From b492fb3ba89205b984c390a39cc9f57d4b2059fa Mon Sep 17 00:00:00 2001 From: samuel niederer Date: Thu, 19 May 2022 18:00:01 +0200 Subject: add tick figure of simple mass system --- buch/papers/kra/images/simple_mass_spring.tex | 64 +++++++++++++++++++++++++++ 1 file changed, 64 insertions(+) create mode 100644 buch/papers/kra/images/simple_mass_spring.tex (limited to 'buch/papers') diff --git a/buch/papers/kra/images/simple_mass_spring.tex b/buch/papers/kra/images/simple_mass_spring.tex new file mode 100644 index 0000000..e98ee3e --- /dev/null +++ b/buch/papers/kra/images/simple_mass_spring.tex @@ -0,0 +1,64 @@ +% create tikz drawing of a simple mass spring system + +\tikzstyle{hmline}=[{Latex[length=3.3,width=2.2]}-{Latex[length=3.3,width=2.2]},line width=0.3] +\tikzstyle{vmline}=[red, dashed,line width=0.4,dash pattern=on 1pt off 1pt] +\tikzstyle{ground}=[pattern=north east lines] +\tikzstyle{mass}=[line width=0.6,red!30!black,fill=red!40!black!10,rounded corners=1,top color=red!40!black!20,bottom color=red!40!black!10,shading angle=20] +\tikzstyle{spring}=[line width=0.8,blue!7!black!80,snake=coil,segment amplitude=5,line cap=round] + +\begin{tikzpicture}[scale=2] + \newcommand{\ticks}[2] + { + % arguments: x, y coordinates + \draw[thick] (#1, #2 - 0.1 / 2) --++ (0, 0.1); + } + + \tikzmath{ + \hWall = 1.5; + \wWall = 0.3; + \lWall = 3.5; + \hMass = 0.6; + \wMass = 1.1; + \xMass1 = 1.2; + \xMass2 = 2.2; + \xAxisYpos = 0; + \originX1 = 0; + \originY1 = 0.5; + \originX2 = 0; + \originY2 = -2; + \springscale=7; + } + + % create x axis + \draw[->,thick] (0,\xAxisYpos) --+ (\lWall, 0) node[right]{$x$}; + + % create ticks on x axis + \ticks{\wWall}{\xAxisYpos} + \ticks{\xMass1}{\xAxisYpos} + \ticks{\xMass2}{\xAxisYpos} + + % create underground + \draw[ground] (\originX1, \originY1) ++ (0, 0) --+(\lWall,0) --+(\lWall, \wWall) --+(\wWall, \wWall) --+(\wWall, \hWall) --+(0, \hWall) -- cycle; + \draw[ground] (\originX2, \originY2) ++ (0, 0) --+(\lWall,0) --+(\lWall, \wWall) --+(\wWall, \wWall) --+(\wWall, \hWall) --+(0, \hWall) -- cycle; + + % create masses + \draw[mass] (\originX1, \originY1) ++ (\xMass1, \wWall) rectangle ++ (\wMass,\hMass) node[midway] {$m$}; + \draw[mass] (\originX2, \originY2) ++ (\xMass2, \wWall) rectangle ++ (\wMass,\hMass) node[midway] {$m$}; + + % create springs + \draw[spring, segment length=(\xMass1 - \wWall) * \springscale] (\originX1, \originY1) ++ (\wWall, \wWall + \hMass / 2) --+ (\xMass1 - \wWall, 0); + \draw[spring, segment length=(\xMass2 - \wWall) * \springscale] (\originX2, \originY2) ++ (\wWall, \wWall + \hMass / 2) --+ (\xMass2 - \wWall, 0); + + % create vertical measurement line + \draw[vmline] (\xMass1, \xAxisYpos) --+(0, \originY1 + \wWall); + \draw[vmline] (\xMass2, \xAxisYpos) --+(0, \originY2 + \hMass+\wWall); + \draw[vmline] (\wWall, \originY1+\wWall) --(\wWall, \originY2 + \hWall); + + % create horizontal measurement line + \draw[hmline] (\wWall, \xAxisYpos + 0.2) -- (\xMass1, \xAxisYpos + 0.2) node[midway,fill=white,inner sep=0] {$\ell_0$}; + \draw[hmline] (\xMass1, \xAxisYpos + 0.2) -- (\xMass2, \xAxisYpos + 0.2) node[midway,fill=white,inner sep=0] {$\Delta_{x}$}; + \draw[hmline] (\wWall, \xAxisYpos - 0.3) -- (\xMass2, \xAxisYpos - 0.3) node[midway,fill=white,inner sep=0] {$\ell_{1}$}; + + % create force arrow + \draw[->,blue, very thick,line cap=round] (\xMass2 + \wMass / 2, \originY2 + \wWall + \hMass + 0.15) node[above] {$\vb{F_{R}}$} --+ (-0.5, 0); +\end{tikzpicture} \ No newline at end of file -- cgit v1.2.1 From ffb50c6574c53b805df068fd2ce2e89726597911 Mon Sep 17 00:00:00 2001 From: samuel niederer Date: Thu, 26 May 2022 11:36:20 +0200 Subject: add spring constant --- buch/papers/kra/images/simple_mass_spring.tex | 6 ++++-- 1 file changed, 4 insertions(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kra/images/simple_mass_spring.tex b/buch/papers/kra/images/simple_mass_spring.tex index e98ee3e..207f1e0 100644 --- a/buch/papers/kra/images/simple_mass_spring.tex +++ b/buch/papers/kra/images/simple_mass_spring.tex @@ -46,8 +46,10 @@ \draw[mass] (\originX2, \originY2) ++ (\xMass2, \wWall) rectangle ++ (\wMass,\hMass) node[midway] {$m$}; % create springs - \draw[spring, segment length=(\xMass1 - \wWall) * \springscale] (\originX1, \originY1) ++ (\wWall, \wWall + \hMass / 2) --+ (\xMass1 - \wWall, 0); - \draw[spring, segment length=(\xMass2 - \wWall) * \springscale] (\originX2, \originY2) ++ (\wWall, \wWall + \hMass / 2) --+ (\xMass2 - \wWall, 0); + \draw[spring, segment length=(\xMass1 - \wWall) * \springscale] (\originX1, \originY1) ++ + (\wWall, \wWall + \hMass / 2) --++ (\xMass1 - \wWall, 0) node[midway,above=0.2] {$k$}; + \draw[spring, segment length=(\xMass2 - \wWall) * \springscale] (\originX2, \originY2) ++ + (\wWall, \wWall + \hMass / 2) --++ (\xMass2 - \wWall, 0) node[midway,above=0.2] {$k$}; % create vertical measurement line \draw[vmline] (\xMass1, \xAxisYpos) --+(0, \originY1 + \wWall); -- cgit v1.2.1 From a0b6394bd559e7d2e1a6d7c028cfc73586503d58 Mon Sep 17 00:00:00 2001 From: samuel niederer Date: Thu, 26 May 2022 11:46:24 +0200 Subject: add drawing --- buch/papers/kra/images/multi_mass_spring.tex | 54 ++++++++++++++++++++++++++++ 1 file changed, 54 insertions(+) create mode 100644 buch/papers/kra/images/multi_mass_spring.tex (limited to 'buch/papers') diff --git a/buch/papers/kra/images/multi_mass_spring.tex b/buch/papers/kra/images/multi_mass_spring.tex new file mode 100644 index 0000000..f255cc8 --- /dev/null +++ b/buch/papers/kra/images/multi_mass_spring.tex @@ -0,0 +1,54 @@ +% create tikz drawing of a multi mass multi spring system + +\tikzstyle{vmline}=[red, dashed,line width=0.4,dash pattern=on 1pt off 1pt] +\tikzstyle{ground}=[pattern=north east lines] +\tikzstyle{mass}=[line width=0.6,red!30!black,fill=red!40!black!10,rounded corners=1,top color=red!40!black!20,bottom color=red!40!black!10,shading angle=20] +\tikzstyle{spring}=[line width=0.8,blue!7!black!80,snake=coil,segment amplitude=5,line cap=round] + +\begin{tikzpicture}[scale=2] + \newcommand{\ticks}[3] + { + % x, y coordinates + \draw[thick] (#1, #2 - 0.1 / 2) --++ (0, 0.1) node[scale=0.8,below=0.2] {#3}; + } + \tikzmath{ + \hWall = 1.2; + \wWall = 0.3; + \lWall = 5; + \hMass = 0.6; + \wMass = 1.1; + \xMass1 = 1.0; + \xMass2 = 3.0; + \xAxisYpos = 0; + \originX1 = 0; + \originY1 = 0.5; + \springscale=7; + } + + % create axis + \draw[->,thick] (0,\xAxisYpos) --+ (\xMass2 + \wMass, 0) node[right]{$q$}; + % create ticks on x / q axis + \ticks{\xMass1}{\xAxisYpos}{$q_{1}$} + \ticks{\xMass2}{\xAxisYpos}{$q_{2}$} + + % create non-moving backgrounds + \draw[ground] (\originX1, \originY1) ++ (0, 0) --+(\lWall,0) --+(\lWall, \hWall) + --+ (\lWall - \wWall, \hWall) --+(\lWall - \wWall, \wWall) --+ (\wWall, \wWall) --+(\wWall, \hWall) --+(0, \hWall) -- cycle; + + % create masses + \draw[mass] (\originX1, \originY1) ++ (\xMass1, \wWall) rectangle ++ (\wMass,\hMass) node[midway] {$m_{1}$}; + \draw[mass] (\originX1, \originY1) ++ (\xMass2, \wWall) rectangle ++ (\wMass,\hMass) node[midway] {$m_{2}$}; + + % create springs + \draw[spring, segment length=(\xMass1 - \wWall) * \springscale] (\originX1, \originY1) ++ + (\wWall, \wWall + \hMass / 2) --++ (\xMass1 - \wWall, 0) node[midway,above=0.2] {$k_1$}; + \draw[spring, segment length=(\xMass1 - \wWall) * \springscale] (\originX1, \originY1) ++ + (\xMass1 + \wMass, \wWall + \hMass / 2) --++ (\xMass2 - \xMass1 - \wMass, 0) node[midway,above=0.2] {$k_c$}; + \draw[spring, segment length=(\xMass1 - \wWall) * \springscale] (\originX1, \originY1) ++ + (\xMass2 + \wMass, \wWall + \hMass / 2) --++ (\lWall - \xMass2 - \wMass - \wWall, 0) node[midway,above=0.2] {$k_2$}; + + % create vertical measurement line + \draw[vmline] (\xMass1, \xAxisYpos) --+(0, \originY1 + \wWall); + \draw[vmline] (\xMass2, \xAxisYpos) --+(0, \originY1 + \wWall); + +\end{tikzpicture} -- cgit v1.2.1 From da8dbf2a727537fbf279268b4a42145677034994 Mon Sep 17 00:00:00 2001 From: runterer Date: Sat, 28 May 2022 18:13:13 +0200 Subject: started with presentation --- buch/papers/zeta/presentation/presentation.tex | 224 +++++++++++++++++++++ .../zeta/presentation/youtube_screenshot.png | Bin 0 -> 378662 bytes 2 files changed, 224 insertions(+) create mode 100644 buch/papers/zeta/presentation/presentation.tex create mode 100644 buch/papers/zeta/presentation/youtube_screenshot.png (limited to 'buch/papers') diff --git a/buch/papers/zeta/presentation/presentation.tex b/buch/papers/zeta/presentation/presentation.tex new file mode 100644 index 0000000..0833f14 --- /dev/null +++ b/buch/papers/zeta/presentation/presentation.tex @@ -0,0 +1,224 @@ +\documentclass[ngerman, aspectratio=169]{beamer} + +%style +\mode{ + \usetheme{Frankfurt} +} +%packages +\usepackage[utf8]{inputenc} +\usepackage[english]{babel} +\usepackage{graphicx} +\usepackage{array} + +\newcolumntype{L}[1]{>{\raggedright\let\newline\\\arraybackslash\hspace{0pt}}m{#1}} +\usepackage{ragged2e} + +\usepackage{bm} % bold math +\usepackage{amsfonts} +\usepackage{amssymb} +\usepackage{mathtools} +\usepackage{amsmath} +\usepackage{multirow} % multi row in tables +\usepackage{scrextend} + +\usepackage{tikz} + +\usepackage{algorithmic} + +%\usepackage{algorithm} % http://ctan.org/pkg/algorithm +%\usepackage{algpseudocode} % http://ctan.org/pkg/algorithmicx + +%\usepackage{algorithmicx} + + +%citations +\usepackage[style=verbose,backend=biber]{biblatex} +\addbibresource{references.bib} + + + +\usefonttheme[onlymath]{serif} + +%Beamer Template modifications +%\definecolor{mainColor}{HTML}{0065A3} % HSR blue +\definecolor{mainColor}{HTML}{D72864} % OST pink +\definecolor{invColor}{HTML}{28d79b} % OST pink +\definecolor{dgreen}{HTML}{38ad36} % Dark green + +%\definecolor{mainColor}{HTML}{000000} % HSR blue +\setbeamercolor{palette primary}{bg=white,fg=mainColor} +\setbeamercolor{palette secondary}{bg=orange,fg=mainColor} +\setbeamercolor{palette tertiary}{bg=yellow,fg=red} +\setbeamercolor{palette quaternary}{bg=mainColor,fg=white} %bg = Top bar, fg = active top bar topic +\setbeamercolor{structure}{fg=black} % itemize, enumerate, etc (bullet points) +\setbeamercolor{section in toc}{fg=black} % TOC sections +\setbeamertemplate{section in toc}[sections numbered] +\setbeamertemplate{subsection in toc}{% + \hspace{1.2em}{$\bullet$}~\inserttocsubsection\par} + +\setbeamertemplate{itemize items}[circle] +\setbeamertemplate{description item}[circle] +\setbeamertemplate{title page}[default][colsep=-4bp,rounded=true] +\beamertemplatenavigationsymbolsempty + +\setbeamercolor{footline}{fg=gray} +\setbeamertemplate{footline}{% + \hfill\usebeamertemplate***{navigation symbols} + \hspace{0.5cm} + \insertframenumber{}\hspace{0.2cm}\vspace{0.2cm} +} + +\usepackage{caption} +\captionsetup{labelformat=empty} + +%Title Page +\title{Riemannsche Zeta Funktion} +\author{Raphael Unterer} +\institute{Mathematisches Seminar 2022: Spezielle Funktionen} + +\newcommand*{\HL}{\textcolor{mainColor}} +\newcommand*{\RD}{\textcolor{red}} +\newcommand*{\BL}{\textcolor{blue}} +\newcommand*{\GN}{\textcolor{dgreen}} + + + + +\makeatletter +\newcount\my@repeat@count +\newcommand{\myrepeat}[2]{% + \begingroup + \my@repeat@count=\z@ + \@whilenum\my@repeat@count<#1\do{#2\advance\my@repeat@count\@ne}% + \endgroup +} +\makeatother + + + + +\usetikzlibrary{automata,arrows,positioning,calc} + + +\begin{document} + + %Titelseite + \begin{frame} + \titlepage + \end{frame} + + %Inhaltsverzeichnis + \begin{frame} + \frametitle{Inhalt} + \tableofcontents + \end{frame} + + \section{Motivation} + + \begin{frame} + \frametitle{Summe aller Natürlichen Zahlen} + \begin{equation*} + \sum_{n=1}^{\infty} n + = + 1 + 2 + 3 + \ldots + \infty + = + - \frac{1}{12} + \end{equation*} + \end{frame} + \begin{frame} + \frametitle{Summe aller Natürlichen Zahlen} + \begin{center} + \includegraphics[width=0.7\textwidth]{youtube_screenshot.png} + \end{center} + \end{frame} + \begin{frame} + \frametitle{Riemannsche Zeta Funktion} + \begin{equation*} + \zeta(s) + = + \sum_{n=1}^{\infty} + \frac{1}{n^s} + \end{equation*} + \pause + \begin{equation*} + \zeta(-1) + = + \sum_{n=1}^{\infty} + \frac{1}{n^{-1}} + = + \sum_{n=1}^{\infty} n + \end{equation*} + \end{frame} + \begin{frame} + \frametitle{Originaler Definitionsbereich} + Wir kennen die divergierende harmonische Reihe + \begin{equation*} + \zeta(1) + = + \sum_{n=1}^{\infty} + \frac{1}{n} + \rightarrow + \infty, + \end{equation*} + und somit ist $\Re(s) > 1$. + \end{frame} + + \section{Analytische Fortsetzung} + \begin{frame} + \frametitle{Plan für die Analytische Fortsetzung von $\zeta(s)$} + \begin{center} + \input{../continuation_overview.tikz.tex} + \end{center} + \end{frame} + \begin{frame} + \frametitle{Fortsetzung auf $\Re(s) > 0$} + Dirichletsche Etafunktion ist + \begin{equation*}\label{zeta:equation:eta} + \eta(s) + = + \sum_{n=1}^{\infty} + \frac{(-1)^{n-1}}{n^s}, + \end{equation*} + und konvergiert im Bereich $\Re(s) > 0$. + \end{frame} + +% Zuerst wiederholen wir zweimal die Definition der Zetafunktion \eqref{zeta:equation1}, wobei wir sie einmal durch $2^{s-1}$ teilen +% \begin{align} +% \zeta(s) +% &= +% \sum_{n=1}^{\infty} +% \frac{1}{n^s} \label{zeta:align1} +% \\ +% \frac{1}{2^{s-1}} +% \zeta(s) +% &= +% \sum_{n=1}^{\infty} +% \frac{2}{(2n)^s}. \label{zeta:align2} +% \end{align} +% Durch Subtraktion der beiden Gleichungen \eqref{zeta:align1} minus \eqref{zeta:align2}, ergibt sich +% \begin{align} +% \left(1 - \frac{1}{2^{s-1}} \right) +% \zeta(s) +% &= +% \frac{1}{1^s} +% \underbrace{-\frac{2}{2^s} + \frac{1}{2^s}}_{-\frac{1}{2^s}} +% + \frac{1}{3^s} +% \underbrace{-\frac{2}{4^s} + \frac{1}{4^s}}_{-\frac{1}{4^s}} +% \ldots +% \\ +% &= \eta(s). +% \end{align} +% Dies ist die Fortsetzung auf den noch unbekannten Bereich $0 < \Re(s) < 1$ +% \begin{equation} \label{zeta:equation:fortsetzung1} +% \zeta(s) +% := +% \left(1 - \frac{1}{2^{s-1}} \right)^{-1} \eta(s). +% \end{equation} +% \section{Euler Produkt} +% +% \section{Weitere Eigenschaften} +% +% + +\end{document} + diff --git a/buch/papers/zeta/presentation/youtube_screenshot.png b/buch/papers/zeta/presentation/youtube_screenshot.png new file mode 100644 index 0000000..434041b Binary files /dev/null and b/buch/papers/zeta/presentation/youtube_screenshot.png differ -- cgit v1.2.1 From bd59e9086178019b48f10db3ad2ca8356c96e2c0 Mon Sep 17 00:00:00 2001 From: runterer Date: Sat, 28 May 2022 19:49:04 +0200 Subject: wip working on presentation --- buch/papers/zeta/presentation/presentation.tex | 111 ++++-- buch/papers/zeta/primzahlfunktion.pgf | 505 +++++++++++++++++++++++++ buch/papers/zeta/python/primzahlfunktion.py | 24 ++ 3 files changed, 603 insertions(+), 37 deletions(-) create mode 100644 buch/papers/zeta/primzahlfunktion.pgf create mode 100644 buch/papers/zeta/python/primzahlfunktion.py (limited to 'buch/papers') diff --git a/buch/papers/zeta/presentation/presentation.tex b/buch/papers/zeta/presentation/presentation.tex index 0833f14..bb6d515 100644 --- a/buch/papers/zeta/presentation/presentation.tex +++ b/buch/papers/zeta/presentation/presentation.tex @@ -181,44 +181,81 @@ \end{equation*} und konvergiert im Bereich $\Re(s) > 0$. \end{frame} + \begin{frame} + \frametitle{Fortsetzung auf $\Re(s) > 0$} + \begin{align} + \zeta(s) + &= + \sum_{n=1}^{\infty} + \frac{1}{n^s} \label{zeta:align1} + \\ + \frac{1}{2^{s-1}} + \zeta(s) + &= + \sum_{n=1}^{\infty} + \frac{2}{(2n)^s} \label{zeta:align2} + \end{align} + \pause + \eqref{zeta:align1} - \eqref{zeta:align2}: + \begin{align*} + \left(1 - \frac{1}{2^{s-1}} \right) + \zeta(s) + &= + \frac{1}{1^s} + \underbrace{-\frac{2}{2^s} + \frac{1}{2^s}}_{-\frac{1}{2^s}} + + \frac{1}{3^s} + \underbrace{-\frac{2}{4^s} + \frac{1}{4^s}}_{-\frac{1}{4^s}} + \ldots + \\ + &= \eta(s) + \end{align*} + \end{frame} + \begin{frame} + \frametitle{Fortsetzung auf $\Re(s) > 0$} + Somit haben wir die Fortsetzung gefunden als + \begin{equation} \label{zeta:equation:fortsetzung1} + \zeta(s) + := + \left(1 - \frac{1}{2^{s-1}} \right)^{-1} \eta(s). + \end{equation} + \end{frame} + \begin{frame} + \frametitle{Spiegelungseigenschaft für $\Re(s) < 0$} + \begin{equation*}\label{zeta:equation:functional} + \frac{\Gamma \left( \frac{s}{2} \right)}{\pi^{\frac{s}{2}}} + \zeta(s) + = + \frac{\Gamma \left( \frac{1-s}{2} \right)}{\pi^{\frac{1-s}{2}}} + \zeta(1-s). + \end{equation*} + \end{frame} + %TODO maybe explain gamma-fct + + \section{Euler Produkt und Primzahlen} + \begin{frame} + \frametitle{Wieso ist die Zeta Funktion so bekannt?} + \begin{itemize} + \item Interessante Funktionswerte z.B. $\zeta(2) = \frac{\pi^2}{6}$ + \item Primzahlenverteilung (Riemannhypothese) + \item Forschungsgebiet der analytischen Zahlentheorie seit dem 18. Jahrhundert + \item ... + \end{itemize} + \end{frame} + \begin{frame} + \frametitle{Primzahlfunktion} + \begin{center} + \scalebox{0.5}{\input{../primzahlfunktion.pgf}} + \end{center} + \end{frame} + \begin{frame} + \frametitle{Zusammenhang Zeta und Primzahlen} + %TODO + \end{frame} + + + \section{Weitere Eigenschaften} + -% Zuerst wiederholen wir zweimal die Definition der Zetafunktion \eqref{zeta:equation1}, wobei wir sie einmal durch $2^{s-1}$ teilen -% \begin{align} -% \zeta(s) -% &= -% \sum_{n=1}^{\infty} -% \frac{1}{n^s} \label{zeta:align1} -% \\ -% \frac{1}{2^{s-1}} -% \zeta(s) -% &= -% \sum_{n=1}^{\infty} -% \frac{2}{(2n)^s}. \label{zeta:align2} -% \end{align} -% Durch Subtraktion der beiden Gleichungen \eqref{zeta:align1} minus \eqref{zeta:align2}, ergibt sich -% \begin{align} -% \left(1 - \frac{1}{2^{s-1}} \right) -% \zeta(s) -% &= -% \frac{1}{1^s} -% \underbrace{-\frac{2}{2^s} + \frac{1}{2^s}}_{-\frac{1}{2^s}} -% + \frac{1}{3^s} -% \underbrace{-\frac{2}{4^s} + \frac{1}{4^s}}_{-\frac{1}{4^s}} -% \ldots -% \\ -% &= \eta(s). -% \end{align} -% Dies ist die Fortsetzung auf den noch unbekannten Bereich $0 < \Re(s) < 1$ -% \begin{equation} \label{zeta:equation:fortsetzung1} -% \zeta(s) -% := -% \left(1 - \frac{1}{2^{s-1}} \right)^{-1} \eta(s). -% \end{equation} -% \section{Euler Produkt} -% -% \section{Weitere Eigenschaften} -% -% \end{document} diff --git a/buch/papers/zeta/primzahlfunktion.pgf b/buch/papers/zeta/primzahlfunktion.pgf new file mode 100644 index 0000000..7d4f4fc --- /dev/null +++ b/buch/papers/zeta/primzahlfunktion.pgf @@ -0,0 +1,505 @@ +%% Creator: Matplotlib, PGF backend +%% +%% To include the figure in your LaTeX document, write +%% \input{.pgf} +%% +%% Make sure the required packages are loaded in your preamble +%% \usepackage{pgf} +%% +%% and, on pdftex +%% \usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-} +%% +%% or, on luatex and xetex +%% \usepackage{unicode-math} +%% +%% Figures using additional raster images can only be included by \input if +%% they are in the same directory as the main LaTeX file. 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+\pgfsetmiterjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.800000in}{0.528000in}}% +\pgfpathlineto{\pgfqpoint{5.760000in}{0.528000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetrectcap% +\pgfsetmiterjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.800000in}{4.224000in}}% +\pgfpathlineto{\pgfqpoint{5.760000in}{4.224000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\end{pgfpicture}% +\makeatother% +\endgroup% diff --git a/buch/papers/zeta/python/primzahlfunktion.py b/buch/papers/zeta/python/primzahlfunktion.py new file mode 100644 index 0000000..9434de9 --- /dev/null +++ b/buch/papers/zeta/python/primzahlfunktion.py @@ -0,0 +1,24 @@ +import matplotlib.pyplot as plt +import numpy as np + +primzahlfunktion = [0, 0, 0, 0] +x = [0, 1-1e-12, 1, 2-1e-12] +x_last = 1 +value = 0 +for i in range(2, 30, 1): + new_value = value + 1 + for j in range(2, i, 1): + if i % j == 0: + new_value = value + value = new_value + primzahlfunktion.append(new_value) + x_last += 1 + x.append(x_last) + primzahlfunktion.append(new_value) + x.append(x_last + 1 - 1e-12) + + +plt.rcParams.update({"pgf.texsystem": "pdflatex"}) +plt.plot(x, primzahlfunktion) +plt.show() + -- cgit v1.2.1 From 052ccbe27e19d53ac7272bf29c9c16e071d7059b Mon Sep 17 00:00:00 2001 From: samuel niederer Date: Wed, 1 Jun 2022 13:34:26 +0200 Subject: add phase space plot --- buch/papers/kra/images/phase_space.tex | 67 ++++++++++++++++++++++++++++++++++ 1 file changed, 67 insertions(+) create mode 100644 buch/papers/kra/images/phase_space.tex (limited to 'buch/papers') diff --git a/buch/papers/kra/images/phase_space.tex b/buch/papers/kra/images/phase_space.tex new file mode 100644 index 0000000..cd51ea4 --- /dev/null +++ b/buch/papers/kra/images/phase_space.tex @@ -0,0 +1,67 @@ +\colorlet{mypurple}{red!50!blue!90!black!80} + +% style to create arrows +\tikzset{ + traj/.style 2 args={thick, postaction={decorate},decoration={markings, + mark=at position #1 with {\arrow{<}}, + mark=at position #2 with {\arrow{<}}} + } +} + +\begin{tikzpicture}[scale=0.6] + % p(t=0) = 0, q(t=0) = A, max(p) = mwA + \tikzmath{ + \axh = 5.2; + \axw1 = 4.2; + \axw2 = 4.8; + \d1 = 0.9; + \a0 = 1; + \b0 = 2; + \a1 = \a0 + \d1; + \b1 = \b0 + \d1; + \a2 = \a1 + \d1; + \b2 = \b1 + \d1; + \a3 = \a2 + \d1; + \b3 = \b2 + \d1; + \d2 = 0.75; + \aa0 = 2; + \bb0 = 1; + \aa1 = \aa0 + \d2; + \bb1 = \bb0 + \d2; + \aa2 = \aa1 + \d2; + \bb2 = \bb1 + \d2; + \aa3 = \aa2 + \d2; + \bb3 = \bb2 + \d2; + } + + \draw[->,thick] (-\axw1,0) -- (\axw1,0) node[right] {$q$}; + \draw[->,thick] (0,-\axh) -- (0,\axh) node[above] {$p$}; + + \draw[traj={0.375}{0.875},darkgreen] ellipse (\a0 and \b0); + \draw[traj={0.375}{0.875},blue] ellipse (\a1 and \b1); + \draw[traj={0.375}{0.875},cyan] ellipse (\a2 and \b2); + \draw[traj={0.375}{0.875},mypurple] ellipse (\a3 and \b3); + + \node[right,darkgreen] at (45:{\a0} and {\b0}) {$E_A$}; + \node[right, blue] at (45:{\a1} and {\b1}) {$E_B$}; + \node[right, cyan] at (45:{\a2} and {\b2}) {$E_C$}; + \node[right, mypurple] at (45:{\a3} and {\b3}) {$E_D$}; + \node[above left] at (110:\b3 + 0.1) {grosses $\omega$}; + + \begin{scope}[xshift=12cm] + \draw[->,thick] (-\axw2,0) -- (\axw2,0) node[right] {$q$}; + \draw[->,thick] (0,-\axh) -- (0,\axh) node[above] {$p$}; + + \draw[traj={0.375}{0.875},darkgreen] ellipse (\aa0 and \bb0); + \draw[traj={0.375}{0.875},blue] ellipse (\aa1 and \bb1); + \draw[traj={0.375}{0.875},cyan] ellipse (\aa2 and \bb2); + \draw[traj={0.375}{0.875},mypurple] ellipse (\aa3 and \bb3); + + \node[above, darkgreen] at (45:{\aa0} and {\bb0}) {$E_A$}; + \node[above, blue] at (45:{\aa1} and {\bb1}) {$E_B$}; + \node[above, cyan] at (45:{\aa2} and {\bb2}) {$E_C$}; + \node[above, mypurple] at (45:{\aa3} and {\bb3}) {$E_D$}; + + \node[above left] at (110:\b3 + 0.1) {kleines $\omega$}; + \end{scope} +\end{tikzpicture} \ No newline at end of file -- cgit v1.2.1 From 45e8902e2409339cfc363033e622980600cbcf41 Mon Sep 17 00:00:00 2001 From: runterer Date: Thu, 2 Jun 2022 00:28:08 +0200 Subject: presentation finished? --- buch/papers/zeta/presentation/presentation.tex | 112 +- .../zeta/presentation/zeta_color_plot-img0.png | Bin 0 -> 37362 bytes buch/papers/zeta/presentation/zeta_color_plot.pgf | 402 +++++++ buch/papers/zeta/python/plot_zeta.py | 39 + buch/papers/zeta/python/plot_zeta2.py | 31 + buch/papers/zeta/zeta_re_-1_plot.pgf | 1147 ++++++++++++++++++ buch/papers/zeta/zeta_re_0.5_plot.pgf | 1206 +++++++++++++++++++ buch/papers/zeta/zeta_re_0_plot.pgf | 1242 ++++++++++++++++++++ 8 files changed, 4174 insertions(+), 5 deletions(-) create mode 100644 buch/papers/zeta/presentation/zeta_color_plot-img0.png create mode 100644 buch/papers/zeta/presentation/zeta_color_plot.pgf create mode 100644 buch/papers/zeta/python/plot_zeta.py create mode 100644 buch/papers/zeta/python/plot_zeta2.py create mode 100644 buch/papers/zeta/zeta_re_-1_plot.pgf create mode 100644 buch/papers/zeta/zeta_re_0.5_plot.pgf create mode 100644 buch/papers/zeta/zeta_re_0_plot.pgf (limited to 'buch/papers') diff --git a/buch/papers/zeta/presentation/presentation.tex b/buch/papers/zeta/presentation/presentation.tex index bb6d515..be3e12c 100644 --- a/buch/papers/zeta/presentation/presentation.tex +++ b/buch/papers/zeta/presentation/presentation.tex @@ -80,6 +80,7 @@ \newcommand*{\RD}{\textcolor{red}} \newcommand*{\BL}{\textcolor{blue}} \newcommand*{\GN}{\textcolor{dgreen}} +\newcommand*{\YE}{\textcolor{violet}} @@ -241,21 +242,122 @@ \item ... \end{itemize} \end{frame} + \begin{frame} + \frametitle{Euler Produkt: Verbindung von Zeta und Primzahlen} + \begin{equation*} + \zeta(s) + = + \sum_{n=1}^\infty + \frac{1}{n^s} + = + \prod_{p \in P} + \frac{1}{1-p^{-s}} + \end{equation*} + \pause + Geometrische Reihe + \begin{equation*} + \prod_{p \in P} + \frac{1}{1-p^{-s}} + = + \prod_{p \in P} + \left( + 1 + + + \frac{1}{p^s} + + + \frac{1}{p^{2s}} + + + \frac{1}{p^{3s}} + + + \ldots + \right) + \end{equation*} + \pause + Erste Terme ausmultiplizieren + \begin{align*} + \left( + 1 + + + \RD{\frac{1}{2^s}} + + + \GN{\frac{1}{2^{2s}}} + + + \frac{1}{2^{3s}} + + + \ldots + \right) + \left( + 1 + + + \BL{\frac{1}{3^s}} + + + \frac{1}{3^{2s}} + + + \frac{1}{3^{3s}} + + + \ldots + \right) + \left( + 1 + + + \YE{\frac{1}{5^s}} + + + \frac{1}{5^{2s}} + + + \frac{1}{5^{3s}} + + + \ldots + \right) + \\ + = + 1 + + + \RD{\frac{1}{2^s}} + + + \BL{\frac{1}{3^s}} + + + \GN{\frac{1}{4^s}} + + + \YE{\frac{1}{5^s}} + + + \ldots + \end{align*} + \end{frame} \begin{frame} \frametitle{Primzahlfunktion} \begin{center} \scalebox{0.5}{\input{../primzahlfunktion.pgf}} \end{center} \end{frame} - \begin{frame} - \frametitle{Zusammenhang Zeta und Primzahlen} - %TODO - \end{frame} - \section{Weitere Eigenschaften} + \section{Darstellungen} + \begin{frame} + \frametitle{Farbcodierung} + \begin{center} + \scalebox{0.6}{\input{zeta_color_plot.pgf}} + \end{center} + \end{frame} + \begin{frame} + \frametitle{Konstanter Realteil $\Re(s)=-1$ und $\Im(s)=0\ldots40$} + \begin{center} + \scalebox{0.6}{\input{../zeta_re_-1_plot.pgf}} + \end{center} + \end{frame} + \begin{frame} + \frametitle{Konstanter Realteil $\Re(s)=0$ und $\Im(s)=0\ldots40$} + \begin{center} + \scalebox{0.6}{\input{../zeta_re_0_plot.pgf}} + \end{center} + \end{frame} + \begin{frame} + \frametitle{Konstanter Realteil $\Re(s)=0.5$ und $\Im(s)=0\ldots40$} + \begin{center} + \scalebox{0.6}{\input{../zeta_re_0.5_plot.pgf}} + \end{center} + \end{frame} \end{document} diff --git a/buch/papers/zeta/presentation/zeta_color_plot-img0.png b/buch/papers/zeta/presentation/zeta_color_plot-img0.png new file mode 100644 index 0000000..b8c7298 Binary files /dev/null and b/buch/papers/zeta/presentation/zeta_color_plot-img0.png differ diff --git a/buch/papers/zeta/presentation/zeta_color_plot.pgf b/buch/papers/zeta/presentation/zeta_color_plot.pgf new file mode 100644 index 0000000..0fd7cb8 --- /dev/null +++ b/buch/papers/zeta/presentation/zeta_color_plot.pgf @@ -0,0 +1,402 @@ +%% Creator: Matplotlib, PGF backend +%% +%% To include the figure in your LaTeX document, write +%% \input{.pgf} +%% +%% Make sure the required packages are loaded in your preamble +%% \usepackage{pgf} +%% +%% and, on pdftex +%% \usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-} +%% +%% or, on luatex and xetex +%% \usepackage{unicode-math} +%% +%% Figures using additional raster images can only be included by \input if +%% they are in the same directory as the main LaTeX file. For loading figures +%% from other directories you can use the `import` package +%% \usepackage{import} +%% +%% and then include the figures with +%% \import{}{.pgf} +%% +%% Matplotlib used the following preamble +%% +\begingroup% +\makeatletter% +\begin{pgfpicture}% +\pgfpathrectangle{\pgfpointorigin}{\pgfqpoint{6.400000in}{4.800000in}}% +\pgfusepath{use as bounding box, clip}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetmiterjoin% +\definecolor{currentfill}{rgb}{1.000000,1.000000,1.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.000000pt}% +\definecolor{currentstroke}{rgb}{1.000000,1.000000,1.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{6.400000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{6.400000in}{4.800000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{4.800000in}}% +\pgfpathclose% +\pgfusepath{fill}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetmiterjoin% 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base]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle {15}\)}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=2.225499in,y=2.376000in,,bottom,rotate=90.000000]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle \Im\)}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetrectcap% +\pgfsetmiterjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{2.588156in}{0.528000in}}% +\pgfpathlineto{\pgfqpoint{2.588156in}{4.224000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetrectcap% +\pgfsetmiterjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{3.971844in}{0.528000in}}% +\pgfpathlineto{\pgfqpoint{3.971844in}{4.224000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetrectcap% +\pgfsetmiterjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{2.588156in}{0.528000in}}% +\pgfpathlineto{\pgfqpoint{3.971844in}{0.528000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetrectcap% +\pgfsetmiterjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{2.588156in}{4.224000in}}% +\pgfpathlineto{\pgfqpoint{3.971844in}{4.224000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\end{pgfpicture}% +\makeatother% +\endgroup% diff --git a/buch/papers/zeta/python/plot_zeta.py b/buch/papers/zeta/python/plot_zeta.py new file mode 100644 index 0000000..53097c5 --- /dev/null +++ b/buch/papers/zeta/python/plot_zeta.py @@ -0,0 +1,39 @@ +import numpy as np +from mpmath import zeta +import matplotlib.pyplot as plt +from matplotlib import colors +import matplotlib +matplotlib.use("pgf") +matplotlib.rcParams.update( + { + "pgf.texsystem": "pdflatex", + "font.family": "serif", + "font.size": 8, + "text.usetex": True, + "pgf.rcfonts": False, + "axes.unicode_minus": False, + } +) + +print(zeta(-1)) +print(zeta(-1 + 2j)) + +re_values = np.arange(-10, 5, 0.04) +im_values = np.arange(-20, 20, 0.04) +plot_matrix = np.zeros((len(im_values), len(re_values), 3)) +for im_i, im in enumerate(im_values): + print(im_i) + for re_i, re in enumerate(re_values): + z = complex(zeta(re + 1j*im)) + h = (np.angle(z) + np.pi) / (2*np.pi) + v = np.abs(z) + s = 1.0 + plot_matrix[im_i, re_i] = [h, s, v] + +log10_v = np.log10(plot_matrix[:, :, 2]) +log10_v += np.abs(np.min(log10_v)) +plot_matrix[:, :, 2] = (log10_v) / np.max(log10_v) +plt.imshow(colors.hsv_to_rgb(plot_matrix), extent=[re_values.min(), re_values.max(), im_values.min(), im_values.max()]) +plt.xlabel("$\Re$") +plt.ylabel("$\Im$") +plt.savefig(f"zeta_color_plot.pgf") diff --git a/buch/papers/zeta/python/plot_zeta2.py b/buch/papers/zeta/python/plot_zeta2.py new file mode 100644 index 0000000..b730703 --- /dev/null +++ b/buch/papers/zeta/python/plot_zeta2.py @@ -0,0 +1,31 @@ +import numpy as np +from mpmath import zeta +import matplotlib.pyplot as plt +import matplotlib +matplotlib.use("pgf") +matplotlib.rcParams.update( + { + "pgf.texsystem": "pdflatex", + "font.family": "serif", + "font.size": 8, + "text.usetex": True, + "pgf.rcfonts": False, + "axes.unicode_minus": False, + } +) +# const re plot +re_values = [-1, 0, 0.5] +im_values = np.arange(0, 40, 0.04) +buf = np.zeros((len(re_values), len(im_values), 2)) +for im_i, im in enumerate(im_values): + print(im_i) + for re_i, re in enumerate(re_values): + z = complex(zeta(re + 1j*im)) + buf[re_i, im_i] = [np.real(z), np.imag(z)] + +for i in range(len(re_values)): + plt.figure() + plt.plot(buf[i,:,0], buf[i,:,1], label=f"$\Re={re_values[i]}$") + plt.xlabel("$\Re$") + plt.ylabel("$\Im$") + plt.savefig(f"zeta_re_{re_values[i]}_plot.pgf") diff --git a/buch/papers/zeta/zeta_re_-1_plot.pgf b/buch/papers/zeta/zeta_re_-1_plot.pgf new file mode 100644 index 0000000..dd15ba1 --- /dev/null +++ b/buch/papers/zeta/zeta_re_-1_plot.pgf @@ -0,0 +1,1147 @@ +%% Creator: Matplotlib, PGF backend +%% +%% To include the figure in your LaTeX document, write +%% \input{.pgf} +%% +%% Make sure the required packages are loaded in your preamble +%% \usepackage{pgf} +%% +%% and, on pdftex +%% \usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-} +%% +%% or, on luatex and xetex +%% \usepackage{unicode-math} +%% +%% Figures using additional raster images can only be included by \input if +%% they are in the same directory as the main LaTeX file. For loading figures +%% from other directories you can use the `import` package +%% \usepackage{import} +%% +%% and then include the figures with +%% \import{}{.pgf} +%% +%% Matplotlib used the following preamble +%% +\begingroup% +\makeatletter% +\begin{pgfpicture}% +\pgfpathrectangle{\pgfpointorigin}{\pgfqpoint{6.400000in}{4.800000in}}% +\pgfusepath{use as bounding box, clip}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetmiterjoin% +\definecolor{currentfill}{rgb}{1.000000,1.000000,1.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.000000pt}% +\definecolor{currentstroke}{rgb}{1.000000,1.000000,1.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{6.400000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{6.400000in}{4.800000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{4.800000in}}% +\pgfpathclose% +\pgfusepath{fill}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetmiterjoin% 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b/buch/papers/zeta/zeta_re_0.5_plot.pgf new file mode 100644 index 0000000..3ac7df8 --- /dev/null +++ b/buch/papers/zeta/zeta_re_0.5_plot.pgf @@ -0,0 +1,1206 @@ +%% Creator: Matplotlib, PGF backend +%% +%% To include the figure in your LaTeX document, write +%% \input{.pgf} +%% +%% Make sure the required packages are loaded in your preamble +%% \usepackage{pgf} +%% +%% and, on pdftex +%% \usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-} +%% +%% or, on luatex and xetex +%% \usepackage{unicode-math} +%% +%% Figures using additional raster images can only be included by \input if +%% they are in the same directory as the main LaTeX file. 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+\pgfpathmoveto{\pgfqpoint{0.800000in}{4.224000in}}% +\pgfpathlineto{\pgfqpoint{5.760000in}{4.224000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\end{pgfpicture}% +\makeatother% +\endgroup% -- cgit v1.2.1 From 6e787a660b0a1a456d42d8a420dfe790431dfc40 Mon Sep 17 00:00:00 2001 From: Nicolas Tobler Date: Thu, 2 Jun 2022 01:28:17 +0200 Subject: working on presentation --- buch/papers/ellfilter/elliptic.tex | 23 +- .../papers/ellfilter/presentation/presentation.tex | 49 ++- buch/papers/ellfilter/python/F_N_elliptic.pgf | 335 ++++++++++----------- buch/papers/ellfilter/python/elliptic.pgf | 232 +++++++------- buch/papers/ellfilter/python/elliptic.py | 4 +- buch/papers/ellfilter/python/elliptic2.py | 38 ++- buch/papers/ellfilter/python/k.pgf | 4 +- buch/papers/ellfilter/tikz/arccos.tikz.tex | 10 +- buch/papers/ellfilter/tikz/cd.tikz.tex | 16 +- buch/papers/ellfilter/tikz/cd2.tikz.tex | 14 +- buch/papers/ellfilter/tikz/cd3.tikz.tex | 84 ++++++ .../ellfilter/tikz/elliptic_transform.tikz.tex | 64 ++++ buch/papers/ellfilter/tikz/sn.tikz.tex | 16 +- 13 files changed, 557 insertions(+), 332 deletions(-) create mode 100644 buch/papers/ellfilter/tikz/cd3.tikz.tex create mode 100644 buch/papers/ellfilter/tikz/elliptic_transform.tikz.tex (limited to 'buch/papers') diff --git a/buch/papers/ellfilter/elliptic.tex b/buch/papers/ellfilter/elliptic.tex index 88bfbfe..96731c8 100644 --- a/buch/papers/ellfilter/elliptic.tex +++ b/buch/papers/ellfilter/elliptic.tex @@ -69,7 +69,15 @@ Analog zu Abbildung \ref{ellfilter:fig:arccos2} können wir auch bei den ellipti \label{ellfilter:fig:elliptic} \end{figure} -\subsection{Degree Equation} + +\begin{figure} + \centering + \input{papers/ellfilter/python/elliptic.pgf} + \caption{Die resultierende frequenzantwort eines elliptischs filter.} + \label{ellfilter:fig:elliptic_freq} +\end{figure} + +\subsection{Gradgleichung} Der $\cd^{-1}$ Term muss so verzogen werden, dass die umgebene $\cd$-Funktion die Nullstellen und Pole trifft. Dies trifft ein wenn die Degree Equation erfüllt ist. @@ -82,6 +90,19 @@ Dies trifft ein wenn die Degree Equation erfüllt ist. Leider ist das lösen dieser Gleichung nicht trivial. Die Rechnung wird in \ref{ellfilter:bib:orfanidis} im Detail angeschaut. +\begin{figure} + \centering + \input{papers/ellfilter/python/k.pgf} + \caption{Die Periodizitäten in realer und imaginärer Richtung in Abhängigkeit vom elliptischen Modul $k$.} +\end{figure} + +\begin{figure} + \centering + \input{papers/ellfilter/tikz/elliptic_transform.tikz} + \caption{Die Gradgleichung als geometrisches Problem.} +\end{figure} + + \subsection{Polynome?} diff --git a/buch/papers/ellfilter/presentation/presentation.tex b/buch/papers/ellfilter/presentation/presentation.tex index 7fdb864..adbf925 100644 --- a/buch/papers/ellfilter/presentation/presentation.tex +++ b/buch/papers/ellfilter/presentation/presentation.tex @@ -117,7 +117,7 @@ \tableofcontents \end{frame} - \section{Linear Filter} + \section{Lineare Filter} \begin{frame} \frametitle{Lineare Filter} @@ -349,6 +349,23 @@ \end{frame} + \section{Elliptisches Filter} + + \begin{frame} + \frametitle{Elliptisches Filter} + + \begin{equation*} + z_1 = N~\frac{K_1}{K}~\cd^{-1}(w, k) + \end{equation*} + + \begin{center} + \scalebox{0.75}{ + \input{../tikz/cd3.tikz.tex} + } + \end{center} + + \end{frame} + \begin{frame} \frametitle{Periodizität in realer und imaginärer Richtung} @@ -357,23 +374,42 @@ \end{center} + \end{frame} + + \begin{frame} + \frametitle{Gradgleichung} + + \begin{equation} + N \frac{K^\prime}{K} = \frac{K^\prime_1}{K_1} + \end{equation} + + \begin{center} + \scalebox{0.95}{ + \input{../tikz/elliptic_transform.tikz} + } + \end{center} + + \end{frame} \begin{frame} \frametitle{Elliptisches Filter} \begin{equation*} + R_N = \cd(z_1, k_1), + \quad z_1 = N~\frac{K_1}{K}~\cd^{-1}(w, k) \end{equation*} \begin{center} - \scalebox{0.8}{ + \scalebox{0.75}{ \input{../tikz/cd2.tikz.tex} } \end{center} \end{frame} + \begin{frame} \frametitle{Elliptisches Filter} @@ -401,13 +437,4 @@ \end{frame} - \begin{frame} - \frametitle{Gradgleichung} - - \begin{equation} - N \frac{K^\prime}{K} = \frac{K^\prime_1}{K_1} - \end{equation} - - \end{frame} - \end{document} diff --git a/buch/papers/ellfilter/python/F_N_elliptic.pgf b/buch/papers/ellfilter/python/F_N_elliptic.pgf index 03084c6..50faaaa 100644 --- a/buch/papers/ellfilter/python/F_N_elliptic.pgf +++ b/buch/papers/ellfilter/python/F_N_elliptic.pgf @@ -94,8 +94,8 @@ \pgfsetstrokeopacity{0.200000}% \pgfsetdash{}{0pt}% \pgfpathmoveto{\pgfqpoint{2.247564in}{1.250043in}}% -\pgfpathlineto{\pgfqpoint{2.262704in}{1.250043in}}% -\pgfpathlineto{\pgfqpoint{2.262704in}{1.600680in}}% +\pgfpathlineto{\pgfqpoint{2.262583in}{1.250043in}}% +\pgfpathlineto{\pgfqpoint{2.262583in}{1.600680in}}% \pgfpathlineto{\pgfqpoint{2.247564in}{1.600680in}}% \pgfpathlineto{\pgfqpoint{2.247564in}{1.250043in}}% \pgfpathclose% @@ -114,11 +114,11 @@ \pgfsetstrokecolor{currentstroke}% \pgfsetstrokeopacity{0.200000}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{2.262704in}{1.600680in}}% -\pgfpathlineto{\pgfqpoint{3.776737in}{1.600680in}}% 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a/buch/papers/ellfilter/python/elliptic.pgf b/buch/papers/ellfilter/python/elliptic.pgf index 31b77d4..89ffb60 100644 --- a/buch/papers/ellfilter/python/elliptic.pgf +++ b/buch/papers/ellfilter/python/elliptic.pgf @@ -94,8 +94,8 @@ \pgfsetstrokeopacity{0.200000}% \pgfsetdash{}{0pt}% \pgfpathmoveto{\pgfqpoint{2.189776in}{0.724087in}}% -\pgfpathlineto{\pgfqpoint{2.205494in}{0.724087in}}% -\pgfpathlineto{\pgfqpoint{2.205494in}{1.788459in}}% +\pgfpathlineto{\pgfqpoint{2.205368in}{0.724087in}}% +\pgfpathlineto{\pgfqpoint{2.205368in}{1.788459in}}% \pgfpathlineto{\pgfqpoint{2.189776in}{1.788459in}}% \pgfpathlineto{\pgfqpoint{2.189776in}{0.724087in}}% \pgfpathclose% @@ -114,11 +114,11 @@ \pgfsetstrokecolor{currentstroke}% \pgfsetstrokeopacity{0.200000}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{2.205494in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{3.777315in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{3.777315in}{0.724087in}}% -\pgfpathlineto{\pgfqpoint{2.205494in}{0.724087in}}% 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29c6f47..cfa16ea 100644 --- a/buch/papers/ellfilter/python/elliptic2.py +++ b/buch/papers/ellfilter/python/elliptic2.py @@ -1,5 +1,6 @@ # %% +import enum import matplotlib.pyplot as plt import scipy.signal import numpy as np @@ -8,7 +9,9 @@ from matplotlib.patches import Rectangle import plot_params -def ellip_filter(N): +N=5 + +def ellip_filter(N, mode=-1): order = N passband_ripple_db = 3 @@ -26,7 +29,16 @@ def ellip_filter(N): fs=None ) - w, mag_db, phase = scipy.signal.bode((a, b), w=np.linspace(0*omega_c,2*omega_c, 4000)) + if mode == 0: + w = np.linspace(0*omega_c,omega_c, 2000) + elif mode == 1: + w = np.linspace(omega_c,1.00992*omega_c, 2000) + elif mode == 2: + w = np.linspace(1.00992*omega_c,2*omega_c, 2000) + else: + w = np.linspace(0*omega_c,2*omega_c, 4000) + + w, mag_db, phase = scipy.signal.bode((a, b), w=w) mag = 10**(mag_db/20) @@ -40,9 +52,9 @@ def ellip_filter(N): plt.figure(figsize=(4,2.5)) -for N in [5]: - w, FN2, mag, a, b = ellip_filter(N) - plt.semilogy(w, FN2, label=f"$N={N}, k=0.1$", linewidth=1) +for mode, c in enumerate(["green", "orange", "red"]): + w, FN2, mag, a, b = ellip_filter(N, mode=mode) + plt.semilogy(w, FN2, label=f"$N={N}, k=0.1$", linewidth=1, color=c) plt.gca().add_patch(Rectangle( (0, 0), @@ -53,21 +65,21 @@ plt.gca().add_patch(Rectangle( )) plt.gca().add_patch(Rectangle( (1, 1), - 0.01, 1e2-1, + 0.00992, 1e2-1, fc ='orange', alpha=0.2, lw = 10, )) plt.gca().add_patch(Rectangle( - (1.01, 100), + (1.00992, 100), 1, 1e6, fc ='red', alpha=0.2, lw = 10, )) -zeros = [0,0.87,1] +zeros = [0,0.87,0.995] poles = [1.01,1.155] import matplotlib.transforms @@ -99,7 +111,7 @@ plt.ylim([1e-4,1e6]) plt.grid() plt.xlabel("$w$") plt.ylabel("$F^2_N(w)$") -plt.legend() +# plt.legend() plt.tight_layout() plt.savefig("F_N_elliptic.pgf") plt.show() @@ -107,7 +119,9 @@ plt.show() plt.figure(figsize=(4,2.5)) -plt.plot(w, mag, linewidth=1) +for mode, c in enumerate(["green", "orange", "red"]): + w, FN2, mag, a, b = ellip_filter(N, mode=mode) + plt.plot(w, mag, linewidth=1, color=c) plt.gca().add_patch(Rectangle( (0, np.sqrt(2)/2), @@ -118,14 +132,14 @@ plt.gca().add_patch(Rectangle( )) plt.gca().add_patch(Rectangle( (1, 0.1), - 0.01, np.sqrt(2)/2 - 0.1, + 0.00992, np.sqrt(2)/2 - 0.1, fc ='orange', alpha=0.2, lw = 10, )) plt.gca().add_patch(Rectangle( - (1.01, 0), + (1.00992, 0), 1, 0.1, fc ='red', alpha=0.2, diff --git a/buch/papers/ellfilter/python/k.pgf b/buch/papers/ellfilter/python/k.pgf index 95d61d4..52dd705 100644 --- a/buch/papers/ellfilter/python/k.pgf +++ b/buch/papers/ellfilter/python/k.pgf @@ -320,7 +320,7 @@ \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% -\pgfsetlinewidth{0.100375pt}% +\pgfsetlinewidth{1.003750pt}% \definecolor{currentstroke}{rgb}{0.121569,0.466667,0.705882}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% @@ -434,7 +434,7 @@ \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% -\pgfsetlinewidth{0.100375pt}% +\pgfsetlinewidth{1.003750pt}% \definecolor{currentstroke}{rgb}{1.000000,0.498039,0.054902}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% diff --git a/buch/papers/ellfilter/tikz/arccos.tikz.tex b/buch/papers/ellfilter/tikz/arccos.tikz.tex index 2772620..987f885 100644 --- a/buch/papers/ellfilter/tikz/arccos.tikz.tex +++ b/buch/papers/ellfilter/tikz/arccos.tikz.tex @@ -10,10 +10,10 @@ \clip(-7.5,-2) rectangle (7.5,2); - \draw[thick, ->, darkgreen] (0, 0) -- (0,1.5); - \draw[thick, ->, orange] (1, 0) -- (0,0); - \draw[thick, ->, red] (2, 0) -- (1,0); - \draw[thick, ->, blue] (2,1.5) -- (2, 0); + \draw[ultra thick, ->, darkgreen] (0, 0) -- (0,1.5); + \draw[ultra thick, ->, orange] (1, 0) -- (0,0); + \draw[ultra thick, ->, red] (2, 0) -- (1,0); + \draw[ultra thick, ->, blue] (2,1.5) -- (2, 0); \foreach \i in {-2,...,1} { \begin{scope}[opacity=0.5, xshift=\i*4cm] @@ -45,7 +45,7 @@ \end{scope} - \begin{scope}[yshift=-2.5cm] + \begin{scope}[yshift=-3cm] \draw[gray, ->] (-5,0) -- (5,0) node[anchor=west]{$w$}; diff --git a/buch/papers/ellfilter/tikz/cd.tikz.tex b/buch/papers/ellfilter/tikz/cd.tikz.tex index 7155a85..7a2767b 100644 --- a/buch/papers/ellfilter/tikz/cd.tikz.tex +++ b/buch/papers/ellfilter/tikz/cd.tikz.tex @@ -4,7 +4,7 @@ \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}} - \begin{scope}[xscale=1, yscale=2] + \begin{scope}[xscale=0.9, yscale=1.8] \draw[gray, ->] (0,-1.5) -- (0,1.5) node[anchor=south]{$\mathrm{Im}~z$}; \draw[gray, ->] (-5,0) -- (5,0) node[anchor=west]{$\mathrm{Re}~z$}; @@ -23,12 +23,12 @@ \fill[yellow!30] (0,0) rectangle (1, 0.5); - \draw[thick, ->, darkgreen] (0, 0) -- (0,0.5); - \draw[thick, ->, orange] (1, 0) -- (0,0); - \draw[thick, ->, red] (2, 0) -- (1,0); - \draw[thick, ->, blue] (2,0.5) -- (2, 0); - \draw[thick, ->, purple] (1, 0.5) -- (2,0.5); - \draw[thick, ->, cyan] (0, 0.5) -- (1,0.5); + \draw[ultra thick, ->, darkgreen] (0, 0) -- (0,0.5); + \draw[ultra thick, ->, orange] (1, 0) -- (0,0); + \draw[ultra thick, ->, red] (2, 0) -- (1,0); + \draw[ultra thick, ->, blue] (2,0.5) -- (2, 0); + \draw[ultra thick, ->, purple] (1, 0.5) -- (2,0.5); + \draw[ultra thick, ->, cyan] (0, 0.5) -- (1,0.5); @@ -63,7 +63,7 @@ \end{scope} - \begin{scope}[yshift=-3.5cm, xscale=0.75] + \begin{scope}[yshift=-4cm, xscale=0.75] \draw[gray, ->] (-6,0) -- (6,0) node[anchor=west]{$w$}; diff --git a/buch/papers/ellfilter/tikz/cd2.tikz.tex b/buch/papers/ellfilter/tikz/cd2.tikz.tex index 0743f7d..425db95 100644 --- a/buch/papers/ellfilter/tikz/cd2.tikz.tex +++ b/buch/papers/ellfilter/tikz/cd2.tikz.tex @@ -5,9 +5,9 @@ \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}} - \begin{scope}[xscale=1.25, yscale=2.5] + \begin{scope}[xscale=1.25, yscale=3.5] - \draw[gray, ->] (0,-0.75) -- (0,1.25) node[anchor=south]{$\mathrm{Im}~z_1$}; + \draw[gray, ->] (0,-0.55) -- (0,1.05) node[anchor=south]{$\mathrm{Im}~z_1$}; \draw[gray, ->] (-1.5,0) -- (6,0) node[anchor=west]{$\mathrm{Re}~z_1$}; \draw[gray] ( 1,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $K_1$}; @@ -35,12 +35,12 @@ % \node[] at (2.5, 0.25) {\small $N=3$}; \fill[orange!30] (0,0) rectangle (5, 0.5); - \fill[yellow!30] (0,0) rectangle (1, 0.5); + % \fill[yellow!30] (0,0) rectangle (1, 0.1); \node[] at (2.5, 0.25) {\small $N=5$}; \draw[decorate,decoration={brace,amplitude=3pt,mirror}, yshift=0.05cm] - (5,0.5) node(t_k_unten){} -- node[above, yshift=0.1cm]{$NK$} + (5,0.5) node(t_k_unten){} -- node[above, yshift=0.1cm]{$NK_1$} (0,0.5) node(t_k_opt_unten){}; \draw[decorate,decoration={brace,amplitude=3pt,mirror}, xshift=0.1cm] @@ -63,9 +63,9 @@ - \draw[thick, ->, darkgreen] (5, 0) -- node[yshift=-0.5cm]{Durchlassbereich} (0,0); - \draw[thick, ->, orange] (-0, 0) -- node[align=center]{Übergangs-\\berech} (0,0.5); - \draw[thick, ->, red] (0,0.5) -- node[align=center, yshift=0.5cm]{Sperrbereich} (5, 0.5); + \draw[ultra thick, ->, darkgreen] (5, 0) -- node[yshift=-0.5cm]{Durchlassbereich} (0,0); + \draw[ultra thick, ->, orange] (-0, 0) -- node[align=center]{Übergangs-\\berech} (0,0.5); + \draw[ultra thick, ->, red] (0,0.5) -- node[align=center, yshift=0.7cm]{Sperrbereich} (5, 0.5); \draw (4,0 ) node[dot]{} node[anchor=south] {\small $1$}; \draw (2,0 ) node[dot]{} node[anchor=south] {\small $-1$}; diff --git a/buch/papers/ellfilter/tikz/cd3.tikz.tex b/buch/papers/ellfilter/tikz/cd3.tikz.tex new file mode 100644 index 0000000..fa9cc08 --- /dev/null +++ b/buch/papers/ellfilter/tikz/cd3.tikz.tex @@ -0,0 +1,84 @@ +\begin{tikzpicture}[>=stealth', auto, node distance=2cm, scale=1.2] + + \tikzstyle{zero} = [draw, circle, inner sep =0, minimum height=0.15cm] + \tikzstyle{dot} = [fill, circle, inner sep =0, minimum height=0.1cm] + + \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}} + + \begin{scope}[xscale=1.25, yscale=2.5] + + \draw[gray, ->] (0,-0.55) -- (0,1.05) node[anchor=south]{$\mathrm{Im}$}; + \draw[gray, ->] (-1.5,0) -- (6,0) node[anchor=west]{$\mathrm{Re}$}; + + % \draw[gray] ( 1,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $K_1$}; + % \draw[gray] ( 5,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $5K_1$}; + % \draw[gray] (0, 0.5) +(0.1, 0) -- +(-0.1, 0) node[inner sep=0, anchor=east]{\small $jK^\prime_1$}; + + \begin{scope} + + \clip(-1.5,-0.75) rectangle (6.8,1.25); + + % \draw[>->, line width=0.05, thick, blue] (1, 0.45) -- (2, 0.45) -- (2, 0.05) -- ( 0.1, 0.05) -- ( 0.1,0.45) -- (1, 0.45); + % \draw[>->, line width=0.05, thick, orange] (2, 0.5 ) -- (4, 0.5 ) -- (4, 0 ) -- ( 0 , 0 ) -- ( 0 ,0.5 ) -- (2, 0.5 ); + % \draw[>->, line width=0.05, thick, red] (3, 0.55) -- (6, 0.55) -- (6,-0.05) -- (-0.1,-0.05) -- (-0.1,0.55) -- (3, 0.55); + % \node[blue] at (1, 0.25) {$N=1$}; + % \node[orange] at (3, 0.25) {$N=2$}; + % \node[red] at (5, 0.25) {$N=3$}; + + + + % \draw[line width=0.1cm, fill, red!50] (0,0) rectangle (3, 0.5); + % \draw[line width=0.05cm, fill, orange!50] (0,0) rectangle (2, 0.5); + % \fill[yellow!50] (0,0) rectangle (1, 0.5); + % \node[] at (0.5, 0.25) {\small $N=1$}; + % \node[] at (1.5, 0.25) {\small $N=2$}; + % \node[] at (2.5, 0.25) {\small $N=3$}; + + % \fill[orange!30] (0,0) rectangle (5, 0.5); + \fill[yellow!30] (0,0) rectangle (1, 0.5); + \node[] at (2.5, 0.25) {\small $N=5$}; + + + % \draw[decorate,decoration={brace,amplitude=3pt,mirror}, yshift=0.05cm] + % (5,0.5) node(t_k_unten){} -- node[above, yshift=0.1cm]{$NK_1$} + % (0,0.5) node(t_k_opt_unten){}; + + % \draw[decorate,decoration={brace,amplitude=3pt,mirror}, xshift=0.1cm] + % (5,0) node(t_k_unten){} -- node[right, xshift=0.1cm]{$K^\prime \frac{K_1N}{K} = K^\prime_1$} + % (5,0.5) node(t_k_opt_unten){}; + + \foreach \i in {-2,...,1} { + \foreach \j in {-2,...,1} { + \begin{scope}[xshift=\i*4cm, yshift=\j*1cm] + + \node[zero] at ( 1, 0) {}; + \node[zero] at ( 3, 0) {}; + \node[pole] at ( 1,0.5) {}; + \node[pole] at ( 3,0.5) {}; + + \end{scope} + } + } + + + + + \draw[ultra thick, ->, darkgreen] (5, 0) -- node[yshift=-0.4cm]{Durchlassbereich} (0,0); + \draw[ultra thick, ->, orange] (-0, 0) -- node[align=center]{Übergangs-\\berech} (0,0.5); + \draw[ultra thick, ->, red] (0,0.5) -- node[align=center, yshift=0.4cm]{Sperrbereich} (5, 0.5); + + \draw (4,0 ) node[dot]{} node[anchor=south] {\small $1$}; + \draw (2,0 ) node[dot]{} node[anchor=south] {\small $-1$}; + \draw (0,0 ) node[dot]{} node[anchor=south west] {\small $1$}; + \draw (0,0.5) node[dot]{} node[anchor=north west] {\small $1/k$}; + \draw (2,0.5) node[dot]{} node[anchor=north] {\small $-1/k$}; + \draw (4,0.5) node[dot]{} node[anchor=north] {\small $1/k$}; + + + + \end{scope} + + + \end{scope} + +\end{tikzpicture} \ No newline at end of file diff --git a/buch/papers/ellfilter/tikz/elliptic_transform.tikz.tex b/buch/papers/ellfilter/tikz/elliptic_transform.tikz.tex new file mode 100644 index 0000000..c91ecf1 --- /dev/null +++ b/buch/papers/ellfilter/tikz/elliptic_transform.tikz.tex @@ -0,0 +1,64 @@ + +\def\d{0.2} +\def\n{3} +\def\nn{2} +\def\a{2.5} + +\begin{tikzpicture}[>=stealth', auto, node distance=2cm, scale=1.2] + + \tikzstyle{zero} = [draw, circle, inner sep =0, minimum height=0.15cm] + \tikzstyle{dot} = [fill, circle, inner sep =0, minimum height=0.1cm] + + \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}} + + \begin{scope}[xscale=3, yscale=3] + + \begin{scope}[] + + \fill[orange!30, scale=1.735] (0,0) rectangle (\d*\a+0.5, \d/\a+0.5); + \fill[yellow!30] (0,0) rectangle (\d*\a+0.5, \d/\a+0.5); + + \begin{scope}[scale=1.735, red] + \draw (0,0) rectangle (\d*\a/\n+0.5/\n, \d/\a+0.5); + \draw[gray] (0,0) -- (\d*\a/\n+0.5/\n, \d/\a+0.5); + + \node[zero] at ( \d*\a/\n+0.5/\n, \d/\a+0.5) {}; + \node[pole, color=red] at ( \d*\a/\n+0.5/\n, 0) {}; + + + \draw[] ( \d*\a/\n+0.5/\n,0) node[anchor=north] {\small $K_1$}; + \draw[] (0, \d/\a+0.5) node[anchor=east]{\small $jK_1^\prime$}; + + \end{scope} + + \begin{scope}[blue] + \draw[] (0,0) rectangle (\d*\a+0.5, \d/\a+0.5); + \foreach \i in {1,...,\nn} { + \draw[gray, dotted] (\i*\d*\a/\n+\i*0.5/\n, 0) -- (\i*\d*\a/\n+\i*0.5/\n, \d/\a+0.5); + } + + \node[zero] at ( \d*\a+0.5, \d/\a+0.5) {}; + \node[pole, color=blue] at ( \d*\a+0.5, 0) {}; + + \draw[] ( \d*\a+0.5,0) node[anchor=north] {\small $K$}; + \draw[] (0, \d/\a+0.5) node[anchor=east]{\small $jK^\prime$}; + + \node[dot, gray] at (\d*\a/\n+0.5/\n, \d/\a+0.5) {}; + \node[above] at (0.5*\d*\a/\n+0.5*0.5/\n, \d/\a+0.5) {\small $K/N$}; + + \end{scope} + + \draw[thick, gray, ->] (0,-0.25) -- (0,1.25) node[anchor=south]{$\mathrm{Im}$}; + \draw[thick, gray, ->] (-0.25,0) -- (2,0) node[anchor=west]{$\mathrm{Re}$}; + + \begin{scope}[] + \clip(0,0) rectangle (2,1.25); + \draw[scale=1, domain=0.1:10, variable=\x, smooth, samples=200] plot ({\d*\x1+0.5}, {\d/\x+0.5}); + + \end{scope} + \end{scope} + + +\end{scope} + +\end{tikzpicture} diff --git a/buch/papers/ellfilter/tikz/sn.tikz.tex b/buch/papers/ellfilter/tikz/sn.tikz.tex index 87c63c0..6ced3c5 100644 --- a/buch/papers/ellfilter/tikz/sn.tikz.tex +++ b/buch/papers/ellfilter/tikz/sn.tikz.tex @@ -4,7 +4,7 @@ \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}} - \begin{scope}[xscale=1, yscale=2] + \begin{scope}[xscale=0.9, yscale=1.8] \draw[gray, ->] (0,-1.5) -- (0,1.5) node[anchor=south]{$\mathrm{Im}~z$}; \draw[gray, ->] (-5,0) -- (5,0) node[anchor=west]{$\mathrm{Re}~z$}; @@ -17,12 +17,12 @@ \begin{scope}[xshift=-1cm] - \draw[thick, ->, darkgreen] (0, 0) -- (0,0.5); - \draw[thick, ->, orange] (1, 0) -- (0,0); - \draw[thick, ->, red] (2, 0) -- (1,0); - \draw[thick, ->, blue] (2,0.5) -- (2, 0); - \draw[thick, ->, purple] (1, 0.5) -- (2,0.5); - \draw[thick, ->, cyan] (0, 0.5) -- (1,0.5); + \draw[ultra thick, ->, darkgreen] (0, 0) -- (0,0.5); + \draw[ultra thick, ->, orange] (1, 0) -- (0,0); + \draw[ultra thick, ->, red] (2, 0) -- (1,0); + \draw[ultra thick, ->, blue] (2,0.5) -- (2, 0); + \draw[ultra thick, ->, purple] (1, 0.5) -- (2,0.5); + \draw[ultra thick, ->, cyan] (0, 0.5) -- (1,0.5); \foreach \i in {-2,...,2} { @@ -61,7 +61,7 @@ \end{scope} - \begin{scope}[yshift=-3.5cm, xscale=0.75] + \begin{scope}[yshift=-4cm, xscale=0.75] \draw[gray, ->] (-6,0) -- (6,0) node[anchor=west]{$w$}; -- cgit v1.2.1 From b3b175c9b728bb9e4224167ad91e34e6b3bd07f6 Mon Sep 17 00:00:00 2001 From: runterer Date: Thu, 2 Jun 2022 21:18:57 +0200 Subject: minor presentation improvements --- buch/papers/zeta/presentation/presentation.tex | 21 +++++++++++++-------- 1 file changed, 13 insertions(+), 8 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/zeta/presentation/presentation.tex b/buch/papers/zeta/presentation/presentation.tex index be3e12c..e106089 100644 --- a/buch/papers/zeta/presentation/presentation.tex +++ b/buch/papers/zeta/presentation/presentation.tex @@ -109,10 +109,10 @@ \end{frame} %Inhaltsverzeichnis - \begin{frame} - \frametitle{Inhalt} - \tableofcontents - \end{frame} +% \begin{frame} +% \frametitle{Inhalt} +% \tableofcontents +% \end{frame} \section{Motivation} @@ -187,14 +187,18 @@ \begin{align} \zeta(s) &= + \RD{ \sum_{n=1}^{\infty} \frac{1}{n^s} \label{zeta:align1} + } \\ \frac{1}{2^{s-1}} \zeta(s) &= + \BL{ \sum_{n=1}^{\infty} \frac{2}{(2n)^s} \label{zeta:align2} + } \end{align} \pause \eqref{zeta:align1} - \eqref{zeta:align2}: @@ -202,10 +206,10 @@ \left(1 - \frac{1}{2^{s-1}} \right) \zeta(s) &= - \frac{1}{1^s} - \underbrace{-\frac{2}{2^s} + \frac{1}{2^s}}_{-\frac{1}{2^s}} - + \frac{1}{3^s} - \underbrace{-\frac{2}{4^s} + \frac{1}{4^s}}_{-\frac{1}{4^s}} + \RD{\frac{1}{1^s}} + \underbrace{-\BL{\frac{2}{2^s}} + \RD{\frac{1}{2^s}}}_{-\frac{1}{2^s}} + + \RD{\frac{1}{3^s}} + \underbrace{-\BL{\frac{2}{4^s}} + \RD{\frac{1}{4^s}}}_{-\frac{1}{4^s}} \ldots \\ &= \eta(s) @@ -308,6 +312,7 @@ + \ldots \right) + \ldots \\ = 1 -- cgit v1.2.1 From 5e25727877da020b0b23132fec8c0ea70288a18b Mon Sep 17 00:00:00 2001 From: samuel niederer Date: Sun, 24 Jul 2022 12:03:25 +0200 Subject: add presentation files --- buch/papers/kra/images/simple_mass_spring.tex | 2 +- buch/papers/kra/presentation/presentation.tex | 491 ++++++++++++++++++++++++++ buch/papers/kra/scripts/animation.py | 243 +++++++++++++ buch/papers/kra/scripts/simulation.py | 40 +++ 4 files changed, 775 insertions(+), 1 deletion(-) create mode 100644 buch/papers/kra/presentation/presentation.tex create mode 100644 buch/papers/kra/scripts/animation.py create mode 100644 buch/papers/kra/scripts/simulation.py (limited to 'buch/papers') diff --git a/buch/papers/kra/images/simple_mass_spring.tex b/buch/papers/kra/images/simple_mass_spring.tex index 207f1e0..e0e869a 100644 --- a/buch/papers/kra/images/simple_mass_spring.tex +++ b/buch/papers/kra/images/simple_mass_spring.tex @@ -14,7 +14,7 @@ } \tikzmath{ - \hWall = 1.5; + \hWall = 1.2; \wWall = 0.3; \lWall = 3.5; \hMass = 0.6; diff --git a/buch/papers/kra/presentation/presentation.tex b/buch/papers/kra/presentation/presentation.tex new file mode 100644 index 0000000..eb6541b --- /dev/null +++ b/buch/papers/kra/presentation/presentation.tex @@ -0,0 +1,491 @@ +\documentclass[ngerman, aspectratio=169, xcolor={rgb}]{beamer} + +% style +\mode{ + \usetheme{Frankfurt} +} +%packages +\usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-} +\usepackage[english]{babel} +\usepackage{graphicx} +\usepackage{array} + +\newcolumntype{L}[1]{>{\raggedright\let\newline\\\arraybackslash\hspace{0pt}}m{#1}} +\usepackage{ragged2e} + +\usepackage{bm} % bold math +\usepackage{amsfonts} +\usepackage{amssymb} +\usepackage{mathtools} +\usepackage{amsmath} +\usepackage{multirow} % multi row in tables +\usepackage{booktabs} %toprule midrule bottomrue in tables +\usepackage{scrextend} +\usepackage{textgreek} +\usepackage[rgb]{xcolor} + +\usepackage[normalem]{ulem} % \sout + +\usepackage{ marvosym } % \Lightning + +\usepackage{multimedia} % embedded videos + +\usepackage{tikz} +\usepackage{pgf} +\usepackage{pgfplots} + +\usepackage{algorithmic} + +%citations +\usepackage[style=verbose,backend=biber]{biblatex} +\addbibresource{references.bib} + + +%math font +\usefonttheme[onlymath]{serif} + +%Beamer Template modifications +%\definecolor{mainColor}{HTML}{0065A3} % HSR blue +\definecolor{mainColor}{HTML}{D72864} % OST pink +\definecolor{invColor}{HTML}{28d79b} % OST pink +\definecolor{dgreen}{HTML}{38ad36} % Dark green + +%\definecolor{mainColor}{HTML}{000000} % HSR blue +\setbeamercolor{palette primary}{bg=white,fg=mainColor} +\setbeamercolor{palette secondary}{bg=orange,fg=mainColor} +\setbeamercolor{palette tertiary}{bg=yellow,fg=red} +\setbeamercolor{palette quaternary}{bg=mainColor,fg=white} %bg = Top bar, fg = active top bar topic +\setbeamercolor{structure}{fg=black} % itemize, enumerate, etc (bullet points) +\setbeamercolor{section in toc}{fg=black} % TOC sections +\setbeamertemplate{section in toc}[sections numbered] +\setbeamertemplate{subsection in toc}{% + \hspace{1.2em}{$\bullet$}~\inserttocsubsection\par} + +\setbeamertemplate{itemize items}[circle] +\setbeamertemplate{description item}[circle] +\setbeamertemplate{title page}[default][colsep=-4bp,rounded=true] +\beamertemplatenavigationsymbolsempty + +\setbeamercolor{footline}{fg=gray} +\setbeamertemplate{footline}{% + \hfill\usebeamertemplate***{navigation symbols} + \hspace{0.5cm} + \insertframenumber{}\hspace{0.2cm}\vspace{0.2cm} +} + +\usepackage{caption} +\captionsetup{labelformat=empty} + +%Title Page +\title{KRA} +\subtitle{Kalman Riccati Abel} +\author{Samuel Niederer} +% \institute{OST Ostschweizer Fachhochschule} +% \institute{\includegraphics[scale=0.3]{../img/ost_logo.png}} +\date{\today} + +\input{../packages.tex} + +\newcommand*{\QED}{\hfill\ensuremath{\blacksquare}}% + +\newcommand*{\HL}{\textcolor{mainColor}} +\newcommand*{\RD}{\textcolor{red}} +\newcommand*{\BL}{\textcolor{blue}} +\newcommand*{\GN}{\textcolor{dgreen}} +\newcommand{\dt}[0]{\frac{d}{dt}} + +\definecolor{darkgreen}{rgb}{0,0.6,0} + + +\makeatletter +\newcount\my@repeat@count +\newcommand{\myrepeat}[2]{% + \begingroup + \my@repeat@count=\z@ + \@whilenum\my@repeat@count<#1\do{#2\advance\my@repeat@count\@ne}% + \endgroup +} +\makeatother + +\usetikzlibrary{automata,arrows,positioning,calc,shapes.geometric, fadings} + +\begin{document} + +\begin{frame} + \titlepage +\end{frame} + +\begin{frame} + \frametitle{Content} + \tableofcontents +\end{frame} + +\section{Einführung} + +\begin{frame} + \begin{itemize} + \item<1|only@1> \textbf{K}alman + \item<1|only@1> \textbf{R}iccati + \item<1|only@1> \textbf{A}bel + + \item<2|only@2> \textcolor{red}{\sout{\textbf{K}alman}} + \item<2|only@2> \textbf{R}iccati + \item<2|only@2> \textbf{A}bel + + \item<3|only@3> \textcolor{red}{\sout{\textbf{K}alman}} \textcolor{green}{Federmassesytem} + \item<3|only@3> \textbf{R}iccati + \item<3|only@3> \textbf{A}bel + + \item<4|only@4> \textcolor{red}{\sout{\textbf{K}alman}} \textcolor{green}{Federmassesytem} + \item<4|only@4> \textbf{R}iccati + \item<4|only@4> \uwave{\textbf{A}bel} + \end{itemize} +\end{frame} + +\section{Riccati} + +\begin{frame} + \frametitle{Riccatische Differentialgleichung} + \begin{equation*} + % y'(x) = f(x)y^2(x) + g(x)y(x) + h(x) + x'(t) = f(t)x^2(t) + g(t)x(t) + h(t) + \end{equation*} + + \pause + + \begin{equation*} + \dot{X}(t) = - X(t)BX(t) - X(t)A + DX(t) + C + \end{equation*} + + % \pause + % Anwendungen + % \begin{itemize} + % \item Zeitkontinuierlicher Kalmanfilter + % \item Regelungstechnik LQ-Regler + % \item Federmassesyteme + % \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{Auftreten der Gleichung} + \begin{columns} + \column{0.4 \textwidth} + \begin{equation*} + \dt + \begin{pmatrix} + X \\ + Y + \end{pmatrix} + = + \underbrace{ + \begin{pmatrix} + A & B \\ + C & D + \end{pmatrix} + }_{H} + \begin{pmatrix} + X \\ + Y + \end{pmatrix} + \end{equation*} + + \pause + + \column{0.4 \textwidth} + \begin{equation*} + U = YX^{-1} \qquad \dt U = ? + \end{equation*} + \end{columns} + + \pause + + \begin{align*} + \dt U & = \dot{Y} X^{-1} + Y \dt X^{-1} \\ + \uncover<4->{ & = (CX + DY) X^{-1} - Y (X^{-1} \dot{X} X^{-1})\\} + \uncover<5->{ & = C\underbrace{XX^{-1}}_\text{I} + D\underbrace{YX^{-1}}_\text{U} - Y(X^{-1} (AX + BY) X^{-1})\\} + \uncover<6->{ & = C + DU - \underbrace{YX^{-1}}_\text{U}(A\underbrace{XX^{-1}}_\text{I} + B\underbrace{YX^{-1}}_\text{U})\\} + \uncover<7->{ & = C + DU - UA - UBU} + \end{align*} +\end{frame} + +\begin{frame} + \frametitle{Lösen der Gleichung} + \begin{equation*} + \begin{pmatrix} + X(t) \\ + Y(t) + \end{pmatrix} + = + \Phi(t_0, t) + \begin{pmatrix} + I(t) \\ + U_0(t) + \end{pmatrix} + = + \begin{pmatrix} + \Phi_{11}(t_0, t) & \Phi_{12}(t_0, t) \\ + \Phi_{21}(t_0, t) & \Phi_{22}(t_0, t) + \end{pmatrix} + \begin{pmatrix} + I(t) \\ + U_0(t) + \end{pmatrix} + \end{equation*} + + \pause + + \begin{equation*} + U(t) = + \begin{pmatrix} + \Phi_{21}(t_0, t) + \Phi_{22}(t_0, t) U_0(t) + \end{pmatrix} + \begin{pmatrix} + \Phi_{11}(t_0, t) + \Phi_{12}(t_0, t) U_0(t) + \end{pmatrix} + ^{-1} + \end{equation*} + + \pause + + % wobei $\Phi(t, t_0)$ die sogennante Zustandsübergangsmatrix ist. + + \begin{equation*} + \Phi(t_0, t) = e^{H(t - t_0)} + \end{equation*} +\end{frame} + +\section{Federmassystem} +\begin{frame} + \frametitle{Federmassesystem} + \begin{columns} + \column{0.5 \textwidth} + \input{../images/simple_mass_spring.tex} + + \column{0.5 \textwidth} + \begin{align*} + \uncover<2->{F_R & = k \Delta_x \\} + \uncover<3->{F_a & = am = \ddot{x} m \\} + \uncover<4->{F_R & = F_a \Leftrightarrow k \Delta_x = \ddot{x} m\\} + \uncover<5->{\ddot{x} & = \frac{k \Delta_x}{m} \\} + \uncover<6->{x(t) & = A \cos(\omega_0 + \Phi), \quad \omega_0 = \sqrt{\frac{k}{m}}} + \end{align*} + \end{columns} +\end{frame} + +\begin{frame} + \frametitle{Phasenraum} + \begin{columns} + \column{0.3 \textwidth} + \begin{tikzpicture}[scale=3] + \draw[->, thick] (0, 0) -- (1,0) node[right] {$q$}; + \draw[->, thick] (0.5, -0.5) -- (0.5,0.5) node[above]{$p$}; + \end{tikzpicture} + \column{0.7 \textwidth} + Impulskoordinaten $p = (p_{1}, p_{2}, ..., p_{n}), \quad p=mv$ \\ + Ortskoordinaten $q = (q_{1}, q_{2}, ..., q_{n})$ \\ + + + + \begin{align*} + \uncover<2->{\mathcal{H}(q, p) & = \underbrace{T(p)}_{E_{kin}} + \underbrace{V(q)}_{E_{pot}} = E_{tot} \\} + \uncover<3->{ & = \frac{p^2}{2m}+ \frac{k q^2}{2}} + \end{align*} + + + + \begin{equation*} + \uncover<4->{ + \dot{q_{k}} = \frac{\partial \mathcal{H}}{\partial p_k} + \qquad + \dot{p_{k}} = -\frac{\partial \mathcal{H}}{\partial q_k} + } + \end{equation*} + + \pause + + \begin{equation*} + \uncover<5->{ + \begin{pmatrix} + \dot{q} \\ + \dot{p} + \end{pmatrix} + = + \begin{pmatrix} + 0 & \frac{1}{m} \\ + -k & 0 + \end{pmatrix} + \begin{pmatrix} + q \\ + p + \end{pmatrix} + } + \end{equation*} + + \end{columns} +\end{frame} + +\begin{frame} + \frametitle{Phasenraum} + \input{../images/phase_space.tex} +\end{frame} + +\begin{frame} + \frametitle{Federmassesystem} + \begin{columns} + \column{0.6 \textwidth} + \scalebox{0.8}{\input{../images/multi_mass_spring.tex}} + \begin{align*} + \uncover<2->{\mathcal{H} & = T + V \\} + \uncover<7->{ & = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2}} + \end{align*} + + \column{0.4 \textwidth} + \begin{align*} + \uncover<3->{T & = T_1 + T_2} \\ + \uncover<5->{ & = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} } \\ + \uncover<4->{V & = V_1 + V_c + V_2 } \\ + \uncover<6->{ & = \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2}} + \end{align*} + \end{columns} +\end{frame} + +\begin{frame} + \frametitle{Federmassesystem} + \begin{equation*} + \begin{pmatrix} + \dot{q_1} \\ + \dot{q_2} \\ + \dot{p_1} \\ + \dot{p_2} \\ + \end{pmatrix} + = + \begin{pmatrix} + 0 & 0 & \frac{1}{2m_1} & 0 \\ + 0 & 0 & 0 & \frac{1}{2m_2} \\ + -(k_1 + k_c) & k_c & 0 & 0 \\ + k_c & -(k_c + k_2) & 0 & 0 \\ + \end{pmatrix} + \begin{pmatrix} + q_1 \\ + q_2 \\ + p_1 \\ + p_2 \\ + \end{pmatrix} + \Leftrightarrow + \dt + \begin{pmatrix} + Q \\ + P \\ + \end{pmatrix} + \underbrace{ + \begin{pmatrix} + 0 & M \\ + K & 0 + \end{pmatrix} + }_{H} + \begin{pmatrix} + Q \\ + P \\ + \end{pmatrix} + \end{equation*} + + \pause + + $U = PQ^{-1} \qquad \dt U = ?$ + + \pause + + \begin{align*} + \dt U & = C + DU - UA - UBU \\ + & = K - UMU + \end{align*} + +\end{frame} + +\begin{frame} + \frametitle{Einfluss der Anfangsbedingung:} + \begin{columns} + \column{0.4 \textwidth} + \begin{equation*} + \uncover<2->{q_0 = + \begin{pmatrix} + q_{10} \\ + q_{20} + \end{pmatrix} + = + \begin{pmatrix} + 3 \\ + 1 + \end{pmatrix} + } + \end{equation*} + \begin{equation*} + \uncover<3->{q_0 = + \begin{pmatrix} + q_{10} \\ + q_{20} + \end{pmatrix} + = + \begin{pmatrix} + 3 \\ + 3 + \end{pmatrix} + } + \end{equation*} + \begin{equation*} + \uncover<4->{q_0 = + \begin{pmatrix} + q_{10} \\ + q_{20} + \end{pmatrix} + = + \begin{pmatrix} + 2 \\ + -2 + \end{pmatrix} + } + \end{equation*} + \column{0.6 \textwidth} + \scalebox{0.8}{\input{../images/multi_mass_spring.tex}} + \end{columns} +\end{frame} + +\section{Schlussteil} +\begin{frame} + \frametitle{Zusammenfassung} + \begin{itemize} + \pause + \item{Riccatische Differentialgleichung} + \pause + \begin{itemize} + \item{Ausgansgleichung} + \pause + \item{Lösung} + \end{itemize} + \pause + \item{Harmonischer Ozillator} + \pause + \begin{itemize} + \item{Hamiltonfunktion} + \pause + \item{Phasenraum} + \end{itemize} + \pause + \item{Gekoppelter harmonischer Ozillator} + \pause + \begin{itemize} + \item{Riccatische Differentialgleichung} + \pause + \item{Einfluss der Anfangsbedingungen} + \end{itemize} + \pause + \item{\uwave{Abel}} + \begin{itemize} + \pause + \item{Nichtlineare Federkonstante} + \end{itemize} + + \end{itemize} +\end{frame} + +\end{document} diff --git a/buch/papers/kra/scripts/animation.py b/buch/papers/kra/scripts/animation.py new file mode 100644 index 0000000..5e805ae --- /dev/null +++ b/buch/papers/kra/scripts/animation.py @@ -0,0 +1,243 @@ +import numpy as np +import matplotlib.pyplot as plt +import matplotlib.patches +import matplotlib.transforms +import matplotlib.text +from matplotlib.animation import FuncAnimation +import imageio + +from simulation import Simulation + + +class Mass: + def __init__(self, x_0, width, height, **kwargs): + self._x_0 = x_0 + xy = (x_0, 0) + self._rect = matplotlib.patches.Rectangle(xy, width, height, **kwargs) + + @property + def patch(self): + return self._rect + + @property + def x(self): + return self._rect.get_x() + + @property + def width(self): + return self._rect.get_width() + + def move(self, x): + self._rect.set_x(self._x_0 + x) + + +class Spring: + def __init__(self, n, height, ax, resolution=1000, **kwargs): + self._n = n + self._height = height + self._N = resolution + (self._line,) = ax.plot([], [], "-", **kwargs) + + def set(self, x_0, x_1): + T = (x_1 - x_0) / self._n + x = np.linspace(x_0, x_1, self._N, endpoint=True) + t = np.linspace(0, x_1 - x_0, self._N) + y = (np.sin(2 * np.pi * t / T) + 1.5) * self._height / 2 + self.line.set_data(x, y) + + @property + def line(self): + return self._line + + +class LinePlot: + def __init__(self, ax, **kwargs): + (self._line,) = ax.plot([], [], "-", **kwargs) + self._x = [] + self._y = [] + + @property + def line(self): + return self._line + + def update(self, x, y): + self._x.append(x) + self._y.append(y) + self._line.set_data(self._x, self._y) + + +class ScatterPlot: + def __init__(self, ax, **kwargs): + self._color = kwargs.get("color", "tab:green") + self._line = ax.scatter([], [], **kwargs) + self._ax = ax + self._x = [] + self._y = [] + + @property + def line(self): + return self._line + + def update(self, x, y, **kwargs): + self._x.append(x) + self._y.append(y) + self._line.remove() + self._line = self._ax.scatter(self._x, self._y, color=self._color, **kwargs) + + +class QuiverPlot: + def __init__(self, ax, **kwargs): + self.x = [] + self.y = [] + self.u = [] + self.v = [] + self.ax = ax + self.ln = self.ax.quiver([], [], [], []) + + def update(self, x, y, u, v): + self.x.append(x) + self.y.append(y) + self.u.append(u) + self.v.append(v) + self.ln.remove() + self.ln = self.ax.quiver(self.x, self.y, self.u, self.v) + + @property + def line(self): + return self.ln + + +anim_folder = "anim_0" +img_counter = 0 + +sim = Simulation() +params = { + "x_0": [2, -2], + "k_1": 1, + "k_c": 2, + "k_2": 1, + "m_1": 0.5, + "m_2": 0.5, +} + +time = 2.1 + + +# create axis +fig = plt.figure(figsize=(20, 15), constrained_layout=True) +fig.suptitle( + " ,".join([f"${key} = {val}$" for (key, val) in params.items()]), fontsize=20 +) +spec = fig.add_gridspec(3, 4) +ax0 = fig.add_subplot(spec[-1, :]) +ax1 = fig.add_subplot(spec[:-1, :2]) +ax2 = fig.add_subplot(spec[:-1, 2:]) + +ax0.set_yticks([]) + +mass_height = 0.5 +spring_height = 0.6 * mass_height +x_max = 21 +y_max = 2 * mass_height + +mass_1 = Mass( + 7, + 2, + mass_height, + color="tab:red", +) +mass_2 = Mass(14, 2, mass_height, color="tab:blue") +masses = [mass_1, mass_2] +patches = [mass.patch for mass in masses] + +spring_1 = Spring(4, spring_height, ax0, color="tab:red", linewidth=10) +spring_2 = Spring(4, spring_height, ax0, color="tab:gray", linewidth=10) +spring_3 = Spring(4, spring_height, ax0, color="tab:blue", linewidth=10) +springs = [spring_1, spring_2, spring_3] + +linePlot_1 = LinePlot(ax1, color="tab:red", label="$m_1$", alpha=1) +linePlot_2 = LinePlot(ax1, color="tab:blue", label="$m_2$", alpha=1) +linePlots = [linePlot_1, linePlot_2] + +# quiverPlot = QuiverPlot(ax2) +scatterPlot = ScatterPlot(ax2) + +lines = [spring.line for spring in springs] +lines.extend([plot.line for plot in linePlots]) +# lines.append(quiverPlot.line) +lines.append(scatterPlot.line) + +objects = lines + patches + +ax0.plot( + np.repeat(mass_1.x, 2), + [0, y_max], + "--", + color="tab:red", + label="Ruhezustand $m_1$", +) +ax0.plot( + np.repeat(mass_2.x, 2), + [0, y_max], + "--", + color="tab:blue", + label="Ruhezustand $m_2$", +) + + +def init(): + ax0.set_xlim(0, x_max) + ax0.set_ylim(0, y_max) + + ax1.set_xlim(0, time) + ax1.set_ylim(-4, 4) + ax1.set_xlabel("time", fontsize=20) + ax1.set_ylabel("$q$", fontsize=20) + + ax2.set_xlim(-4, 4) + ax2.set_ylim(-4, 4) + ax2.set_xlabel("$q_1$", fontsize=20) + ax2.set_ylabel("$q_2$", fontsize=20) + + for patch in patches: + ax0.add_patch(patch) + + spring_1.set(0, mass_1.x) + spring_2.set(mass_1.x + mass_1.width, mass_2.x) + spring_2.set(mass_2.x + mass_2.width, x_max) + + return objects + + +def update(frame): + global img_counter + x_1, x_2 = sim(frame, **params) + + mass_1.move(x_1) + mass_2.move(x_2) + + spring_1.set(0, mass_1.x) + spring_2.set(mass_1.x + mass_1.width, mass_2.x) + spring_3.set(mass_2.x + mass_2.width, x_max) + + linePlot_1.update(frame, x_1) + linePlot_2.update(frame, x_2) + + scatterPlot.update(x_1, x_2, alpha=0.25) + + img_counter += 1 + return objects + + +anim = FuncAnimation( + fig, + update, + frames=np.linspace(0, time, int(time * 30)), + init_func=init, + blit=False, +) + +ax0.legend(fontsize=20) +ax1.legend(fontsize=20) +FFwriter = matplotlib.animation.FFMpegWriter(fps=30) +anim.save("animation.mp4", writer=FFwriter) diff --git a/buch/papers/kra/scripts/simulation.py b/buch/papers/kra/scripts/simulation.py new file mode 100644 index 0000000..8bccb6a --- /dev/null +++ b/buch/papers/kra/scripts/simulation.py @@ -0,0 +1,40 @@ +import sympy as sp + + +class Simulation: + def __init__(self): + self.k_1, self.k_2, self.k_c = sp.symbols("k_1 k_2 k_c") + self.m_1, self.m_2 = sp.symbols("m_1 m_2") + self.t = sp.symbols("t") + K = sp.Matrix( + [[-(self.k_1 + self.k_c), self.k_c], [self.k_c, -(self.k_2 + self.k_c)]] + ) + M = sp.Matrix([[1 / self.m_1, 0], [0, 1 / self.m_2]]) + A = M * K + + self.eigenvecs = [] + self.eigenvals = [] + for ev, mult, vecs in A.eigenvects(): + self.eigenvecs.append(sp.Matrix(vecs)) + self.eigenvals.extend([ev] * mult) + + def __call__(self, t, x_0, k_1, k_c, k_2, m_1, m_2): + params = { + self.k_1: k_1, + self.k_c: k_c, + self.k_2: k_2, + self.m_1: m_1, + self.m_2: m_2, + } + x_0 = sp.Matrix(x_0) + eig_mat = sp.Matrix.hstack(*self.eigenvecs).subs(params) + g = eig_mat.inv() * x_0 + L = sp.Matrix( + [ + g[0] * sp.cos(self.eigenvals[0].subs(params) * self.t), + g[1] * sp.cos(self.eigenvals[1].subs(params) * self.t), + ] + ) + x = eig_mat * L + f = sp.lambdify(self.t, x, "numpy") + return f(t).squeeze() -- cgit v1.2.1 From 433b838cd7a68bb45abf0023edfb5439097693f2 Mon Sep 17 00:00:00 2001 From: samuel niederer Date: Sun, 24 Jul 2022 12:03:46 +0200 Subject: update packages --- buch/papers/kra/packages.tex | 12 ++++++++++++ 1 file changed, 12 insertions(+) (limited to 'buch/papers') diff --git a/buch/papers/kra/packages.tex b/buch/papers/kra/packages.tex index df34dcf..b16f074 100644 --- a/buch/papers/kra/packages.tex +++ b/buch/papers/kra/packages.tex @@ -8,3 +8,15 @@ % following example %\usepackage{packagename} +\usepackage{physics} +\usepackage{pgfplots} +\usepackage{tikz} +\usepackage[outline]{contour} +\pgfplotsset{compat=1.16} +\usetikzlibrary{patterns} +\usetikzlibrary{snakes} +\usetikzlibrary{math} +\usetikzlibrary{arrows.meta} +\usetikzlibrary{decorations} +\usetikzlibrary{decorations.markings} +\usetikzlibrary{calc} \ No newline at end of file -- cgit v1.2.1 From c5a26d2d7bde694d08bff948c48b2615a7e2e973 Mon Sep 17 00:00:00 2001 From: samuel niederer Date: Sun, 24 Jul 2022 12:16:02 +0200 Subject: add current work --- buch/papers/kra/hamilton.tex | 185 +++++++++++++++++++++++++++++++++++++++++ buch/papers/kra/main.tex | 32 ++----- buch/papers/kra/references.bib | 41 ++++----- buch/papers/kra/riccati.tex | 93 +++++++++++++++++++++ buch/papers/kra/teil0.tex | 22 ----- buch/papers/kra/teil1.tex | 55 ------------ buch/papers/kra/teil2.tex | 40 --------- buch/papers/kra/teil3.tex | 40 --------- buch/papers/kra/test.tex | 12 +++ 9 files changed, 313 insertions(+), 207 deletions(-) create mode 100644 buch/papers/kra/hamilton.tex create mode 100644 buch/papers/kra/riccati.tex delete mode 100644 buch/papers/kra/teil0.tex delete mode 100644 buch/papers/kra/teil1.tex delete mode 100644 buch/papers/kra/teil2.tex delete mode 100644 buch/papers/kra/teil3.tex create mode 100644 buch/papers/kra/test.tex (limited to 'buch/papers') diff --git a/buch/papers/kra/hamilton.tex b/buch/papers/kra/hamilton.tex new file mode 100644 index 0000000..14a5e8c --- /dev/null +++ b/buch/papers/kra/hamilton.tex @@ -0,0 +1,185 @@ +\newcommand{\dt}[0]{\frac{d}{dt}} + +\section{Teil abc\label{kra:section:teilabc}} +\rhead{Teil abc} + +\subsection{Hamilton-Funktion} +Die Bewegung der Masse $m$ kann mit Hilfe der hamiltonschen Mechanik im Phasenraum untersucht werden. +Die hamiltonschen Gleichungen verwenden dafür die veralgemeinerten Ortskoordinaten +$q = (q_{1}, q_{2}, ..., q_{n})$ und die verallgemeinerten Impulskoordinaten $p = (p_{1}, p_{2}, ..., p_{n})$, +wobei der Impuls definiert ist als $p_k = m_k \cdot v_k$. +Liegen keine zeitabhängigen Zwangsbedingungen vor, so entspricht die Hamitlon-Funktion der Gesamtenergie des Systems \cite{kra:hamilton}. +Im Falle des einfachen Federmassesystems, Abbildung \ref{kra:fig:simple_spring_mass}, +setzt sich die Hamilton-Funktion aus kinetischer und potentieller Energie zusammen. + +\begin{equation} + \label{hamilton} + \begin{split} + \mathcal{H}(q, p) &= T(p) + V(q) = E \\ + &= \underbrace{\frac{p^2}{2m}}_{E_{kin}} + \underbrace{\frac{k q^2}{2}}_{E_{pot}} + \end{split} +\end{equation} + +Die Hamiltonschen Bewegungsgleichungen liefern \cite{kra:kanonischegleichungen} +\begin{equation} + \label{kra:hamilton:bewegungsgleichung} + \dot{q_{k}} = \frac{\partial \mathcal{H}}{\partial p_k} + \qquad + \dot{p_{k}} = -\frac{\partial \mathcal{H}}{\partial q_k} +\end{equation} + +daraus folgt + +\[ + \dot{q} = \frac{p}{m} + \qquad + \dot{p} = -kq +\] + +in Matrixschreibweise erhalten wir also + +\[ + \begin{pmatrix} + \dot{q} \\ + \dot{p} + \end{pmatrix} + = + \begin{pmatrix} + 0 & \frac{1}{m} \\ + -k & 0 + \end{pmatrix} + \begin{pmatrix} + q \\ + p + \end{pmatrix} +\] + +Für das erweiterte Federmassesystem, Abbildung \ref{kra:fig:multi_spring_mass}, können wir analog vorgehen. +Die kinetische Energie setzt sich nun aus den kinetischen Energien der einzelnen Massen $m_1$ und $m_2$ zusammen. +Die Potentielle Energie erhalten wir aus der Summe der kinetischen Energien der einzelnen Federn mit den Federkonstanten $k_1$, $k_c$ und $k_2$. + +\begin{align*} + \begin{split} + T &= T_1 + T_2 \\ + &= \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \end{split} + \\ + \begin{split} + V &= V_1 + V_c + V_2 \\ + &= \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2} + \end{split} +\end{align*} + +Die Hamilton-Funktion ist also + +\begin{align*} + \begin{split} + \mathcal{H} &= T + V \\ + &= \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2} + \end{split} +\end{align*} + +Die Bewegungsgleichungen \ref{kra:hamilton:bewegungsgleichung} liefern +\begin{align*} + \frac{\partial \mathcal{H}}{\partial p_k} & = \dot{q_k} + \Rightarrow + \left\{ + \begin{alignedat}{2} + \dot{q_1} &= \frac{2p_1}{2m_1} &&= \frac{p_1}{m_1}\\ + \dot{q_2} &= \frac{2p_2}{2m_2} &&= \frac{p_2}{m_2} + \end{alignedat} + \right. + \\ + -\frac{\partial \mathcal{H}}{\partial q_k} & = \dot{p_k} + \Rightarrow + \left\{ + \begin{alignedat}{2} + \dot{p_1} &= -(\frac{2k_1q_1}{2} - \frac{2k_c(q_2-q_1)}{2}) &&= -q_1(k_1+k_c) + q_2k_c \\ + \dot{p_1} &= -(\frac{2k_c(q_2-q_1)}{2} - \frac{2k_2q_2}{2}) &&= q_1k_c - (k_c + k_2) + \end{alignedat} + \right. +\end{align*} + +In Matrixschreibweise erhalten wir + +\begin{equation} + \label{kra:hamilton:multispringmass} + \begin{pmatrix} + \dot{q_1} \\ + \dot{q_2} \\ + \dot{p_1} \\ + \dot{p_2} \\ + \end{pmatrix} + = + \begin{pmatrix} + 0 & 0 & \frac{1}{2m_1} & 0 \\ + 0 & 0 & 0 & \frac{1}{2m_2} \\ + -(k_1 + k_c) & k_c & 0 & 0 \\ + k_c & -(k_c + k_2) & 0 & 0 \\ + \end{pmatrix} + \begin{pmatrix} + q_1 \\ + q_2 \\ + p_1 \\ + p_2 \\ + \end{pmatrix} + \Leftrightarrow + \dt + \begin{pmatrix} + Q \\ + P \\ + \end{pmatrix} + \underbrace{ + \begin{pmatrix} + 0 & M \\ + K & 0 + \end{pmatrix} + }_{G} + \begin{pmatrix} + Q \\ + P \\ + \end{pmatrix} +\end{equation} + + +Wir intressieren uns nun dafür wie der Phasenwinkel $U = PQ^{-1}$ von der Zeit abhängt, +wir suchen also die Grösse $\Theta = \dt U$. + +Ersetzten wir in der Gleichung \ref{kra:hamilton:multispringmass} die Matrix $G$ mit $\tilde{G}$ so erhalten wir +\begin{equation} + \dt + \begin{pmatrix} + Q \\ + P + \end{pmatrix} + = + \underbrace{ + \begin{pmatrix} + A & B \\ + C & D + \end{pmatrix} + }_{\tilde{G}} + \begin{pmatrix} + Q \\ + P + \end{pmatrix} +\end{equation} + +Mit einsetzten folgt + +\begin{align*} + \dot{Q} = AQ + BP \\ + \dot{P} = CQ + DP +\end{align*} +\begin{equation} + \begin{split} + \dt U &= \dot{P} Q^{-1} + P \dt Q^{-1} \\ + &= (CQ + DP) Q^{-1} - P (Q^{-1} \dot{Q} Q^{-1}) \\ + &= C\underbrace{QQ^{-1}}_\text{I} + D\underbrace{PQ^{-1}}_\text{U} - P(Q^{-1} (AQ + BP) Q^{-1}) \\ + &= C + DU - \underbrace{PQ^{-1}}_\text{U}(A\underbrace{QQ^{-1}}_\text{I} + B\underbrace{PQ^{-1}}_\text{U}) \\ + &= C + DU - UA - UBU + \end{split} +\end{equation} + +was uns auf die zeitkontinuierliche Matrix-Riccati-Gleichung führt. + diff --git a/buch/papers/kra/main.tex b/buch/papers/kra/main.tex index fcee25b..456b6ee 100644 --- a/buch/papers/kra/main.tex +++ b/buch/papers/kra/main.tex @@ -6,31 +6,9 @@ \chapter{Kalman, Riccati und Abel\label{chapter:kra}} \lhead{Kalman, Riccati und Abel} \begin{refsection} - \chapterauthor{Samuel Niederer} - - Ein paar Hinweise für die korrekte Formatierung des Textes - \begin{itemize} - \item - Absätze werden gebildet, indem man eine Leerzeile einfügt. - Die Verwendung von \verb+\\+ ist nur in Tabellen und Arrays gestattet. - \item - Die explizite Platzierung von Bildern ist nicht erlaubt, entsprechende - Optionen werden gelöscht. - Verwenden Sie Labels und Verweise, um auf Bilder hinzuweisen. - \item - Beginnen Sie jeden Satz auf einer neuen Zeile. - Damit ermöglichen Sie dem Versionsverwaltungssysteme, Änderungen - in verschiedenen Sätzen von verschiedenen Autoren ohne Konflikt - anzuwenden. - \item - Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren - Übersicht wegen, aber auch um GIT die Arbeit zu erleichtern. - \end{itemize} - - \input{papers/kra/teil0.tex} - \input{papers/kra/teil1.tex} - \input{papers/kra/teil2.tex} - \input{papers/kra/teil3.tex} - - \printbibliography[heading=subbibliography] + \chapterauthor{Samuel Niederer} + \input{papers/kra/hamilton.tex} + \newpage + \input{papers/kra/riccati.tex} + \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/kra/references.bib b/buch/papers/kra/references.bib index f13c3d8..7f972ec 100644 --- a/buch/papers/kra/references.bib +++ b/buch/papers/kra/references.bib @@ -4,32 +4,27 @@ % (c) 2020 Autor, Hochschule Rapperswil % -@online{kra:bibtex, - title = {BibTeX}, - url = {https://de.wikipedia.org/wiki/BibTeX}, - date = {2020-02-06}, - year = {2020}, - month = {2}, - day = {6} +@online{kra:hamilton, + title = {Hamilton-Funktion}, + url = {https://de.wikipedia.org/wiki/Hamilton-Funktion}, + date = {2022-05-26} } -@book{kra:numerical-analysis, - title = {Numerical Analysis}, - author = {David Kincaid and Ward Cheney}, - publisher = {American Mathematical Society}, - year = {2002}, - isbn = {978-8-8218-4788-6}, - inseries = {Pure and applied undegraduate texts}, - volume = {2} +@misc{kra:kanonischegleichungen, + title = {Kanonische Gleichungen}, + url = {https://de.wikipedia.org/wiki/Kanonische_Gleichungen}, + date = {2022-05-26} } -@article{kra:mendezmueller, - author = { Tabea Méndez and Andreas Müller }, - title = { Noncommutative harmonic analysis and image registration }, - journal = { Appl. Comput. Harmon. Anal.}, - year = 2019, - volume = 47, - pages = {607--627}, - url = {https://doi.org/10.1016/j.acha.2017.11.004} +@misc{kra:newton, + title = {Newtonsche Gesetze}, + url = {https://de.wikipedia.org/wiki/Newtonsche_Gesetze}, + date = {2022-05-26} } +@misc{kra:kalmanisae, + author = {D.Alazard}, + title = {Introduction to Kalman filtering}, + url = {https://pagespro.isae-supaero.fr/IMG/pdf/introKalman_e_151211.pdf}, + date = {2022-05-26} +} diff --git a/buch/papers/kra/riccati.tex b/buch/papers/kra/riccati.tex new file mode 100644 index 0000000..df2921d --- /dev/null +++ b/buch/papers/kra/riccati.tex @@ -0,0 +1,93 @@ +\section{Riccati + \label{kra:section:riccati}} +\rhead{Riccati} + +\begin{equation} + y'(x) = f(x)y^2(x) + g(x)y(x) + h(x) +\end{equation} +% einfache (normale riccati gleichung und ihre loesung) +% (kann man diese bei einfachem federmasse system benutzten?) +% matrix riccati gleichung + + +Die zeitkontinuierliche Riccati-Matrix-Gleichung hat die Form +\begin{equation} + \label{kra:riccati:riccatiequation} + \dot{U(t)} = DU(t) - UA(t) - U(t)BU(t) +\end{equation} + +Betrachten wir das Differentialgleichungssystem \ref{kra:riccati:derivation} + +\begin{equation} + \label{kra:riccati:derivation} + \dt + \begin{pmatrix} + X \\ + Y + \end{pmatrix} + = + \underbrace{ + \begin{pmatrix} + A & B \\ + C & D + \end{pmatrix} + }_{H} + \begin{pmatrix} + X \\ + Y + \end{pmatrix} +\end{equation} + +interessieren wir uns für die zeitliche Änderung der Grösse $U = YX^{-1}$, so erhalten wir durch einsetzten + +\begin{align*} + \dt U & = \dot{Y} X^{-1} + Y \dt X^{-1} \\ + & = (CX + DY) X^{-1} - Y (X^{-1} \dot{X} X^{-1}) \\ + & = C\underbrace{XX^{-1}}_\text{I} + D\underbrace{YX^{-1}}_\text{U} - Y(X^{-1} (AX + BY) X^{-1}) \\ + & = C + DU - \underbrace{YX^{-1}}_\text{U}(A\underbrace{XX^{-1}}_\text{I} + B\underbrace{YX^{-1}}_\text{U}) \\ + & = C + DU - UA - UBU +\end{align*} + +was uns auf die Riccati-Matrix-Gleichung \ref{kra:riccati:riccatiequation} führt. +Die Lösung dieser Gleichung erhalten wir nach \cite{kra:kalmanisae} folgendermassen +\begin{equation} + \begin{pmatrix} + X(t) \\ + Y(t) + \end{pmatrix} + = + \Phi(t_0, t) + \begin{pmatrix} + I(t) \\ + U_0(t) + \end{pmatrix} + = + \begin{pmatrix} + \Phi_{11}(t_0, t) & \Phi_{12}(t_0, t) \\ + \Phi_{21}(t_0, t) & \Phi_{22}(t_0, t) + \end{pmatrix} + \begin{pmatrix} + I(t) \\ + U_0(t) + \end{pmatrix} +\end{equation} + +\begin{equation} + U(t) = + \begin{pmatrix} + \Phi_{21}(t_0, t) + \Phi_{22}(t_0, t) + \end{pmatrix} + \begin{pmatrix} + \Phi_{11}(t_0, t) + \Phi_{12}(t_0, t) + \end{pmatrix} + ^{-1} +\end{equation} + +wobei $\Phi(t, t_0)$ die sogennante Zustandsübergangsmatrix ist. + +\begin{equation} + \Phi(t_0, t) = e^{H(t - t_0)} +\end{equation} + + + diff --git a/buch/papers/kra/teil0.tex b/buch/papers/kra/teil0.tex deleted file mode 100644 index d06a055..0000000 --- a/buch/papers/kra/teil0.tex +++ /dev/null @@ -1,22 +0,0 @@ -% -% einleitung.tex -- Beispiel-File für die Einleitung -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 0\label{kra:section:teil0}} -\rhead{Teil 0} -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua \cite{kra:bibtex}. -At vero eos et accusam et justo duo dolores et ea rebum. -Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum -dolor sit amet. - -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua. -At vero eos et accusam et justo duo dolores et ea rebum. Stet clita -kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit -amet. - - diff --git a/buch/papers/kra/teil1.tex b/buch/papers/kra/teil1.tex deleted file mode 100644 index 0c0977d..0000000 --- a/buch/papers/kra/teil1.tex +++ /dev/null @@ -1,55 +0,0 @@ -% -% teil1.tex -- Beispiel-File für das Paper -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 1 -\label{kra:section:teil1}} -\rhead{Problemstellung} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. -Nemo enim ipsam voluptatem quia voluptas sit aspernatur aut odit -aut fugit, sed quia consequuntur magni dolores eos qui ratione -voluptatem sequi nesciunt -\begin{equation} -\int_a^b x^2\, dx -= -\left[ \frac13 x^3 \right]_a^b -= -\frac{b^3-a^3}3. -\label{kra:equation1} -\end{equation} -Neque porro quisquam est, qui dolorem ipsum quia dolor sit amet, -consectetur, adipisci velit, sed quia non numquam eius modi tempora -incidunt ut labore et dolore magnam aliquam quaerat voluptatem. - -Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis -suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur? -Quis autem vel eum iure reprehenderit qui in ea voluptate velit -esse quam nihil molestiae consequatur, vel illum qui dolorem eum -fugiat quo voluptas nulla pariatur? - -\subsection{De finibus bonorum et malorum -\label{kra:subsection:finibus}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga \eqref{000tempmlate:equation1}. - -Et harum quidem rerum facilis est et expedita distinctio -\ref{kra:section:loesung}. -Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil -impedit quo minus id quod maxime placeat facere possimus, omnis -voluptas assumenda est, omnis dolor repellendus -\ref{kra:section:folgerung}. -Temporibus autem quibusdam et aut officiis debitis aut rerum -necessitatibus saepe eveniet ut et voluptates repudiandae sint et -molestiae non recusandae. -Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis -voluptatibus maiores alias consequatur aut perferendis doloribus -asperiores repellat. - - diff --git a/buch/papers/kra/teil2.tex b/buch/papers/kra/teil2.tex deleted file mode 100644 index 249f078..0000000 --- a/buch/papers/kra/teil2.tex +++ /dev/null @@ -1,40 +0,0 @@ -% -% teil2.tex -- Beispiel-File für teil2 -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 2 -\label{kra:section:teil2}} -\rhead{Teil 2} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{kra:subsection:bonorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. - - diff --git a/buch/papers/kra/teil3.tex b/buch/papers/kra/teil3.tex deleted file mode 100644 index 2515c7d..0000000 --- a/buch/papers/kra/teil3.tex +++ /dev/null @@ -1,40 +0,0 @@ -% -% teil3.tex -- Beispiel-File für Teil 3 -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 3 -\label{kra:section:teil3}} -\rhead{Teil 3} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{kra:subsection:malorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. - - diff --git a/buch/papers/kra/test.tex b/buch/papers/kra/test.tex new file mode 100644 index 0000000..ebe0aa0 --- /dev/null +++ b/buch/papers/kra/test.tex @@ -0,0 +1,12 @@ +\begin{figure} + \input{papers/kra/images/phase_space.tex} + % \begin{minipage}{.45\textwidth} + % \input{papers/kra/images/phase_space_small_omega.tex} + % \end{minipage} + % \begin{minipage}{.45\textwidth} + % \input{papers/kra/images/phase_space_large_omega.tex} + % \end{minipage} + % \begin{minipage}[.5\textwidth] + % \input{papers/kra/images/phase_space_large_omega.tex} + % \end{minipage} +\end{figure} \ No newline at end of file -- cgit v1.2.1 From 1d78360ee72a8d0d6cd4b440a2244624c284887f Mon Sep 17 00:00:00 2001 From: samuel niederer Date: Sun, 24 Jul 2022 17:12:49 +0200 Subject: update paper --- buch/papers/kra/Makefile.inc | 11 +- buch/papers/kra/anwendung.tex | 235 +++++++++++++++++++++++++++++++++++++++++ buch/papers/kra/einleitung.tex | 14 +++ buch/papers/kra/hamilton.tex | 185 -------------------------------- buch/papers/kra/loesung.tex | 47 +++++++++ buch/papers/kra/main.tex | 10 +- buch/papers/kra/riccati.tex | 93 ---------------- 7 files changed, 306 insertions(+), 289 deletions(-) create mode 100644 buch/papers/kra/anwendung.tex create mode 100644 buch/papers/kra/einleitung.tex delete mode 100644 buch/papers/kra/hamilton.tex create mode 100644 buch/papers/kra/loesung.tex delete mode 100644 buch/papers/kra/riccati.tex (limited to 'buch/papers') diff --git a/buch/papers/kra/Makefile.inc b/buch/papers/kra/Makefile.inc index f453e6e..a521e4b 100644 --- a/buch/papers/kra/Makefile.inc +++ b/buch/papers/kra/Makefile.inc @@ -4,11 +4,10 @@ # (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule # dependencies-kra = \ - papers/kra/packages.tex \ + papers/kra/packages.tex \ papers/kra/main.tex \ - papers/kra/references.bib \ - papers/kra/teil0.tex \ - papers/kra/teil1.tex \ - papers/kra/teil2.tex \ - papers/kra/teil3.tex + papers/kra/references.bib \ + papers/kra/einleitung.tex \ + papers/kra/loesung.tex \ + papers/kra/anwendung.tex \ diff --git a/buch/papers/kra/anwendung.tex b/buch/papers/kra/anwendung.tex new file mode 100644 index 0000000..4d4d351 --- /dev/null +++ b/buch/papers/kra/anwendung.tex @@ -0,0 +1,235 @@ +\section{Anwendungen \label{kra:section:anwendung}} +\rhead{Anwendungen} +\newcommand{\dt}[0]{\frac{d}{dt}} + +Die Matrix-Riccati Differentialgleichung findet unter anderem Anwendung in der Regelungstechnik beim RQ- und RQG-Regler oder aber auch beim Kalmanfilter. +Im folgenden Abschnitt möchten wir uns an einem Beispiel anschauen wie wir mit Hilfe der Matrix-Riccati Differentialgleichung (\ref{kra:matrixriccati}) ein Feder-Masse-System untersuchen können. + +\subsection{Feder-Masse-System} +Die Einfachste Form eines Feder-Masse-Systems ist dargestellt in Abbildung \ref{kra:fig:simple_mass_spring}. +Es besteht aus einer Masse $m$ welche reibungsfrei gelagert ist und einer Feder mit der Federkonstante $k$. +Die im System wirkenden Kräfte teilen sich auf in die auf dem hookeschen Gesetz basierenden Rückstellkraft $F_R = k \Delta_x$ und der auf dem Aktionsprinzip basierenden Kraft $F_a = am = \ddot{x} m$. +Das Kräftegleichgewicht fordert $F_R = F_a$ woraus folgt, dass + +\begin{equation*} + k \Delta_x = \ddot{x} m \Leftrightarrow \ddot{x} = \frac{k \Delta_x}{m} +\end{equation*} +Die funktion die diese Differentialgleichung löst ist die harmonische Schwingung +\begin{equation} + x(t) = A \cos(\omega_0 t + \Phi), \quad \omega_0 = \sqrt{\frac{k}{m}} +\end{equation} + + +\begin{figure} + \input{papers/kra/images/simple_mass_spring.tex} + \caption{Einfaches Feder-Masse-System.} + \label{kra:fig:simple_mass_spring} +\end{figure} + +\begin{figure} + \input{papers/kra/images/multi_mass_spring.tex} + \caption{Feder-Masse-System mit zwei Massen und drei Federn.} + \label{kra:fig:multi_mass_spring} +\end{figure} + + +\subsection{Hamilton-Funktion} +Die Bewegung der Masse $m$ kann mit Hilfe der hamiltonschen Mechanik im Phasenraum untersucht werden. +Die hamiltonschen Gleichungen verwenden dafür die veralgemeinerten Ortskoordinaten +$q = (q_{1}, q_{2}, ..., q_{n})$ und die verallgemeinerten Impulskoordinaten $p = (p_{1}, p_{2}, ..., p_{n})$, wobei der Impuls definiert ist als $p_k = m_k \cdot v_k$. +Liegen keine zeitabhängigen Zwangsbedingungen vor, so entspricht die Hamitlon-Funktion der Gesamtenergie des Systems \cite{kra:hamilton}. +Im Falle des einfachen Feder-Masse-Systems, Abbildung \ref{kra:fig:simple_mass_spring}, setzt sich die Hamilton-Funktion aus kinetischer und potentieller Energie zusammen. + +\begin{equation} + \label{kra:harmonischer_oszillator} + \begin{split} + \mathcal{H}(q, p) &= T(p) + V(q) = E \\ + &= \underbrace{\frac{p^2}{2m}}_{E_{kin}} + \underbrace{\frac{k q^2}{2}}_{E_{pot}} + \end{split} +\end{equation} + +Die Hamiltonschen Bewegungsgleichungen liefern \cite{kra:kanonischegleichungen} +\begin{equation} + \label{kra:hamilton:bewegungsgleichung} + \dot{q_{k}} = \frac{\partial \mathcal{H}}{\partial p_k} + \qquad + \dot{p_{k}} = -\frac{\partial \mathcal{H}}{\partial q_k} +\end{equation} + +daraus folgt + +\[ + \dot{q} = \frac{p}{m} + \qquad + \dot{p} = -kq +\] + +in Matrixschreibweise erhalten wir also + +\[ + \begin{pmatrix} + \dot{q} \\ + \dot{p} + \end{pmatrix} + = + \begin{pmatrix} + 0 & \frac{1}{m} \\ + -k & 0 + \end{pmatrix} + \begin{pmatrix} + q \\ + p + \end{pmatrix} +\] + +Für das erweiterte Federmassesystem, Abbildung \ref{kra:fig:multi_mass_spring}, können wir analog vorgehen. +Die kinetische Energie setzt sich nun aus den kinetischen Energien der einzelnen Massen $m_1$ und $m_2$ zusammen. +Die Potentielle Energie erhalten wir aus der Summe der kinetischen Energien der einzelnen Federn mit den Federkonstanten $k_1$, $k_c$ und $k_2$. + +\begin{align*} + \begin{split} + T &= T_1 + T_2 \\ + &= \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \end{split} + \\ + \begin{split} + V &= V_1 + V_c + V_2 \\ + &= \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2} + \end{split} +\end{align*} + +Die Hamilton-Funktion ist also + +\begin{align*} + \begin{split} + \mathcal{H} &= T + V \\ + &= \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2} + \end{split} +\end{align*} + +Die Bewegungsgleichungen \ref{kra:hamilton:bewegungsgleichung} liefern +\begin{align*} + \frac{\partial \mathcal{H}}{\partial p_k} & = \dot{q_k} + \Rightarrow + \left\{ + \begin{alignedat}{2} + \dot{q_1} &= \frac{2p_1}{2m_1} &&= \frac{p_1}{m_1}\\ + \dot{q_2} &= \frac{2p_2}{2m_2} &&= \frac{p_2}{m_2} + \end{alignedat} + \right. + \\ + -\frac{\partial \mathcal{H}}{\partial q_k} & = \dot{p_k} + \Rightarrow + \left\{ + \begin{alignedat}{2} + \dot{p_1} &= -(\frac{2k_1q_1}{2} - \frac{2k_c(q_2-q_1)}{2}) &&= -q_1(k_1+k_c) + q_2k_c \\ + \dot{p_1} &= -(\frac{2k_c(q_2-q_1)}{2} - \frac{2k_2q_2}{2}) &&= q_1k_c - (k_c + k_2) + \end{alignedat} + \right. +\end{align*} + +In Matrixschreibweise erhalten wir + +\begin{equation} + \label{kra:hamilton:multispringmass} + \begin{pmatrix} + \dot{q_1} \\ + \dot{q_2} \\ + \dot{p_1} \\ + \dot{p_2} \\ + \end{pmatrix} + = + \begin{pmatrix} + 0 & 0 & \frac{1}{2m_1} & 0 \\ + 0 & 0 & 0 & \frac{1}{2m_2} \\ + -(k_1 + k_c) & k_c & 0 & 0 \\ + k_c & -(k_c + k_2) & 0 & 0 \\ + \end{pmatrix} + \begin{pmatrix} + q_1 \\ + q_2 \\ + p_1 \\ + p_2 \\ + \end{pmatrix} + \Leftrightarrow + \dt + \begin{pmatrix} + Q \\ + P \\ + \end{pmatrix} + = + \underbrace{ + \begin{pmatrix} + 0 & M \\ + K & 0 + \end{pmatrix} + }_{G} + \begin{pmatrix} + Q \\ + P \\ + \end{pmatrix} +\end{equation} + +\subsection{Phasenraum} +Der Phasenraum erlaubt die eindeutige Beschreibung aller möglichen Bewegungszustände eines mechanischen System durch einen Punkt. +Die Phasenraumdarstellung eignet sich somit sehr gut für die systematische Untersuchung der Feder-Masse-Systeme. + +\subsubsection{Harmonischer Oszillator} +Die Hamiltonfunktion des harmonischen Oszillators \ref{kra:harmonischer_oszillator} führt auf eine Lösung der Form +\begin{equation*} + q(t) = A \cos(\omega_0 T + \Phi), \quad p(t) = -m \omega_0 A \sin(\omega_0 t + \Phi) +\end{equation*} +die Phasenraumtrajektorien bilden also Ellipsen mit Zentrum $q=0, p=0$ und Halbachsen $A$ und $m \omega A$. +Abbildung \ref{kra:fig:phasenraum} zeigt Phasenraumtrajektorien mit den Energien $E_{x \in \{A, B, C, D\}}$ und verschiedenen Werten von $\omega$. + +\begin{figure} + \input{papers/kra/images/phase_space.tex} + \caption{Phasenraumdarstellung des einfachen Feder-Masse-Systems.} + \label{kra:fig:phasenraum} +\end{figure} + +\subsubsection{Erweitertes Feder-Masse-System} +Wir intressieren uns nun dafür wie der Phasenwinkel $U = PQ^{-1}$ von der Zeit abhängt, +wir suchen also die Grösse $\Theta = \dt U$. + +Ersetzten wir in der Gleichung \ref{kra:hamilton:multispringmass} die Matrix $G$ mit $\tilde{G}$ so erhalten wir +\begin{equation} + \dt + \begin{pmatrix} + Q \\ + P + \end{pmatrix} + = + \underbrace{ + \begin{pmatrix} + A & B \\ + C & D + \end{pmatrix} + }_{\tilde{G}} + \begin{pmatrix} + Q \\ + P + \end{pmatrix} +\end{equation} + +Mit einsetzten folgt + +\begin{align*} + \dot{Q} = AQ + BP \\ + \dot{P} = CQ + DP +\end{align*} +\begin{equation} + \begin{split} + \dt U &= \dot{P} Q^{-1} + P \dt Q^{-1} \\ + &= (CQ + DP) Q^{-1} - P (Q^{-1} \dot{Q} Q^{-1}) \\ + &= C\underbrace{QQ^{-1}}_\text{I} + D\underbrace{PQ^{-1}}_\text{U} - P(Q^{-1} (AQ + BP) Q^{-1}) \\ + &= C + DU - \underbrace{PQ^{-1}}_\text{U}(A\underbrace{QQ^{-1}}_\text{I} + B\underbrace{PQ^{-1}}_\text{U}) \\ + &= C + DU - UA - UBU + \end{split} +\end{equation} + +was uns auf die Matrix-Riccati Gleichung \ref{kra:matrixriccati} führt. + + +\subsection{Fazit} +% @TODO diff --git a/buch/papers/kra/einleitung.tex b/buch/papers/kra/einleitung.tex new file mode 100644 index 0000000..1a347a8 --- /dev/null +++ b/buch/papers/kra/einleitung.tex @@ -0,0 +1,14 @@ +\section{Einleitung} \label{kra:section:einleitung} +\rhead{Einleitung} +Die riccatische Differentialgleichung ist eine nichtlineare gewöhnliche Differentialgleichunge erster Ordnung der form +\begin{equation} + \label{kra:riccati} + y'(x) = f(x)y^2(x) + g(x)y(x) + h(x) +\end{equation} +Sie ist bennant nach dem italienischen Grafen Jacopo Francesco Riccati (1676–1754) der sich mit der Klassifizierung von Differentialgleichungen befasste und Methoden zur Verringerung der Ordnung von Gleichungen entwickelte. +Als Riccati Gleichung werden auch Matrixgleichugen der Form +\begin{equation} + \label{kra:matrixriccati} + \dot{U}(t) = DU(t) - UA(t) - U(t)BU(t) % +Q ? +\end{equation} +bezeichnet, welche aufgrund ihres quadratischen Terms eine gewisse ähnlichkeit aufweisen. \ No newline at end of file diff --git a/buch/papers/kra/hamilton.tex b/buch/papers/kra/hamilton.tex deleted file mode 100644 index 14a5e8c..0000000 --- a/buch/papers/kra/hamilton.tex +++ /dev/null @@ -1,185 +0,0 @@ -\newcommand{\dt}[0]{\frac{d}{dt}} - -\section{Teil abc\label{kra:section:teilabc}} -\rhead{Teil abc} - -\subsection{Hamilton-Funktion} -Die Bewegung der Masse $m$ kann mit Hilfe der hamiltonschen Mechanik im Phasenraum untersucht werden. -Die hamiltonschen Gleichungen verwenden dafür die veralgemeinerten Ortskoordinaten -$q = (q_{1}, q_{2}, ..., q_{n})$ und die verallgemeinerten Impulskoordinaten $p = (p_{1}, p_{2}, ..., p_{n})$, -wobei der Impuls definiert ist als $p_k = m_k \cdot v_k$. -Liegen keine zeitabhängigen Zwangsbedingungen vor, so entspricht die Hamitlon-Funktion der Gesamtenergie des Systems \cite{kra:hamilton}. -Im Falle des einfachen Federmassesystems, Abbildung \ref{kra:fig:simple_spring_mass}, -setzt sich die Hamilton-Funktion aus kinetischer und potentieller Energie zusammen. - -\begin{equation} - \label{hamilton} - \begin{split} - \mathcal{H}(q, p) &= T(p) + V(q) = E \\ - &= \underbrace{\frac{p^2}{2m}}_{E_{kin}} + \underbrace{\frac{k q^2}{2}}_{E_{pot}} - \end{split} -\end{equation} - -Die Hamiltonschen Bewegungsgleichungen liefern \cite{kra:kanonischegleichungen} -\begin{equation} - \label{kra:hamilton:bewegungsgleichung} - \dot{q_{k}} = \frac{\partial \mathcal{H}}{\partial p_k} - \qquad - \dot{p_{k}} = -\frac{\partial \mathcal{H}}{\partial q_k} -\end{equation} - -daraus folgt - -\[ - \dot{q} = \frac{p}{m} - \qquad - \dot{p} = -kq -\] - -in Matrixschreibweise erhalten wir also - -\[ - \begin{pmatrix} - \dot{q} \\ - \dot{p} - \end{pmatrix} - = - \begin{pmatrix} - 0 & \frac{1}{m} \\ - -k & 0 - \end{pmatrix} - \begin{pmatrix} - q \\ - p - \end{pmatrix} -\] - -Für das erweiterte Federmassesystem, Abbildung \ref{kra:fig:multi_spring_mass}, können wir analog vorgehen. -Die kinetische Energie setzt sich nun aus den kinetischen Energien der einzelnen Massen $m_1$ und $m_2$ zusammen. -Die Potentielle Energie erhalten wir aus der Summe der kinetischen Energien der einzelnen Federn mit den Federkonstanten $k_1$, $k_c$ und $k_2$. - -\begin{align*} - \begin{split} - T &= T_1 + T_2 \\ - &= \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} - \end{split} - \\ - \begin{split} - V &= V_1 + V_c + V_2 \\ - &= \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2} - \end{split} -\end{align*} - -Die Hamilton-Funktion ist also - -\begin{align*} - \begin{split} - \mathcal{H} &= T + V \\ - &= \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2} - \end{split} -\end{align*} - -Die Bewegungsgleichungen \ref{kra:hamilton:bewegungsgleichung} liefern -\begin{align*} - \frac{\partial \mathcal{H}}{\partial p_k} & = \dot{q_k} - \Rightarrow - \left\{ - \begin{alignedat}{2} - \dot{q_1} &= \frac{2p_1}{2m_1} &&= \frac{p_1}{m_1}\\ - \dot{q_2} &= \frac{2p_2}{2m_2} &&= \frac{p_2}{m_2} - \end{alignedat} - \right. - \\ - -\frac{\partial \mathcal{H}}{\partial q_k} & = \dot{p_k} - \Rightarrow - \left\{ - \begin{alignedat}{2} - \dot{p_1} &= -(\frac{2k_1q_1}{2} - \frac{2k_c(q_2-q_1)}{2}) &&= -q_1(k_1+k_c) + q_2k_c \\ - \dot{p_1} &= -(\frac{2k_c(q_2-q_1)}{2} - \frac{2k_2q_2}{2}) &&= q_1k_c - (k_c + k_2) - \end{alignedat} - \right. -\end{align*} - -In Matrixschreibweise erhalten wir - -\begin{equation} - \label{kra:hamilton:multispringmass} - \begin{pmatrix} - \dot{q_1} \\ - \dot{q_2} \\ - \dot{p_1} \\ - \dot{p_2} \\ - \end{pmatrix} - = - \begin{pmatrix} - 0 & 0 & \frac{1}{2m_1} & 0 \\ - 0 & 0 & 0 & \frac{1}{2m_2} \\ - -(k_1 + k_c) & k_c & 0 & 0 \\ - k_c & -(k_c + k_2) & 0 & 0 \\ - \end{pmatrix} - \begin{pmatrix} - q_1 \\ - q_2 \\ - p_1 \\ - p_2 \\ - \end{pmatrix} - \Leftrightarrow - \dt - \begin{pmatrix} - Q \\ - P \\ - \end{pmatrix} - \underbrace{ - \begin{pmatrix} - 0 & M \\ - K & 0 - \end{pmatrix} - }_{G} - \begin{pmatrix} - Q \\ - P \\ - \end{pmatrix} -\end{equation} - - -Wir intressieren uns nun dafür wie der Phasenwinkel $U = PQ^{-1}$ von der Zeit abhängt, -wir suchen also die Grösse $\Theta = \dt U$. - -Ersetzten wir in der Gleichung \ref{kra:hamilton:multispringmass} die Matrix $G$ mit $\tilde{G}$ so erhalten wir -\begin{equation} - \dt - \begin{pmatrix} - Q \\ - P - \end{pmatrix} - = - \underbrace{ - \begin{pmatrix} - A & B \\ - C & D - \end{pmatrix} - }_{\tilde{G}} - \begin{pmatrix} - Q \\ - P - \end{pmatrix} -\end{equation} - -Mit einsetzten folgt - -\begin{align*} - \dot{Q} = AQ + BP \\ - \dot{P} = CQ + DP -\end{align*} -\begin{equation} - \begin{split} - \dt U &= \dot{P} Q^{-1} + P \dt Q^{-1} \\ - &= (CQ + DP) Q^{-1} - P (Q^{-1} \dot{Q} Q^{-1}) \\ - &= C\underbrace{QQ^{-1}}_\text{I} + D\underbrace{PQ^{-1}}_\text{U} - P(Q^{-1} (AQ + BP) Q^{-1}) \\ - &= C + DU - \underbrace{PQ^{-1}}_\text{U}(A\underbrace{QQ^{-1}}_\text{I} + B\underbrace{PQ^{-1}}_\text{U}) \\ - &= C + DU - UA - UBU - \end{split} -\end{equation} - -was uns auf die zeitkontinuierliche Matrix-Riccati-Gleichung führt. - diff --git a/buch/papers/kra/loesung.tex b/buch/papers/kra/loesung.tex new file mode 100644 index 0000000..ece0f15 --- /dev/null +++ b/buch/papers/kra/loesung.tex @@ -0,0 +1,47 @@ +\section{Lösungsmethoden} \label{kra:section:loesung} +\rhead{Lösungsmethoden} +% @TODO Lösung normal riccati +Lösung der Riccatischen Differentialgleichung \ref{kra:riccati}. + + +% Lösung matrix riccati +Die Lösung der Matrix-Riccati Gleichung \ref{kra:matrixriccati} erhalten wir nach \cite{kra:kalmanisae} folgendermassen +\begin{equation} + \label{kra:matrixriccati-solution} + \begin{pmatrix} + X(t) \\ + Y(t) + \end{pmatrix} + = + \Phi(t_0, t) + \begin{pmatrix} + I(t) \\ + U_0(t) + \end{pmatrix} + = + \begin{pmatrix} + \Phi_{11}(t_0, t) & \Phi_{12}(t_0, t) \\ + \Phi_{21}(t_0, t) & \Phi_{22}(t_0, t) + \end{pmatrix} + \begin{pmatrix} + I(t) \\ + U_0(t) + \end{pmatrix} +\end{equation} + +\begin{equation} + U(t) = + \begin{pmatrix} + \Phi_{21}(t_0, t) + \Phi_{22}(t_0, t) + \end{pmatrix} + \begin{pmatrix} + \Phi_{11}(t_0, t) + \Phi_{12}(t_0, t) + \end{pmatrix} + ^{-1} +\end{equation} + +wobei $\Phi(t, t_0)$ die sogennante Zustandsübergangsmatrix ist. + +\begin{equation} + \Phi(t_0, t) = e^{H(t - t_0)} +\end{equation} diff --git a/buch/papers/kra/main.tex b/buch/papers/kra/main.tex index 456b6ee..a84ebaf 100644 --- a/buch/papers/kra/main.tex +++ b/buch/papers/kra/main.tex @@ -3,12 +3,12 @@ % % (c) 2020 Hochschule Rapperswil % -\chapter{Kalman, Riccati und Abel\label{chapter:kra}} -\lhead{Kalman, Riccati und Abel} +\chapter{Riccati Differentialgleichung\label{chapter:kra}} +\lhead{Riccati Differentialgleichung} \begin{refsection} \chapterauthor{Samuel Niederer} - \input{papers/kra/hamilton.tex} - \newpage - \input{papers/kra/riccati.tex} + \input{papers/kra/einleitung.tex} + \input{papers/kra/loesung.tex} + \input{papers/kra/anwendung.tex} \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/kra/riccati.tex b/buch/papers/kra/riccati.tex deleted file mode 100644 index df2921d..0000000 --- a/buch/papers/kra/riccati.tex +++ /dev/null @@ -1,93 +0,0 @@ -\section{Riccati - \label{kra:section:riccati}} -\rhead{Riccati} - -\begin{equation} - y'(x) = f(x)y^2(x) + g(x)y(x) + h(x) -\end{equation} -% einfache (normale riccati gleichung und ihre loesung) -% (kann man diese bei einfachem federmasse system benutzten?) -% matrix riccati gleichung - - -Die zeitkontinuierliche Riccati-Matrix-Gleichung hat die Form -\begin{equation} - \label{kra:riccati:riccatiequation} - \dot{U(t)} = DU(t) - UA(t) - U(t)BU(t) -\end{equation} - -Betrachten wir das Differentialgleichungssystem \ref{kra:riccati:derivation} - -\begin{equation} - \label{kra:riccati:derivation} - \dt - \begin{pmatrix} - X \\ - Y - \end{pmatrix} - = - \underbrace{ - \begin{pmatrix} - A & B \\ - C & D - \end{pmatrix} - }_{H} - \begin{pmatrix} - X \\ - Y - \end{pmatrix} -\end{equation} - -interessieren wir uns für die zeitliche Änderung der Grösse $U = YX^{-1}$, so erhalten wir durch einsetzten - -\begin{align*} - \dt U & = \dot{Y} X^{-1} + Y \dt X^{-1} \\ - & = (CX + DY) X^{-1} - Y (X^{-1} \dot{X} X^{-1}) \\ - & = C\underbrace{XX^{-1}}_\text{I} + D\underbrace{YX^{-1}}_\text{U} - Y(X^{-1} (AX + BY) X^{-1}) \\ - & = C + DU - \underbrace{YX^{-1}}_\text{U}(A\underbrace{XX^{-1}}_\text{I} + B\underbrace{YX^{-1}}_\text{U}) \\ - & = C + DU - UA - UBU -\end{align*} - -was uns auf die Riccati-Matrix-Gleichung \ref{kra:riccati:riccatiequation} führt. -Die Lösung dieser Gleichung erhalten wir nach \cite{kra:kalmanisae} folgendermassen -\begin{equation} - \begin{pmatrix} - X(t) \\ - Y(t) - \end{pmatrix} - = - \Phi(t_0, t) - \begin{pmatrix} - I(t) \\ - U_0(t) - \end{pmatrix} - = - \begin{pmatrix} - \Phi_{11}(t_0, t) & \Phi_{12}(t_0, t) \\ - \Phi_{21}(t_0, t) & \Phi_{22}(t_0, t) - \end{pmatrix} - \begin{pmatrix} - I(t) \\ - U_0(t) - \end{pmatrix} -\end{equation} - -\begin{equation} - U(t) = - \begin{pmatrix} - \Phi_{21}(t_0, t) + \Phi_{22}(t_0, t) - \end{pmatrix} - \begin{pmatrix} - \Phi_{11}(t_0, t) + \Phi_{12}(t_0, t) - \end{pmatrix} - ^{-1} -\end{equation} - -wobei $\Phi(t, t_0)$ die sogennante Zustandsübergangsmatrix ist. - -\begin{equation} - \Phi(t_0, t) = e^{H(t - t_0)} -\end{equation} - - - -- cgit v1.2.1 From 022616988c2a0ad10d83133a330e4194f4d7d4a3 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Tue, 26 Jul 2022 13:04:26 +0200 Subject: Changed Authors of Sturm-Liouville chapter. --- buch/papers/sturmliouville/main.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/main.tex b/buch/papers/sturmliouville/main.tex index a7d2857..f1a500e 100644 --- a/buch/papers/sturmliouville/main.tex +++ b/buch/papers/sturmliouville/main.tex @@ -6,7 +6,7 @@ \chapter{Thema\label{chapter:sturmliouville}} \lhead{Thema} \begin{refsection} -\chapterauthor{Hans Muster} +\chapterauthor{Réda Haddouche und Erik Löffler} Ein paar Hinweise für die korrekte Formatierung des Textes \begin{itemize} -- cgit v1.2.1 From 1073ef257d9f511a1d5b6c733401390933c7566a Mon Sep 17 00:00:00 2001 From: haddoucher Date: Tue, 26 Jul 2022 13:12:42 +0200 Subject: change title --- buch/papers/sturmliouville/main.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/main.tex b/buch/papers/sturmliouville/main.tex index f1a500e..9b04219 100644 --- a/buch/papers/sturmliouville/main.tex +++ b/buch/papers/sturmliouville/main.tex @@ -3,7 +3,7 @@ % % (c) 2020 Hochschule Rapperswil % -\chapter{Thema\label{chapter:sturmliouville}} +\chapter{Sturm-Liouville-Problem\label{chapter:sturmliouville}} \lhead{Thema} \begin{refsection} \chapterauthor{Réda Haddouche und Erik Löffler} -- cgit v1.2.1 From 8d3f5416af1f8a5ce30db4eb275be3cdae67c8eb Mon Sep 17 00:00:00 2001 From: haddoucher Date: Tue, 26 Jul 2022 13:53:12 +0200 Subject: makefile makefile --- buch/papers/sturmliouville/Makefile | 34 +++++++++++++++++++-- buch/papers/sturmliouville/einleitung.tex | 22 +++++++++++++ buch/papers/sturmliouville/main.tex | 2 +- buch/papers/sturmliouville/standalone.tex | 31 +++++++++++++++++++ .../sturmliouville/standalone/standalone.pdf | Bin 0 -> 77574 bytes buch/papers/sturmliouville/teil0.tex | 22 ------------- 6 files changed, 85 insertions(+), 26 deletions(-) create mode 100644 buch/papers/sturmliouville/einleitung.tex create mode 100644 buch/papers/sturmliouville/standalone.tex create mode 100644 buch/papers/sturmliouville/standalone/standalone.pdf delete mode 100644 buch/papers/sturmliouville/teil0.tex (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/Makefile b/buch/papers/sturmliouville/Makefile index da902e7..70de9fc 100644 --- a/buch/papers/sturmliouville/Makefile +++ b/buch/papers/sturmliouville/Makefile @@ -1,9 +1,37 @@ # -# Makefile -- make file for the paper sturmliouville +# Makefile -- make file for the paper fm # # (c) 2020 Prof Dr Andreas Mueller # -images: - @echo "no images to be created in sturmliouville" +SOURCES := \ + einleitung.tex\ + teil1.tex \ + teil2.tex \ + teil3.tex \ + main.tex +#TIKZFIGURES := \ + tikz/atoms-grid-still.tex \ + +#FIGURES := $(patsubst tikz/%.tex, figures/%.pdf, $(TIKZFIGURES)) + +#.PHONY: images +#images: $(FIGURES) + +#figures/%.pdf: tikz/%.tex +# mkdir -p figures +# pdflatex --output-directory=figures $< + +.PHONY: standalone +standalone: standalone.tex $(SOURCES) #$(FIGURES) + mkdir -p standalone + cd ../..; \ + pdflatex \ + --halt-on-error \ + --shell-escape \ + --output-directory=papers/sturmliouville/standalone \ + papers/sturmliouville/standalone.tex; + cd standalone; \ + bibtex standalone; \ + makeindex standalone; \ No newline at end of file diff --git a/buch/papers/sturmliouville/einleitung.tex b/buch/papers/sturmliouville/einleitung.tex new file mode 100644 index 0000000..ffcb8f3 --- /dev/null +++ b/buch/papers/sturmliouville/einleitung.tex @@ -0,0 +1,22 @@ +% +% einleitung.tex -- Beispiel-File für die Einleitung +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Teil 0\label{sturmliouville:section:teil0}} +\rhead{Teil 0} +Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam +nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam +erat, sed diam voluptua \cite{sturmliouville:bibtex}. +At vero eos et accusam et justo duo dolores et ea rebum. +Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum +dolor sit amet. + +Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam +nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam +erat, sed diam voluptua. +At vero eos et accusam et justo duo dolores et ea rebum. Stet clita +kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit +amet. + + diff --git a/buch/papers/sturmliouville/main.tex b/buch/papers/sturmliouville/main.tex index 9b04219..2a779db 100644 --- a/buch/papers/sturmliouville/main.tex +++ b/buch/papers/sturmliouville/main.tex @@ -27,7 +27,7 @@ Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren Übersicht wegen, aber auch um GIT die Arbeit zu erleichtern. \end{itemize} -\input{papers/sturmliouville/teil0.tex} +\input{papers/sturmliouville/einleitung.tex} \input{papers/sturmliouville/teil1.tex} \input{papers/sturmliouville/teil2.tex} \input{papers/sturmliouville/teil3.tex} diff --git a/buch/papers/sturmliouville/standalone.tex b/buch/papers/sturmliouville/standalone.tex new file mode 100644 index 0000000..cd0e8dc --- /dev/null +++ b/buch/papers/sturmliouville/standalone.tex @@ -0,0 +1,31 @@ +\documentclass{book} + +\def\IncludeBookCover{0} +\input{common/packages.tex} + +% additional packages used by the individual papers, add a line for +% each paper +\input{papers/common/addpackages.tex} + +% workaround for biblatex bug +\makeatletter +\def\blx@maxline{77} +\makeatother +\addbibresource{chapters/references.bib} + +% Bibresources for each article +\input{papers/common/addbibresources.tex} + +% make sure the last index starts on an odd page +\AtEndDocument{\clearpage\ifodd\value{page}\else\null\clearpage\fi} +\makeindex + +%\pgfplotsset{compat=1.12} +\setlength{\headheight}{15pt} % fix headheight warning +\DeclareGraphicsRule{*}{mps}{*}{} + +\begin{document} + \input{common/macros.tex} + \def\chapterauthor#1{{\large #1}\bigskip\bigskip} + \input{papers/sturmliouville/main.tex} +\end{document} diff --git a/buch/papers/sturmliouville/standalone/standalone.pdf b/buch/papers/sturmliouville/standalone/standalone.pdf new file mode 100644 index 0000000..1b5acdb Binary files /dev/null and b/buch/papers/sturmliouville/standalone/standalone.pdf differ diff --git a/buch/papers/sturmliouville/teil0.tex b/buch/papers/sturmliouville/teil0.tex deleted file mode 100644 index ffcb8f3..0000000 --- a/buch/papers/sturmliouville/teil0.tex +++ /dev/null @@ -1,22 +0,0 @@ -% -% einleitung.tex -- Beispiel-File für die Einleitung -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 0\label{sturmliouville:section:teil0}} -\rhead{Teil 0} -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua \cite{sturmliouville:bibtex}. -At vero eos et accusam et justo duo dolores et ea rebum. -Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum -dolor sit amet. - -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua. -At vero eos et accusam et justo duo dolores et ea rebum. Stet clita -kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit -amet. - - -- cgit v1.2.1 From 08931fd248fc0c14173b5ee9bb34e545d7d4bf03 Mon Sep 17 00:00:00 2001 From: haddoucher Date: Tue, 26 Jul 2022 15:05:25 +0200 Subject: struktur verbessert --- buch/papers/sturmliouville/Makefile | 1 + buch/papers/sturmliouville/beispiele.tex | 40 +++++++++++++++ buch/papers/sturmliouville/eigenschaften.tex | 55 +++++++++++++++++++++ buch/papers/sturmliouville/einleitung.tex | 2 +- buch/papers/sturmliouville/main.tex | 8 +-- ...2008_Chapter_Sturm-Liouville-ProblemeUndSpe.pdf | Bin 0 -> 548591 bytes buch/papers/sturmliouville/teil1.tex | 55 --------------------- buch/papers/sturmliouville/teil2.tex | 40 --------------- buch/papers/sturmliouville/teil3.tex | 40 --------------- 9 files changed, 102 insertions(+), 139 deletions(-) create mode 100644 buch/papers/sturmliouville/beispiele.tex create mode 100644 buch/papers/sturmliouville/eigenschaften.tex create mode 100644 buch/papers/sturmliouville/papers/sturmliouville/2008_Chapter_Sturm-Liouville-ProblemeUndSpe.pdf delete mode 100644 buch/papers/sturmliouville/teil1.tex delete mode 100644 buch/papers/sturmliouville/teil2.tex delete mode 100644 buch/papers/sturmliouville/teil3.tex (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/Makefile b/buch/papers/sturmliouville/Makefile index 70de9fc..23214a2 100644 --- a/buch/papers/sturmliouville/Makefile +++ b/buch/papers/sturmliouville/Makefile @@ -31,6 +31,7 @@ standalone: standalone.tex $(SOURCES) #$(FIGURES) --halt-on-error \ --shell-escape \ --output-directory=papers/sturmliouville/standalone \ + --extra-mem-top=10000000 \ papers/sturmliouville/standalone.tex; cd standalone; \ bibtex standalone; \ diff --git a/buch/papers/sturmliouville/beispiele.tex b/buch/papers/sturmliouville/beispiele.tex new file mode 100644 index 0000000..7fc3d2c --- /dev/null +++ b/buch/papers/sturmliouville/beispiele.tex @@ -0,0 +1,40 @@ +% +% teil2.tex -- Beispiel-File für teil2 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Teil 2 +\label{sturmliouville:section:teil2}} +\rhead{Teil 2} +Sed ut perspiciatis unde omnis iste natus error sit voluptatem +accusantium doloremque laudantium, totam rem aperiam, eaque ipsa +quae ab illo inventore veritatis et quasi architecto beatae vitae +dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit +aspernatur aut odit aut fugit, sed quia consequuntur magni dolores +eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam +est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci +velit, sed quia non numquam eius modi tempora incidunt ut labore +et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima +veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, +nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure +reprehenderit qui in ea voluptate velit esse quam nihil molestiae +consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla +pariatur? + +\subsection{De finibus bonorum et malorum +\label{sturmliouville:subsection:bonorum}} +At vero eos et accusamus et iusto odio dignissimos ducimus qui +blanditiis praesentium voluptatum deleniti atque corrupti quos +dolores et quas molestias excepturi sint occaecati cupiditate non +provident, similique sunt in culpa qui officia deserunt mollitia +animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis +est et expedita distinctio. Nam libero tempore, cum soluta nobis +est eligendi optio cumque nihil impedit quo minus id quod maxime +placeat facere possimus, omnis voluptas assumenda est, omnis dolor +repellendus. Temporibus autem quibusdam et aut officiis debitis aut +rerum necessitatibus saepe eveniet ut et voluptates repudiandae +sint et molestiae non recusandae. Itaque earum rerum hic tenetur a +sapiente delectus, ut aut reiciendis voluptatibus maiores alias +consequatur aut perferendis doloribus asperiores repellat. + + diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex new file mode 100644 index 0000000..c23c1d6 --- /dev/null +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -0,0 +1,55 @@ +% +% teil1.tex -- Beispiel-File für das Paper +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Teil 1 +\label{sturmliouville:section:teil1}} +\rhead{Problemstellung} +Sed ut perspiciatis unde omnis iste natus error sit voluptatem +accusantium doloremque laudantium, totam rem aperiam, eaque ipsa +quae ab illo inventore veritatis et quasi architecto beatae vitae +dicta sunt explicabo. +Nemo enim ipsam voluptatem quia voluptas sit aspernatur aut odit +aut fugit, sed quia consequuntur magni dolores eos qui ratione +voluptatem sequi nesciunt +\begin{equation} +\int_a^b x^2\, dx += +\left[ \frac13 x^3 \right]_a^b += +\frac{b^3-a^3}3. +\label{sturmliouville:equation1} +\end{equation} +Neque porro quisquam est, qui dolorem ipsum quia dolor sit amet, +consectetur, adipisci velit, sed quia non numquam eius modi tempora +incidunt ut labore et dolore magnam aliquam quaerat voluptatem. + +Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis +suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur? +Quis autem vel eum iure reprehenderit qui in ea voluptate velit +esse quam nihil molestiae consequatur, vel illum qui dolorem eum +fugiat quo voluptas nulla pariatur? + +\subsection{De finibus bonorum et malorum +\label{sturmliouville:subsection:finibus}} +At vero eos et accusamus et iusto odio dignissimos ducimus qui +blanditiis praesentium voluptatum deleniti atque corrupti quos +dolores et quas molestias excepturi sint occaecati cupiditate non +provident, similique sunt in culpa qui officia deserunt mollitia +animi, id est laborum et dolorum fuga \eqref{000tempmlate:equation1}. + +Et harum quidem rerum facilis est et expedita distinctio +\ref{sturmliouville:section:loesung}. +Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil +impedit quo minus id quod maxime placeat facere possimus, omnis +voluptas assumenda est, omnis dolor repellendus +\ref{sturmliouville:section:folgerung}. +Temporibus autem quibusdam et aut officiis debitis aut rerum +necessitatibus saepe eveniet ut et voluptates repudiandae sint et +molestiae non recusandae. +Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis +voluptatibus maiores alias consequatur aut perferendis doloribus +asperiores repellat. + + diff --git a/buch/papers/sturmliouville/einleitung.tex b/buch/papers/sturmliouville/einleitung.tex index ffcb8f3..073ba6e 100644 --- a/buch/papers/sturmliouville/einleitung.tex +++ b/buch/papers/sturmliouville/einleitung.tex @@ -3,7 +3,7 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Teil 0\label{sturmliouville:section:teil0}} +\section{Was ist Sturm-Liouville-Problem\label{sturmliouville:section:wa}} \rhead{Teil 0} Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam diff --git a/buch/papers/sturmliouville/main.tex b/buch/papers/sturmliouville/main.tex index 2a779db..0dd3ca5 100644 --- a/buch/papers/sturmliouville/main.tex +++ b/buch/papers/sturmliouville/main.tex @@ -28,9 +28,11 @@ Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren \end{itemize} \input{papers/sturmliouville/einleitung.tex} -\input{papers/sturmliouville/teil1.tex} -\input{papers/sturmliouville/teil2.tex} -\input{papers/sturmliouville/teil3.tex} +%einleitung "was ist das sturm-liouville-problem" +ng\input{papers/sturmliouville/eigenschaften.tex} +%Eigenschaften von Lösungen eines solchen Problems +\input{papers/sturmliouville/beispiele.tex} +%Beispiele sind: Wärmeleitung in einem Stab, Tschebyscheff-Polynome \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/sturmliouville/papers/sturmliouville/2008_Chapter_Sturm-Liouville-ProblemeUndSpe.pdf b/buch/papers/sturmliouville/papers/sturmliouville/2008_Chapter_Sturm-Liouville-ProblemeUndSpe.pdf new file mode 100644 index 0000000..2237e55 Binary files /dev/null and b/buch/papers/sturmliouville/papers/sturmliouville/2008_Chapter_Sturm-Liouville-ProblemeUndSpe.pdf differ diff --git a/buch/papers/sturmliouville/teil1.tex b/buch/papers/sturmliouville/teil1.tex deleted file mode 100644 index c23c1d6..0000000 --- a/buch/papers/sturmliouville/teil1.tex +++ /dev/null @@ -1,55 +0,0 @@ -% -% teil1.tex -- Beispiel-File für das Paper -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 1 -\label{sturmliouville:section:teil1}} -\rhead{Problemstellung} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. -Nemo enim ipsam voluptatem quia voluptas sit aspernatur aut odit -aut fugit, sed quia consequuntur magni dolores eos qui ratione -voluptatem sequi nesciunt -\begin{equation} -\int_a^b x^2\, dx -= -\left[ \frac13 x^3 \right]_a^b -= -\frac{b^3-a^3}3. -\label{sturmliouville:equation1} -\end{equation} -Neque porro quisquam est, qui dolorem ipsum quia dolor sit amet, -consectetur, adipisci velit, sed quia non numquam eius modi tempora -incidunt ut labore et dolore magnam aliquam quaerat voluptatem. - -Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis -suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur? -Quis autem vel eum iure reprehenderit qui in ea voluptate velit -esse quam nihil molestiae consequatur, vel illum qui dolorem eum -fugiat quo voluptas nulla pariatur? - -\subsection{De finibus bonorum et malorum -\label{sturmliouville:subsection:finibus}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga \eqref{000tempmlate:equation1}. - -Et harum quidem rerum facilis est et expedita distinctio -\ref{sturmliouville:section:loesung}. -Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil -impedit quo minus id quod maxime placeat facere possimus, omnis -voluptas assumenda est, omnis dolor repellendus -\ref{sturmliouville:section:folgerung}. -Temporibus autem quibusdam et aut officiis debitis aut rerum -necessitatibus saepe eveniet ut et voluptates repudiandae sint et -molestiae non recusandae. -Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis -voluptatibus maiores alias consequatur aut perferendis doloribus -asperiores repellat. - - diff --git a/buch/papers/sturmliouville/teil2.tex b/buch/papers/sturmliouville/teil2.tex deleted file mode 100644 index 7fc3d2c..0000000 --- a/buch/papers/sturmliouville/teil2.tex +++ /dev/null @@ -1,40 +0,0 @@ -% -% teil2.tex -- Beispiel-File für teil2 -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 2 -\label{sturmliouville:section:teil2}} -\rhead{Teil 2} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{sturmliouville:subsection:bonorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. - - diff --git a/buch/papers/sturmliouville/teil3.tex b/buch/papers/sturmliouville/teil3.tex deleted file mode 100644 index 3aa1b40..0000000 --- a/buch/papers/sturmliouville/teil3.tex +++ /dev/null @@ -1,40 +0,0 @@ -% -% teil3.tex -- Beispiel-File für Teil 3 -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 3 -\label{sturmliouville:section:teil3}} -\rhead{Teil 3} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{sturmliouville:subsection:malorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. - - -- cgit v1.2.1 From 12b32e8ba83f426f96364327e013517f3356723a Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Tue, 26 Jul 2022 15:12:58 +0200 Subject: added .gitignore to sturm liouville folder --- buch/papers/sturmliouville/.gitignore | 1 + 1 file changed, 1 insertion(+) create mode 100644 buch/papers/sturmliouville/.gitignore (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/.gitignore b/buch/papers/sturmliouville/.gitignore new file mode 100644 index 0000000..f08278d --- /dev/null +++ b/buch/papers/sturmliouville/.gitignore @@ -0,0 +1 @@ +*.pdf \ No newline at end of file -- cgit v1.2.1 From 355193f2047a9c34e6a96281c73ed04cc8287c1e Mon Sep 17 00:00:00 2001 From: haddoucher Date: Tue, 26 Jul 2022 15:36:01 +0200 Subject: =?UTF-8?q?=C3=A4nderungen?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- buch/papers/sturmliouville/beispiele.tex | 35 +-------------- buch/papers/sturmliouville/eigenschaften.tex | 49 +-------------------- buch/papers/sturmliouville/einleitung.tex | 17 +------ buch/papers/sturmliouville/main.tex | 19 +------- ...2008_Chapter_Sturm-Liouville-ProblemeUndSpe.pdf | Bin 548591 -> 0 bytes 5 files changed, 7 insertions(+), 113 deletions(-) delete mode 100644 buch/papers/sturmliouville/papers/sturmliouville/2008_Chapter_Sturm-Liouville-ProblemeUndSpe.pdf (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/beispiele.tex b/buch/papers/sturmliouville/beispiele.tex index 7fc3d2c..d5ec3f9 100644 --- a/buch/papers/sturmliouville/beispiele.tex +++ b/buch/papers/sturmliouville/beispiele.tex @@ -3,38 +3,7 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Teil 2 +\section{Beispiele \label{sturmliouville:section:teil2}} -\rhead{Teil 2} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{sturmliouville:subsection:bonorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. - +\rhead{Beispiele} diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex index c23c1d6..6d37612 100644 --- a/buch/papers/sturmliouville/eigenschaften.tex +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -3,53 +3,8 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Teil 1 +\section{Eigenschaften von Lösungen \label{sturmliouville:section:teil1}} -\rhead{Problemstellung} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. -Nemo enim ipsam voluptatem quia voluptas sit aspernatur aut odit -aut fugit, sed quia consequuntur magni dolores eos qui ratione -voluptatem sequi nesciunt -\begin{equation} -\int_a^b x^2\, dx -= -\left[ \frac13 x^3 \right]_a^b -= -\frac{b^3-a^3}3. -\label{sturmliouville:equation1} -\end{equation} -Neque porro quisquam est, qui dolorem ipsum quia dolor sit amet, -consectetur, adipisci velit, sed quia non numquam eius modi tempora -incidunt ut labore et dolore magnam aliquam quaerat voluptatem. - -Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis -suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur? -Quis autem vel eum iure reprehenderit qui in ea voluptate velit -esse quam nihil molestiae consequatur, vel illum qui dolorem eum -fugiat quo voluptas nulla pariatur? - -\subsection{De finibus bonorum et malorum -\label{sturmliouville:subsection:finibus}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga \eqref{000tempmlate:equation1}. - -Et harum quidem rerum facilis est et expedita distinctio -\ref{sturmliouville:section:loesung}. -Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil -impedit quo minus id quod maxime placeat facere possimus, omnis -voluptas assumenda est, omnis dolor repellendus -\ref{sturmliouville:section:folgerung}. -Temporibus autem quibusdam et aut officiis debitis aut rerum -necessitatibus saepe eveniet ut et voluptates repudiandae sint et -molestiae non recusandae. -Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis -voluptatibus maiores alias consequatur aut perferendis doloribus -asperiores repellat. +\rhead{Eigenschaften von Lösungen} diff --git a/buch/papers/sturmliouville/einleitung.tex b/buch/papers/sturmliouville/einleitung.tex index 073ba6e..384a642 100644 --- a/buch/papers/sturmliouville/einleitung.tex +++ b/buch/papers/sturmliouville/einleitung.tex @@ -3,20 +3,7 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Was ist Sturm-Liouville-Problem\label{sturmliouville:section:wa}} -\rhead{Teil 0} -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua \cite{sturmliouville:bibtex}. -At vero eos et accusam et justo duo dolores et ea rebum. -Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum -dolor sit amet. - -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua. -At vero eos et accusam et justo duo dolores et ea rebum. Stet clita -kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit -amet. +\section{Was ist Sturm-Liouville-Problem\label{sturmliouville:section:teil0}} +\rhead{Einleitung} diff --git a/buch/papers/sturmliouville/main.tex b/buch/papers/sturmliouville/main.tex index 0dd3ca5..dfd2c38 100644 --- a/buch/papers/sturmliouville/main.tex +++ b/buch/papers/sturmliouville/main.tex @@ -8,24 +8,7 @@ \begin{refsection} \chapterauthor{Réda Haddouche und Erik Löffler} -Ein paar Hinweise für die korrekte Formatierung des Textes -\begin{itemize} -\item -Absätze werden gebildet, indem man eine Leerzeile einfügt. -Die Verwendung von \verb+\\+ ist nur in Tabellen und Arrays gestattet. -\item -Die explizite Platzierung von Bildern ist nicht erlaubt, entsprechende -Optionen werden gelöscht. -Verwenden Sie Labels und Verweise, um auf Bilder hinzuweisen. -\item -Beginnen Sie jeden Satz auf einer neuen Zeile. -Damit ermöglichen Sie dem Versionsverwaltungssysteme, Änderungen -in verschiedenen Sätzen von verschiedenen Autoren ohne Konflikt -anzuwenden. -\item -Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren -Übersicht wegen, aber auch um GIT die Arbeit zu erleichtern. -\end{itemize} + \input{papers/sturmliouville/einleitung.tex} %einleitung "was ist das sturm-liouville-problem" diff --git a/buch/papers/sturmliouville/papers/sturmliouville/2008_Chapter_Sturm-Liouville-ProblemeUndSpe.pdf b/buch/papers/sturmliouville/papers/sturmliouville/2008_Chapter_Sturm-Liouville-ProblemeUndSpe.pdf deleted file mode 100644 index 2237e55..0000000 Binary files a/buch/papers/sturmliouville/papers/sturmliouville/2008_Chapter_Sturm-Liouville-ProblemeUndSpe.pdf and /dev/null differ -- cgit v1.2.1 From 188c4356ae2e9431ce68b2f6332256da915b6ce9 Mon Sep 17 00:00:00 2001 From: haddoucher Date: Tue, 26 Jul 2022 15:37:27 +0200 Subject: Delete standalone.pdf --- buch/papers/sturmliouville/standalone/standalone.pdf | Bin 77574 -> 0 bytes 1 file changed, 0 insertions(+), 0 deletions(-) delete mode 100644 buch/papers/sturmliouville/standalone/standalone.pdf (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/standalone/standalone.pdf b/buch/papers/sturmliouville/standalone/standalone.pdf deleted file mode 100644 index 1b5acdb..0000000 Binary files a/buch/papers/sturmliouville/standalone/standalone.pdf and /dev/null differ -- cgit v1.2.1 From b781675d44f18b172ebdd24c8c011f75d1d30e6c Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Tue, 26 Jul 2022 15:40:05 +0200 Subject: Added missing new line in .gitignore file. --- buch/papers/sturmliouville/.gitignore | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/.gitignore b/buch/papers/sturmliouville/.gitignore index f08278d..a136337 100644 --- a/buch/papers/sturmliouville/.gitignore +++ b/buch/papers/sturmliouville/.gitignore @@ -1 +1 @@ -*.pdf \ No newline at end of file +*.pdf -- cgit v1.2.1 From 26ed6c0f968b723821c92606a1c5aa53fa274754 Mon Sep 17 00:00:00 2001 From: haddoucher Date: Tue, 26 Jul 2022 15:44:27 +0200 Subject: Update Makefile --- buch/papers/sturmliouville/Makefile | 5 ++--- 1 file changed, 2 insertions(+), 3 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/Makefile b/buch/papers/sturmliouville/Makefile index 23214a2..8d3e0af 100644 --- a/buch/papers/sturmliouville/Makefile +++ b/buch/papers/sturmliouville/Makefile @@ -6,9 +6,8 @@ SOURCES := \ einleitung.tex\ - teil1.tex \ - teil2.tex \ - teil3.tex \ + eigenschaften.tex \ + beispiele.tex \ main.tex #TIKZFIGURES := \ -- cgit v1.2.1 From 796f2b607d90c7d2aed4ac38b39830bb2a93cfea Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Tue, 26 Jul 2022 16:04:10 +0200 Subject: Added comments on what to work on. --- buch/papers/sturmliouville/beispiele.tex | 2 +- buch/papers/sturmliouville/eigenschaften.tex | 3 +-- 2 files changed, 2 insertions(+), 3 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/beispiele.tex b/buch/papers/sturmliouville/beispiele.tex index d5ec3f9..49703c9 100644 --- a/buch/papers/sturmliouville/beispiele.tex +++ b/buch/papers/sturmliouville/beispiele.tex @@ -6,4 +6,4 @@ \section{Beispiele \label{sturmliouville:section:teil2}} \rhead{Beispiele} - +% Fourier: Erik work diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex index 6d37612..a397dcc 100644 --- a/buch/papers/sturmliouville/eigenschaften.tex +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -6,5 +6,4 @@ \section{Eigenschaften von Lösungen \label{sturmliouville:section:teil1}} \rhead{Eigenschaften von Lösungen} - - +% Erik work -- cgit v1.2.1 From 250488bcb7e08beeb0d2a8b8b50c917aa12fd2a4 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Wed, 27 Jul 2022 14:08:52 +0200 Subject: Added comment to main.tex pointing to buch.tex in order to compile from sturmlouville folder directly. --- buch/papers/sturmliouville/main.tex | 1 + 1 file changed, 1 insertion(+) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/main.tex b/buch/papers/sturmliouville/main.tex index dfd2c38..4956664 100644 --- a/buch/papers/sturmliouville/main.tex +++ b/buch/papers/sturmliouville/main.tex @@ -1,3 +1,4 @@ +% !TeX root = ../../buch.tex % % main.tex -- Paper zum Thema % -- cgit v1.2.1 From c97459b91cd980d3db65c3ca1944d8998ccf7006 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Wed, 27 Jul 2022 14:46:02 +0200 Subject: Added file for fourier example. --- buch/papers/sturmliouville/waermeleitung_beispiel.tex | 8 ++++++++ 1 file changed, 8 insertions(+) create mode 100644 buch/papers/sturmliouville/waermeleitung_beispiel.tex (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex new file mode 100644 index 0000000..6cfb50f --- /dev/null +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -0,0 +1,8 @@ +% +% waermeleitung_beispiel.tex -- Beispiel Wärmeleitung in homogenem Stab. +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% + +\subsubsection{Wärmeleitung in einem Homogenen Stab} + -- cgit v1.2.1 From d9bbd9cc6541847425f1fced501b5865e2ba282e Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Wed, 27 Jul 2022 14:48:48 +0200 Subject: Adjusted labels and included new file. --- buch/papers/sturmliouville/beispiele.tex | 4 +++- buch/papers/sturmliouville/eigenschaften.tex | 2 +- buch/papers/sturmliouville/main.tex | 4 +--- 3 files changed, 5 insertions(+), 5 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/beispiele.tex b/buch/papers/sturmliouville/beispiele.tex index 49703c9..b23593e 100644 --- a/buch/papers/sturmliouville/beispiele.tex +++ b/buch/papers/sturmliouville/beispiele.tex @@ -4,6 +4,8 @@ % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % \section{Beispiele -\label{sturmliouville:section:teil2}} +\label{sturmliouville:section:examples}} \rhead{Beispiele} + % Fourier: Erik work +\input{papers/sturmliouville/waermeleitung_beispiel.tex} \ No newline at end of file diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex index a397dcc..9f20070 100644 --- a/buch/papers/sturmliouville/eigenschaften.tex +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -4,6 +4,6 @@ % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % \section{Eigenschaften von Lösungen -\label{sturmliouville:section:teil1}} +\label{sturmliouville:section:solution-properties}} \rhead{Eigenschaften von Lösungen} % Erik work diff --git a/buch/papers/sturmliouville/main.tex b/buch/papers/sturmliouville/main.tex index 4956664..d21b013 100644 --- a/buch/papers/sturmliouville/main.tex +++ b/buch/papers/sturmliouville/main.tex @@ -9,11 +9,9 @@ \begin{refsection} \chapterauthor{Réda Haddouche und Erik Löffler} - - \input{papers/sturmliouville/einleitung.tex} %einleitung "was ist das sturm-liouville-problem" -ng\input{papers/sturmliouville/eigenschaften.tex} +\input{papers/sturmliouville/eigenschaften.tex} %Eigenschaften von Lösungen eines solchen Problems \input{papers/sturmliouville/beispiele.tex} %Beispiele sind: Wärmeleitung in einem Stab, Tschebyscheff-Polynome -- cgit v1.2.1 From 6b0cb2b62e6d5da19dffc90c49d11dea48f5cdbb Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Wed, 27 Jul 2022 15:12:50 +0200 Subject: Added intro and differential equation to fourier example. --- buch/papers/sturmliouville/main.tex | 2 +- buch/papers/sturmliouville/waermeleitung_beispiel.tex | 12 ++++++++++++ 2 files changed, 13 insertions(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/main.tex b/buch/papers/sturmliouville/main.tex index d21b013..4b5b8af 100644 --- a/buch/papers/sturmliouville/main.tex +++ b/buch/papers/sturmliouville/main.tex @@ -5,7 +5,7 @@ % (c) 2020 Hochschule Rapperswil % \chapter{Sturm-Liouville-Problem\label{chapter:sturmliouville}} -\lhead{Thema} +\lhead{Sturm-Liouville-Problem} \begin{refsection} \chapterauthor{Réda Haddouche und Erik Löffler} diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 6cfb50f..64bf974 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -6,3 +6,15 @@ \subsubsection{Wärmeleitung in einem Homogenen Stab} +In diesem Abschnitt betrachten wir das Problem der Wärmeleitung in einem +homogenen Stab und wie das Sturm-Liouville-Problem bei der Beschreibung dieses +physikalischen Phänomenes auftritt. + +Zunächst wird ein eindimensionaler homogener Stab der Länge $l$ und +Wärmeleitkoeffizient $\kappa$. Somit ergibt sich für das Wärmeleitungsproblem +die partielle Differentialgleichung + +\[ + \frac{\partial u}{\partial t} = + \kappa \frac{\partial^{2}u}{{\partial x}^{2}}. +\] \ No newline at end of file -- cgit v1.2.1 From 3e57ab690350ad4ab447cdd0d263d87c414c96b5 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Wed, 27 Jul 2022 15:53:20 +0200 Subject: Added boundary condiutions for fourier example. --- .../sturmliouville/waermeleitung_beispiel.tex | 54 ++++++++++++++++++++-- 1 file changed, 49 insertions(+), 5 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 64bf974..243d0e1 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -1,10 +1,11 @@ % % waermeleitung_beispiel.tex -- Beispiel Wärmeleitung in homogenem Stab. +%%%%%%%%%%%%%%%%%%%%%%%%%%% Erster Entwurf %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\subsubsection{Wärmeleitung in einem Homogenen Stab} +\subsection{Wärmeleitung in einem Homogenen Stab} In diesem Abschnitt betrachten wir das Problem der Wärmeleitung in einem homogenen Stab und wie das Sturm-Liouville-Problem bei der Beschreibung dieses @@ -12,9 +13,52 @@ physikalischen Phänomenes auftritt. Zunächst wird ein eindimensionaler homogener Stab der Länge $l$ und Wärmeleitkoeffizient $\kappa$. Somit ergibt sich für das Wärmeleitungsproblem -die partielle Differentialgleichung - +die partielle Differentialgleichung \[ \frac{\partial u}{\partial t} = - \kappa \frac{\partial^{2}u}{{\partial x}^{2}}. -\] \ No newline at end of file + \kappa \frac{\partial^{2}u}{{\partial x}^{2}} +\] +wobei der Stab in diesem Fall auf der X-Achse im Intervall $[0,l]$ liegt. + +Da diese Differentialgleichung das Problem allgemein für einen homogenen +Stab beschreibt, werden zusätzliche Bedingungen benötigt, um beispielsweise +die Lösung für einen Stab zu finden, bei dem die Enden auf konstanter +Tempreatur gehalten werden. + +%%%%%%%%%%%%% Randbedingungen für Stab mit konstanten Endtemperaturen %%%%%%%%% + +\subsubsection{Stab mit Enden auf konstanter Temperatur} + +Die Enden des Stabes auf konstanter Temperatur zu halten bedeutet, dass die +Lösungsfunktion $u(t,x)$ bei $x = 0$ und $x = l$ nur die vorgegebene +Temperatur zurückgeben darf. Es folgen nun +\[ + u(t,0) + = + u(t,l) + = + 0 +\] +als Randbedingungen. + +%%%%%%%%%%%%% Randbedingungen für Stab mit isolierten Enden %%%%%%%%%%%%%%%%%%% + +\subsubsection{Stab mit isolierten Enden} + +Bei isolierten Enden des Stabes können belibige Temperaturen für $x = 0$ und +$x = l$ auftreten. In diesem Fall nicht erlaubt ist es, dass Wärme vom Stab +an die Umgebung oder von der Umgebung an den Stab abgegeben wird. + +Aus der Physik ist bekannt, dass Wärme immer von der höheren zur tieferen +Temperatur fliesst. Um Wärmefluss zu unterdrücken, muss also dafür gesorgt +werden, dass am Rand des Stabes keine Temperaturdifferenz existiert oder +indem die partielle Ableitung von $u(t,x)$ nach $x$ bei $x = 0$ und $x = l$ +verschwinden. Somit folgen +\[ + \frac{\partial}{\partial x} u(t, 0) + = + \frac{\partial}{\partial x} u(t, l) + = + 0 +\] +als Randbedingungen. \ No newline at end of file -- cgit v1.2.1 From d71e2db54a66ac9233757253b85eb678cc3e5f78 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Wed, 27 Jul 2022 16:19:37 +0200 Subject: Added separation for diff. eq. in fourier example. --- .../sturmliouville/waermeleitung_beispiel.tex | 48 +++++++++++++++++++++- 1 file changed, 46 insertions(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 243d0e1..cd7a620 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -52,7 +52,7 @@ an die Umgebung oder von der Umgebung an den Stab abgegeben wird. Aus der Physik ist bekannt, dass Wärme immer von der höheren zur tieferen Temperatur fliesst. Um Wärmefluss zu unterdrücken, muss also dafür gesorgt werden, dass am Rand des Stabes keine Temperaturdifferenz existiert oder -indem die partielle Ableitung von $u(t,x)$ nach $x$ bei $x = 0$ und $x = l$ +dass die partiellen Ableitungen von $u(t,x)$ nach $x$ bei $x = 0$ und $x = l$ verschwinden. Somit folgen \[ \frac{\partial}{\partial x} u(t, 0) @@ -61,4 +61,48 @@ verschwinden. Somit folgen = 0 \] -als Randbedingungen. \ No newline at end of file +als Randbedingungen. + +%%%%%%%%%%% Lösung der Differenzialgleichung %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +\subsubsection{Lösung der Differenzialgleichung} + +% TODO: Referenz Separationsmethode +% TODO: Formeln sauber in Text einbinden. + +Da die Lösungsfunktion von zwei Variablen abhängig ist, wird als Lösungsansatz +die Separationsmethode verwendet. + +\[ + u(t,x) + = + T(t)X(x) +\] +Dieser Ausdruck wird in die ursprüngliche Differenzialgleichung eingesetzt: +\[ + T^{\prime}(t)X(x) + = + \kappa T(t)X^{\prime \prime}(x) +\] +Nun können alle von $t$ abhängigen Ausdrücke auf die linke Seite, sowie alle +von $x$ abhängigen Ausdrücke auf die rechte Seite gebracht werden und mittels +der neuen Variablen $\mu$ gekoppelt werden: +\[ + \frac{T^{\prime}(t)}{\kappa T(t)} + = + \frac{X^{\prime \prime}(x)}{X(x)} + = + \mu +\] +Durch die Einführung von $\mu$ kann das Problem nun in zwei separate +Differenzialgleichungen aufgeteilt werden: +\[ + T^{\prime}(t) - \kappa \mu T(t) + = + 0 +\] +\[ + X^{\prime \prime}(x) - \mu X(x) + = + 0 +\] -- cgit v1.2.1 From 29fd344738894593ae434a271613815d1aa563ac Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Thu, 28 Jul 2022 12:56:49 +0200 Subject: Added solutions for heat conduction. --- .../sturmliouville/waermeleitung_beispiel.tex | 32 ++++++++++++++++++++++ 1 file changed, 32 insertions(+) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index cd7a620..a493749 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -106,3 +106,35 @@ Differenzialgleichungen aufgeteilt werden: = 0 \] + +% TODO: Rechenweg +TODO: Rechenweg... Enden auf konstanter Temperatur: +\[ + u(t,x) + = + \sum_{n=1}^{\infty}a_{n}e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t} + \sin\left(\frac{n\pi}{l}x\right) +\] +\[ + a_{n} + = + \frac{2}{l}\int_{0}^{l}u(0,x)sin\left(\frac{n\pi}{l}x\right) dx +\] + +TODO: Rechenweg... Enden isoliert: +\[ + u(t,x) + = + a_{0} + \sum_{n=1}^{\infty}a_{n}e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t} + \cos\left(\frac{n\pi}{l}x\right) +\] +\[ + a_{0} + = + \frac{1}{l}\int_{0}^{l}u(0,x) dx +\] +\[ + a_{n} + = + \frac{2}{l}\int_{0}^{l}u(0,x)sin\left(\frac{n\pi}{l}x\right) dx +\] -- cgit v1.2.1 From 2fa5e32a5bbb88cb0f676ac080f0bef54623599e Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Thu, 28 Jul 2022 16:22:07 +0200 Subject: Added solution for T(t) in fourier example. --- .../sturmliouville/waermeleitung_beispiel.tex | 31 +++++++++++++++++++--- 1 file changed, 28 insertions(+), 3 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index a493749..b25fc89 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -71,19 +71,20 @@ als Randbedingungen. % TODO: Formeln sauber in Text einbinden. Da die Lösungsfunktion von zwei Variablen abhängig ist, wird als Lösungsansatz -die Separationsmethode verwendet. - +die Separationsmethode verwendet. Dazu wird \[ u(t,x) = T(t)X(x) \] -Dieser Ausdruck wird in die ursprüngliche Differenzialgleichung eingesetzt: +in die ursprüngliche Differenzialgleichung eingesetzt. Daraus ergibt sich \[ T^{\prime}(t)X(x) = \kappa T(t)X^{\prime \prime}(x) \] +als neue Form. + Nun können alle von $t$ abhängigen Ausdrücke auf die linke Seite, sowie alle von $x$ abhängigen Ausdrücke auf die rechte Seite gebracht werden und mittels der neuen Variablen $\mu$ gekoppelt werden: @@ -107,6 +108,30 @@ Differenzialgleichungen aufgeteilt werden: 0 \] +Es ist an dieser Stelle zu bemerken, dass die Gleichung in $x$ in +Sturm-Liouville-Form ist. Erfüllen die Randbedingungen des Stab-Problems auch +die Randbedingungen des Sturm-Liouville-Problems, kann bereits die Aussage +getroffen werden, dass alle Lösungen für die Gleichung in $x$ orthogonal sein +werden. + +Widmen wir uns zunächst der ersten Gleichung. Diese Lösen wir über das +charakteristische Polynom +\[ + \lambda - \kappa \mu + = + 0. +\] +Es ist direkt ersichtlich, dass $\lambda = \kappa \mu$ gelten muss, was zur +Lösung +\[ + T(t) + = + e^{\kappa \mu t} +\] +führt. + +Etwas aufwändiger wird es, die zweite Gleichung zu lösen. + % TODO: Rechenweg TODO: Rechenweg... Enden auf konstanter Temperatur: \[ -- cgit v1.2.1 From 02ad63db71adf381e21c0230c502c3ead7e11ecc Mon Sep 17 00:00:00 2001 From: haddoucher Date: Fri, 29 Jul 2022 16:49:36 +0200 Subject: erste Variante Einleitung Kapitel "Was ist das Sturm-Liouville-Problem" --- buch/papers/sturmliouville/einleitung.tex | 58 ++++++++++++++++++++++++++++++- buch/papers/sturmliouville/main.tex | 1 + 2 files changed, 58 insertions(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/einleitung.tex b/buch/papers/sturmliouville/einleitung.tex index 384a642..ec37a3f 100644 --- a/buch/papers/sturmliouville/einleitung.tex +++ b/buch/papers/sturmliouville/einleitung.tex @@ -3,7 +3,63 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Was ist Sturm-Liouville-Problem\label{sturmliouville:section:teil0}} +\section{Was ist das Sturm-Liouville-Problem\label{sturmliouville:section:teil0}} \rhead{Einleitung} +Das Sturm-Liouville-Problem wurde benannt nach dem schweizerisch-französischer Mathematiker und Physiker Jacques Charles Fran\c{c}ois Sturm und dem französischer Mathematiker Joseph Liouville. +Gemeinsam haben sie in der mathematischen Physik die Sturm-Liouville-Theorie entwickelt und gilt für die Lösung von gewohnlichen Differentialgleichungen, jedoch verwendet man die Theorie öfters bei der Lösung von partiellen Differentialgleichungen. +Normalerweise betrachtet man für das Strum-Liouville-Problem eine gewöhnliche Differentialgleichung 2. Ordnung, und wenn es sich um eine partielle Differentialgleichung handelt, kann man sie mit Hilfe einiger Methoden in mehrere gewöhnliche Differentialgleichungen umwandeln, wie z. B. den Separationsansatz, die partielle Differentialgleichung mit mehreren Variablen. +Angenommen man hat die lineare homogene Differentialgleichung + +\begin{equation} + \frac{d^2y}{dx^2} + a(x)\frac{dy}{dx} + b(x)y = 0 +\end{equation} + +und schreibt die Gleichung um in: + +\begin{equation} + \label{eq:sturm-liouville-equation} + \frac{d}{dx}\lbrack p(x) \frac{dy}{dx} \rbrack + \lbrack q(x) + \lambda w(x) \rbrack y = 0 +\end{equation}, + +diese Gleichung wird dann Sturm-liouville-Gleichung bezeichnet. +Alle homogene 2.Ordnung lineare gewöhnliche Differentialgleichungen können in die Form der Gleichung \ref{eq:sturm-liouville-equation} umgeformt werden. +Die Sturm-Liouville-Theorie besagt, dass, wenn man die Sturm-Liouville-Gleichung mit den homogenen Randbedingungen + +\begin{equation} +\begin{aligned} + \label{ali:randbedingungen} + k_a y(a) + h_a p(a) y'(a) &= 0 \\ + k_b y(b) + h_b p(b) y'(b) &= 0 +\end{aligned} +\end{equation} + +kombiniert, wie schon im Kapitel \ref{sub:differentailgleichung} erwähnt, auf dem Intervall (a,b), dann bekommt man das klassische Sturm-Liouville-Problem. +Lösungen die nicht Null sind, werden nicht betrachtet und diese zwei Gleichungen (\ref{eq:sturm-liouville-equation} und \ref{ali:randbedingungen}) kombiniert, nennt man Eigenfunktionen. +Wenn bei der Sturm-Liouville-Gleichung \ref{eq:sturm-liouville-equation} alles konstant bleibt, aber der Wert von $\lambda$ sich ändert, erhält man eine andere Eigenfunktion, weil man eine andere gewöhnliche Differentialgleichung löst; +der Parameter $\lambda$ wird als Eigenwert bezeichnet. +Es ist genau das gleiche Prinzip wie bei den Matrizen, andere Eigenwerte ergeben andere Eigenvektoren. +Es besteht eine Korrespondenz zwischen den Eigenwerten und den Eigenvektoren. +Das gleiche gilt auch beim Sturm-Liouville-Problem, und zwar + +\begin{equation} + \lambda \overset{Korrespondenz}\leftrightarrow y +\end{equation}. + +Die Theorie besagt, wenn $y_m$, $y_n$ Eigenfuktionen des Sturm-Liouville-Problems sind, die verschiedene Eigenwerte $\lambda_m$, $\lambda_n$ ($\lambda_m \neq \lambda_n$) entsprechen, so sind $y_m$, $y_n$ orthogonal zu y - +dies gilt für das Intervall (a,b). +Somit ergibt die Gleichung + +\begin{equation} + \int_{a}^{b} w(x)y_m y_n = 0 +\end{equation}. + +Die Funktionen $p(x)$, $q(x)$ und $w(x)$ werden als Koeffizientenfunktionen mit ihren freien Variablen $x$ bezeichnet. Die Funktion $w(x)$ (manchmal auch $r(x)$ genannt) wird als Gewichtsfunktion oder Dichtefunktion bezeichnet. + + + + + + + diff --git a/buch/papers/sturmliouville/main.tex b/buch/papers/sturmliouville/main.tex index dfd2c38..4c25843 100644 --- a/buch/papers/sturmliouville/main.tex +++ b/buch/papers/sturmliouville/main.tex @@ -10,6 +10,7 @@ + \input{papers/sturmliouville/einleitung.tex} %einleitung "was ist das sturm-liouville-problem" ng\input{papers/sturmliouville/eigenschaften.tex} -- cgit v1.2.1 From 5c71b098ca50b4bb11f273f8c78279c8ce23ef02 Mon Sep 17 00:00:00 2001 From: daHugen Date: Sun, 31 Jul 2022 18:09:55 +0200 Subject: Update to next version includes changes in syntax and structure --- .../papers/lambertw/Bilder/VerfolgungskurveBsp.png | Bin 297455 -> 356399 bytes buch/papers/lambertw/teil4.tex | 168 +++++++++++---------- 2 files changed, 91 insertions(+), 77 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png b/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png index 90758cd..e6e7c1e 100644 Binary files a/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png and b/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png differ diff --git a/buch/papers/lambertw/teil4.tex b/buch/papers/lambertw/teil4.tex index c79aa0c..0050b61 100644 --- a/buch/papers/lambertw/teil4.tex +++ b/buch/papers/lambertw/teil4.tex @@ -6,15 +6,15 @@ \section{Beispiel einer Verfolgungskurve \label{lambertw:section:teil4}} \rhead{Beispiel einer Verfolgungskurve} -In diesem Abschnitt wird rechnerisch das Beispiel einer Verfolgungskurve mit der Verfolgungsstrategie 1 beschreiben. Dafür werden zuerst Bewegungsraum, Anfangspositionen und Bewegungsverhalten definiert, in einem nächsten Schritt soll eine Differentialgleichung dafür aufgestellt und anschliessend gelöst werden. +In diesem Abschnitt wird rechnerisch das Beispiel einer Verfolgungskurve mit der Verfolgungsstrategie ``Jagd'' beschreiben. Dafür werden zuerst Bewegungsraum, Anfangspositionen und Bewegungsverhalten definiert, in einem nächsten Schritt soll eine Differentialgleichung dafür aufgestellt und anschliessend gelöst werden. \subsection{Anfangsbedingungen definieren und einsetzen \label{lambertw:subsection:Anfangsbedingungen}} -Das zu verfolgende Ziel \(Z\) bewegt sich entlang der \(y\)-Achse mit konstanter Geschwindigkeit \(v = 1\), beginnend beim Ursprung des Kartesischen Koordinatensystems. Der Verfolger \(V\) startet auf einem beliebigen Punkt im ersten Quadranten und bewegt sich auch mit konstanter Geschwindigkeit \(|\dot{V}| = 1\) in Richtung Ziel. Diese Anfangspunkte oder Anfangsbedingungen können wie folgt formuliert werden: +Das zu verfolgende Ziel \(Z\) bewegt sich entlang der \(y\)-Achse mit konstanter Geschwindigkeit \(|\dot{z}| = 1\), beginnend beim Ursprung des Kartesischen Koordinatensystems. Der Verfolger \(V\) startet auf einem beliebigen Punkt im ersten Quadranten und bewegt sich auch mit konstanter Geschwindigkeit \(|\dot{v}| = 1\) in Richtung Ziel. Diese Anfangspunkte oder Anfangsbedingungen können wie folgt formuliert werden: \begin{equation} Z = - \left( \begin{array}{c} 0 \\ v \cdot t \end{array} \right) + \left( \begin{array}{c} 0 \\ |\dot{z}| \cdot t \end{array} \right) = \left( \begin{array}{c} 0 \\ t \end{array} \right) ,\: @@ -22,13 +22,13 @@ Das zu verfolgende Ziel \(Z\) bewegt sich entlang der \(y\)-Achse mit konstanter = \left( \begin{array}{c} x \\ y \end{array} \right) \:\text{und}\:\: - \bigl| \dot{V} \bigl| + |\dot{v}| = 1. \label{lambertw:Anfangsbed} \end{equation} Wir haben nun die Anfangsbedingungen definiert, jetzt fehlt nur noch eine DGL, welche die fortlaufende Änderung der Position und Bewegungsrichtung des Verfolgers beschreibt. -Diese DGL haben wir bereits in Kapitel \ref{lambertw:subsection:Verfolger} definiert, und zwar Gleichung \eqref{lambertw:pursuerDGL}. Wenn man die Startpunkte einfügt, ergibt sich folgender Ausdruck: +Diese DGL haben wir bereits in Kapitel \ref{lambertw:subsection:Verfolger} definiert, und zwar Gleichung \eqref{lambertw:pursuerDGL}. Wenn man die Startpunkte einfügt, ergibt sich der Ausdruck \begin{equation} \frac{\left( \begin{array}{c} 0-x \\ t-y \end{array} \right)}{\sqrt{x^2 + (t-y)^2}} \cdot @@ -42,37 +42,38 @@ Diese DGL haben wir bereits in Kapitel \ref{lambertw:subsection:Verfolger} defin \label{lambertw:subsection:DGLvereinfach}} Nun haben wir eine Gleichung, es stellt sich aber die Frage, ob es überhaupt eine geschlossene Lösung dafür gibt. Eine Funktion welche die Beziehung \(y(x)\) beschreibt oder sogar \(x(t)\) und \(y(t)\) liefert. Zum jetzigen Zeitpunkt mag es nicht trivial scheinen, aber mit den gewählten Anfangsbedingungen \eqref{lambertw:Anfangsbed} ist es möglich eine geschlossene Lösung für die Gleichung \eqref{lambertw:eqMitAnfangsbed} zu finden. -Auf dem Weg dahin muss die definierte DGL zuerst wesentlich vereinfacht werden, sei es mittels algebraischer Umformungen oder mit den Tools aus der Analysis. Da die nächsten Schritte sehr algebralastig sind und sie das Lesen dieses Papers einfach nur mühsam machen würden, werden wir uns hier nur die wesentlichsten Schritte konzentrieren, welche notwendig sind, um den Lösungsweg nachvollziehen zu können. +Auf dem Weg dahin muss die definierte DGL zuerst wesentlich vereinfacht werden, sei es mittels algebraischer Umformungen oder mit den Tools aus der Analysis. Da die nächsten Schritte sehr algebralastig sind und sie das Lesen dieses Papers träge machen würden, werden wir uns hier nur auf die wesentlichsten Schritte konzentrieren, welche notwendig sind, um den Lösungsweg nachvollziehen zu können. \subsubsection{Skalarprodukt auflösen \label{lambertw:subsubsection:SkalProdAufl}} -Zuerst müssen wir den Bruch und das Skalarprodukt in \eqref{lambertw:eqMitAnfangsbed} wegbringen, damit wir eine. Dies führt zu: +Zuerst müssen wir den Bruch und das Skalarprodukt in \eqref{lambertw:eqMitAnfangsbed} wegbringen, damit wir eine viel handlichere Differentialgleichung erhalten. Dies führt zu \begin{equation} -x \cdot \dot{x} + (t-y) \cdot \dot{y} = \sqrt{x^2 + (t-y)^2}. \label{lambertw:eqOhneSkalarprod} \end{equation} -Im letzten Schritt, fällt die Nützlichkeit des Skalarproduktes in der Verfolgungsgleichung \eqref{lambertw:pursuerDGL} markant auf. Anstatt zwei gekoppelte Differentialgleichungen zu erhalten, eine für die \(x\) und die andere für die \(y\)-Komponente, erhält man einen einzigen Ausdruck, was in der Regel mit weniger Lösungsaufwand verbunden ist. +Im letzten Schritt, fällt die Nützlichkeit des Skalarproduktes in der Verfolgungsgleichung \eqref{lambertw:pursuerDGL} markant auf. Anstatt zwei gekoppelte Differentialgleichungen zu erhalten, eine für die \(x\)- und die andere für die \(y\)-Komponente, erhält man einen einzigen Ausdruck, was in der Regel mit weniger Lösungsaufwand verbunden ist. \subsubsection{Quadrieren und Gruppieren \label{lambertw:subsubsection:QuadUndGrup}} -Mit der Quadratwurzel in \ref{lambertw:eqOhneSkalarprod} kann man nichts anfangen, sie steht nur im Weg, also muss man sie loswerden. Wenn man dies macht, kann \eqref{lambertw:eqOhneSkalarprod} auf folgende Form gebracht werden: +Mit der Quadratwurzel in \eqref{lambertw:eqOhneSkalarprod} kann man nichts anfangen, sie steht nur im Weg, also muss man sie loswerden. Wenn man dies macht, kann \eqref{lambertw:eqOhneSkalarprod} auf die Form \begin{equation} \left(\dot{x}^2-1\right) \cdot x^2 -2x \left(t-y\right) \dot{x}\dot{y} + \left(\dot{y}^2-1\right) \cdot \left(t-y\right)^2 - =0. + =0 \label{lambertw:eqOhneWurzel} \end{equation} +gebracht werden. Diese Form mag auf den ersten Blick nicht gerade nützlich sein, aber man kann sie mit einer Substitution weiter vereinfachen. \subsubsection{Wichtige Substitution \label{lambertw:subsubsection:WichtSubst}} -Wenn man beachtet, dass die Geschwindigkeit des Verfolgers konstant und gleich 1 ist, dann kann man folgende Gleichung aufstellen: +Wenn man beachtet, dass die Geschwindigkeit des Verfolgers konstant und gleich 1 ist, dann ergibt sich die Beziehung \begin{equation} \dot{x}^2 + \dot{y}^2 = 1. \label{lambertw:eqGeschwSubst} \end{equation} -Umformungen der Gleichung \eqref{lambertw:eqGeschwSubst} können in \eqref{lambertw:eqOhneWurzel} erkannt werden. Ersetzt führen sie zu folgendem Ausdruck: +Umformungen der Gleichung \eqref{lambertw:eqGeschwSubst} können in \eqref{lambertw:eqOhneWurzel} erkannt werden. Wenn man sie ersetzt, erhält man \begin{equation} \dot{y}^2 \cdot x^2 +2x \left(t-y\right) \dot{x}\dot{y} + \dot{x}^2 \cdot \left(t-y\right)^2 =0. @@ -82,27 +83,31 @@ Diese unscheinbare Substitution führt dazu, dass weitere Vereinfachungen durchg \subsubsection{Binom erkennen und vereinfachen \label{lambertw:subsubsection:BinomVereinfach}} -Versteckt im Ausdruck \eqref{lambertw:eqGeschwSubstituiert} befindet sich die erste binomische Formel, welche zu folgender Gleichung führt: +Versteckt im Ausdruck \eqref{lambertw:eqGeschwSubstituiert} befindet sich die erste binomische Formel, wobei \begin{equation} (x \dot{y} + (t-y) \dot{x})^2 - = 0. + = 0 \label{lambertw:eqAlgVerinfacht} \end{equation} -Da der linke Term gleich Null ist, muss auch der Inhalt des Quadrates gleich Null sein, somit folgt eine weitere Vereinfachung, welche zu einer im Vergleich zu \eqref{lambertw:eqOhneSkalarprod} wesentlich einfacheren DGL führt: +die faktorisierte Darstellung davon ist. +Da der linke Term gleich Null ist, muss auch der Inhalt des Quadrates gleich Null sein. Es ergibt sich eine weitere Vereinfachung, welche zu der im Vergleich zu \eqref{lambertw:eqOhneSkalarprod} wesentlich einfacheren DGL \begin{equation} x \dot{y} + (t-y) \dot{x} - = 0. + = 0 \label{lambertw:eqGanzVerinfacht} \end{equation} -Kompakt, ohne Wurzelterme und Quadrate, nur elementare Operationen und Ableitungen. Nun stellt sich die Frage wie es weiter gehen soll, bei der Gleichung \eqref{lambertw:eqGanzVerinfacht} scheinen keine weiteren Vereinfachungen möglich zu sein. Wir brauchen einen neuen Ansatz, um unser Ziel einer möglichen Lösung zu verfolgen. +führt. +Kompakt, ohne Wurzelterme und Quadrate, nur elementare Operationen und Ableitungen. + +Nun stellt sich die Frage wie es weiter gehen soll, bei der Gleichung \eqref{lambertw:eqGanzVerinfacht} scheinen keine weiteren Vereinfachungen möglich zu sein. Wir brauchen einen neuen Ansatz, um unser Ziel einer möglichen Lösung zu verfolgen. \subsection{Zeitabhängigkeit loswerden \label{lambertw:subsection:ZeitabhLoswerden}} -Der nächste logischer Schritt scheint irgendwie die Zeitabhängigkeit in der Gleichung \eqref{lambertw:eqGanzVerinfacht} loszuwerden, aber wieso? Nun, wie am Anfang von Abschnitt \ref{lambertw:subsection:DGLvereinfach} beschrieben, suchen wir eine Lösung der Art \(y(x)\), dies ist natürlich erst möglich wenn wir die Abhängigkeit nach \(t\) eliminieren können. +Der nächste logische Schritt scheint irgendwie die Zeitabhängigkeit in der Gleichung \eqref{lambertw:eqGanzVerinfacht} loszuwerden, aber wieso? Nun, wie am Anfang von Abschnitt \ref{lambertw:subsection:DGLvereinfach} beschrieben, suchen wir eine Lösung der Art \(y(x)\), dies ist natürlich erst möglich wenn wir die Abhängigkeit nach \(t\) eliminieren können. \subsubsection{Zeitliche Ableitungen loswerden \label{lambertw:subsubsection:ZeitAbleit}} -Der erste Schritt auf dem Weg zur Funktion \(y(x)\), ist es die zeitlichen Ableitungen los zu werden, dafür wird \eqref{lambertw:eqGanzVerinfacht} beidseitig mit \(\dot{x}\) dividiert, was erlaubt ist, weil diese Änderung ungleich Null ist: +Der erste Schritt auf dem Weg zur Funktion \(y(x)\) ist, die zeitlichen Ableitungen los zu werden, dafür wird \eqref{lambertw:eqGanzVerinfacht} beidseitig durch \(\dot{x}\) dividiert, was erlaubt ist, weil diese Änderung ungleich Null ist: \begin{equation} x \frac{\dot{y}}{\dot{x}} + (t-y) \frac{\dot{x}}{\dot{x}} = 0. @@ -126,30 +131,31 @@ Nach dem die Eigenschaft \eqref{lambertw:eqQuotZeitAbleit} in \eqref{lambertw:eq \subsubsection{Variable \(t\) eliminieren \label{lambertw:subsubsection:ZeitAbleit}} -Hier wäre es natürlich passend, wenn man die Abhängigkeit nach \(t\) komplett wegbringen könnte. Um dies zu erreichen, muss man auf die Definition der Bogenlänge zurückgreifen. -Die Strecke \(s\) entspricht +Hier wäre es natürlich passend, wenn man die Abhängigkeit nach \(t\) komplett wegbringen könnte, aber wie? +Wir wissen, dass sich der Verfolger mit Geschwindigkeit 1 bewegt, also legt er in der Zeit \(t\) die Strecke \(1\cdot t = t\) zurück. Längen und Strecken können auch mit der Bogenlänge repräsentiert werden, somit kann Zeit und zurückgelegte Strecke in der Gleichung \begin{equation} s = - v \cdot t + |\dot{v}| \cdot t = 1 \cdot t = t = - \int_{\displaystyle x_0}^{\displaystyle x_{\text{end}}}\sqrt{1+y^{\prime\, 2}} \: dx. + \int_{\displaystyle x_0}^{\displaystyle x_{\text{end}}}\sqrt{1+y^{\prime\, 2}} \: dx \label{lambertw:eqZuBogenlaenge} \end{equation} - +verbunden werden. + Nicht gerade auffällig ist die Richtung, in welche hier integriert wird. Wenn der Verfolger sich wie vorgesehen am Anfang im ersten Quadranten befindet, dann muss sich dieser nach links bewegen, was nicht der üblichen Integrationsrichtung entspricht. Um eine Integration wie üblich von links nach rechts ausführen zu können, müssen die Integrationsgenerzen vertauscht werden, was in einem Vorzeichenwechsel resultiert. -Wenn man nun \eqref{lambertw:eqZuBogenlaenge} in die DGL \eqref{lambertw:DGLmitT} einfügt, dann ergibt sich folgender Ausdruck: +Wenn man nun \eqref{lambertw:eqZuBogenlaenge} in die DGL \eqref{lambertw:DGLmitT} einfügt, dann ergibt sich der neue Ausdruck \begin{equation} x y^{\prime} - \int\sqrt{1+y^{\prime\, 2}} \: dx - y = 0. \label{lambertw:DGLohneT} \end{equation} -Um das Integral los zu werden, leitet man den vorherigen Ausdruck \eqref{lambertw:DGLohneT} nach \(x\) ab und erhaltet folgende DGL zweiter Ordnung \eqref{lambertw:DGLohneInt}: +Um das Integral los zu werden, leitet man \eqref{lambertw:DGLohneT} nach \(x\) ab und erhält die DGL zweiter Ordnung \begin{align} y^{\prime}+ xy^{\prime\prime} - \sqrt{1+y^{\prime\, 2}} - y^{\prime} &= 0, \\ @@ -157,16 +163,17 @@ Um das Integral los zu werden, leitet man den vorherigen Ausdruck \eqref{lambert &= 0. \label{lambertw:DGLohneInt} \end{align} -Nun sind wir unserem Ziel einen weiteren Schritt näher. Die Gleichung \eqref{lambertw:DGLohneInt} mag auf den ersten Blick nicht gerade einfach sein, aber im Nächsten Abschnitt werden wir sehen, dass sie relativ einfach zu lösen ist. +Nun sind wir unserem Ziel einen weiteren Schritt näher. Die Gleichung \eqref{lambertw:DGLohneInt} mag auf den ersten Blick nicht gerade einfach sein, aber im nächsten Abschnitt werden wir sehen, dass sie relativ einfach zu lösen ist. \subsection{Differentialgleichung lösen \label{lambertw:subsection:DGLloes}} -Die Gleichung \eqref{lambertw:DGLohneInt} ist eine DGL zweiter Ordnung, in der \(y\) nicht vorkommt. Sie kann mittels der Substitution \(y^{\prime} = u\) in eine DGL erster Ordnung umgewandelt werden: +Die Gleichung \eqref{lambertw:DGLohneInt} ist eine DGL zweiter Ordnung, in der \(y\) nicht vorkommt. Sie kann mittels der Substitution \(y^{\prime} = u\) in die DGL \begin{equation} xu^{\prime} - \sqrt{1+u^2} - = 0. + = 0 \label{lambertw:DGLmitU} \end{equation} +erster Ordnung umgewandelt werden. Diese Gleichung ist separierbar, was sie viel handlicher macht. In der separierten Form \begin{equation} \int{\frac{1}{\sqrt{1+u^2}}\:du} @@ -174,7 +181,7 @@ Diese Gleichung ist separierbar, was sie viel handlicher macht. In der separiert \int{\frac{1}{x}\:dx}, \end{equation} lässt sich die Gleichung mittels einer Integrationstabelle sehr rasch lösen. -Mit dem Ergebnis: +Das Ergebnis ist \begin{align} \operatorname{arsinh}(u) &= @@ -184,20 +191,23 @@ Mit dem Ergebnis: \operatorname{sinh}(\operatorname{ln}(x) + C). \label{lambertw:loesDGLmitU} \end{align} -Wenn man in \eqref{lambertw:loesDGLmitU} die Substitution rückgängig macht, erhält man folgende DGL erster Ordnung, die bereits separiert ist: +Wenn man in \eqref{lambertw:loesDGLmitU} die Substitution rückgängig macht, erhält man die DGL \begin{equation} y^{\prime} = - \operatorname{sinh}(\operatorname{ln}(x) + C). + \operatorname{sinh}(\operatorname{ln}(x) + C) \label{lambertw:loesDGLmitY} \end{equation} -Ersetzt man den \(\operatorname{sinh}\) mit seiner exponentiellen Definition \(\operatorname{sinh}(x)=\frac{1}{2}(e^x-e^{-x})\), so resultiert auf sehr einfache Art folgende Lösung für \eqref{lambertw:loesDGLmitY}: +erster Ordnung, die bereits separiert ist. +Ersetzt man den \(\operatorname{sinh}\) durch seine exponentiellen Definition \(\operatorname{sinh}(x)=\frac{1}{2}(e^x-e^{-x})\), so resultiert auf sehr einfache Art die Lösung \begin{equation} y = - C_1 + C_2 x^2 - \frac{\operatorname{ln}(x)}{8 \cdot C_2}. + C_1 + C_2 x^2 - \frac{\operatorname{ln}(x)}{8 \cdot C_2} \end{equation} -Nun haben wir eine Lösung, aber wie es immer mit Lösungen ist, stellt sich die Frage, ob sie überhaupt plausibel ist. Dieser Frage werden wir im nächsten Abschnitt nachgehen. +für \eqref{lambertw:loesDGLmitY}. + +Nun haben wir eine Lösung, aber wie es immer mit Lösungen ist, stellt sich die Frage, ob sie überhaupt plausibel ist. \subsection{Lösung analysieren \label{lambertw:subsection:LoesAnalys}} @@ -210,37 +220,34 @@ Nun haben wir eine Lösung, aber wie es immer mit Lösungen ist, stellt sich die } \end{figure} -Das Resultat, wie ersichtlich, ist folgende Funktion \eqref{lambertw:funkLoes} welche mittels Anfangsbedingungen parametrisiert werden kann: +Das Resultat, wie ersichtlich, ist die Funktion \begin{equation} {\color{red}{y(x)}} = - C_1 + C_2 {\color{darkgreen}{x^2}} {\color{blue}{-}} \frac{\color{blue}{\operatorname{ln}(x)}}{8 \cdot C_2}. + C_1 + C_2 {\color{darkgreen}{x^2}} {\color{blue}{-}} \frac{\color{blue}{\operatorname{ln}(x)}}{8 \cdot C_2}, \label{lambertw:funkLoes} \end{equation} -Für die Koeffizienten \(C_1\) und \(C_2\) ergibt sich ein Anfangswertproblem, welches für deren Bestimmung gelöst werden muss. Zuerst soll aber eine qualitative Intuition oder Idee für das Aussehen der Funktion \(y(x)\) geschaffen werden: +für welche die Koeffizienten \(C_1\) und \(C_2\) aus den Anfangsbedingungen bestimmt werden können. Zuerst soll aber eine qualitative Intuition oder Idee für das Aussehen der Funktion \(y(x)\) geschaffen werden: \begin{itemize} \item Für grosse \(x\)-Werte, welche in der Regel in der Nähe von \(x_0\) sein sollten, ist der quadratisch Term in der Funktion \eqref{lambertw:funkLoes} dominant. \item - Für immer kleiner werdende \(x\) geht der Verfolger in Richtung \(y\)-Achse, wobei seine Steigung stetig sinkt, was Sinn macht wenn der Verfolgte entlang der \(y\)-Achse steigt. Irgendwann werden Verfolger und Ziel auf gleicher Höhe sein, also gleiche \(y\) aber verschiedene \(x\)-Koordinate besitzen. + Für immer kleiner werdende \(x\) geht der Verfolger in Richtung \(y\)-Achse, wobei seine Steigung stetig sinkt, was Sinn macht wenn der Verfolgte entlang der \(y\)-Achse steigt. Irgendwann werden Verfolger und Ziel auf gleicher Höhe sein, also gleiche \(y\)- aber verschiedene \(x\)-Koordinate besitzen. + In diesem Punkt findet ein Monotoniewechsel in der Kurve \eqref{lambertw:funkLoes} statt, was zu einem Minimum führt. \item Für \(x\)-Werte in der Nähe von \(0\) ist das asymptotische Verhalten des Logarithmus dominant, dies macht auch Sinn, da sich der Verfolgte auf der \(y\)-Achse bewegt und der Verfolger ihm nachgeht. - \item - Aufgrund des Monotoniewechsels in der Kurve \eqref{lambertw:funkLoes} muss diese auch ein Minimum aufweisen. Es stellt sich nun die Frage: Wo befindet sich dieser Punkt? - Eine Abschätzung darüber kann getroffen werden und zwar, dass dieser dann entsteht, wenn \(A\) und \(P\) die gleiche \(y\)-Koordinaten besitzen. In diesem Moment ändert die Richtung der \(y\)-Komponente der Geschwindigkeit des Verfolgers, somit auch sein Vorzeichen und dadurch entsteht auch das Minimum. \end{itemize} Alle diese Eigenschaften stimmen mit dem überein, was man von einer Kurve dieser Art erwarten würde, welche durch die Grafik \ref{lambertw:BildFunkLoes} repräsentiert wurde. \subsection{Anfangswertproblem \label{lambertw:subsection:AllgLoes}} -In diesem Abschnitt soll eine Parameterfunktion hergeleitet werden, bei der jeder beliebige Anfangspunkt im ersten Quadranten eingesetzt werden kann, ausser der Ursprung im Koordinatensystem. Diese Aufgabe erfordert ein Anfangswertproblem. +In diesem Abschnitt soll eine Parameterfunktion hergeleitet werden, bei der jeder beliebige Anfangspunkt im ersten Quadranten eingesetzt werden kann, ausser der Ursprung im Koordinatensystem. Diese Aufgabe ist ein Anfangswertproblem für \(y(x)\). -Das Lösen des Anfangswertproblems ist ein Problem aus der Algebra, auf welches hier nicht explizit eingegangen wird. Zur Vollständigkeit und Nachvollziehbarkeit, wird aber das Gleichungssystem präsentiert, welches notwendig ist, um das Anfangswertproblem zu lösen. +Das Lösen des Anfangswertproblems ist ein Problem aus der Analysis, auf welches hier nicht explizit eingegangen wird. Zur Vollständigkeit und Nachvollziehbarkeit, wird aber das Gleichungssystem präsentiert, welches notwendig ist, um das Anfangswertproblem zu lösen. \subsubsection{Anfangswerte bestimmen \label{lambertw:subsubsection:Anfangswerte}} -Der erste Schritt auf dem Weg zur gesuchten Parameterfunktion ist, die Anfangswerte \eqref{lambertw:eq1Anfangswert} zu definieren. -Die Anfangswerte sind: +Der erste Schritt auf dem Weg zur gesuchten Parameterfunktion ist, die Anfangswerte \begin{equation} y(x)\big \vert_{t=0} = @@ -255,15 +262,17 @@ und = y^{\prime}(x_0) = - \frac{y_0}{x_0}. + \frac{y_0}{x_0} \label{lambertw:eq2Anfangswert} \end{equation} +zu definieren. Der zweite Anfangswert \eqref{lambertw:eq2Anfangswert} mag nicht grade offensichtlich sein. Die Erklärung dafür ist aber simpel: Der Verfolger wird sich zum Zeitpunkt \(t=0\) in Richtung Koordinatenursprung bewegen wollen, wo sich das Ziel befindet. Somit entsteht das Steigungsdreieck mit \(\Delta x = x_0\) und \(\Delta y = y_0\). \subsubsection{Gleichungssystem aufstellen und lösen \label{lambertw:subsubsection:GlSys}} -Wenn man die Anfangswerte \eqref{lambertw:eq1Anfangswert} und \eqref{lambertw:eq2Anfangswert} in die Gleichung \eqref{lambertw:funkLoes} und deren Ableitung \(y^{\prime}(x)\) einsetzt, dann ergibt sich folgendes Gleichungssystem: +Wenn man die Anfangswerte \eqref{lambertw:eq1Anfangswert} und \eqref{lambertw:eq2Anfangswert} in die Gleichung \eqref{lambertw:funkLoes} und deren Ableitung \(y^{\prime}(x)\) einsetzt, dann ergibt sich das Gleichungssystem \begin{subequations} + \label{lambertw:eqGleichungssystem} \begin{align} y_0 &= @@ -272,9 +281,8 @@ Wenn man die Anfangswerte \eqref{lambertw:eq1Anfangswert} und \eqref{lambertw:eq &= 2 \cdot C_2 x_0 - \frac{1}{8 \cdot C_2 \cdot x_0}. \end{align} - \label{lambertw:eqGleichungssystem} \end{subequations} -Damit die gesuchte Funktion im ersten Quadranten bleibt, werden nur die positiven Lösungen des Gleichungssystems gewählt, welche wie folgt aussehen: +Damit die gesuchte Funktion im ersten Quadranten bleibt, werden nur die positiven Lösungen \begin{subequations} \begin{align} \label{lambertw:eqKoeff1} @@ -284,16 +292,17 @@ Damit die gesuchte Funktion im ersten Quadranten bleibt, werden nur die positive \label{lambertw:eqKoeff2} C_2 &= - \frac{\sqrt{x_0^2 + y_0^2} + y_0}{4x_0^2}. + \frac{\sqrt{x_0^2 + y_0^2} + y_0}{4x_0^2} \end{align} \end{subequations} +des Gleichungssystems gewählt. \subsubsection{Gesuchte Parameterfunktion aufstellen \label{lambertw:subsubsection:ParamFunk}} -Wenn man die Koeffizienten \eqref{lambertw:eqKoeff1} und \eqref{lambertw:eqKoeff2} in die Funktion \eqref{lambertw:funkLoes} einsetzt, dann ergibt sich nach dem Vereinfachen die gesuchte Parameterfunktion: +Wenn man die Koeffizienten \eqref{lambertw:eqKoeff1} und \eqref{lambertw:eqKoeff2} in die Funktion \eqref{lambertw:funkLoes} einsetzt, dann ergibt sich beim Vereinfachen die gesuchte Parameterfunktion \begin{equation} y(x) = - \frac{1}{4}\left(\left(y_0+r_0\right)\eta+\left(r_0-y_0\right)\operatorname{ln}\left(\eta\right)-r_0+3y_0\right). + \frac{1}{4}\left(\left(y_0+r_0\right)\eta+\left(y_0-r_0\right)\operatorname{ln}\left(\eta\right)-r_0+3y_0\right). \label{lambertw:eqAllgLoes} \end{equation} Damit die Funktion \eqref{lambertw:eqAllgLoes} trotzdem übersichtlich bleibt, wurden Anfangssteigung \(\eta\) und Anfangsentfernung \(r_0\) wie folgt definiert: @@ -316,27 +325,28 @@ In diesem Abschnitt werden algebraischen Umformungen ein wenig detaillierter als \subsubsection{Zeitabhängigkeit wiederherstellen \label{lambertw:subsubsection:ZeitabhWiederherst}} -Der erste Schritt ist es herauszufinden, wie die Zeitabhängigkeit wieder hineingebracht werden kann. Dafür greifen wir auf die letzte Gleichung zu, in welcher \(t\) noch enthalten war, und zwar DGL \eqref{lambertw:DGLmitT}, welche zur Übersichtlichkeit hier nochmals aufgeführt wird: +Der erste Schritt ist es herauszufinden, wie die Zeitabhängigkeit wieder hineingebracht werden kann. Dafür greifen wir auf die letzte Gleichung zu, in welcher \(t\) noch enthalten war, und zwar DGL \begin{equation} x y^{\prime} + t - y - = 0. + = 0 \label{lambertw:eqDGLmitTnochmals} \end{equation} +aus dem Abschnitt \eqref{lambertw:subsection:ZeitabhLoswerden}, welche zur Übersichtlichkeit hier nochmals aufgeführt wurde. Wie in \eqref{lambertw:eqDGLmitTnochmals} zu sehen ist, werden \(y\) und deren Ableitung \(y^{\prime}\) benötigt, diese sind: \begin{subequations} + \label{lambertw:eqFunkUndAbleit} \begin{align} + \label{lambertw:eqFunkUndAbleit1} y &= - \frac{1}{4}\left(\left(y_0+r_0\right)\eta+\left(r_0-y_0\right)\operatorname{ln}\left(\eta\right)-r_0+3y_0\right), \\ - \label{lambertw:eqFunkUndAbleit1} + \frac{1}{4}\left(\left(y_0+r_0\right)\eta+\left(y_0-r_0\right)\operatorname{ln}\left(\eta\right)-r_0+3y_0\right), \\ y^\prime &= - \frac{1}{2}\left(\left(y_0+r_0\right)\frac{x}{x_0^2}+\left(r_0-y_0\right)\frac{1}{x}\right). + \frac{1}{2}\left(\left(y_0+r_0\right)\frac{x}{x_0^2}+\left(y_0-r_0\right)\frac{1}{x}\right). \end{align} - \label{lambertw:eqFunkUndAbleit} \end{subequations} -Wenn man diese Gleichungen \ref{lambertw:eqFunkUndAbleit} in die DGL \label{lambertw:eqDGLmitTnochmals} einfügt, vereinfacht und nach \(t\) auflöst, dann ergibt sich folgenden Ausdruck: +Wenn man diese Gleichungen \eqref{lambertw:eqFunkUndAbleit} in die DGL \eqref{lambertw:eqDGLmitTnochmals} einfügt, vereinfacht und nach \(t\) auflöst, dann ergibt sich der Ausdruck \begin{equation} -4t = @@ -348,17 +358,20 @@ Wenn man diese Gleichungen \ref{lambertw:eqFunkUndAbleit} in die DGL \label{lamb \label{lambertw:subsubsection:UmformBisZumZiel}} Mit dem Ausdruck \eqref{lambertw:eqFunkUndAbleitEingefuegt}, welcher Terme mit \(x\) und \(t\) verbindet, kann nun nach der gesuchten Variable \(x\) aufgelöst werden. - -In einem nächsten Schritt wird alles mit \(x\) auf die eine Seite gebracht, der Rest auf die andere Seite und anschliessend beidseitig exponentiert, was wie folgt aussieht: -\begin{align} +In einem nächsten Schritt wird alles mit \(x\) auf die eine Seite gebracht, der Rest auf die andere Seite und anschliessend beidseitig exponenziert, sodass man +\begin{equation} -4t+\left(y_0+r_0\right) - &= - \left(y_0+r_0\right)\eta+\left(r_0-y_0\right)\operatorname{ln}\left(\eta\right), \\ + = + \left(y_0+r_0\right)\eta+\left(r_0-y_0\right)\operatorname{ln}\left(\eta\right) +\end{equation} +und anschliessend +\begin{equation} e^{\displaystyle -4t+\left(y_0+r_0\right)} - &= - e^{\displaystyle \left(y_0+r_0\right)\eta}\cdot\eta^{\displaystyle \left(r_0-y_0\right)}. + = + e^{\displaystyle \left(y_0+r_0\right)\eta}\cdot\eta^{\displaystyle \left(r_0-y_0\right)} \label{lambertw:eqMitExp} -\end{align} +\end{equation} +erhält. Auf dem rechten Term von \eqref{lambertw:eqMitExp} beginnen wir langsam eine ähnliche Struktur wie \(\eta e^\eta\) zu erkennen, dies schreit nach der Struktur die benötigt wird um \(\eta\) mittels der Lambert-\(W\)-Funktion \(W(x)\) zu erhalten. Dies macht durchaus Sinn, wenn wir die Funktion \(x(t)\) finden wollen und \(W(x)\) die Umkehrfunktion von \(x e^x\) ist. Die erste Sache die uns in \eqref{lambertw:eqMitExp} stört ist, dass \(\eta\) als Potenz da steht. Dieses Problem können wir loswerden, indem wir beidseitig mit \(\:\displaystyle \frac{1}{r_0-y_0}\:\) potenzieren: @@ -368,30 +381,32 @@ Die erste Sache die uns in \eqref{lambertw:eqMitExp} stört ist, dass \(\eta\) a \eta\cdot \operatorname{exp}\left(\displaystyle \frac{y_0+r_0}{r_0-y_0}\eta\right). \label{lambertw:eqOhnePotenz} \end{equation} -Das nächste Problem auf welches wir in \eqref{lambertw:eqOhnePotenz} treffen ist, dass \(\eta\) nicht alleine im Exponent steht. Dies kann elegant mit folgender Substitution gelöst werden: +Das nächste Problem auf welches wir in \eqref{lambertw:eqOhnePotenz} treffen ist, dass \(\eta\) nicht alleine im Exponent steht. Dies kann elegant mit der Substitution \begin{equation} \chi = - \frac{y_0+r_0}{r_0-y_0}. + \frac{y_0+r_0}{r_0-y_0} \label{lambertw:eqChiSubst} \end{equation} +gelöst werden. Es gäbe natürlich andere Substitutionen wie z.B. \[\displaystyle \chi=\frac{y_0+r_0}{r_0-y_0}\cdot\eta,\] -die auf dasselbe Ergebnis führen würden, aber \eqref{lambertw:eqChiSubst} liefert in einem Schritt die kompakteste Lösung. Also fahren wir mit der Substitution \eqref{lambertw:eqChiSubst} weiter, setzen diese in die Gleichung \eqref{lambertw:eqOhnePotenz} ein und multiplizieren beidseitig mit \(\chi\). Daraus erhalten wir folgende Gleichung: +die auf dasselbe Ergebnis führen würden, aber \eqref{lambertw:eqChiSubst} liefert in einem Schritt die kompakteste Lösung. Also fahren wir mit der Substitution \eqref{lambertw:eqChiSubst} weiter, setzen diese in die Gleichung \eqref{lambertw:eqOhnePotenz} ein und multiplizieren beidseitig mit \(\chi\). Daraus erhalten wir die Gleichung \begin{equation} \chi\cdot \operatorname{exp}\left(\displaystyle \chi-\frac{4t}{r_0-y_0}\right) = \chi\eta\cdot e^{\displaystyle \chi\eta}. \label{lambertw:eqNachSubst} \end{equation} -Nun sind wir endlich soweit, dass wir die angedeutete Lambert-\(W\)-Funktion \(W(x)\)einsetzen können. Wenn wir beidseitig \(W(x)\) anwenden, dann erhalten wir folgenden Ausdruck: +Nun sind wir endlich soweit, dass wir die angedeutete Lambert-\(W\)-Funktion \(W(x)\) einsetzen können. Wenn wir beidseitig \(W(x)\) anwenden, dann erhalten wir den Ausdruck \begin{equation} W\left(\chi\cdot \operatorname{exp}\left(\displaystyle \chi-\frac{4t}{r_0-y_0}\right)\right) = \chi\eta. \end{equation} -Nach dem Auflösen nach \(x\) welches in \(\eta\) enthalten ist, erhalten wir die gesuchte \(x(t)\)-Funktion \eqref{lambertw:eqFunkXNachT}. Dieses \(x(t)\) in Kombination mit \eqref{lambertw:eqFunkUndAbleit1} liefert die Position des Verfolgers zu jedem Zeitpunkt. Das Gleichungspaar \eqref{lambertw:eqFunktionenNachT}, besteht aus folgenden Gleichungen: +Nach dem Auflösen nach \(x\) welches in \(\eta\) enthalten ist, erhalten wir die gesuchte \(x(t)\)-Funktion \eqref{lambertw:eqFunkXNachT}. Dieses \(x(t)\) in Kombination mit \eqref{lambertw:eqFunkUndAbleit1} liefert die Position des Verfolgers zu jedem Zeitpunkt. Das Gleichungspaar besteht also aus den Gleichungen \begin{subequations} + \label{lambertw:eqFunktionenNachT} \begin{align} \label{lambertw:eqFunkXNachT} x(t) @@ -402,15 +417,14 @@ Nach dem Auflösen nach \(x\) welches in \(\eta\) enthalten ist, erhalten wir di = y(t) &= - \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(r_0-y_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right). + \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right). \end{align} - \label{lambertw:eqFunktionenNachT} \end{subequations} Nun haben wir unser letztes Ziel erreicht und sind in der Lage eine Verfolgung rechnerisch sowie graphisch zu repräsentieren. \subsubsection{Hinweise zur Lambert-\(W\)-Funktion \label{lambertw:subsubsection:HinwLambertW}} -Wir sind aber noch nicht ganz fertig, eine Frage muss noch beantwortet werden. Und zwar wieso, dass man schon bei der Gleichung \eqref{lambertw:eqFunkUndAbleitEingefuegt} weiss, dass die Lambert-\(W\)-Funktion zum Einsatz kommen wird. +Wir sind aber noch nicht ganz fertig, eine Frage muss noch beantwortet werden. Und zwar wieso, man schon bei der Gleichung \eqref{lambertw:eqFunkUndAbleitEingefuegt} weiss, dass die Lambert-\(W\)-Funktion zum Einsatz kommen wird. Nun, der Grund dafür ist die Struktur \begin{equation} y @@ -420,4 +434,4 @@ Nun, der Grund dafür ist die Struktur \end{equation} bei welcher \(p(x)\) eine beliebige Potenz von \(x\) darstellt. -Jedes Mal wenn \(x\) gesucht ist und in einer Struktur der Art \eqref{lambertw:eqEinsatzLambW} vorkommt, dann kann mit ein paar Umformungen die Struktur \(f(x)e^{f(x)}\) erzielt werden. Wie bereits in diesem Abschnitt \ref{lambertw:subsection:FunkNachT} gezeigt wurde, kann \(x\) nun mittels der \(W(x)\)-Funktion aufgelöst werden. Erstaunlicherweise ist \eqref{lambertw:eqEinsatzLambW} eine Struktur die oftmals vorkommt, was die Lambert-\(W\)-Funktion so wichtig macht. \ No newline at end of file +Jedes Mal wenn \(x\) gesucht ist und in einer Struktur der Art \eqref{lambertw:eqEinsatzLambW} vorkommt, dann kann mit ein paar Umformungen die Struktur \(f(x)e^{f(x)}\) erzielt werden. Wie bereits in diesem Abschnitt \ref{lambertw:subsection:FunkNachT} gezeigt wurde, kann \(x\) nun mittels der \(W(x)\)-Funktion aufgelöst werden. Erstaunlicherweise ist \eqref{lambertw:eqEinsatzLambW} eine Struktur die oft vorkommt, was die Lambert-\(W\)-Funktion so wichtig macht. \ No newline at end of file -- cgit v1.2.1 From 435a9b21bad8244ea81f63cf4254d85212942436 Mon Sep 17 00:00:00 2001 From: daHugen Date: Sun, 31 Jul 2022 18:14:56 +0200 Subject: Update also includes some changes in the pursuitcurve-picture --- buch/papers/lambertw/teil4.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/lambertw/teil4.tex b/buch/papers/lambertw/teil4.tex index 0050b61..1053dd1 100644 --- a/buch/papers/lambertw/teil4.tex +++ b/buch/papers/lambertw/teil4.tex @@ -434,4 +434,4 @@ Nun, der Grund dafür ist die Struktur \end{equation} bei welcher \(p(x)\) eine beliebige Potenz von \(x\) darstellt. -Jedes Mal wenn \(x\) gesucht ist und in einer Struktur der Art \eqref{lambertw:eqEinsatzLambW} vorkommt, dann kann mit ein paar Umformungen die Struktur \(f(x)e^{f(x)}\) erzielt werden. Wie bereits in diesem Abschnitt \ref{lambertw:subsection:FunkNachT} gezeigt wurde, kann \(x\) nun mittels der \(W(x)\)-Funktion aufgelöst werden. Erstaunlicherweise ist \eqref{lambertw:eqEinsatzLambW} eine Struktur die oft vorkommt, was die Lambert-\(W\)-Funktion so wichtig macht. \ No newline at end of file +Jedes Mal wenn \(x\) gesucht ist und in einer Struktur der Art \eqref{lambertw:eqEinsatzLambW} vorkommt, dann kann mit ein paar Umformungen die Struktur \(f(x)e^{f(x)}\) erzielt werden. Wie bereits in diesem Abschnitt \ref{lambertw:subsection:FunkNachT} gezeigt wurde, kann \(x\) nun mittels der \(W(x)\)-Funktion aufgelöst werden. Erstaunlicherweise ist \eqref{lambertw:eqEinsatzLambW} eine Struktur die oft vorkommt, was die Lambert-\(W\)-Funktion so wichtig macht. \ No newline at end of file -- cgit v1.2.1 From 3ccdc3ec4dcc7d33b16fc1469b0c95c0e8def66d Mon Sep 17 00:00:00 2001 From: Andrea Mozzini Vellen Date: Tue, 2 Aug 2022 14:51:41 +0200 Subject: =?UTF-8?q?=C3=A4nderungen=2002.08.2022=20andrea?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- buch/papers/kreismembran/teil1.tex | 54 +++++++++++++++++++------------------- buch/papers/kreismembran/teil2.tex | 45 +++++++++++++------------------ buch/papers/kreismembran/teil3.tex | 8 +++--- 3 files changed, 49 insertions(+), 58 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil1.tex b/buch/papers/kreismembran/teil1.tex index 39ca598..377ba48 100644 --- a/buch/papers/kreismembran/teil1.tex +++ b/buch/papers/kreismembran/teil1.tex @@ -30,37 +30,33 @@ Da es sich um eine Kreisscheibe handelt, werden Polarkoordinaten verwendet, so d ergibt. Es wird eine runde elastische Membran berücksichtigt, die das Gebiet $\Omega$ abdeckt und am Rand $\Gamma$ befestigt ist. -Es wird daher davon ausgegangen, dass die Membran aus einem homogenen Material von vernachlässigbarer Dicke gefertigt ist. -Die Membran kann verformt werden, aber innere elastische Kräfte wirken den Verformungen entgegen. Es wirken keine äusseren Kräfte. Es handelt sich somit von einer kreisförmligen eingespannten homogenen schwingenden Membran. +Es wirken keine äusseren Kräfte. Es handelt sich somit von einer kreisförmligen eingespannten homogenen schwingenden Membran. Daher ist die Membranabweichung im Punkt $(r,\varphi)$ $\in$ $\overline{\rm \Omega}$ zum Zeitpunkt $t$: \begin{align*} u: \overline{\rm \Omega} \times \mathbb{R}_{\geq 0} &\longrightarrow \mathbb{R}\\ (r,\varphi,t) &\longmapsto u(r,\varphi,t) \end{align*} -Da die Membran am Rand befestigt ist, kann es keine Schwingungen geben, so dass die \textit{Dirichlet-Randbedingung} \cite{prof_dr_horst_knorrer_kreisformige_2013} -\begin{equation*} - u\big|_{\Gamma} = 0 \quad \text{für} \quad 0 \leq \varphi \leq 2\pi,\quad t \geq 0 -\end{equation*} -gilt. +Um die Vergleichbarkeit der beiden nachfolgend vorgestellten Lösungsverfahren in Abschnitt \ref{kreismembran:vergleich} zu vereinfachen, werden keine Randbedingungen vorgegeben. -Um eine eindeutige Lösung bestimmen zu können, werden die folgenden Anfangsbedingungen festgelegt: +Um eine eindeutige Lösung bestimmen zu können, werden die folgenden Anfangsbedingungen festgelegt zur zeit $t = \text{0}$: \begin{align*} u(r,\varphi, 0) &= f(r,\varphi)\\ u_t(r,\varphi, 0) &= g(r,\varphi). \end{align*} \subsection{Lösung\label{sub:lösung1}} +Nun wird das in Abschnitt \ref{sub:aufgabestellung} vorgestellte Problem mit Hilfe der varibalen Trennungsmethode gelöst. \subsubsection{Ansatz der Separation der Variablen\label{subsub:ansatz_separation}} -Daher muss an dieser Stelle von einer Separation der Variablen ausgegangen werden: +Bezug muss an dieser Stelle von einer Separation der Variablen ausgegangen werden: \begin{equation*} u(r,\varphi, t) = F(r)G(\varphi)T(t) \end{equation*} -Dank der Randbedingungen kann also gefordert werden, dass $F(R)=0$ ist, und natürlich, dass $G(\varphi)$ $2\pi$ periodisch ist. Eingesetzt in der Differenzialgleichung ergibt sich: +Dank der Randbedingungen kann gefordert werden, dass $F(R)=0$ ist, und natürlich, dass $G(\varphi)$ $2\pi$ periodisch ist. Eingesetzt in der Differenzialgleichung ergibt sich: \begin{equation*} - \frac{1}{c^2}\frac{T''(t)}{T(t)}=\frac{F''(r)}{F(r)}+\frac{1}{r}\frac{F'(r)}{F(r)}+\frac{1}{r^2}\frac{G''(\varphi)}{G(\varphi)}. + \frac{1}{c^2}\frac{T''(t)}{T(t)}=-\kappa^2=\frac{F''(r)}{F(r)}+\frac{1}{r}\frac{F'(r)}{F(r)}+\frac{1}{r^2}\frac{G''(\varphi)}{G(\varphi)}. \end{equation*} -Da die linke Seite nur von $t$ und die rechte Seite nur von $r$ und $\varphi$ abhängt, müssen sie gleich einer reellen Zahl sein. Aus physikalischen Gründen suchen wir nach Lösungen, die weder exponentiell in der Zeit wachsen noch exponentiell abklingen. Dies bedeutet, dass die Konstante negativ sein muss, also schreibt man $k=-k^2$. Daraus ergeben sich die folgenden zwei Gleichungen: +Da die linke Seite nur von $t$ und die rechte Seite nur von $r$ und $\varphi$ abhängt, müssen sie gleich einer reellen Zahl sein. Aus physikalischen Gründen suchen wir nach Lösungen, die weder exponentiell in der Zeit wachsen noch exponentiell abklingen. Dies bedeutet, dass die Konstante negativ sein muss, also schreibt man $-\kappa^2$. Daraus ergeben sich die folgenden zwei Gleichungen: \begin{align*} T''(t) + c^2\kappa^2T(t) &= 0\\ r^2\frac{F''(r)}{F(r)} + r \frac{F'(r)}{F(r)} +\kappa^2 r^2 &= - \frac{G''(\varphi)}{G(\varphi)}. @@ -72,14 +68,14 @@ In der zweiten Gleichung hängt die linke Seite nur von $r$ ab, während die rec \end{align*} \subsubsection{Lösung für $G(\varphi)$\label{subsub:lösung_G}} -Da für die Zweite Gelichung Lösungen von Schwingungen erwartet werden, für die $G''(\varphi)=-\omega^2 G(\varphi)$ gilt, schreibt die gemeinsame Konstante als $-\nu^2$, was die Formeln später vereinfacht. Also: +Da für die zweite Gleichung Lösungen von Schwingungen erwartet werden, für die $G''(\varphi)=-\omega^2 G(\varphi)$ gilt, schreibt man die gemeinsame Konstante als $\nu=-\omega^2$, was die Formeln später vereinfacht. Also: \begin{equation*} G(\varphi) = C_n \cos(\varphi) + D_n \sin(\varphi) \label{eq:cos_sin_überlagerung} \end{equation*} \subsubsection{Lösung für $F(r)$\label{subsub:lösung_F}} -Die Gleichung für $F$ hat die Gestalt +Die Gleichung für $F$ hat die Gestalt (verweis auf \ref{buch:differentialgleichungen:bessel-operator}) \begin{align} r^2F''(r) + rF'(r) + (\kappa^2 r^2 - n^2)F(r) = 0 \label{eq:2nd_degree_PDE} @@ -90,19 +86,9 @@ Wir bereits in Kapitel \ref{buch:differntialgleichungen:section:bessel} gezeigt, \end{equation*} Lösungen der Besselschen Differenzialgleichung \begin{equation*} - x^2 y'' + xy' + (x^2 - \nu^2)y = 0 -\end{equation*} -Die Funktionen $F(r) = J_n(\kappa r)$ lösen also die Differentialgleichung \eqref{eq:2nd_degree_PDE}. Die -Randbedingung $F(R)=0$ impliziert, dass $\kappa R$ eine Nullstelle der Besselfunktion -$J_n$ sein muss. Man kann zeigen, dass die Besselfunktionen $J_n, n \geq 0$, alle unendlich -viele Nullstellen -\begin{equation*} - \alpha_{1n} < \alpha_{2n} < ... -\end{equation*} -haben, und dass $\underset{\substack{m\to\infty}}{\text{lim}} \alpha_{mn}=\infty$. Somit ergibt sich, dass $\kappa = \frac{\alpha_{mn}}{R}$ für ein $m\geq 1$, und dass -\begin{equation*} - F(r) = J_n (\kappa_{mn}r) \quad \text{mit} \quad \kappa_{mn}=\frac{\alpha_{mn}}{R} + x^2 y'' + xy' + (\kappa^2 - \nu^2)y = 0 \end{equation*} +Die Funktionen $F(r) = J_n(\kappa r)$ lösen die Differentialgleichung \eqref{eq:2nd_degree_PDE}. \subsubsection{Lösung für $T(t)$\label{subsub:lösung_T}} Die Differenzialgleichung $T''(t) + c^2\kappa^2T(t) = 0$, wird auf ähnliche Weise gelöst wie $G(\varphi)$. @@ -115,7 +101,21 @@ Durch Überlagerung aller Ergebnisse erhält man die Lösung \end{align} Dabei sind $m$ und $n$ ganze Zahlen, wobei $m$ für die Anzahl der Knotenkreise und $n$ -für die Anzahl der Knotenlinien steht. Es gibt bestimmte Bereiche auf der Membran, in denen es keine Bewegung oder Vibration gibt. Wenn der nicht schwingende Bereich ein Kreis ist, nennt man ihn einen Knotenkreis, und wenn er eine Linie ist, nennt man ihn ebenfalls eine Knotenlinie. $Jn(\kappa_{mn}r)$ ist die Besselfunktion $n$-ter Ordnung, wobei $\kappa mn$ die Wellenzahl und $r$ der Radius ist. $a_{mn}$ und $b_{mn}$ sind die zu bestimmenden Konstanten. +für die Anzahl der Knotenlinien steht. Es gibt bestimmte Bereiche auf der Membran, in denen es keine Bewegung oder Vibration gibt. Wenn der nicht schwingende Bereich ein Kreis ist, nennt man ihn einen Knotenkreis, und wenn er eine Linie ist, nennt man ihn ebenfalls eine Knotenlinie; siehe Abbildung \ref{buch:pde:kreis:fig:pauke}. $Jn(\kappa_{mn}r)$ ist die Besselfunktion $n$-ter Ordnung, wobei $\kappa mn$ die Wellenzahl und $r$ der Radius ist. $a_{mn}$ und $b_{mn}$ sind die zu bestimmenden Konstanten. + +\begin{figure} + \centering + \includegraphics[width=\textwidth]{chapters/090-pde/bessel/pauke.pdf} + %\includegraphics{chapters/090-pde/bessel/pauke.pdf} + \caption{Vorzeichen der Lösungsfunktionen und Knotenlinien + für verschiedene Werte von $\mu$ und $k$. + Die Bereiche, in denen die Lösungsfunktion positiv sind, ist + rot dargestellt, die negativen Bereiche blau. + In jeder Darstellung gibt es genau $k+\mu$ Knotenlinien. + Die Radien der kreisförmigen Knotenlinien müssen aus den Nullstellen + der Besselfunktionen berechnet werden. + \label{buch:pde:kreis:fig:pauke}} +\end{figure} An diesem Punkt stellte sich die Frage, ob es möglich wäre, die partielle Differentialgleichung mit einer anderen Methode als der der Trennung der Variablen zu lösen. Nach einer kurzen Recherche wurde festgestellt, dass die beste Methode die Transformationsmethode ist, genauer gesagt die Anwendung der Hankel-Transformation. Im nächsten Kapitel wird daher diese Integraltransformation vorgestellt und entwickelt, und es wird erläutert, warum sie für diese Art von Problem geeignet ist. diff --git a/buch/papers/kreismembran/teil2.tex b/buch/papers/kreismembran/teil2.tex index 6efda49..4fb139c 100644 --- a/buch/papers/kreismembran/teil2.tex +++ b/buch/papers/kreismembran/teil2.tex @@ -11,30 +11,30 @@ Er studierte auch Funktionen, jetzt Hankel- oder Bessel- Funktionen genannt, der Die Hankel-Transformation, die die Bessel-Funktion enthält, taucht natürlich bei achsensymmetrischen Problemen auf, die in zylindrischen Polarkoordinaten formuliert sind. In diesem Abschnitt werden die Theorie der Transformation und einige Eigenschaften der Grundoperationen erläutert. -\subsubsection{Hankel-Transformation \label{subsub:hankel_tansformation}} +\subsubsection{Definition der Hankel-Transformation \label{subsub:hankel_tansformation}} Wir führen die Definition der Hankel-Transformation \cite{lokenath_debnath_integral_2015} aus der zweidimensionalen Fourier-Transformation und ihrer Umkehrung ein, die durch: \begin{align} - \mathscr{F}\{f(x,y)\} & = F(k,l)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i( \bm{\kappa}\cdot \mathbf{r})}f(x,y) \; dx dy,\label{equation:fourier_transform}\\ - \mathscr{F}^{-1}\{F(x,y)\} & = f(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{i(\bm{\kappa}\cdot \mathbf{r}))}F(k,l) \; dx dy \label{equation:inv_fourier_transform} + \mathscr{F}\{f(x,y)\} & = F(k,l)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i( \bm{\kappa}\cdot \mathbf{r})}f(x,y) \; dx \; dy,\label{equation:fourier_transform}\\ + \mathscr{F}^{-1}\{F(x,y)\} & = f(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{i(\bm{\kappa}\cdot \mathbf{r})}F(k,l) \; dx \; dy \label{equation:inv_fourier_transform} \end{align} -wo $\mathbf{r}=(x,y)$ und $\bm{\kappa}=(k,l)$. Polarkoordinaten sind für diese Art von Problemen am besten geeignet, mit $(x,y)=r(\cos\theta,\sin\theta)$ und $(k,l)=\kappa(\cos\phi,\sin\phi)$ findet man $\bm{\kappa}\cdot\mathbf{r}=\kappa r(\cos(\theta-\phi))$ und danach: +wo $\mathbf{r}=(x,y)$ und $\bm{\kappa}=(k,l)$. Polarkoordinaten sind für diese Art von Problem am besten geeignet, mit $(x,y)=r(\cos\theta,\sin\theta)$ und $(k,l)=\kappa(\cos\phi,\sin\phi)$ findet man $\bm{\kappa}\cdot\mathbf{r}=\kappa r(\cos(\theta-\phi))$ und danach: \begin{align} F(k,\phi)=\frac{1}{2\pi}\int_{0}^{\infty}r \; dr \int_{0}^{2\pi}e^{-ikr\cos(\theta-\phi)}f(r,\theta) \; d\phi. \label{equation:F_ohne_variable_wechsel} \end{align} -Dann wird angenommen dass, $f(r,\theta)=e^{in\theta}f(r)$, was keine strenge Einschränkung ist, und es wird eine Änderung der Variabeln vorgenommen $\theta-\phi=\alpha-\frac{\pi}{2}$, um \eqref{equation:F_ohne_variable_wechsel} zu reduzieren: +Dann wird angenommen dass, $f(r,\theta)=e^{in\theta}f(r)$, was keine strenge Einschränkung ist, weil die \textit{Fourier-Theorie} besagt, dass sich jede Funktion durch Überlagerung solcher Terme darstellen lässt. Es wird auch eine Änderung der Variabeln vorgenommen $\theta-\phi=\alpha-\frac{\pi}{2}$, um \eqref{equation:F_ohne_variable_wechsel} zu reduzieren: \begin{align} F(k,\phi)=\frac{1}{2\pi}\int_{0}^{\infty}rf(r) \; dr \int_{\phi_{0}}^{2\pi+\phi_{0}}e^{in(\phi-\frac{\pi}{2})+i(n\alpha-kr\sin\alpha)} \; d\alpha, \label{equation:F_ohne_bessel} \end{align} wo $\phi_{0}=(\frac{\pi}{2}-\phi)$. -Unter Verwendung der Integraldarstellung der Besselfunktion vom Ordnung $n$ \eqref{buch:fourier:eqn:bessel-integraldarstellung} +Unter Verwendung der Integraldarstellung \begin{equation*} J_n(\kappa r)=\frac{1}{2\pi}\int_{\phi_{0}}^{2\pi + \phi_{0}}e^{i(n\alpha-\kappa r \sin \alpha)} \; d\alpha \label{equation:bessel_n_ordnung} \end{equation*} -\eqref{equation:F_ohne_bessel} wird sie zu: + der Besselfunktion vom Ordnung $n$ \eqref{buch:fourier:eqn:bessel-integraldarstellung} wird \eqref{equation:F_ohne_bessel} zu: \begin{align} F(k,\phi)&=e^{in(\phi-\frac{\pi}{2})}\int_{0}^{\infty}rJ_n(\kappa r) f(r) \; dr \nonumber \\ &=e^{in(\phi-\frac{\pi}{2})}\tilde{f}_n(\kappa), @@ -47,37 +47,28 @@ wo $\tilde{f}_n(\kappa)$ ist die \textit{Hankel-Transformation} von $f(r)$ und i \end{align} \subsubsection{Inverse Hankel-Transformation \label{subsub:inverse_hankel_tansformation}} -Ähnlich verhält es sich mit der inversen Fourier Transformation in Form von polaren Koordinaten unter der Annahme $f(r,\theta)=e^{in\theta}f(r)$ mit \eqref{equation:F_mit_bessel_step_2}, wird die inverse Fourier Transformation \eqref{equation:inv_fourier_transform}: +Wie bei der Entwicklung der Hankel-Transformation können auch für die Umkehrformel Analogien zur Fourier-Transformation hergestellt werden. Vergleicht man die beiden Transformationen, so stellt man fest, dass sie sehr ähnlich sind, wenn man den Term $J_n(\kappa r)$ der Hankel-Transformation durch $e^{-i( \bm{\kappa}\cdot \mathbf{r})}$ der Fourier-Transformation ersetzt. Diese beide Funktionen sind orthogonal, und bei orthogonalen Matrizen genügt bekanntlich die Transponierung, um sie zu invertieren. Da das Skalarprodukt der Bessel-Funktionen jedoch nicht dasselbe ist wie das der Exponentialfunktionen, muss man durch $\kappa\; d\kappa$ statt nur durch $d\kappa$ integrieren, um die Umkehrfunktion zu erhalten. -\begin{align*} - e^{in\theta}f(r)&=\frac{1}{2\pi}\int_{0}^{\infty}\kappa \; d\kappa \int_{0}^{2\pi}e^{i\kappa r \cos (\theta - \phi)}F(\kappa,\phi) \; d\phi \\ - &= \frac{1}{2\pi}\int_{0}^{\infty}\kappa \tilde{f}_n(\kappa) \; d\kappa \int_{0}^{2\pi}e^{in(\phi - \frac{\pi}{2})- i\kappa r \cos (\theta - \phi)} \; d\phi, -\end{align*} -was durch den Wechsel der Variablen $\theta-\phi=-(\alpha+\frac{\pi}{2})$ und $\theta_0=-(\theta+\frac{\pi}{2})$, - -\begin{align*} - &= \frac{1}{2\pi}\int_{0}^{\infty}\kappa \tilde{f}_n(\kappa) \; d\kappa \int_{\theta_0}^{2\pi+\theta_0}e^{in(\theta + \alpha - i\kappa r \sin\alpha)} \; d\alpha \\ - &= e^{in\theta}\int_{0}^{\infty}\kappa J_n(\kappa r) \tilde{f}_n(\kappa) \; d\kappa, -\end{align*} - -von \eqref{equation:bessel_n_ordnung} also ist, die inverse \textit{Hankel-Transformation} so definiert: +Von \eqref{equation:hankel} also ist, die inverse \textit{Hankel-Transformation} so definiert: \begin{align} \mathscr{H}^{-1}_n\{\tilde{f}_n(\kappa)\}=f(r)=\int_{0}^{\infty}\kappa J_n(\kappa r) \tilde{f}_n(\kappa) \; d\kappa. \label{equation:inv_hankel} \end{align} -Anstelle von $\tilde{f}_n(\kappa)$, wird häufig für die Hankel-Transformation verwendet, indem die Ordnung angegeben wird. -\eqref{equation:hankel} und \eqref{equation:inv_hankel} Integralen existieren für eine grosse Klasse von Funktionen, die normalerweise in physikalischen Anwendungen benötigt werden. -Alternativ kann auch die berühmte Hankel-Transformationsformel verwendet werden, +Anstelle von $\tilde{f}_n(\kappa)$, wird häufig einfach $\tilde{f}(\kappa)$ für die Hankel-Transformation verwendet, indem die Ordnung angegeben wird. +Die Integrale \eqref{equation:hankel} und \eqref{equation:inv_hankel} existieren für bestimmte grosse Klassen von Funktionen, die normalerweise in physikalischen Anwendungen vorkommen. + +Alternativ dazu kann die berühmte Hankel-Integralformel \begin{align*} f(r) = \int_{0}^{\infty}\kappa J_n(\kappa r) \; d\kappa \int_{0}^{\infty} p J_n(\kappa p)f(p) \; dp, \label{equation:hankel_integral_formula} \end{align*} -um die Hankel-Transformation \eqref{equation:hankel} und ihre Inverse \eqref{equation:inv_hankel} zu definieren. +verwendet werden, um die Hankel-Transformation \eqref{equation:hankel} und ihre Umkehrung \eqref{equation:inv_hankel} zu definieren. + Insbesondere die Hankel-Transformation der nullten Ordnung ($n=0$) und der ersten Ordnung ($n=1$) sind häufig nützlich, um Lösungen für Probleme mit der Laplace Gleichung in einer achsensymmetrischen zylindrischen Geometrie zu finden. -\subsection{Operative Eigenschaften der Hankel-Transformation\label{sub:op_properties_hankel}} +\subsection{Operatoreigenschaften der Hankel-Transformation \label{sub:op_properties_hankel}} In diesem Kapitel werden die operativen Eigenschaften der Hankel-Transformation aufgeführt. Der Beweis für ihre Gültigkeit wird jedoch nicht analysiert. \begin{satz}{Skalierung:} @@ -88,7 +79,7 @@ In diesem Kapitel werden die operativen Eigenschaften der Hankel-Transformation \end{equation*} \end{satz} -\begin{satz}{Persevalsche Relation (Skalarprodukt bleibt erhalten):} +\begin{satz}{Parsevalsche Relation:} Wenn $\tilde{f}(\kappa)=\mathscr{H}_n\{f(r)\}$ und $\tilde{g}(\kappa)=\mathscr{H}_n\{g(r)\}$, dann: \begin{equation*} @@ -103,7 +94,7 @@ Wenn $\tilde{f}_n(\kappa)=\mathscr{H}_n\{f(r)\}$, dann: &\mathscr{H}_n\{f'(r)\}=\frac{\kappa}{2n}\left[(n-1)\tilde{f}_{n+1}(\kappa)-(n+1)\tilde{f}_{n-1}(\kappa)\right], \quad n\geq1, \\ &\mathscr{H}_1\{f'(r)\}=-\kappa \tilde{f}_0(\kappa), \end{align*} -bereitgestellt dass $[rf(r)]$ verschwindet als $r\to0$ und $r\to\infty$. +vorausgesetzt dass $[rf(r)]$ verschwindet wenn $r\to0$ und $r\to\infty$. \end{satz} \begin{satz} diff --git a/buch/papers/kreismembran/teil3.tex b/buch/papers/kreismembran/teil3.tex index 7d5648a..014b6e6 100644 --- a/buch/papers/kreismembran/teil3.tex +++ b/buch/papers/kreismembran/teil3.tex @@ -40,7 +40,7 @@ bekommt man: \tilde{u}(\kappa,0)=\tilde{f}(\kappa), \quad \tilde{u}_t(\kappa,0)=\tilde{g}(\kappa). \end{equation*} -Die allgemeine Lösung für diese Transformation lautet, wie in Gleighung \eqref{eq:cos_sin_überlagerung} gesehen, wie folgt +Die allgemeine Lösung für diese Gleichung lautet, wie in Abschnitt \eqref{eq:cos_sin_überlagerung} gesehen, wie folgt \begin{equation*} \tilde{u}(\kappa,t)=\tilde{f}(\kappa)\cos(c\kappa t) + \frac{1}{c\kappa}\tilde{g}(\kappa)\sin(c\kappa t). @@ -60,7 +60,7 @@ Es wird in Folgenden davon ausgegangen, dass sich die Membran verformt und zum Z \end{equation*} so dass $\tilde{g}(\kappa)\equiv 0$ und \begin{equation*} - \tilde{f}(\kappa)=Aa\int_{0}^{\infty}r(a^2 + r^2)^{-\frac{1}{2}} J_0 (\kappa r) \; dr=\frac{Aa}{\kappa}e^{-a\kappa} + \tilde{f}(\kappa)=Aa\int_{0}^{\infty}r(a^2 + r^2)^{-\frac{1}{2}} J_0 (\kappa r) \; dr=\frac{Aa}{\kappa}e^{-a\kappa}. \end{equation*} Die formale Lösung \eqref{eq:formale_lösung} lautet also \begin{align*} @@ -68,7 +68,7 @@ Die formale Lösung \eqref{eq:formale_lösung} lautet also &=AaRe\left\{r^2+\left(a+ict\right)^2\right\}^{-\frac{1}{2}} \end{align*} -Nimmt man jedoch die allgemeine Lösung mit Summationen, +Nimmt man jedoch die allgemeine Lösung durch Überlagerung, \begin{align} u(r, t) = \displaystyle\sum_{m=1}^{\infty} J_0 (k_{m}r)[a_{m}\cos(c \kappa_{m} t)+b_{m}\sin(c \kappa_{m} t)] @@ -78,7 +78,7 @@ kann man die Lösungsmethoden 1 und 2 vergleichen. \subsection{Vergleich der Analytischen Lösungen \label{kreismembran:vergleich}} -Bei der Analyse der Gleichungen \eqref{eq:lösung_endliche_generelle} und \eqref{eq:lösung_unendliche_generelle} fällt sofort auf, dass die Gleichung \eqref{eq:lösung_unendliche_generelle} nicht mehr von $m$ und $n$ abhängt, sondern nur noch von $n$ \cite{nishanth_p_vibrations_2018}. Das macht Sinn, denn $n$ beschreibt die Anzahl der Knotenlinien, und in einer unendlichen Membran gibt es keine. Tatsächlich werden $a_{m0}$, $b_{m0}$ und $\kappa_{m0}$ in $a_m$, $b_m$ bzw. $\kappa_m$ umbenannt. Die beiden Termen $\cos(n\varphi)$ und $\sin(n\varphi)$ verschwinden ebenfalls, da für $n=0$ der $\cos(n\varphi)$ gleich 1 und der $\sin(n \varphi)$ gleich 0 ist. +Bei der Analyse der Gleichungen \eqref{eq:lösung_endliche_generelle} und \eqref{eq:lösung_unendliche_generelle} fällt sofort auf, dass die Gleichung \eqref{eq:lösung_unendliche_generelle} nicht mehr von $m$ und $n$ abhängt, sondern nur noch von $n$ \cite{nishanth_p_vibrations_2018}. Das macht Sinn, denn $n$ beschreibt die Anzahl der Knotenlinien, und in einer unendlichen Membran gibt es keine, dato che abbiamo assunto che la soluzione è rotationssymmetrisch. Tatsächlich werden $a_{m0}$, $b_{m0}$ und $\kappa_{m0}$ in $a_m$, $b_m$ bzw. $\kappa_m$ umbenannt. Die beiden Termen $\cos(n\varphi)$ und $\sin(n\varphi)$ verschwinden ebenfalls, da für $n=0$ der $\cos(n\varphi)$ gleich 1 und der $\sin(n \varphi)$ gleich 0 ist. Die Funktion hängt also nicht mehr von der Besselfunktionen $n$-ter Ordnung ab, sondern nur von der $0$-ter Ordnung. -- cgit v1.2.1 From de76ac03caa4e7a09a99fe1271fb6a22a809ade2 Mon Sep 17 00:00:00 2001 From: haddoucher Date: Tue, 2 Aug 2022 17:38:32 +0200 Subject: Was ist das Sturm-Liouville-Problem erste Version --- buch/papers/sturmliouville/einleitung.tex | 48 ++++++++++++++++++++++++++++--- 1 file changed, 44 insertions(+), 4 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/einleitung.tex b/buch/papers/sturmliouville/einleitung.tex index ec37a3f..7d39cf4 100644 --- a/buch/papers/sturmliouville/einleitung.tex +++ b/buch/papers/sturmliouville/einleitung.tex @@ -23,17 +23,30 @@ und schreibt die Gleichung um in: diese Gleichung wird dann Sturm-liouville-Gleichung bezeichnet. Alle homogene 2.Ordnung lineare gewöhnliche Differentialgleichungen können in die Form der Gleichung \ref{eq:sturm-liouville-equation} umgeformt werden. -Die Sturm-Liouville-Theorie besagt, dass, wenn man die Sturm-Liouville-Gleichung mit den homogenen Randbedingungen +Die Sturm-Liouville-Theorie besagt, dass, wenn man die Sturm-Liouville-Gleichung mit den homogenen Randbedingungen des dritten Typs\footnote{Die Randbedingung des dritten Typs, oder Robin-Randbedingungen (benannt nach dem französischen mathematischen Analytiker und angewandten Mathematiker Victor Gustave Robin), wird genannt, wenn sie einer gewöhnlichen oder partiellen Differentialgleichung auferlegt wird, so sind die Spezifikationen einer Linearkombination der Werte einer Funktion sowie die Werte ihrer Ableitung am Rande des Bereichs} \begin{equation} \begin{aligned} - \label{ali:randbedingungen} + \label{eq:randbedingungen} k_a y(a) + h_a p(a) y'(a) &= 0 \\ k_b y(b) + h_b p(b) y'(b) &= 0 \end{aligned} \end{equation} - + kombiniert, wie schon im Kapitel \ref{sub:differentailgleichung} erwähnt, auf dem Intervall (a,b), dann bekommt man das klassische Sturm-Liouville-Problem. +Wenn von der Funktion $y(x)$ die Werte $x$ des jeweiligen Randes des Definitionsbereiches anzunehmen sind,also + +\begin{equation} + y(a) = y(b) = 0 +\end{equation} + +, so spricht man von einer Dirichlet-Randbedingung, und von einer Neumann-Randbedingung spricht man, wenn + +\begin{equation} + y'(a) = y'(b) = 0 +\end{equation} + +ergibt - die Existenz und Eindeutigkeit der Lösung kann mit den zwei Randbedingungen sichergestellt werden Lösungen die nicht Null sind, werden nicht betrachtet und diese zwei Gleichungen (\ref{eq:sturm-liouville-equation} und \ref{ali:randbedingungen}) kombiniert, nennt man Eigenfunktionen. Wenn bei der Sturm-Liouville-Gleichung \ref{eq:sturm-liouville-equation} alles konstant bleibt, aber der Wert von $\lambda$ sich ändert, erhält man eine andere Eigenfunktion, weil man eine andere gewöhnliche Differentialgleichung löst; der Parameter $\lambda$ wird als Eigenwert bezeichnet. @@ -53,10 +66,37 @@ Somit ergibt die Gleichung \int_{a}^{b} w(x)y_m y_n = 0 \end{equation}. -Die Funktionen $p(x)$, $q(x)$ und $w(x)$ werden als Koeffizientenfunktionen mit ihren freien Variablen $x$ bezeichnet. Die Funktion $w(x)$ (manchmal auch $r(x)$ genannt) wird als Gewichtsfunktion oder Dichtefunktion bezeichnet. +Die Funktionen $p(x)$, $q(x)$ und $w(x)$ werden als Koeffizientenfunktionen mit ihren freien Variablen $x$ bezeichnet. +Die Funktion $w(x)$ (manchmal auch $r(x)$ genannt) wird als Gewichtsfunktion oder Dichtefunktion bezeichnet. +Es gibt zwei verschiedene Sturm-Liouville-Probleme: das reguläre Sturm-Liouville-Problem und das singuläre Sturm-Liouville-Problem. +Die Funktionen für das reguläre und das singuläre Sturm-Liouville-Problem sind nicht dieselben. + +\subsection{Das reguläre Sturm-Liouville-Problem\label{sub:reguläre_sturm_liouville_problem}} +Damit es sich um ein reguläres Sturm-Liouville-Problem handelt, müssen einige Bedingungen beachtet werden. + +\begin{itemize} + \item Die Funktionen $p(x), p'(x), q(x)$ und $w(x)$ müssen stetig und reell sein. + \item sowie müssen in einem Endlichen Intervall $[ \ a,b] \ $ integrierbar sein. + \item $p(x)^{-1}$ und $w(x)$ sind $>0$. + \item Es gelten die Randbedingungen \ref{eq:randbedingungen}, wobei $|k_i|^2 + |h_i|^2\ne 0$ mit $i=a,b$. +\end{itemize} + +Bei einem regulären Sturm-Liouville-Problem geht es darum, ohne genaue Kenntnis der Eigenfunktionen diese dennoch beschreiben zu können. + +\subsection{Das singuläre Sturm-Liouville-Problem\label{sub:singuläre_sturm_liouville_problem}} +Von einem singulären Sturm-Liouville-Problem spricht man, wenn die oben genannten Bedingungen nicht erfüllt sind, d.h: +\begin{itemize} + \item wenn sein Definitionsbereich auf dem Intervall $[ \ a,b] \ $ unbeschränkt ist oder + \item wenn die Koeffizienten an den Randpunkten Singularitäten haben. +\end{itemize} +Allerdings kann nur eine der Bedingungen nicht erfüllt sein, so dass es sich bereits um ein singuläres Sturm-Liouville-Problem handelt. +Verwendet man das reguläre Sturm-Liouville-Problem, obwohl eine oder beide Bedingungen nicht erfüllt sind, dann ist es schwierig zu sagen, ob die Lösung fundierte Ergebnisse hat. +Es ist schwierig, bestehende Kriterien anzuwenden, da die Formulierungen z.B. in der Lösungsfunktion liegen. +Das Spektrum besteht im singulärem Problem nicht mehr nur aus Eigenwerte, sondern kann auch einen stetigen Anteil enthalten. +Ähnlich wie bei der Fourier-Reihe gegenüber der Fourier-Transformation gibt es immer noch eine zugehörige Eigenfunktionsentwicklung, und zwar die Integraltransformation sowie gibt es weiterhin eine verallgemeinerte Eigenfunktionen. -- cgit v1.2.1 From 8dc531ac53ae1b085482c9f1bda6001ca803c164 Mon Sep 17 00:00:00 2001 From: Kuster Yanik Date: Tue, 2 Aug 2022 21:14:53 +0200 Subject: Created python files for graphics. Created addtional subsection verlockende Intuition --- buch/papers/lambertw/Bilder/Abstand.py | 18 +++++ buch/papers/lambertw/Bilder/Intuition.pdf | Bin 0 -> 186972 bytes buch/papers/lambertw/Bilder/Strategie.pdf | Bin 120904 -> 151640 bytes buch/papers/lambertw/Bilder/Strategie.py | 2 +- buch/papers/lambertw/Bilder/konvergenz.py | 20 ++++++ .../Bilder/lambertAbstandBauchgef\303\274hl.py" | 58 ++++++++++++++++ buch/papers/lambertw/teil1.tex | 76 ++++++++++++++++++++- 7 files changed, 170 insertions(+), 4 deletions(-) create mode 100644 buch/papers/lambertw/Bilder/Abstand.py create mode 100644 buch/papers/lambertw/Bilder/Intuition.pdf create mode 100644 buch/papers/lambertw/Bilder/konvergenz.py create mode 100644 "buch/papers/lambertw/Bilder/lambertAbstandBauchgef\303\274hl.py" (limited to 'buch/papers') diff --git a/buch/papers/lambertw/Bilder/Abstand.py b/buch/papers/lambertw/Bilder/Abstand.py new file mode 100644 index 0000000..d787c34 --- /dev/null +++ b/buch/papers/lambertw/Bilder/Abstand.py @@ -0,0 +1,18 @@ +# -*- coding: utf-8 -*- +""" +Created on Sat Jul 30 23:09:33 2022 + +@author: yanik +""" + +import numpy as np +import matplotlib.pyplot as plt + +phi = np.pi/2 +t = np.linspace(0, 10, 10**5) +x0 = 1 + +def D(t): + return np.sqrt(x0**2+2*x0*t*np.cos(phi)+2*t**2-2*t**2*np.sin(phi)) + +plt.plot(t, D(t)) diff --git a/buch/papers/lambertw/Bilder/Intuition.pdf b/buch/papers/lambertw/Bilder/Intuition.pdf new file mode 100644 index 0000000..236212a Binary files /dev/null and b/buch/papers/lambertw/Bilder/Intuition.pdf differ diff --git a/buch/papers/lambertw/Bilder/Strategie.pdf b/buch/papers/lambertw/Bilder/Strategie.pdf index 0de3001..91442cc 100644 Binary files a/buch/papers/lambertw/Bilder/Strategie.pdf and b/buch/papers/lambertw/Bilder/Strategie.pdf differ diff --git a/buch/papers/lambertw/Bilder/Strategie.py b/buch/papers/lambertw/Bilder/Strategie.py index b9b41bf..28f7bcd 100644 --- a/buch/papers/lambertw/Bilder/Strategie.py +++ b/buch/papers/lambertw/Bilder/Strategie.py @@ -44,7 +44,7 @@ plt.rcParams.update({ "font.serif": ["New Century Schoolbook"], }) -ax.text(1.6, 4.3, r"$\vec{v}$", size=30) +ax.text(1.6, 4.3, r"$\dot{v}$", size=30) ax.text(0.6, 3.9, r"$V$", size=30, c='b') ax.text(5.1, 4.77, r"$Z$", size=30, c='b') diff --git a/buch/papers/lambertw/Bilder/konvergenz.py b/buch/papers/lambertw/Bilder/konvergenz.py new file mode 100644 index 0000000..dac99a7 --- /dev/null +++ b/buch/papers/lambertw/Bilder/konvergenz.py @@ -0,0 +1,20 @@ +# -*- coding: utf-8 -*- +""" +Created on Sun Jul 31 14:34:13 2022 + +@author: yanik +""" + +import numpy as np +import matplotlib.pyplot as plt + +t = 0 +phi = np.linspace(np.pi/2, 3*np.pi/2, 10**5) +x0 = 1 +y0 = -2 + +def D(t): + return (x0+t*np.cos(phi))*np.cos(phi)+(y0+t*(np.sin(phi)-1))*(np.sin(phi)-1)/(np.sqrt((x0+t*np.cos(phi))**2+(y0+t*(np.sin(phi)-1))**2)) + + +plt.plot(phi, D(t)) \ No newline at end of file diff --git "a/buch/papers/lambertw/Bilder/lambertAbstandBauchgef\303\274hl.py" "b/buch/papers/lambertw/Bilder/lambertAbstandBauchgef\303\274hl.py" new file mode 100644 index 0000000..9031bfc --- /dev/null +++ "b/buch/papers/lambertw/Bilder/lambertAbstandBauchgef\303\274hl.py" @@ -0,0 +1,58 @@ +# -*- coding: utf-8 -*- +""" +Created on Sun Jul 31 13:32:53 2022 + +@author: yanik +""" + +import numpy as np +import matplotlib.pyplot as plt +import scipy.special as sci + +W = sci.lambertw + + +t = np.linspace(0, 1.2, 1000) +x0 = 1 +y0 = 1 + +r0 = np.sqrt(x0**2+y0**2) +chi = (r0+y0)/(r0-y0) + +x = x0*np.sqrt(1/chi*W(chi*np.exp(chi-4*t/(r0-y0)))) +eta = (x/x0)**2 +y = 1/4*((y0+r0)*eta+(y0-r0)*np.log(eta)-r0+3*y0) + +ymin= (min(y)).real +xmin = (x[np.where(y == ymin)][0]).real + + +#Verfolger +plt.plot(x, y, 'r--') +plt.plot(xmin, ymin, 'bo', markersize=10) + +#Ziel +plt.plot(np.zeros_like(t), t, 'g--') +plt.plot(0, ymin, 'bo', markersize=10) + + +plt.plot([0, xmin], [ymin, ymin], 'k--') +#plt.xlim(-0.1, 1) +#plt.ylim(1, 2) +#plt.ylabel("y") +#plt.xlabel("x") +plt.grid(True) +plt.quiver(xmin, ymin, -0.2, 0, scale=1) + +plt.text(xmin+0.1, ymin-0.1, "Verfolgungskurve", size=20, rotation=20, color='r') +plt.text(0.01, 0.02, "Fluchtkurve", size=20, rotation=90, color='g') + +plt.rcParams.update({ + "text.usetex": True, + "font.family": "serif", + "font.serif": ["New Century Schoolbook"], +}) + +plt.text(xmin-0.11, ymin-0.12, r"$\dot{v}$", size=30) +plt.text(xmin-0.02, ymin+0.05, r"$V$", size=30, c='b') +plt.text(0.02, ymin+0.05, r"$Z$", size=30, c='b') \ No newline at end of file diff --git a/buch/papers/lambertw/teil1.tex b/buch/papers/lambertw/teil1.tex index 2733759..2da07db 100644 --- a/buch/papers/lambertw/teil1.tex +++ b/buch/papers/lambertw/teil1.tex @@ -205,8 +205,78 @@ Durch quadrieren verschwindet die Wurzel des Betrages, womit % die neue Bedingung ist. Da sowohl der Betrag als auch $a_{min}$ grösser null sind, bleibt die Aussage unverändert. - - - +% +\subsection{verleitende/trügerisch/verführerisch Intuition} +In der Grafik \ref{lambertw:grafic:intuition} ist eine Mögliche Verfolgungskurve dargestellt, wobei für die Startbedingung der erste-Quadrant verwendet wurde. +Als erste Intuition bietet sich der tiefste Punkt der Verfolgungskurve an, bei dem der y-Anteil des Richtungsvektors null entspricht. +Wenn sich der Verfolger an diesem Punkt befindet, muss zwingend das Ziel auf gleicher Höhe sein. +Es lässt sich vermuten, dass bei diesem Punkt der Abstand zum Ziel minimal sein könnte. +\begin{figure} + \centering + \includegraphics[scale=0.4]{./papers/lambertw/Bilder/Intuition.pdf} + \caption{Intuition} + \label{lambertw:grafic:intuition} +\end{figure} +% +Dies kann leicht überprüft werden, indem wir lokal alle relevanten benachbarten Punkte betrachten und das Vorzeichen der Änderung des Abstandes prüfen. +Dafür wird ein Ausdruck benötigt, der den Abstand und die benachbarten Punkte beschreibt. +Der Richtungsvektor wird allgemein mit dem Winkel $\alpha \in[ 0, 2\pi)$ +Die Ortsvektoren der Punkte können wiederum mit +\begin{align} + v + &= + t\cdot\left(\begin{array}{c} \cos (\alpha) \\ \sin (\alpha) \end{array}\right) +\left(\begin{array}{c} x_0 \\ y_0 \end{array}\right) + \\ + z + &= + \left(\begin{array}{c} 0 \\ t \end{array}\right) +\end{align} +beschrieben werden. Der Verfolger wurde allgemein für jede Richtung $\alpha$ definiert, um alle unmittelbar benachbarten Punkte beschreiben zu können. +Da der Abstand +\begin{equation} + a + = + |v-z| + \geq + 0 +\end{equation} +ist, kann durch quadrieren ohne Informationsverlust die Rechnung vereinfacht werden zu +\begin{equation} + a^2 + = + |v-z|^2 + = + (t\cdot\cos(\alpha)+x_0)^2+t^2(\sin(\alpha)-1)^2 + \text{.} +\end{equation} +Der Abstand im Quadrat abgeleitet nach der Zeit ist +\begin{equation} + \frac{d a^2}{d t} + = + 2(t\cdot\cos (\alpha)+x_0)\cdot\cos(\alpha)(\alpha)+2t(\sin(\alpha)-1)^2 + \text{.} +\end{equation} +Da nur die unmittelbar benachbarten Punkten von Interesse sind, wird die Ableitung für $t=0$ untersucht. Dabei kann die Ableitung in +\begin{align} + \frac{d a^2}{d t} + &= + 2x_0\cos(\alpha) + \\ + \frac{d a^2}{d t} + &< + 0\Leftrightarrow\alpha\in\left( \frac{\pi}{2}, \frac{3\pi}{2}\right) + \\ + \frac{d a^2}{d t} + &> + 0\Leftrightarrow\alpha\in\left[0, \frac{\pi}{2}\right)\cup\left(\frac{3\pi}{2}, 2\pi\right) + \\ + \frac{d a^2}{d t} + &= + 0\Leftrightarrow\alpha\in\left\{ \frac{\pi}{2}, \frac{3\pi}{2}\right\} +\end{align} +unterteilt werden. +Von Interesse ist lediglich das Intervall $\alpha\in\left( \frac{\pi}{2}, \frac{3\pi}{2}\right)$, da der Verfolger sich stets in die negative $y$-Richtung bewegt. +In diesem Intervall ist die Ableitung negativ, woraus folgt, dass jeglicher unmittelbar benachbarte Punkt, den der Verfolger als nächstes begehen könnte, stets näher am Ziel ist als zuvor. +Dies bedeutet, dass der Scheitelpunkt der Verfolgungskurve nie ein lokales Minimum bezüglich des Abstandes sein kann. -- cgit v1.2.1 From 796815b4b22a3cae2db58125be8045a72fe30471 Mon Sep 17 00:00:00 2001 From: haddoucher Date: Tue, 2 Aug 2022 21:17:50 +0200 Subject: Update einleitung.tex Korrektur der Einleitung --- buch/papers/sturmliouville/einleitung.tex | 84 ++++++++++++++++++++----------- 1 file changed, 54 insertions(+), 30 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/einleitung.tex b/buch/papers/sturmliouville/einleitung.tex index 7d39cf4..44c3192 100644 --- a/buch/papers/sturmliouville/einleitung.tex +++ b/buch/papers/sturmliouville/einleitung.tex @@ -8,23 +8,23 @@ Das Sturm-Liouville-Problem wurde benannt nach dem schweizerisch-französischer Mathematiker und Physiker Jacques Charles Fran\c{c}ois Sturm und dem französischer Mathematiker Joseph Liouville. Gemeinsam haben sie in der mathematischen Physik die Sturm-Liouville-Theorie entwickelt und gilt für die Lösung von gewohnlichen Differentialgleichungen, jedoch verwendet man die Theorie öfters bei der Lösung von partiellen Differentialgleichungen. Normalerweise betrachtet man für das Strum-Liouville-Problem eine gewöhnliche Differentialgleichung 2. Ordnung, und wenn es sich um eine partielle Differentialgleichung handelt, kann man sie mit Hilfe einiger Methoden in mehrere gewöhnliche Differentialgleichungen umwandeln, wie z. B. den Separationsansatz, die partielle Differentialgleichung mit mehreren Variablen. -Angenommen man hat die lineare homogene Differentialgleichung +\begin{definition} + \index{Sturm-Liouville-Gleichung} +Angenommen man hat die lineare homogene Differentialgleichung \begin{equation} \frac{d^2y}{dx^2} + a(x)\frac{dy}{dx} + b(x)y = 0 \end{equation} - und schreibt die Gleichung um in: - \begin{equation} \label{eq:sturm-liouville-equation} \frac{d}{dx}\lbrack p(x) \frac{dy}{dx} \rbrack + \lbrack q(x) + \lambda w(x) \rbrack y = 0 -\end{equation}, +\end{equation} +, diese Gleichung wird dann Sturm-liouville-Gleichung bezeichnet. +\end{definition} -diese Gleichung wird dann Sturm-liouville-Gleichung bezeichnet. Alle homogene 2.Ordnung lineare gewöhnliche Differentialgleichungen können in die Form der Gleichung \ref{eq:sturm-liouville-equation} umgeformt werden. Die Sturm-Liouville-Theorie besagt, dass, wenn man die Sturm-Liouville-Gleichung mit den homogenen Randbedingungen des dritten Typs\footnote{Die Randbedingung des dritten Typs, oder Robin-Randbedingungen (benannt nach dem französischen mathematischen Analytiker und angewandten Mathematiker Victor Gustave Robin), wird genannt, wenn sie einer gewöhnlichen oder partiellen Differentialgleichung auferlegt wird, so sind die Spezifikationen einer Linearkombination der Werte einer Funktion sowie die Werte ihrer Ableitung am Rande des Bereichs} - \begin{equation} \begin{aligned} \label{eq:randbedingungen} @@ -32,28 +32,22 @@ Die Sturm-Liouville-Theorie besagt, dass, wenn man die Sturm-Liouville-Gleichung k_b y(b) + h_b p(b) y'(b) &= 0 \end{aligned} \end{equation} - kombiniert, wie schon im Kapitel \ref{sub:differentailgleichung} erwähnt, auf dem Intervall (a,b), dann bekommt man das klassische Sturm-Liouville-Problem. -Wenn von der Funktion $y(x)$ die Werte $x$ des jeweiligen Randes des Definitionsbereiches anzunehmen sind,also - +Wenn von der Funktion $y(x)$ die Werte $x$ des jeweiligen Randes des Definitionsbereiches anzunehmen sind, also \begin{equation} y(a) = y(b) = 0 \end{equation} - , so spricht man von einer Dirichlet-Randbedingung, und von einer Neumann-Randbedingung spricht man, wenn - \begin{equation} y'(a) = y'(b) = 0 \end{equation} - -ergibt - die Existenz und Eindeutigkeit der Lösung kann mit den zwei Randbedingungen sichergestellt werden -Lösungen die nicht Null sind, werden nicht betrachtet und diese zwei Gleichungen (\ref{eq:sturm-liouville-equation} und \ref{ali:randbedingungen}) kombiniert, nennt man Eigenfunktionen. +ergibt - die Existenz und Eindeutigkeit der Lösung kann mit den zwei Randbedingungen sichergestellt werden. +Lösungen die nicht Null sind, werden nicht betrachtet und diese zwei Gleichungen (\ref{eq:sturm-liouville-equation} und \ref{eq:randbedingungen}) kombiniert, nennt man Eigenfunktionen. Wenn bei der Sturm-Liouville-Gleichung \ref{eq:sturm-liouville-equation} alles konstant bleibt, aber der Wert von $\lambda$ sich ändert, erhält man eine andere Eigenfunktion, weil man eine andere gewöhnliche Differentialgleichung löst; der Parameter $\lambda$ wird als Eigenwert bezeichnet. Es ist genau das gleiche Prinzip wie bei den Matrizen, andere Eigenwerte ergeben andere Eigenvektoren. Es besteht eine Korrespondenz zwischen den Eigenwerten und den Eigenvektoren. Das gleiche gilt auch beim Sturm-Liouville-Problem, und zwar - \begin{equation} \lambda \overset{Korrespondenz}\leftrightarrow y \end{equation}. @@ -61,7 +55,6 @@ Das gleiche gilt auch beim Sturm-Liouville-Problem, und zwar Die Theorie besagt, wenn $y_m$, $y_n$ Eigenfuktionen des Sturm-Liouville-Problems sind, die verschiedene Eigenwerte $\lambda_m$, $\lambda_n$ ($\lambda_m \neq \lambda_n$) entsprechen, so sind $y_m$, $y_n$ orthogonal zu y - dies gilt für das Intervall (a,b). Somit ergibt die Gleichung - \begin{equation} \int_{a}^{b} w(x)y_m y_n = 0 \end{equation}. @@ -71,28 +64,60 @@ Die Funktion $w(x)$ (manchmal auch $r(x)$ genannt) wird als Gewichtsfunktion ode Es gibt zwei verschiedene Sturm-Liouville-Probleme: das reguläre Sturm-Liouville-Problem und das singuläre Sturm-Liouville-Problem. Die Funktionen für das reguläre und das singuläre Sturm-Liouville-Problem sind nicht dieselben. +% +%Kapitel mit "Das reguläre Sturm-Liouville-Problem" +% + \subsection{Das reguläre Sturm-Liouville-Problem\label{sub:reguläre_sturm_liouville_problem}} Damit es sich um ein reguläres Sturm-Liouville-Problem handelt, müssen einige Bedingungen beachtet werden. +\begin{definition} + \index{regläres Sturm-Liouville-Problem} + Die Bedingungen für ein reguläres Sturm-Liouville-Problem sind: + \begin{itemize} + \item Die Funktionen $p(x), p'(x), q(x)$ und $w(x)$ müssen stetig und reell sein. + \item sowie müssen in einem Endlichen Intervall $[ \ a,b] \ $ integrierbar sein. + \item $p(x)^{-1}$ und $w(x)$ sind $>0$. + \item Es gelten die Randbedingungen \ref{eq:randbedingungen}, wobei $|k_i|^2 + |h_i|^2\ne 0$ mit $i=a,b$. + \end{itemize} +\end{definition} +Bei einem regulären Sturm-Liouville-Problem geht es darum, ohne genaue Kenntnis der Eigenfunktionen diese dennoch beschreiben zu können. -\begin{itemize} - \item Die Funktionen $p(x), p'(x), q(x)$ und $w(x)$ müssen stetig und reell sein. - \item sowie müssen in einem Endlichen Intervall $[ \ a,b] \ $ integrierbar sein. - \item $p(x)^{-1}$ und $w(x)$ sind $>0$. - \item Es gelten die Randbedingungen \ref{eq:randbedingungen}, wobei $|k_i|^2 + |h_i|^2\ne 0$ mit $i=a,b$. -\end{itemize} -Bei einem regulären Sturm-Liouville-Problem geht es darum, ohne genaue Kenntnis der Eigenfunktionen diese dennoch beschreiben zu können. +% +%Kapitel mit "Das singuläre Sturm-Liouville-Problem" +% \subsection{Das singuläre Sturm-Liouville-Problem\label{sub:singuläre_sturm_liouville_problem}} -Von einem singulären Sturm-Liouville-Problem spricht man, wenn die oben genannten Bedingungen nicht erfüllt sind, d.h: +Von einem singulären Sturm-Liouville-Problem spricht man, wenn die Bedingungen des regulärem Problem nicht erfüllt sind. +\begin{definition} + \index{singuläres Sturm-Liouville-Problem} +Es handelt sich um ein singuläres Sturm-Liouville-Problem, wenn: + \begin{itemize} + \item wenn sein Definitionsbereich auf dem Intervall $[ \ a,b] \ $ unbeschränkt ist oder + \item wenn die Koeffizienten an den Randpunkten Singularitäten haben. + \end{itemize} +\end{definition} +Allerdings kann nur eine der Bedingungen nicht erfüllt sein, so dass es sich bereits um ein singuläres Sturm-Liouville-Problem handelt. -\begin{itemize} - \item wenn sein Definitionsbereich auf dem Intervall $[ \ a,b] \ $ unbeschränkt ist oder - \item wenn die Koeffizienten an den Randpunkten Singularitäten haben. -\end{itemize} +\begin{beispiel} + Das Randwertproblem + \begin{equation} + \begin{aligned} + x^2y'' + xy' + (\lambda^2x^2 - m^2)y &= 0, 0 Date: Tue, 2 Aug 2022 23:54:02 +0200 Subject: improved Einleitung --- buch/papers/ellfilter/einleitung.tex | 74 ++++--- buch/papers/ellfilter/elliptic.tex | 14 +- .../papers/ellfilter/presentation/presentation.tex | 239 ++++++++++++++++----- buch/papers/ellfilter/tikz/arccos.tikz.tex | 55 +++-- buch/papers/ellfilter/tikz/arccos2.tikz.tex | 5 +- buch/papers/ellfilter/tikz/cd.tikz.tex | 5 + buch/papers/ellfilter/tikz/cd2.tikz.tex | 27 +-- buch/papers/ellfilter/tikz/cd3.tikz.tex | 8 +- .../ellfilter/tikz/elliptic_transform.tikz.tex | 64 ------ .../ellfilter/tikz/elliptic_transform1.tikz.tex | 76 +++++++ .../ellfilter/tikz/elliptic_transform2.tikz.tex | 75 +++++++ buch/papers/ellfilter/tikz/filter.tikz.tex | 26 +++ buch/papers/ellfilter/tikz/sn.tikz.tex | 49 +++-- 13 files changed, 496 insertions(+), 221 deletions(-) delete mode 100644 buch/papers/ellfilter/tikz/elliptic_transform.tikz.tex create mode 100644 buch/papers/ellfilter/tikz/elliptic_transform1.tikz.tex create mode 100644 buch/papers/ellfilter/tikz/elliptic_transform2.tikz.tex create mode 100644 buch/papers/ellfilter/tikz/filter.tikz.tex (limited to 'buch/papers') diff --git a/buch/papers/ellfilter/einleitung.tex b/buch/papers/ellfilter/einleitung.tex index 37fd89f..18913fb 100644 --- a/buch/papers/ellfilter/einleitung.tex +++ b/buch/papers/ellfilter/einleitung.tex @@ -1,44 +1,45 @@ \section{Einleitung} -% Lineare filter - -% Filter, Signalverarbeitung - - -Der womöglich wichtigste Filtertyp ist das Tiefpassfilter. -Dieses soll im Durchlassbereich unter der Grenzfrequenz $\Omega_p$ unverstärkt durchlassen und alle anderen Frequenzen vollständig auslöschen. - -% Bei der Implementierung von Filtern - -In der Elektrotechnik führen Schaltungen mit linearen Bauelementen wie Kondensatoren, Spulen und Widerständen immer zu linearen zeitinvarianten Systemen (LTI-System von englich \textit{time-invariant system}). -Die Übertragungsfunktion im Frequenzbereich $|H(\Omega)|$ eines solchen Systems ist dabei immer eine rationale Funktion, also eine Division von zwei Polynomen. -Die Polynome habe dabei immer reelle oder komplex-konjugierte Nullstellen. - - +Filter sind womöglich eines der wichtigsten Element in der Signalverarbeitung und finden Anwendungen in der digitalen und analogen Elektrotechnik. +Besonders hilfreich ist die Untergruppe der linearen Filter. +Elektronische Schaltungen mit linearen Bauelementen wie Kondensatoren, Spulen und Widerständen führen immer zu linearen zeitinvarianten Systemen (LTI-System von englich \textit{time-invariant system}). +Durch die Linearität werden beim das Filtern keine neuen Frequenzanteile erzeugt, was es erlaubt, einen Frequenzanteil eines Signals verzerrungsfrei herauszufiltern. %TODO review sentence +Diese Eigenschaft macht es Sinnvoll, lineare Filter im Frequenzbereich zu beschreiben. +Die Übertragungsfunktion eines linearen Filters im Frequenzbereich $H(\Omega)$ ist dabei immer eine rationale Funktion, also eine Division von zwei Polynomen. +Dabei ist $\Omega = 2 \pi f$ die analoge Frequenzeinheit. +Die Polynome haben dabei immer reelle oder komplex-konjugierte Nullstellen. + +Ein breit angewendeter Filtertyp ist das Tiefpassfilter, welches beabsichtigt alle Frequenzen eines Signals über der Grenzfrequenz $\Omega_p$ auszulöschen. +Der Rest soll dabei unverändert passieren. +Ein solches Filter hat idealerweise eine Frequenzantwort \begin{equation} \label{ellfilter:eq:h_omega} - | H(\Omega)|^2 = \frac{1}{1 + \varepsilon_p^2 F_N^2(w)}, \quad w=\frac{\Omega}{\Omega_p} + H(\Omega) = + \begin{cases} + 1 & \Omega < \Omega_p \\ + 0 & \Omega < \Omega_p + \end{cases}. \end{equation} - -$\Omega = 2 \pi f$ ist die analoge Frequenz - - -% Linear filter -Damit das Filter implementierbar und stabil ist, muss $H(\Omega)^2$ eine rationale Funktion sein, deren Nullstellen und Pole auf der linken Halbebene liegen. - -$N \in \mathbb{N} $ gibt dabei die Ordnung des Filters vor, also die maximale Anzahl Pole oder Nullstellen. - -Damit ein Filter die Passband Kondition erfüllt muss $|F_N(w)| \leq 1 \forall |w| \leq 1$ und für $|w| \geq 1$ sollte die Funktion möglichst schnell divergieren. -Eine einfaches Polynom, dass das erfüllt, erhalten wir wenn $F_N(w) = w^N$. +Leider ist eine solche Funktion nicht als rationale Funktion darstellbar. +Aus diesem Grund sind realisierbare Approximationen gesucht. +Jede Approximation wird einen kontinuierlichen übergang zwischen Durchlassbereich und Sperrbereich aufweisen. +Oft wird dabei der Faktor $1/\sqrt{2}$ als Schwelle zwischen den beiden Bereichen gewählt. +Somit lassen sich lineare Tiefpassfilter mit folgender Funktion zusammenfassen: +\begin{equation} \label{ellfilter:eq:h_omega} + | H(\Omega)|^2 = \frac{1}{1 + \varepsilon_p^2 F_N^2(w)}, \quad w=\frac{\Omega}{\Omega_p}, +\end{equation} +%TODO figure? +wobei $F_N(w)$ eine rationale Funktion ist, $|F_N(w)| \leq 1 ~\forall~ |w| \leq 1$ erfüllt und für $|w| \geq 1$ möglichst schnell divergiert. +Des weiteren müssen alle Nullstellen und Pole von $F_N$ auf der linken Halbebene liegen, damit das Filter implementierbar und stabil ist. +$N \in \mathbb{N} $ gibt dabei die Ordnung des Filters vor, also die maximale Anzahl Pole oder Nullstellen, die zur Komplexitätsmilderung klein gehalten werden soll. +Eine einfache Funktion für $F_N$ ist das Polynom $w^N$. Tatsächlich erhalten wir damit das Butterworth Filter, wie in Abbildung \ref{ellfilter:fig:butterworth} ersichtlich. \begin{figure} \centering \input{papers/ellfilter/python/F_N_butterworth.pgf} - \caption{$F_N$ für Butterworth filter. Der grüne Bereich definiert die erlaubten Werte für alle $F_N$-Funktionen.} + \caption{$F_N$ für Butterworth filter. Der grüne und gelbe Bereich definiert die erlaubten Werte für alle $F_N$-Funktionen.} \label{ellfilter:fig:butterworth} \end{figure} - -wenn $F_N(w)$ eine rationale Funktion ist, ist auch $H(\Omega)$ eine rationale Funktion und daher ein lineares Filter. %proof? - +Eine Reihe von rationalen Funktionen können für $F_N$ eingesetzt werden, um Tiefpassfilter\-approximationen mit unterschiedlichen Eigenschaften zu erhalten: \begin{align} F_N(w) & = \begin{cases} @@ -48,9 +49,14 @@ wenn $F_N(w)$ eine rationale Funktion ist, ist auch $H(\Omega)$ eine rationale F R_N(w, \xi) & \text{Elliptisch (Cauer)} \\ \end{cases} \end{align} - Mit der Ausnahme vom Butterworth filter sind alle Filter nach speziellen Funktionen benannt. -Alle diese Filter sind optimal für unterschiedliche Anwendungsgebiete. +Alle diese Filter sind optimal hinsichtlich einer Eigenschaft. Das Butterworth-Filter, zum Beispiel, ist maximal flach im Durchlassbereich. -Das Tschebyscheff-1 Filter sind maximal steil für eine definierte Welligkeit im Durchlassbereich, währendem es im Sperrbereich monoton abfallend ist. +Das Tschebyscheff-1 Filter ist maximal steil für eine definierte Welligkeit im Durchlassbereich, währendem es im Sperrbereich monoton abfallend ist. Es scheint so als sind gewisse Eigenschaften dieser speziellen Funktionen verantwortlich für die Optimalität dieser Filter. + +Dieses Paper betrachtet die Theorie hinter dem elliptischen Filter, dem wohl exotischsten dieser Auswahl. +Es weist sich aus durch den Steilsten Übergangsbereich für eine gegebene Filterdesignspezifikation. +Des weiteren kann es als Verallgemeinerung des Tschebyscheff-Filters angesehen werden. + +% wenn $F_N(w)$ eine rationale Funktion ist, ist auch $H(\Omega)$ eine rationale Funktion und daher ein lineares Filter. %proof? diff --git a/buch/papers/ellfilter/elliptic.tex b/buch/papers/ellfilter/elliptic.tex index 96731c8..861600b 100644 --- a/buch/papers/ellfilter/elliptic.tex +++ b/buch/papers/ellfilter/elliptic.tex @@ -31,13 +31,13 @@ Die $\cd^{-1}(w, k)$-Funktion ist um $K$ verschoben zur $\sn^{-1}(w, k)$-Funktio \end{figure} Auffallend ist, dass sich alle Nullstellen und Polstellen um $K$ verschoben haben. -Durch das Konzept vom fundamentalen Rechteck, siehe Abbildung \ref{ellfilter:fig:fundamental_rectangle} können für alle inversen Jaccobi elliptischen Funktionen die Positionen der Null- und Polstellen anhand eines Diagramms ermittelt werden. +Durch das Konzept vom fundamentalen Rechteck, siehe Abbildung \ref{ellfilter:fig:fundamental_rectangle} können für alle inversen Jacobi elliptischen Funktionen die Positionen der Null- und Polstellen anhand eines Diagramms ermittelt werden. Der erste Buchstabe bestimmt die Position der Nullstelle und der zweite Buchstabe die Polstelle. \begin{figure} \centering \input{papers/ellfilter/tikz/fundamental_rectangle.tikz.tex} \caption{ - Fundamentales Rechteck der inversen Jaccobi elliptischen Funktionen. + Fundamentales Rechteck der inversen Jacobi elliptischen Funktionen. } \label{ellfilter:fig:fundamental_rectangle} \end{figure} @@ -80,7 +80,7 @@ Analog zu Abbildung \ref{ellfilter:fig:arccos2} können wir auch bei den ellipti \subsection{Gradgleichung} Der $\cd^{-1}$ Term muss so verzogen werden, dass die umgebene $\cd$-Funktion die Nullstellen und Pole trifft. -Dies trifft ein wenn die Degree Equation erfüllt ist. +Dies trifft ein wenn die Gradengleichung erfüllt ist. \begin{equation} N \frac{K^\prime}{K} = \frac{K^\prime_1}{K_1} @@ -96,9 +96,15 @@ Die Rechnung wird in \ref{ellfilter:bib:orfanidis} im Detail angeschaut. \caption{Die Periodizitäten in realer und imaginärer Richtung in Abhängigkeit vom elliptischen Modul $k$.} \end{figure} +%TODO combine figures? \begin{figure} \centering - \input{papers/ellfilter/tikz/elliptic_transform.tikz} + \input{papers/ellfilter/tikz/elliptic_transform1.tikz} + \caption{Die Gradgleichung als geometrisches Problem.} +\end{figure} +\begin{figure} + \centering + \input{papers/ellfilter/tikz/elliptic_transform2.tikz} \caption{Die Gradgleichung als geometrisches Problem.} \end{figure} diff --git a/buch/papers/ellfilter/presentation/presentation.tex b/buch/papers/ellfilter/presentation/presentation.tex index adbf925..96bdfd3 100644 --- a/buch/papers/ellfilter/presentation/presentation.tex +++ b/buch/papers/ellfilter/presentation/presentation.tex @@ -76,9 +76,9 @@ %Title Page \title{Elliptische Filter} -\subtitle{Eine Anwendung der Jaccobi elliptischen Funktionen} +\subtitle{Eine Anwendung der Jacobi elliptischen Funktionen} \author{Nicolas Tobler} -% \institute{OST Ostschweizer Fachhochschule} +\institute{Mathematisches Seminar 2022 | Spezielle Funktionen} % \institute{\includegraphics[scale=0.3]{../img/ost_logo.png}} \date{\today} @@ -113,7 +113,7 @@ \end{frame} \begin{frame} - \frametitle{Content} + \frametitle{Inhalt} \tableofcontents \end{frame} @@ -122,16 +122,29 @@ \begin{frame} \frametitle{Lineare Filter} + \begin{center} + \scalebox{0.75}{ + \input{../tikz/filter.tikz.tex} + } + \end{center} - \begin{equation} + + \begin{equation*} | H(\Omega)|^2 = \frac{1}{1 + \varepsilon_p^2 F_N^2(w)}, \quad w=\frac{\Omega}{\Omega_p} - \end{equation} + \end{equation*} \pause - \begin{equation} + \begin{align*} + |F_N(w)| &< 1 \quad \forall \quad |w| < 1 \\ + |F_N(w)| &= 1 \quad \forall \quad |w| = 1 \\ + |F_N(w)| &> 1 \quad \forall \quad |w| > 1 + \end{align*} + + + \begin{equation*} F_N(w) = w^N - \end{equation} + \end{equation*} \end{frame} @@ -218,10 +231,36 @@ Darstellung mit trigonometrischen Funktionen: - \begin{align} \label{ellfilter:eq:chebychef_polynomials} + \begin{align*} T_N(w) &= \cos \left( N \cos^{-1}(w) \right) \\ &= \cos \left(N~z \right), \quad w= \cos(z) - \end{align} + \end{align*} + + \pause + + \begin{align*} + \cos^{-1}(x) + &= + \int_{x}^{1} + \frac{ + dz + }{ + \sqrt{ + 1-z^2 + } + }\\ + &= + \int_{0}^{x} + \frac{ + -1 + }{ + \sqrt{ + 1-z^2 + } + } + ~dz + + \frac{\pi}{2} + \end{align*} \end{frame} @@ -229,15 +268,41 @@ \begin{frame} \frametitle{Tschebyscheff-Filter} - \begin{equation*} - z = \cos^{-1}(w) - \end{equation*} + \begin{columns} + + \begin{column}{0.2\textwidth} + + \begin{equation*} + z = \cos^{-1}(w) + \end{equation*} + + \vspace{0.5cm} + + Integrand: + \begin{equation*} + \frac{ + -1 + }{ + \sqrt{ + 1-z^2 + } + } + \end{equation*} + + \end{column} + \begin{column}{0.8\textwidth} + + + \begin{center} + \scalebox{0.7}{ + \input{../tikz/arccos.tikz.tex} + } + \end{center} + + \end{column} + \end{columns} + - \begin{center} - \scalebox{0.85}{ - \input{../tikz/arccos.tikz.tex} - } - \end{center} \end{frame} @@ -245,7 +310,7 @@ \frametitle{Tschebyscheff-Filter} \begin{equation*} - z_1 = N~\cos^{-1}(w) + T_N(w) = \cos \left(z_1 \right), \quad z_1 = N~\cos^{-1}(w) \end{equation*} \begin{center} @@ -257,15 +322,14 @@ \end{frame} - \section{Jaccobi elliptische Funktionen} + \section{Jacobi elliptische Funktionen} \begin{frame} - \frametitle{Jaccobi elliptische Funktionen} + \frametitle{Jacobi elliptische Funktionen} + Elliptisches Integral erster Art - \begin{equation} - z - = + \begin{equation*} F(\phi, k) = \int_{0}^{\phi} @@ -276,18 +340,18 @@ 1-k^2 \sin^2 \theta } } - = - \int_{0}^{\phi} - \frac{ - dt - }{ - \sqrt{ - (1-t^2)(1-k^2 t^2) - } - } - \end{equation} + % = + % \int_{0}^{\phi} + % \frac{ + % dt + % }{ + % \sqrt{ + % (1-t^2)(1-k^2 t^2) + % } + % } + \end{equation*} - \begin{equation} + \begin{equation*} K(k) = \int_{0}^{\pi / 2} @@ -298,24 +362,88 @@ 1-k^2 \sin^2 \theta } } - \end{equation} + \end{equation*} \end{frame} + + + + \begin{frame} - \frametitle{Jaccobi elliptische Funktionen} + \frametitle{Jacobi elliptische Funktionen} + + \begin{equation*} + \sn^{-1}(w, k) + = + F(\phi, k), + \quad + \phi = \sin^{-1}(w) + \end{equation*} + + \begin{align*} + \sn^{-1}(w, k) + & = + \int_{0}^{\phi} + \frac{ + d\theta + }{ + \sqrt{ + 1-k^2 \sin^2 \theta + } + }, + \quad + \phi = \sin^{-1}(w) + \\ + & = + \int_{0}^{w} + \frac{ + dt + }{ + \sqrt{ + (1-t^2)(1-k^2 t^2) + } + } + \end{align*} - \begin{equation*} - z = \sn^{-1}(w, k) - \end{equation*} - \begin{center} - \scalebox{0.7}{ - \input{../tikz/sn.tikz.tex} - } - \end{center} + + \end{frame} + + \begin{frame} + \frametitle{Jacobi elliptische Funktionen} + \begin{columns} + \begin{column}{0.2\textwidth} + + \begin{equation*} + z = \sn^{-1}(w, k) + \end{equation*} + + \vspace{0.5cm} + + Integrand: + \begin{equation*} + \frac{ + 1 + }{ + \sqrt{ + (1-t^2)(1-k^2 t^2) + } + } + \end{equation*} + + \end{column} + \begin{column}{0.8\textwidth} + \begin{center} + \scalebox{0.75}{ + \input{../tikz/sn.tikz.tex} + } + \end{center} + \end{column} + \end{columns} + \end{frame} @@ -334,7 +462,7 @@ \begin{frame} - \frametitle{Jaccobi elliptische Funktionen} + \frametitle{Jacobi elliptische Funktionen} \begin{equation*} z = \cd^{-1}(w, k) @@ -354,9 +482,9 @@ \begin{frame} \frametitle{Elliptisches Filter} - \begin{equation*} - z_1 = N~\frac{K_1}{K}~\cd^{-1}(w, k) - \end{equation*} + % \begin{equation*} + % z_1 = N~\frac{K_1}{K}~\cd^{-1}(w, k) + % \end{equation*} \begin{center} \scalebox{0.75}{ @@ -379,16 +507,17 @@ \begin{frame} \frametitle{Gradgleichung} - \begin{equation} - N \frac{K^\prime}{K} = \frac{K^\prime_1}{K_1} - \end{equation} - \begin{center} \scalebox{0.95}{ - \input{../tikz/elliptic_transform.tikz} + \input{../tikz/elliptic_transform2.tikz} } \end{center} + \onslide<5->{ + \begin{equation*} + N \frac{K^\prime}{K} = \frac{K^\prime_1}{K_1} + \end{equation*} + } \end{frame} @@ -398,7 +527,9 @@ \begin{equation*} R_N = \cd(z_1, k_1), \quad - z_1 = N~\frac{K_1}{K}~\cd^{-1}(w, k) + z_1 = N~\frac{K_1}{K}~\cd^{-1}(w, k), + \quad + N \frac{K^\prime}{K} = \frac{K^\prime_1}{K_1} \end{equation*} \begin{center} diff --git a/buch/papers/ellfilter/tikz/arccos.tikz.tex b/buch/papers/ellfilter/tikz/arccos.tikz.tex index 987f885..4211053 100644 --- a/buch/papers/ellfilter/tikz/arccos.tikz.tex +++ b/buch/papers/ellfilter/tikz/arccos.tikz.tex @@ -8,29 +8,6 @@ \begin{scope}[xscale=0.6] - \clip(-7.5,-2) rectangle (7.5,2); - - \draw[ultra thick, ->, darkgreen] (0, 0) -- (0,1.5); - \draw[ultra thick, ->, orange] (1, 0) -- (0,0); - \draw[ultra thick, ->, red] (2, 0) -- (1,0); - \draw[ultra thick, ->, blue] (2,1.5) -- (2, 0); - - \foreach \i in {-2,...,1} { - \begin{scope}[opacity=0.5, xshift=\i*4cm] - \draw[->, orange] (-1, 0) -- (0,0); - \draw[->, darkgreen] (0, 0) -- (0,1.5); - \draw[->, darkgreen] (0, 0) -- (0,-1.5); - \draw[->, orange] (1, 0) -- (0,0); - \draw[->, red] (2, 0) -- (1,0); - \draw[->, blue] (2,1.5) -- (2, 0); - \draw[->, blue] (2,-1.5) -- (2, 0); - \draw[->, red] (2, 0) -- (3,0); - - \node[zero] at (1,0) {}; - \node[zero] at (3,0) {}; - \end{scope} - } - \node[gray, anchor=north] at (-6,0) {$-3\pi$}; \node[gray, anchor=north] at (-4,0) {$-2\pi$}; \node[gray, anchor=north] at (-2,0) {$-\pi$}; @@ -43,8 +20,40 @@ % \node[gray, anchor=south east] at (0, 0) {$0$}; \node[gray, anchor=east] at (0, 1.5) {$\infty$}; + \clip(-7.5,-2) rectangle (7.5,2); + + % \pause + \draw[ultra thick, ->, orange] (1, 0) -- (0,0); + % \pause + \draw[ultra thick, ->, darkgreen] (0, 0) -- (0,1.5); + % \pause + \draw[ultra thick, ->, red] (2, 0) -- (1,0); + \draw[ultra thick, ->, blue] (2,1.5) -- (2, 0); + + % \pause + + \foreach \i in {-2,...,1} { + \begin{scope}[xshift=\i*4cm] + \begin{scope}[opacity=0.5] + \draw[->, orange] (-1, 0) -- (0,0); + \draw[->, darkgreen] (0, 0) -- (0,1.5); + \draw[->, darkgreen] (0, 0) -- (0,-1.5); + \draw[->, orange] (1, 0) -- (0,0); + \draw[->, red] (2, 0) -- (1,0); + \draw[->, blue] (2,1.5) -- (2, 0); + \draw[->, blue] (2,-1.5) -- (2, 0); + \draw[->, red] (2, 0) -- (3,0); + \end{scope} + \node[zero] at (1,0) {}; + \node[zero] at (3,0) {}; + \end{scope} + } + \end{scope} + \node[zero] at (4,2) (n) {}; + \node[anchor=west] at (n.east) {Zero}; + \begin{scope}[yshift=-3cm] \draw[gray, ->] (-5,0) -- (5,0) node[anchor=west]{$w$}; diff --git a/buch/papers/ellfilter/tikz/arccos2.tikz.tex b/buch/papers/ellfilter/tikz/arccos2.tikz.tex index 3fc3cc6..755e8a0 100644 --- a/buch/papers/ellfilter/tikz/arccos2.tikz.tex +++ b/buch/papers/ellfilter/tikz/arccos2.tikz.tex @@ -14,7 +14,6 @@ \draw[>->, line width=0.05, thick, orange] (4, 1.5) -- (4,0) -- node[anchor=south, pos=0.25]{$N=2$} (0,0) -- (0,1.5); \draw[>->, line width=0.05, thick, red] (6, 1.5) node[anchor=north west]{$-\infty$} -- (6,-0.05) node[anchor=west]{$-1$} -- node[anchor=north]{$0$} node[anchor=south, pos=0.1666]{$N=3$} (-0.1,-0.05) node[anchor=east]{$1$} -- (-0.1,1.5) node[anchor=north east]{$\infty$}; - \node[zero] at (-7,0) {}; \node[zero] at (-5,0) {}; \node[zero] at (-3,0) {}; @@ -24,7 +23,6 @@ \node[zero] at (5,0) {}; \node[zero] at (7,0) {}; - \end{scope} \node[gray, anchor=north] at (-8,0) {$-4\pi$}; @@ -42,4 +40,7 @@ \end{scope} + \node[zero] at (4,2) (n) {}; + \node[anchor=west] at (n.east) {Zero}; + \end{tikzpicture} \ No newline at end of file diff --git a/buch/papers/ellfilter/tikz/cd.tikz.tex b/buch/papers/ellfilter/tikz/cd.tikz.tex index 7a2767b..b2b0090 100644 --- a/buch/papers/ellfilter/tikz/cd.tikz.tex +++ b/buch/papers/ellfilter/tikz/cd.tikz.tex @@ -63,6 +63,11 @@ \end{scope} + \node[zero] at (4,3) (n) {}; + \node[anchor=west] at (n.east) {Zero}; + \node[pole, below=0.25cm of n] (n) {}; + \node[anchor=west] at (n.east) {Pole}; + \begin{scope}[yshift=-4cm, xscale=0.75] \draw[gray, ->] (-6,0) -- (6,0) node[anchor=west]{$w$}; diff --git a/buch/papers/ellfilter/tikz/cd2.tikz.tex b/buch/papers/ellfilter/tikz/cd2.tikz.tex index 425db95..bba5789 100644 --- a/buch/papers/ellfilter/tikz/cd2.tikz.tex +++ b/buch/papers/ellfilter/tikz/cd2.tikz.tex @@ -47,21 +47,6 @@ (5,0) node(t_k_unten){} -- node[right, xshift=0.1cm]{$K^\prime \frac{K_1N}{K} = K^\prime_1$} (5,0.5) node(t_k_opt_unten){}; - \foreach \i in {-2,...,1} { - \foreach \j in {-2,...,1} { - \begin{scope}[xshift=\i*4cm, yshift=\j*1cm] - - \node[zero] at ( 1, 0) {}; - \node[zero] at ( 3, 0) {}; - \node[pole] at ( 1,0.5) {}; - \node[pole] at ( 3,0.5) {}; - - \end{scope} - } - } - - - \draw[ultra thick, ->, darkgreen] (5, 0) -- node[yshift=-0.5cm]{Durchlassbereich} (0,0); \draw[ultra thick, ->, orange] (-0, 0) -- node[align=center]{Übergangs-\\berech} (0,0.5); @@ -74,10 +59,20 @@ \draw (2,0.5) node[dot]{} node[anchor=north] {\small $-1/k$}; \draw (4,0.5) node[dot]{} node[anchor=north] {\small $1/k$}; + \foreach \i in {-2,...,1} { + \foreach \j in {-2,...,1} { + \begin{scope}[xshift=\i*4cm, yshift=\j*1cm] + \node[zero] at ( 1, 0) {}; + \node[zero] at ( 3, 0) {}; + \node[pole] at ( 1,0.5) {}; + \node[pole] at ( 3,0.5) {}; - \end{scope} + \end{scope} + } + } + \end{scope} \end{scope} diff --git a/buch/papers/ellfilter/tikz/cd3.tikz.tex b/buch/papers/ellfilter/tikz/cd3.tikz.tex index fa9cc08..ae18519 100644 --- a/buch/papers/ellfilter/tikz/cd3.tikz.tex +++ b/buch/papers/ellfilter/tikz/cd3.tikz.tex @@ -36,7 +36,6 @@ % \fill[orange!30] (0,0) rectangle (5, 0.5); \fill[yellow!30] (0,0) rectangle (1, 0.5); - \node[] at (2.5, 0.25) {\small $N=5$}; % \draw[decorate,decoration={brace,amplitude=3pt,mirror}, yshift=0.05cm] @@ -62,11 +61,13 @@ - + \onslide<2->{ \draw[ultra thick, ->, darkgreen] (5, 0) -- node[yshift=-0.4cm]{Durchlassbereich} (0,0); \draw[ultra thick, ->, orange] (-0, 0) -- node[align=center]{Übergangs-\\berech} (0,0.5); \draw[ultra thick, ->, red] (0,0.5) -- node[align=center, yshift=0.4cm]{Sperrbereich} (5, 0.5); - + \node[] at (2.5, 0.25) {\small $N=5$}; + } + \onslide<1->{ \draw (4,0 ) node[dot]{} node[anchor=south] {\small $1$}; \draw (2,0 ) node[dot]{} node[anchor=south] {\small $-1$}; \draw (0,0 ) node[dot]{} node[anchor=south west] {\small $1$}; @@ -74,6 +75,7 @@ \draw (2,0.5) node[dot]{} node[anchor=north] {\small $-1/k$}; \draw (4,0.5) node[dot]{} node[anchor=north] {\small $1/k$}; + } \end{scope} diff --git a/buch/papers/ellfilter/tikz/elliptic_transform.tikz.tex b/buch/papers/ellfilter/tikz/elliptic_transform.tikz.tex deleted file mode 100644 index c91ecf1..0000000 --- a/buch/papers/ellfilter/tikz/elliptic_transform.tikz.tex +++ /dev/null @@ -1,64 +0,0 @@ - -\def\d{0.2} -\def\n{3} -\def\nn{2} -\def\a{2.5} - -\begin{tikzpicture}[>=stealth', auto, node distance=2cm, scale=1.2] - - \tikzstyle{zero} = [draw, circle, inner sep =0, minimum height=0.15cm] - \tikzstyle{dot} = [fill, circle, inner sep =0, minimum height=0.1cm] - - \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}} - - \begin{scope}[xscale=3, yscale=3] - - \begin{scope}[] - - \fill[orange!30, scale=1.735] (0,0) rectangle (\d*\a+0.5, \d/\a+0.5); - \fill[yellow!30] (0,0) rectangle (\d*\a+0.5, \d/\a+0.5); - - \begin{scope}[scale=1.735, red] - \draw (0,0) rectangle (\d*\a/\n+0.5/\n, \d/\a+0.5); - \draw[gray] (0,0) -- (\d*\a/\n+0.5/\n, \d/\a+0.5); - - \node[zero] at ( \d*\a/\n+0.5/\n, \d/\a+0.5) {}; - \node[pole, color=red] at ( \d*\a/\n+0.5/\n, 0) {}; - - - \draw[] ( \d*\a/\n+0.5/\n,0) node[anchor=north] {\small $K_1$}; - \draw[] (0, \d/\a+0.5) node[anchor=east]{\small $jK_1^\prime$}; - - \end{scope} - - \begin{scope}[blue] - \draw[] (0,0) rectangle (\d*\a+0.5, \d/\a+0.5); - \foreach \i in {1,...,\nn} { - \draw[gray, dotted] (\i*\d*\a/\n+\i*0.5/\n, 0) -- (\i*\d*\a/\n+\i*0.5/\n, \d/\a+0.5); - } - - \node[zero] at ( \d*\a+0.5, \d/\a+0.5) {}; - \node[pole, color=blue] at ( \d*\a+0.5, 0) {}; - - \draw[] ( \d*\a+0.5,0) node[anchor=north] {\small $K$}; - \draw[] (0, \d/\a+0.5) node[anchor=east]{\small $jK^\prime$}; - - \node[dot, gray] at (\d*\a/\n+0.5/\n, \d/\a+0.5) {}; - \node[above] at (0.5*\d*\a/\n+0.5*0.5/\n, \d/\a+0.5) {\small $K/N$}; - - \end{scope} - - \draw[thick, gray, ->] (0,-0.25) -- (0,1.25) node[anchor=south]{$\mathrm{Im}$}; - \draw[thick, gray, ->] (-0.25,0) -- (2,0) node[anchor=west]{$\mathrm{Re}$}; - - \begin{scope}[] - \clip(0,0) rectangle (2,1.25); - \draw[scale=1, domain=0.1:10, variable=\x, smooth, samples=200] plot ({\d*\x1+0.5}, {\d/\x+0.5}); - - \end{scope} - \end{scope} - - -\end{scope} - -\end{tikzpicture} diff --git a/buch/papers/ellfilter/tikz/elliptic_transform1.tikz.tex b/buch/papers/ellfilter/tikz/elliptic_transform1.tikz.tex new file mode 100644 index 0000000..2a36ee0 --- /dev/null +++ b/buch/papers/ellfilter/tikz/elliptic_transform1.tikz.tex @@ -0,0 +1,76 @@ +\begin{tikzpicture}[>=stealth', auto, node distance=2cm, scale=1.2] + + \tikzstyle{zero} = [draw, circle, inner sep =0, minimum height=0.15cm] + + \tikzset{pole/.style={cross out, draw, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}} + + \begin{scope}[xscale=1, yscale=1.5] + + \begin{scope}[] + + \fill[orange!25] (0,0) rectangle (1.5, 0.75); + \fill[yellow!50] (0,0) rectangle (0.5, 0.25); + + \draw[gray, ->] (0,-0.75) -- (0,1.25) node[anchor=south]{$\mathrm{Im}~z$}; + \draw[gray, ->] (-1.75,0) -- (1.75,0) node[anchor=west]{$\mathrm{Re}~z$}; + + \draw[gray] ( 0.5,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $K$}; + \draw[gray] (0, 0.25) +(0.05, 0) -- +(-0.05, 0) node[inner sep=0, anchor=east]{\small $jK^\prime$}; + + % \draw[gray] ( 1.5,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $K_1$}; + % \draw[gray] (0, 0.75) +(0.05, 0) -- +(-0.05, 0) node[inner sep=0, anchor=east]{\small $jK_1^\prime$}; + + \clip(-1.6,-0.6) rectangle (1.6,1.6); + \begin{scope}[xscale=0.5, yscale=0.25, blue] + \foreach \i in {-1,...,1} { + \foreach \j in {-1,...,2} { + \begin{scope}[xshift=\i*2cm, yshift=\j*2cm] + \node[zero] at ( 1, 0) {}; + \node[zero] at ( -1, 0) {}; + \node[pole] at ( 1,1) {}; + \node[pole] at ( -1,1) {}; + \end{scope} + } + } + \end{scope} + + \node at (0,2) {$\cd \left(N~K_1~z , k_1 \right)$}; + \node at (0,2) {$w= \cd(z K, k)$}; + + \draw[scale=0.2, domain=0.02:5, variable=\x, red] plot ({\x1+3}, {1/\x+2}); + + \end{scope} + + \begin{scope}[xshift=5cm] + + \fill[orange!50] (0,0) rectangle (1.5, 0.75); + \fill[yellow!25] (0,0) rectangle (0.5, 0.25); + + \draw[gray, ->] (0,-0.75) -- (0,1.25) node[anchor=south]{$\mathrm{Im}~z$}; + \draw[gray, ->] (-1.75,0) -- (1.75,0) node[anchor=west]{$\mathrm{Re}~z$}; + + % \draw[gray] ( 0.5,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $K$}; + % \draw[gray] (0, 0.25) +(0.05, 0) -- +(-0.05, 0) node[inner sep=0, anchor=east]{\small $jK^\prime$}; + + \draw[gray] ( 0.5,0) +(0,0.05) -- +(0, -0.05) node[inner sep=0, anchor=north] {\small $K_1$}; + \draw[gray] (0, 0.75) +(0.05, 0) -- +(-0.05, 0) node[inner sep=0, anchor=east]{\small $jK_1^\prime$}; + + \clip(-1.6,-0.6) rectangle (1.6,1.6); + \begin{scope}[xscale=0.5, yscale=0.75, red] + \foreach \i in {-1,...,1} { + \foreach \j in {-1,...,0} { + \begin{scope}[xshift=\i*2cm, yshift=\j*2cm] + \node[zero] at ( 1, 0) {}; + \node[zero] at ( -1, 0) {}; + \node[pole] at ( 1,1) {}; + \node[pole] at ( -1,1) {}; + \end{scope} + } + } + \end{scope} + + \end{scope} + +\end{scope} + +\end{tikzpicture} diff --git a/buch/papers/ellfilter/tikz/elliptic_transform2.tikz.tex b/buch/papers/ellfilter/tikz/elliptic_transform2.tikz.tex new file mode 100644 index 0000000..20c2d82 --- /dev/null +++ b/buch/papers/ellfilter/tikz/elliptic_transform2.tikz.tex @@ -0,0 +1,75 @@ + +\def\d{0.2} +\def\n{3} +\def\nn{2} +\def\a{2.5} + +\begin{tikzpicture}[>=stealth', auto, node distance=2cm, scale=1.2] + + \tikzstyle{zero} = [draw, circle, inner sep =0, minimum height=0.15cm] + \tikzstyle{dot} = [fill, circle, inner sep =0, minimum height=0.1cm] + + \tikzset{pole/.style={cross out, draw, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}} + + \begin{scope}[xscale=3, yscale=3] + + \begin{scope}[] + % \onslide<4->{ + \fill[orange!30, scale=1.735] (0,0) rectangle (\d*\a+0.5, \d/\a+0.5); + % } + % \onslide<2->{ + \fill[yellow!30] (0,0) rectangle (\d*\a+0.5, \d/\a+0.5); + % } + + \begin{scope}[] + \clip(0,0) rectangle (2,1.25); + \draw[thick, scale=1, domain=0.1:10, variable=\x, smooth, samples=200] plot ({\d*\x1+0.5}, {\d/\x+0.5}); + \node at(1.25,0.7) {$K + jK^\prime$ Ortskurve}; + \end{scope} + + % \onslide<2->{ + \begin{scope}[blue] + \draw[] (0,0) rectangle (\d*\a+0.5, \d/\a+0.5); + + + \node[pole] at ( \d*\a+0.5, \d/\a+0.5) {}; + \node[zero] at ( \d*\a+0.5, 0) {}; + + \draw[] ( \d*\a+0.5,0) node[anchor=north] {\small $K$}; + \draw[] (0, \d/\a+0.5) node[anchor=east]{\small $jK^\prime$}; + + % \onslide<3->{ + + \foreach \i in {1,...,\nn} { + \draw[gray, dotted] (\i*\d*\a/\n+\i*0.5/\n, 0) -- (\i*\d*\a/\n+\i*0.5/\n, \d/\a+0.5); + } + + \node[dot, gray] at (\d*\a/\n+0.5/\n, \d/\a+0.5) {}; + \node[above] at (0.5*\d*\a/\n+0.5*0.5/\n, \d/\a+0.5) {\small $K/N$}; + % } + \end{scope} + % } + + % \onslide<4->{ + \begin{scope}[scale=1.735, red] + \draw (0,0) rectangle (\d*\a/\n+0.5/\n, \d/\a+0.5); + \draw[gray] (0,0) -- (\d*\a/\n+0.5/\n, \d/\a+0.5); + + \node[pole] at ( \d*\a/\n+0.5/\n, \d/\a+0.5) {}; + \node[zero] at ( \d*\a/\n+0.5/\n, 0) {}; + + + \draw[] ( \d*\a/\n+0.5/\n,0) node[anchor=north] {\small $K_1$}; + \draw[] (0, \d/\a+0.5) node[anchor=east]{\small $jK_1^\prime$}; + + \end{scope} + % } + + \draw[gray, ->] (0,-0.25) -- (0,1.25) node[anchor=south]{$\mathrm{Im}$}; + \draw[gray, ->] (-0.25,0) -- (2,0) node[anchor=west]{$\mathrm{Re}$}; + + \end{scope} + +\end{scope} + +\end{tikzpicture} diff --git a/buch/papers/ellfilter/tikz/filter.tikz.tex b/buch/papers/ellfilter/tikz/filter.tikz.tex new file mode 100644 index 0000000..05b59b9 --- /dev/null +++ b/buch/papers/ellfilter/tikz/filter.tikz.tex @@ -0,0 +1,26 @@ +\begin{tikzpicture}[>=stealth', auto, node distance=2cm, scale=1.2] + + \tikzstyle{zero} = [draw, circle, inner sep =0, minimum height=0.15cm] + + \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}} + + \begin{scope}[xscale=2, yscale=2] + + \fill[ gray!20] (0,0) rectangle (1,0.707); + + \draw[gray, ->] (0,-0.25) -- (0,1.25) node[anchor=south]{$|H(\Omega)|$}; + \draw[gray, ->] (-0.25,0) -- (3,0) node[anchor=west]{$\Omega$}; + + \draw[fill = gray!20] (0,0.707) node[left] {$\sqrt{\frac{1}{1+\varepsilon^2}}$} -| (1,0) node[below] {$\Omega_p$}; + + \draw[fill = gray!20] (0,0.707) node[left] {$\sqrt{\frac{1}{1+\varepsilon^2}}$} -| (1,0) node[below] {$\Omega_p$}; + + \begin{scope}[] + \draw[thick, domain=0:2.5, variable=\x, smooth, samples=200] plot + ({\x}, {sqrt(abs(1/ (1 + \x^10)))}); + + \end{scope} + + \end{scope} + +\end{tikzpicture} diff --git a/buch/papers/ellfilter/tikz/sn.tikz.tex b/buch/papers/ellfilter/tikz/sn.tikz.tex index 6ced3c5..8e4d223 100644 --- a/buch/papers/ellfilter/tikz/sn.tikz.tex +++ b/buch/papers/ellfilter/tikz/sn.tikz.tex @@ -17,29 +17,33 @@ \begin{scope}[xshift=-1cm] - \draw[ultra thick, ->, darkgreen] (0, 0) -- (0,0.5); - \draw[ultra thick, ->, orange] (1, 0) -- (0,0); - \draw[ultra thick, ->, red] (2, 0) -- (1,0); - \draw[ultra thick, ->, blue] (2,0.5) -- (2, 0); - \draw[ultra thick, ->, purple] (1, 0.5) -- (2,0.5); - \draw[ultra thick, ->, cyan] (0, 0.5) -- (1,0.5); - + % \pause + \draw[ultra thick, <-, orange] (2, 0) -- (1,0); + % \pause + \draw[ultra thick, <-, darkgreen] (2,0.5) -- (2, 0); + % \pause + \draw[ultra thick, <-, cyan] (1, 0.5) -- (2,0.5); + % \pause + \draw[ultra thick, <-, blue] (0, 0) -- (0,0.5); + \draw[ultra thick, <-, purple] (0, 0.5) -- (1,0.5); + \draw[ultra thick, <-, red] (1, 0) -- (0,0); + % \pause \foreach \i in {-2,...,2} { \foreach \j in {-2,...,1} { \begin{scope}[xshift=\i*4cm, yshift=\j*1cm] - \draw[opacity=0.5, ->, darkgreen] (0, 0) -- (0,0.5); - \draw[opacity=0.5, ->, orange] (1, 0) -- (0,0); - \draw[opacity=0.5, ->, red] (2, 0) -- (1,0); - \draw[opacity=0.5, ->, blue] (2,0.5) -- (2, 0); - \draw[opacity=0.5, ->, purple] (1, 0.5) -- (2,0.5); - \draw[opacity=0.5, ->, cyan] (0, 0.5) -- (1,0.5); - \draw[opacity=0.5, ->, darkgreen] (0,1) -- (0,0.5); - \draw[opacity=0.5, ->, blue] (2,0.5) -- (2, 1); - \draw[opacity=0.5, ->, purple] (3, 0.5) -- (2,0.5); - \draw[opacity=0.5, ->, cyan] (4, 0.5) -- (3,0.5); - \draw[opacity=0.5, ->, red] (2, 0) -- (3,0); - \draw[opacity=0.5, ->, orange] (3, 0) -- (4,0); + \draw[opacity=0.5, <-, blue] (0, 0) -- (0,0.5); + \draw[opacity=0.5, <-, red] (1, 0) -- (0,0); + \draw[opacity=0.5, <-, orange] (2, 0) -- (1,0); + \draw[opacity=0.5, <-, darkgreen] (2,0.5) -- (2, 0); + \draw[opacity=0.5, <-, cyan] (1, 0.5) -- (2,0.5); + \draw[opacity=0.5, <-, purple] (0, 0.5) -- (1,0.5); + \draw[opacity=0.5, <-, blue] (0,1) -- (0,0.5); + \draw[opacity=0.5, <-, darkgreen] (2,0.5) -- (2, 1); + \draw[opacity=0.5, <-, cyan] (3, 0.5) -- (2,0.5); + \draw[opacity=0.5, <-, purple] (4, 0.5) -- (3,0.5); + \draw[opacity=0.5, <-, orange] (2, 0) -- (3,0); + \draw[opacity=0.5, <-, red] (3, 0) -- (4,0); \node[zero] at ( 1, 0) {}; \node[zero] at ( 3, 0) {}; @@ -57,10 +61,13 @@ \draw[gray] ( 1,0) +(0,0.1) -- +(0, -0.1) node[inner sep=0, anchor=north] {\small $K$}; \draw[gray] (0, 0.5) +(0.1, 0) -- +(-0.1, 0) node[inner sep=0, anchor=east]{\small $jK^\prime$}; - - \end{scope} + \node[zero] at (4,3) (n) {}; + \node[anchor=west] at (n.east) {Zero}; + \node[pole, below=0.25cm of n] (n) {}; + \node[anchor=west] at (n.east) {Pole}; + \begin{scope}[yshift=-4cm, xscale=0.75] \draw[gray, ->] (-6,0) -- (6,0) node[anchor=west]{$w$}; -- cgit v1.2.1 From f8ac7479589ae069c7a509cf9908f8e3dddd8451 Mon Sep 17 00:00:00 2001 From: Joshua Baer Date: Wed, 3 Aug 2022 19:45:04 +0200 Subject: bessel labeled --- buch/papers/fm/03_bessel.tex | 65 +- buch/papers/fm/Python animation/Bessel-FM.ipynb | 50 +- buch/papers/fm/Python animation/bessel.pgf | 2057 +++++++++++++++++++++++ buch/papers/fm/packages.tex | 1 + 4 files changed, 2114 insertions(+), 59 deletions(-) create mode 100644 buch/papers/fm/Python animation/bessel.pgf (limited to 'buch/papers') diff --git a/buch/papers/fm/03_bessel.tex b/buch/papers/fm/03_bessel.tex index 760cdc4..eec64f2 100644 --- a/buch/papers/fm/03_bessel.tex +++ b/buch/papers/fm/03_bessel.tex @@ -24,6 +24,7 @@ Das Ziel ist es unser moduliertes Signal mit der Besselfunktion so auszudrücken \sum_{k= -\infty}^\infty J_{k}(\beta) \cos((\omega_c+k\omega_m)t) \label{fm:eq:proof} \end{align} + \subsubsection{Hilfsmittel} Doch dazu brauchen wir die Hilfe der Additionsthoerme \begin{align} @@ -46,18 +47,18 @@ und die drei Besselfunktions indentitäten, \begin{align} \cos(\beta\sin\phi) &= - J_0(\beta) + 2\sum_{k=1}^\infty J_{2k}(\beta) \cos(2k\phi) + J_0(\beta) + 2\sum_{k=1}^\infty(-1)^k \cdot J_{2k}(\beta) \cos(2k\phi) \label{fm:eq:besselid1} \\ \sin(\beta\sin\phi) &= - J_0(\beta) + 2\sum_{k=1}^\infty J_{2k+1}(\beta) \cos((2k+1)\phi) + 2\sum_{k=0}^\infty (-1)^k J_{2k+1}(\beta) \cos((2k+1)\phi) \label{fm:eq:besselid2} \\ J_{-n}(\beta) &= (-1)^n J_n(\beta) \label{fm:eq:besselid3} \end{align} -welche man im Kapitel (ref), ref, ref findet. +welche man im Kapitel \eqref{buch:fourier:eqn:expinphireal}, \eqref{buch:fourier:eqn:expinphiimaginary}, \eqref{buch:fourier:eqn:symetrie}. \subsubsection{Anwenden des Additionstheorem} Mit dem \eqref{fm:eq:addth1} wird aus dem modulierten Signal @@ -66,26 +67,31 @@ Mit dem \eqref{fm:eq:addth1} wird aus dem modulierten Signal = \cos(\omega_c t + \beta\sin(\omega_mt)) = - \cos(\omega_c t)\cos(\beta\sin(\omega_m t))-\sin(\omega_c)\sin(\beta\sin(\omega_m t)). + \cos(\omega_c t)\cos(\beta\sin(\omega_m t))-\sin(\omega_ct)\sin(\beta\sin(\omega_m t)). \label{fm:eq:start} \] + \subsubsection{Cos-Teil} Zu beginn wird der Cos-Teil \[ - \cos(\omega_c)\cos(\beta\sin(\omega_mt)) + \cos(\omega_c t)\cdot\cos(\beta\sin(\omega_mt)) \] mit hilfe der Besselindentität \eqref{fm:eq:besselid1} zum \begin{align*} - \cos(\omega_c t) \cdot \bigg[\, J_0(\beta) + 2\sum_{k=1}^\infty J_{2k}(\beta) \cos( 2k \omega_m t)\, \bigg] - &=\\ - J_0(\beta)\cos(\omega_c t) + \sum_{k=1}^\infty J_{2k}(\beta) - \underbrace{2\cos(\omega_c t)\cos(2k\omega_m t)}_{\text{Additionstheorem}} + \cos(\omega_c t) \cdot \bigg[ J_0(\beta) + 2\sum_{k=1}^\infty(-1)^k \cdot J_{2k}(\beta) \cos( 2k \omega_m t)\, \bigg] + &= + (-1)^k \cdot \sum_{k=1}^\infty J_{2k}(\beta) \underbrace{2\cos(\omega_c t)\cos(2k\omega_m t)}_{\text{Additionstheorem}} \end{align*} wobei mit dem Additionstheorem \eqref{fm:eq:addth2} \(A = \omega_c t\) und \(B = 2k\omega_m t \) zum -\[ - J_0(\beta)\cdot \cos(\omega_c t) + \sum_{k=1}^\infty J_{2k}(\beta) \{ \cos((\omega_c - 2k \omega_m) t)+\cos((\omega_c + 2k \omega_m) t) \} -\] -wird. +\begin{align*} + J_0(\beta) \cdot \cos(\omega_c t) +(-1)^k \cdot \sum_{k=1}^\infty J_{2k}(\beta) \{ \underbrace{\cos((\omega_c - 2k \omega_m) t)} \,+\, \cos((\omega_c + 2k \omega_m) t) \} + \\ + = + (-1)^k \cdot \sum_{k=-\infty}^{-1} J_{2k}(\beta) \overbrace{\cos((\omega_c +2k \omega_m) t)} + \,+\,J_0(\beta)\cdot \cos(\omega_c t+ 2\cdot0 \omega_m) + \,+\, (-1)^k \cdot\sum_{k=1}^\infty J_{2k}(\beta)\cos((\omega_c + 2k \omega_m) t) +\end{align*} + Wenn dabei \(2k\) durch alle geraden Zahlen von \(-\infty \to \infty\) mit \(n\) substituiert erhält man den vereinfachten Term \[ \sum_{n\, \text{gerade}} J_{n}(\beta) \cos((\omega_c + n\omega_m) t), @@ -96,22 +102,32 @@ dabei gehen nun die Terme von \(-\infty \to \infty\), dabei bleibt n Ganzzahlig. \subsubsection{Sin-Teil} Nun zum zweiten Teil des Term \eqref{fm:eq:start}, den Sin-Teil \[ - \sin(\omega_c)\sin(\beta\sin(\omega_m t)). + -\sin(\omega_c t)\cdot\sin(\beta\sin(\omega_m t)). \] Dieser wird mit der \eqref{fm:eq:besselid2} Besselindentität zu \begin{align*} - \sin(\omega_c t) \cdot \bigg[ J_0(\beta) + 2 \sum_{k=1}^\infty J_{ 2k + 1}(\beta) \cos(( 2k + 1) \omega_m t) \bigg] - &=\\ - J_0(\beta) \cdot \sin(\omega_c t) + \sum_{k=1}^\infty J_{2k+1}(\beta) \underbrace{2\sin(\omega_c t)\cos((2k+1)\omega_m t)}_{\text{Additionstheorem}}. + -\sin(\omega_c t) \cdot \bigg[ 2 \sum_{k=0}^\infty(-1)^k \cdot J_{ 2k + 1}(\beta) \cos(( 2k + 1) \omega_m t) \bigg] + \\ + = + (-1)^k \cdot -\sum_{k=0}^\infty J_{2k+1}(\beta) \underbrace{2\sin(\omega_c t)\cos((2k+1)\omega_m t)}_{\text{Additionstheorem}}. \end{align*} Auch hier wird ein Additionstheorem \eqref{fm:eq:addth3} gebraucht, dabei ist \(A = \omega_c t\) und \(B = (2k+1)\omega_m t \), somit wird daraus -\[ - J_0(\beta) \cdot \sin(\omega_c) + \sum_{k=1}^\infty J_{2k+1}(\beta) \{ \underbrace{\cos((\omega_c-(2k+1)\omega_m) t)}_{\text{neg.Teil}} - \cos((\omega_c+(2k+1)\omega_m) t) \} -\]dieser Term. -Wenn dabei \(2k +1\) durch alle ungeraden Zahlen von \(-\infty \to \infty\) mit \(n\) substituiert. +\begin{align*} + (-1)^k \cdot -\sum_{k=0}^\infty J_{2k+1}(\beta) \{ \underbrace{\cos((\omega_c - (2k+1)\omega_m) t)} \,-\, \cos((\omega_c+(2k+1)\omega_m) t) \} + \\ + = + (-1)^k \cdot -\sum_{k=- \infty}^{-1} J_{2k+1}(\beta) \overbrace{\cos((\omega_c + (2k+1)\omega_m) t)} + \,-\, (-1)^k \cdot -\sum_{k=0}^\infty J_{2k+1}(\beta) \cos((\omega_c + (2k+1)\omega_m) t) +\end{align*} +dieser Term. Zusätzlich dabei noch die letzte Besselindentität \eqref{fm:eq:besselid3} brauchen, ist bei allen ungeraden negativen \(n : J_{-n}(\beta) = -1\cdot J_n(\beta)\). -Somit wird neg.Teil zum Term \(-\cos((\omega_c+(2k+1)\omega_m) t)\) und die Summe vereinfacht sich zu +Somit wird neg.Teil zum Term +\[ + (-1)^k \cdot \sum_{k= \infty}^{1} -1 \cdot J_{2k+1}(\beta) \cos((\omega_c+(2k+1)\omega_m) t). +\] +TODO (jetzt habe ich zwei Summen die immer positiv sind? ) +Wenn dabei \(2k +1\) durch alle ungeraden Zahlen von \(-\infty \to \infty\) mit \(n\) substituiert vereinfacht sich die Summe zu \[ \sum_{n\, \text{ungerade}} -1 \cdot J_{n}(\beta) \cos((\omega_c + n\omega_m) t). \label{fm:eq:ungerade} @@ -122,7 +138,8 @@ Substituiert man nun noch \(n \text{mit} -n \) so fällt das \(-1\) weg. Beide Teile \eqref{fm:eq:gerade} Gerade \[ \sum_{n\, \text{gerade}} J_{n}(\beta) \cos((\omega_c + n\omega_m) t) -\]und \eqref{fm:eq:ungerade} Ungerade +\] +und \eqref{fm:eq:ungerade} Ungerade \[ \sum_{n\, \text{ungerade}} J_{n}(\beta) \cos((\omega_c + n\omega_m) t) \] @@ -140,7 +157,7 @@ Somit ist \eqref{fm:eq:proof} bewiesen. Um sich das ganze noch einwenig Bildlicher vorzustellenhier einmal die Besselfunktion \(J_{k}(\beta)\) in geplottet. \begin{figure} \centering -% \input{./PyPython animation/bessel.pgf} + \input{papers/fm/Python animation/bessel.pgf} \caption{Bessle Funktion \(J_{k}(\beta)\)} \label{fig:bessel} \end{figure} diff --git a/buch/papers/fm/Python animation/Bessel-FM.ipynb b/buch/papers/fm/Python animation/Bessel-FM.ipynb index 6f099a7..74f1011 100644 --- a/buch/papers/fm/Python animation/Bessel-FM.ipynb +++ b/buch/papers/fm/Python animation/Bessel-FM.ipynb @@ -2,7 +2,7 @@ "cells": [ { "cell_type": "code", - "execution_count": 1, + "execution_count": 4, "metadata": {}, "outputs": [], "source": [ @@ -13,7 +13,7 @@ "import scipy.fftpack\n", "import matplotlib as mpl\n", "# Use the pgf backend (must be set before pyplot imported)\n", - "#mpl.use('pgf')\n", + "mpl.use('pgf')\n", "import matplotlib.pyplot as plt\n", "from matplotlib.widgets import Slider\n", "def fm(beta):\n", @@ -70,39 +70,26 @@ "xf = fftfreq(N, 1 / 1000)\n", "plt.plot(xf, np.abs(yf_old))\n", "#plt.xlim(-150, 150)\n", - "plt.show()" - ] - }, - { - "cell_type": "code", - "execution_count": 118, - "metadata": {}, - "outputs": [ - { - "data": { - "image/png": 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", + "image/png": 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", "text/plain": [ "
" ] @@ -111,13 +98,6 @@ "needs_background": "light" }, "output_type": "display_data" - }, - { - "name": "stdout", - "output_type": "stream", - "text": [ - "0.7651976865579666\n" - ] } ], "source": [ @@ -133,8 +113,8 @@ "plt.xlabel(' $ \\\\beta $ ')\n", "plt.plot(x, y)\n", "plt.legend()\n", - "plt.show()\n", - "#plt.savefig('bessel.pgf', format='pgf')\n", + "#plt.show()\n", + "plt.savefig('bessel.pgf', format='pgf')\n", "print(sc.jv(0,1))" ] }, diff --git a/buch/papers/fm/Python animation/bessel.pgf b/buch/papers/fm/Python animation/bessel.pgf new file mode 100644 index 0000000..cc7af1e --- /dev/null +++ b/buch/papers/fm/Python animation/bessel.pgf @@ -0,0 +1,2057 @@ +%% Creator: Matplotlib, PGF backend +%% +%% To include the figure in your LaTeX document, write +%% \input{.pgf} +%% +%% Make sure the required packages are loaded in your preamble +%% \usepackage{pgf} +%% +%% Also ensure that all the required font packages are loaded; for instance, +%% the lmodern package is sometimes necessary when using math font. +%% \usepackage{lmodern} +%% +%% Figures using additional raster images can only be included by \input if +%% they are in the same directory as the main LaTeX file. For loading figures +%% from other directories you can use the `import` package +%% \usepackage{import} +%% +%% and then include the figures with +%% \import{}{.pgf} +%% +%% Matplotlib used the following preamble +%% \usepackage{fontspec} +%% \setmainfont{DejaVuSerif.ttf}[Path=\detokenize{/home/joshua/.local/lib/python3.8/site-packages/matplotlib/mpl-data/fonts/ttf/}] +%% \setsansfont{DejaVuSans.ttf}[Path=\detokenize{/home/joshua/.local/lib/python3.8/site-packages/matplotlib/mpl-data/fonts/ttf/}] +%% \setmonofont{DejaVuSansMono.ttf}[Path=\detokenize{/home/joshua/.local/lib/python3.8/site-packages/matplotlib/mpl-data/fonts/ttf/}] +%% +\begingroup% +\makeatletter% +\begin{pgfpicture}% +\pgfpathrectangle{\pgfpointorigin}{\pgfqpoint{6.000000in}{4.000000in}}% +\pgfusepath{use as bounding box, clip}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetmiterjoin% +\pgfsetlinewidth{0.000000pt}% +\definecolor{currentstroke}{rgb}{1.000000,1.000000,1.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetstrokeopacity{0.000000}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{6.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{6.000000in}{4.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{4.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathclose% +\pgfusepath{}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetmiterjoin% +\definecolor{currentfill}{rgb}{1.000000,1.000000,1.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.000000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetstrokeopacity{0.000000}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.750000in}{0.500000in}}% +\pgfpathlineto{\pgfqpoint{5.400000in}{0.500000in}}% +\pgfpathlineto{\pgfqpoint{5.400000in}{3.520000in}}% +\pgfpathlineto{\pgfqpoint{0.750000in}{3.520000in}}% +\pgfpathlineto{\pgfqpoint{0.750000in}{0.500000in}}% +\pgfpathclose% +\pgfusepath{fill}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.750000in}{0.500000in}}{\pgfqpoint{4.650000in}{3.020000in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.750000in}{0.500000in}}% +\pgfpathlineto{\pgfqpoint{0.750000in}{3.520000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{0.750000in}{0.500000in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=0.750000in,y=0.402778in,,top]{\color{textcolor}\sffamily\fontsize{10.000000}{12.000000}\selectfont \ensuremath{-}10.0}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.750000in}{0.500000in}}{\pgfqpoint{4.650000in}{3.020000in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{1.331250in}{0.500000in}}% +\pgfpathlineto{\pgfqpoint{1.331250in}{3.520000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% 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+\endgroup% diff --git a/buch/papers/fm/packages.tex b/buch/papers/fm/packages.tex index f0ca8cc..7bbbe35 100644 --- a/buch/papers/fm/packages.tex +++ b/buch/papers/fm/packages.tex @@ -8,3 +8,4 @@ % following example %\usepackage{packagename} \usepackage{xcolor} +\usepackage{pgf} -- cgit v1.2.1 From d9a24e13322ba877475300514412d74ad2e3a238 Mon Sep 17 00:00:00 2001 From: tim30b Date: Wed, 3 Aug 2022 20:07:27 +0200 Subject: Simulation: weitergeschrieben --- buch/papers/kreismembran/teil4.tex | 23 +++++++++++++++++------ 1 file changed, 17 insertions(+), 6 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil4.tex b/buch/papers/kreismembran/teil4.tex index c124354..62a34c5 100644 --- a/buch/papers/kreismembran/teil4.tex +++ b/buch/papers/kreismembran/teil4.tex @@ -5,12 +5,23 @@ % \section{Lösungsmethode 3: Simulation \label{kreismembran:section:teil4}} -\paragraph{TODO Einleitung} Um numerisch das Verhalten einer Membran zu ermitteln, muss eine numerische Darstellung definiert werden. -Die Membran wird hier in Form der Matrix $ A $ digitalisiert. -Jedes Element $ A_{ij} $ steh für die Auslenkung der Membran $ u(x,y,t) $ an der Stelle $ \{x,y\}=\{i,j\} $. -Die zeitliche Dimension wird in Form des Array $ X[] $ aus $ v \times A $ Matrizen dargestellt. -Das Element auf Zeile $ i $, Spalte $ j $ der $ w $-ten Matrix von $ X[] $ also $ X[w]_{ij} $ entspricht der Auslenkung $ u(i,j,w) $. +Die Membran wird hier in Form der Matrix $ U $ digitalisiert. +Jedes Element $ U_{ij} $ steh für die Auslenkung der Membran $ u(x,y,t) $ an der Stelle $ \{x,y\}=\{i,j\} $. +Die zeitliche Dimension wird in Form des Array $ U[] $ aus $ z \times U $ Matrizen dargestellt, wobei $ z $ der Anzahl Zeitschritten entspricht. +Das Element auf Zeile $ i $, Spalte $ j $ der $ w $-ten Matrix von $ U[] $ also $ U[w]_{ij} $ entspricht somit der Auslenkung $ u(i,j,w) $. +Da die DGL von Zweiter Ordnung ist, reicht eine Zustandsvariabel pro Membran-Element nicht aus. +Es wird neben der Auslenkung auch die Geschwindigkeit jedes Membran-Elementes benötigt um den Zustand eindeutig zu beschreiben. +Dazu existiert neben $ U[] $ ein analoger Array $ V[] $ welcher die Geschwindigkeiten aller Membran-Elementen repräsentiert. +$ V[w]_{ij} $ entspricht also $ \dot{u}(i,j,w) $. +Der Zustand einer Membran zum Zeitpunkt $ w $ wird mit $ X[w] $ beschrieben, was $ U[w] $ und $ V[w] $ beinhaltet. -\paragraph{title} \ No newline at end of file +\subsection{Propagation} +Um das Verhalten der Membran zu berechnen, muss aus einem gegebenen Zustand $ X[w] $ der Folgezustand $ X[w+1] $ gerechnet werden können, wobei dazwischen ein Zeitintervall $ dt $ vergeht. +Die Berechnung von Folgezuständen kann anschliessend repetiert werden über das zu untersuchende Zeitfenster. +Da die Digitale Membran sich wie die analytisch untersuchte verhalten soll, muss auch sie +\begin{equation*} + \frac{1}{c^2}\frac{\partial^2u}{\partial t^2} = \Delta u +\end{equation*} +erfüllen. -- cgit v1.2.1 From 05b1350074c1c62340c7c32f240cb46078c152e7 Mon Sep 17 00:00:00 2001 From: Kuster Yanik Date: Thu, 4 Aug 2022 17:31:48 +0200 Subject: changed textsize in Strategie.pdf. Did minor changes in Teil0 and Teil1 --- buch/papers/lambertw/Bilder/Strategie.pdf | Bin 151640 -> 151684 bytes buch/papers/lambertw/Bilder/Strategie.py | 9 +++--- buch/papers/lambertw/teil0.tex | 47 ++++++++++++++++++++---------- buch/papers/lambertw/teil1.tex | 36 ++++++++++++----------- 4 files changed, 55 insertions(+), 37 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/lambertw/Bilder/Strategie.pdf b/buch/papers/lambertw/Bilder/Strategie.pdf index 91442cc..b5428f5 100644 Binary files a/buch/papers/lambertw/Bilder/Strategie.pdf and b/buch/papers/lambertw/Bilder/Strategie.pdf differ diff --git a/buch/papers/lambertw/Bilder/Strategie.py b/buch/papers/lambertw/Bilder/Strategie.py index 28f7bcd..d7d06cb 100644 --- a/buch/papers/lambertw/Bilder/Strategie.py +++ b/buch/papers/lambertw/Bilder/Strategie.py @@ -34,7 +34,8 @@ ax.quiver(X, Y, U, W, angles='xy', scale_units='xy', scale=1, headwidth=5, headl ax.plot([V[0], (VZ+V)[0]], [V[1], (VZ+V)[1]], 'k--') ax.plot(np.vstack([V, Z])[:, 0], np.vstack([V, Z])[:,1], 'bo', markersize=10) - +ax.set_xlabel("x") +ax.set_ylabel("y") ax.text(2.5, 4.5, "Visierlinie", size=20, rotation=10) @@ -44,9 +45,9 @@ plt.rcParams.update({ "font.serif": ["New Century Schoolbook"], }) -ax.text(1.6, 4.3, r"$\dot{v}$", size=30) -ax.text(0.6, 3.9, r"$V$", size=30, c='b') -ax.text(5.1, 4.77, r"$Z$", size=30, c='b') +ax.text(1.6, 4.3, r"$\dot{v}$", size=20) +ax.text(0.65, 3.9, r"$V$", size=20, c='b') +ax.text(5.15, 4.85, r"$Z$", size=20, c='b') diff --git a/buch/papers/lambertw/teil0.tex b/buch/papers/lambertw/teil0.tex index 8fa8f9b..088cb7b 100644 --- a/buch/papers/lambertw/teil0.tex +++ b/buch/papers/lambertw/teil0.tex @@ -7,7 +7,7 @@ \label{lambertw:section:Was_sind_Verfolgungskurven}} \rhead{Was sind Verfolgungskurven?} % -Verfolgungskurven tauchen oft auf bei Fragen wie "Welchen Pfad begeht ein Hund während er einer Katze nachrennt?". +Verfolgungskurven tauchen oft auf bei Fragen wie ``Welchen Pfad begeht ein Hund während er einer Katze nachrennt?''. Ein solches Problem hat im Kern immer ein Verfolger und sein Ziel. Der Verfolger verfolgt sein Ziel, das versucht zu entkommen. Der Pfad, den der Verfolger während der Verfolgung begeht, wird Verfolgungskurve genannt. @@ -27,15 +27,15 @@ Daraus folgt, dass eine Strategie zwei dieser drei Parameter festlegen muss, um % \begin{table} \centering - \begin{tabular}{|>{$}c<{$}|>{$}c<{$}|>{$}c<{$}|>{$}c<{$}|} + \begin{tabular}{|>{$}l<{$}|>{$}c<{$}|>{$}c<{$}|>{$}c<{$}|} \hline \text{Strategie}&\text{Geschwindigkeit}&\text{Abstand}&\text{Richtung}\\ \hline \text{Jagd} - & \text{konstant} & \text{-} & \text{direkt auf Ziel hinzu}\\ + & \text{konstant} & \text{-} & \text{direkt auf Ziel zu}\\ \text{Beschattung} - & \text{-} & \text{konstant} & \text{direkt auf Ziel hinzu}\\ + & \text{-} & \text{konstant} & \text{direkt auf Ziel zu}\\ \text{Vorhalt} & \text{konstant} & \text{-} & \text{etwas voraus Zielen}\\ @@ -59,7 +59,7 @@ Der Verfolger und sein Ziel werden als Punkte $V$ und $Z$ modelliert. In der Abbildung \ref{lambertw:grafic:pursuerDGL2} ist das Problem dargestellt, wobei $v$ der Ortsvektor des Verfolgers, $z$ der Ortsvektor des Ziels und $\dot{v}$ der Geschwindigkeitsvektor des Verfolgers ist. Der Geschwindigkeitsvektor entspricht dem Richtungsvektors des Verfolgers. -Die konstante Geschwindigkeit kann man mit der Gleichung +Die konstante Geschwindigkeit kann man mit % \begin{equation} |\dot{v}| @@ -67,38 +67,53 @@ Die konstante Geschwindigkeit kann man mit der Gleichung \text{,}\quad A\in\mathbb{R}^+ \end{equation} % -darstellen. Der Geschwindigkeitsvektor kann mit der Gleichung -% +darstellen. Der Geschwindigkeitsvektor muss auf das Ziel zeigen, woraus folgt +\begin{equation} + \dot{v} + \quad||\quad + z-v + \text{.} +\end{equation} +Um den Richtungsvektor zu konstruieren kann der Einheitsvektor parallel zu $z-v$ um $\dot{v}$ gestreckt werden, was zu \begin{equation} - \frac{z-v}{|z-v|}\cdot|\dot{v}| + \dot{v} = + |\dot{v}|\cdot e_{z-v} +\end{equation} +führt. Dies kann noch ausgeschrieben werden zu +\begin{equation} \dot{v} + = + |\dot{v}|\cdot\frac{z-v}{|z-v|} + \text{.} \end{equation} % -beschrieben werden, wenn die Jagdstrategie verwendet wird. -Die Differenz der Ortsvektoren $v$ und $z$ ist ein Vektor der vom Punkt $V$ auf $Z$ zeigt. -Da die Länge dieses Vektors beliebig sein kann, wird durch Division durch den Betrag, ein Einheitsvektor erzeugt. Aus dem Verfolgungsproblem ist auch ersichtlich, dass die Punkte $V$ und $Z$ nicht am gleichen Ort starten und so eine Division durch Null ausgeschlossen ist. Wenn die Punkte $V$ und $Z$ trotzdem am gleichen Ort starten, ist die Lösung trivial. -% -Nun wird die Gleichung mit $\dot{v}$ skalar multipliziert, um das Gleichungssystem von zwei auf eine Gleichung zu reduzieren. Somit ergeben sich + +Nun wird die Gleichung mit $\dot{v}$ skalar multipliziert, um das Gleichungssystem von zwei auf eine Gleichung zu reduzieren. Somit ergibt sich \begin{align} \frac{z-v}{|z-v|}\cdot|\dot{v}|\cdot\dot{v} &= |\dot{v}|^2 - \\ +\end{align} +was algebraisch zu +\begin{align} \label{lambertw:pursuerDGL} \frac{z-v}{|z-v|}\cdot \frac{\dot{v}}{|\dot{v}|} &= - 1 \text{.} + 1 \end{align} +umgeformt werden kann. Die Lösungen dieser Differentialgleichung sind die gesuchten Verfolgungskurven, insofern der Verfolger die Jagdstrategie verwendet. % \subsection{Ziel \label{lambertw:subsection:Ziel}} Als nächstes gehen wir auf das Ziel ein. Wie der Verfolger wird auch unser Ziel sich strikt an eine Fluchtstrategie halten, welche von Anfang an bekannt ist. -Diese Strategie kann als Parameterdarstellung der Position nach der Zeit beschrieben werden. +Als Strategie eignet sich eine definierte Fluchtkurve oder ähnlich wie beim Verfolger ein Verhalten, das vom Verfolger abhängig ist. +Ein vom Verfolger abhängiges Verhalten führt zu einem gekoppeltem DGL-System, das schwierig zu lösen sein wird. +Eine definierte Fluchtkurve kann mit einer Parameterdarstellung der Position nach der Zeit beschrieben werden. Zum Beispiel könnte ein Ziel auf einer Geraden flüchten, welches auf einer Ebene mit der Parametrisierung % \begin{equation} diff --git a/buch/papers/lambertw/teil1.tex b/buch/papers/lambertw/teil1.tex index 2da07db..0fd0108 100644 --- a/buch/papers/lambertw/teil1.tex +++ b/buch/papers/lambertw/teil1.tex @@ -17,9 +17,10 @@ Nun gilt es zu definieren, wann das Ziel erreicht wird. Da sowohl Ziel und Verfolger als Punkte modelliert wurden, gilt das Ziel als erreicht, wenn die Koordinaten des Verfolgers mit denen des Ziels bei einem diskreten Zeitpunkt $t_1$ übereinstimmen. Somit gilt es % -\begin{equation*} +\begin{equation} z(t_1)=v(t_1) -\end{equation*} + \label{bedingung_treffer} +\end{equation} % zu lösen. Die Parametrisierung von $z(t)$ ist im Beispiel definiert als @@ -29,12 +30,12 @@ Die Parametrisierung von $z(t)$ ist im Beispiel definiert als \left( \begin{array}{c} 0 \\ t \end{array} \right)\text{.} \end{equation} % -Die Parametrisierung von $v(t)$ ist von den Startbedingungen abhängig. Deshalb wird die obige Bedingung jeweils für die unterschiedlichen Startbedingungen separat analysiert. +Die Parametrisierung von $v(t)$ ist von den Startbedingungen abhängig. Deshalb wird \eqref{bedingung_treffer} jeweils für die unterschiedlichen Startbedingungen separat analysiert. % -\subsection{Anfangsbedingung im \RN{1}-Quadranten} +\subsection{Anfangsbedingung im ersten Quadranten} % -Wenn der Verfolger im \RN{1}-Quadranten startet, dann kann $v(t)$ mit den Gleichungen aus \eqref{lambertw:eqFunkXNachT}, welche -\begin{align*} +Wenn der Verfolger im ersten Quadranten startet, dann kann $v(t)$ mit den Gleichungen aus \eqref{lambertw:eqFunkXNachT}, welche +\begin{align} x\left(t\right) &= x_0\cdot\sqrt{\frac{1}{\chi}W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right) \right)} \\ @@ -50,7 +51,8 @@ Wenn der Verfolger im \RN{1}-Quadranten startet, dann kann $v(t)$ mit den Gleich r_0 = \sqrt{x_0^2+y_0^2} -\end{align*} + \text{.} +\end{align} % Der Folger ist durch \begin{equation} @@ -61,9 +63,9 @@ Der Folger ist durch \end{equation} % parametrisiert, wobei $y(t)$ viel komplexer ist als $x(t)$. -Daher wird das Problem in zwei einzelne Teilprobleme zerlegt, wodurch die Bedingung der $x$- und $y$-Koordinaten einzeln überprüft werden müssen. Es entstehen daher folgende Bedingungen +Daher wird das Problem in zwei einzelne Teilprobleme zerlegt, wodurch die Bedingung der $x$- und $y$-Koordinaten einzeln überprüft werden müssen. Es entstehen daher die Bedingungen % -\begin{align*} +\begin{align} 0 &= x(t) @@ -75,7 +77,7 @@ Daher wird das Problem in zwei einzelne Teilprobleme zerlegt, wodurch die Beding y(t) = \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(r_0-y_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right)\text{,} -\end{align*} +\end{align} % welche Beide gleichzeitig erfüllt sein müssen, damit das Ziel erreicht wurde. Zuerst wird die Bedingung der $x$-Koordinate betrachtet. @@ -101,7 +103,7 @@ Diese Gleichung entspricht genau den Nullstellen der Lambert W-Funktion. Da die % Da $\chi\neq0$ und die Exponentialfunktion nie null sein kann, ist diese Bedingung unmöglich zu erfüllen. Beim Grenzwert für $t\rightarrow\infty$ geht die Exponentialfunktion gegen null. -Dies nützt nicht viel, da unendlich viel Zeit vergehen müsste damit ein Einholen möglich wäre. +Dies nützt nicht viel, da unendlich viel Zeit vergehen müsste, damit ein Einholen möglich wäre. Somit kann nach den gestellten Bedingungen das Ziel nie erreicht werden. % % @@ -136,7 +138,7 @@ Somit kann nach den gestellten Bedingungen das Ziel nie erreicht werden. %Somit kann nach den gestellten Bedingungen das Ziel nie erreicht werden. % \subsection{Anfangsbedingung $y_0<0$} -Da die Geschwindigkeit des Verfolgers und des Ziels übereinstimmen, kann der Verfolgers niemals das Ziel einholen. +Da die Geschwindigkeit des Verfolgers und des Ziels übereinstimmen, kann der Verfolger niemals das Ziel einholen. Dies kann veranschaulicht werden anhand % \begin{equation} @@ -184,7 +186,7 @@ was aufgelöst zu führt. Somit wird das Ziel immer erreicht bei $t_1$, wenn der Verfolger auf der positiven $y$-Achse startet. \subsection{Fazit} -Durch die Symmetrie der Fluchtkurve an der $y$-Achse führen die Anfangsbedingungen in den Quadranten \RN{1} und \RN{2} zu den gleichen Ergebnissen. Nun ist klar, dass lediglich Anfangspunkte auf der positiven $y$-Achse oder direkt auf dem Ziel dazu führen, dass der Verfolger das Ziel bei $t_1$ einholt. +Durch die Symmetrie der Fluchtkurve an der $y$-Achse führen die Anfangsbedingungen im ersten und zweiten Quadranten zu den gleichen Ergebnissen. Nun ist klar, dass lediglich Anfangspunkte auf der positiven $y$-Achse oder direkt auf dem Ziel dazu führen, dass der Verfolger das Ziel bei $t_1$ einholt. Bei allen anderen Anfangspunkten wird der Verfolger das Ziel nie erreichen. Dieses Resultat ist aber eher akademischer Natur, weil der Verfolger und das Ziel als Punkt betrachtet wurden. Wobei aber in Realität nicht von Punkten sondern von Objekten mit einer räumlichen Ausdehnung gesprochen werden kann. @@ -193,18 +195,18 @@ Falls dies stattfinden sollte, wird dies als Treffer interpretiert. Mathematisch kann dies mit % \begin{equation} - |v-z| Date: Thu, 4 Aug 2022 18:04:11 +0200 Subject: Herleitung fix --- buch/papers/fm/00_modulation.tex | 12 ++-- buch/papers/fm/01_AM.tex | 4 +- buch/papers/fm/03_bessel.tex | 135 ++++++++++++++++++++++++--------------- 3 files changed, 93 insertions(+), 58 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/fm/00_modulation.tex b/buch/papers/fm/00_modulation.tex index dc99b40..e2ba39f 100644 --- a/buch/papers/fm/00_modulation.tex +++ b/buch/papers/fm/00_modulation.tex @@ -18,10 +18,14 @@ Mathematisch wird dann daraus \omega_i = \omega_c + \frac{d \varphi(t)}{dt} \] mit der Ableitung der Phase\cite{fm:NAT}. -Mit diesen drei parameter ergeben sich auch drei modulationsarten, die Amplitudenmodulation welche \(A_c\) benutzt, -die Phasenmodulation \(\varphi\) und dann noch die Momentankreisfrequenz \(\omega_i\): -\newline -\newline +Mit diesen drei Parameter ergeben sich auch drei Modulationsarten, die Amplitudenmodulation, welche \(A_c\) benutzt, +die Phasenmodulation \(\varphi\) und dann noch die Momentankreisfrequenz \(\omega_i\): +\begin{itemize} + \item AM + \item PM + \item FM +\end{itemize} + To do: Bilder jeder Modulationsart diff --git a/buch/papers/fm/01_AM.tex b/buch/papers/fm/01_AM.tex index 921fcf2..21927f5 100644 --- a/buch/papers/fm/01_AM.tex +++ b/buch/papers/fm/01_AM.tex @@ -17,8 +17,8 @@ Dies sieht man besonders in der Eulerischen Formel \[ x_c(t) = \frac{A_c}{2} \cdot e^{j\omega_ct}\;+\;\frac{A_c}{2} \cdot e^{-j\omega_ct}. \] -Dabei ist die negative Frequenz der zweiten komplexen Schwingung zwingend erforderlich, damit in der Summe immer ein reelwertiges Trägersignal ergibt. -Nun wird der parameter \(A_c\) durch das Moduierende Signal \(m(t)\) ersetzt, wobei so \(m(t) \leqslant |1|\) normiert wurde. +Dabei ist die negative Frequenz der zweiten komplexen Schwingung zwingend erforderlich, damit in der Summe immer ein reellwertiges Trägersignal ergibt. +Nun wird der Parameter \(A_c\) durch das Modulierende Signal \(m(t)\) ersetzt, wobei so \(m(t) \leqslant |1|\) normiert wurde. \newline \newline TODO: diff --git a/buch/papers/fm/03_bessel.tex b/buch/papers/fm/03_bessel.tex index eec64f2..5f85dc6 100644 --- a/buch/papers/fm/03_bessel.tex +++ b/buch/papers/fm/03_bessel.tex @@ -3,11 +3,11 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{FM und Besselfunktion +\section{FM und Bessel-Funktion \label{fm:section:proof}} \rhead{Herleitung} -Die momentane Trägerkreisfrequenz \(\omega_i\) wie schon in (ref) beschrieben ist, bringt die Vorigen Kapittel beschreiben. (Ableitung \(\frac{d \varphi(t)}{dt}\) mit sich). -Diese wiederum kann durch \(\beta\sin(\omega_mt)\) ausgedrückt werden, wobei es das Modulierende Signal \(m(t)\) ist. +Die momentane Trägerkreisfrequenz \(\omega_i\), wie schon in (ref) beschrieben ist, bringt die Ableitung \(\frac{d \varphi(t)}{dt}\) mit sich. +Diese wiederum kann durch \(\beta\sin(\omega_mt)\) ausgedrückt werden, wobei es das modulierende Signal \(m(t)\) ist. Somit haben wir unser \(x_c\) welches \[ \cos(\omega_c t+\beta\sin(\omega_mt)) @@ -15,7 +15,7 @@ Somit haben wir unser \(x_c\) welches ist. \subsection{Herleitung} -Das Ziel ist es unser moduliertes Signal mit der Besselfunktion so auszudrücken: +Das Ziel ist, unser moduliertes Signal mit der Bessel-Funktion so auszudrücken: \begin{align} x_c(t) = @@ -43,22 +43,22 @@ Doch dazu brauchen wir die Hilfe der Additionsthoerme \cos(A-B)-\cos(A+B) \label{fm:eq:addth3} \end{align} -und die drei Besselfunktions indentitäten, +und die drei Bessel-Funktionsindentitäten, \begin{align} \cos(\beta\sin\phi) &= - J_0(\beta) + 2\sum_{k=1}^\infty(-1)^k \cdot J_{2k}(\beta) \cos(2k\phi) + J_0(\beta) + 2\sum_{k=1}^\infty J_{2k}(\beta) \cos(2k\phi) \label{fm:eq:besselid1} \\ \sin(\beta\sin\phi) &= - 2\sum_{k=0}^\infty (-1)^k J_{2k+1}(\beta) \cos((2k+1)\phi) + 2\sum_{k=0}^\infty J_{2k+1}(\beta) \cos((2k+1)\phi) \label{fm:eq:besselid2} \\ J_{-n}(\beta) &= (-1)^n J_n(\beta) \label{fm:eq:besselid3} \end{align} -welche man im Kapitel \eqref{buch:fourier:eqn:expinphireal}, \eqref{buch:fourier:eqn:expinphiimaginary}, \eqref{buch:fourier:eqn:symetrie}. +welche man im Kapitel \eqref{buch:fourier:eqn:expinphireal}, \eqref{buch:fourier:eqn:expinphiimaginary}, \eqref{buch:fourier:eqn:symetrie} findet. \subsubsection{Anwenden des Additionstheorem} Mit dem \eqref{fm:eq:addth1} wird aus dem modulierten Signal @@ -70,70 +70,102 @@ Mit dem \eqref{fm:eq:addth1} wird aus dem modulierten Signal \cos(\omega_c t)\cos(\beta\sin(\omega_m t))-\sin(\omega_ct)\sin(\beta\sin(\omega_m t)). \label{fm:eq:start} \] - +%----------------------------------------------------------------------------------------------------------- \subsubsection{Cos-Teil} Zu beginn wird der Cos-Teil -\[ +\begin{align*} + c(t) + &= \cos(\omega_c t)\cdot\cos(\beta\sin(\omega_mt)) -\] +\end{align*} mit hilfe der Besselindentität \eqref{fm:eq:besselid1} zum \begin{align*} - \cos(\omega_c t) \cdot \bigg[ J_0(\beta) + 2\sum_{k=1}^\infty(-1)^k \cdot J_{2k}(\beta) \cos( 2k \omega_m t)\, \bigg] + c(t) &= - (-1)^k \cdot \sum_{k=1}^\infty J_{2k}(\beta) \underbrace{2\cos(\omega_c t)\cos(2k\omega_m t)}_{\text{Additionstheorem}} + \cos(\omega_c t) \cdot \bigg[ J_0(\beta) + 2\sum_{k=1}^\infty J_{2k}(\beta) \cos( 2k \omega_m t)\, \bigg] + \\ + &= + J_0(\beta) \cdot \cos(\omega_c t) + \sum_{k=1}^\infty J_{2k}(\beta) \underbrace{2\cos(\omega_c t)\cos(2k\omega_m t)}_{\text{Additionstheorem \eqref{fm:eq:addth2}}} \end{align*} -wobei mit dem Additionstheorem \eqref{fm:eq:addth2} \(A = \omega_c t\) und \(B = 2k\omega_m t \) zum +%intertext{} Funktioniert nicht. +wobei mit dem Additionstheorem \eqref{fm:eq:addth2} \(A = \omega_c t\) und \(B = 2k\omega_m t \) ersetzt wurden. \begin{align*} - J_0(\beta) \cdot \cos(\omega_c t) +(-1)^k \cdot \sum_{k=1}^\infty J_{2k}(\beta) \{ \underbrace{\cos((\omega_c - 2k \omega_m) t)} \,+\, \cos((\omega_c + 2k \omega_m) t) \} + c(t) + &= + J_0(\beta) \cdot \cos(\omega_c t) + \sum_{k=1}^\infty J_{2k}(\beta) \{ \underbrace{\cos((\omega_c - 2k \omega_m) t)} \,+\, \cos((\omega_c + 2k \omega_m) t) \} \\ - = - (-1)^k \cdot \sum_{k=-\infty}^{-1} J_{2k}(\beta) \overbrace{\cos((\omega_c +2k \omega_m) t)} + &= + \sum_{k=-\infty}^{-1} J_{2k}(\beta) \overbrace{\cos((\omega_c +2k \omega_m) t)} \,+\,J_0(\beta)\cdot \cos(\omega_c t+ 2\cdot0 \omega_m) - \,+\, (-1)^k \cdot\sum_{k=1}^\infty J_{2k}(\beta)\cos((\omega_c + 2k \omega_m) t) + \,+\, \sum_{k=1}^\infty J_{2k}(\beta)\cos((\omega_c + 2k \omega_m) t) \end{align*} - -Wenn dabei \(2k\) durch alle geraden Zahlen von \(-\infty \to \infty\) mit \(n\) substituiert erhält man den vereinfachten Term -\[ - \sum_{n\, \text{gerade}} J_{n}(\beta) \cos((\omega_c + n\omega_m) t), +wird. +Das Minus im Ersten Term wird zur negativen Summe \(\sum_{-\infty}^{-1}\) ersetzt. +Da \(2k\) immer gerade ist, wird es durch alle negativen und positiven Ganzzahlen \(n\) ersetzt: +\begin{align*} + \sum_{n\, \text{gerade}} J_{n}(\beta) \cos((\omega_c + n \omega_m) t), \label{fm:eq:gerade} -\] -dabei gehen nun die Terme von \(-\infty \to \infty\), dabei bleibt n Ganzzahlig. - +\end{align*} +%---------------------------------------------------------------------------------------------------------------- \subsubsection{Sin-Teil} Nun zum zweiten Teil des Term \eqref{fm:eq:start}, den Sin-Teil -\[ +\begin{align*} + s(t) + &= -\sin(\omega_c t)\cdot\sin(\beta\sin(\omega_m t)). -\] +\end{align*} Dieser wird mit der \eqref{fm:eq:besselid2} Besselindentität zu \begin{align*} - -\sin(\omega_c t) \cdot \bigg[ 2 \sum_{k=0}^\infty(-1)^k \cdot J_{ 2k + 1}(\beta) \cos(( 2k + 1) \omega_m t) \bigg] + s(t) + &= + -\sin(\omega_c t) \cdot \bigg[ 2 \sum_{k=0}^\infty J_{ 2k + 1}(\beta) \cos(( 2k + 1) \omega_m t) \bigg] \\ - = - (-1)^k \cdot -\sum_{k=0}^\infty J_{2k+1}(\beta) \underbrace{2\sin(\omega_c t)\cos((2k+1)\omega_m t)}_{\text{Additionstheorem}}. + &= + \sum_{k=0}^\infty -1 \cdot J_{2k+1}(\beta) 2\sin(\omega_c t)\cos((2k+1)\omega_m t). +\end{align*} +Da \(2k + 1\) alle ungeraden positiven Ganzzahlen entspricht wird es durch \(n\) ersetzt. +Wird die Besselindentität \eqref{fm:eq:besselid3} gebraucht, so ersetzten wird \(J_{-n}(\beta) = -1\cdot J_n(\beta)\) ersetzt: +\begin{align*} + s(t) + &= + \sum_{n=0}^\infty J_{-n}(\beta) \underbrace{2\sin(\omega_c t)\cos(n \omega_m t)}_{\text{Additionstheorem \eqref{fm:eq:addth3}}}. \end{align*} -Auch hier wird ein Additionstheorem \eqref{fm:eq:addth3} gebraucht, dabei ist \(A = \omega_c t\) und \(B = (2k+1)\omega_m t \), -somit wird daraus +Auch hier wird ein Additionstheorem \eqref{fm:eq:addth3} gebraucht, dabei ist \(A = \omega_c t\) und \(B = n \omega_m t \), +somit wird daraus: \begin{align*} - (-1)^k \cdot -\sum_{k=0}^\infty J_{2k+1}(\beta) \{ \underbrace{\cos((\omega_c - (2k+1)\omega_m) t)} \,-\, \cos((\omega_c+(2k+1)\omega_m) t) \} + s(t) + &= + \sum_{n=0}^\infty J_{-n}(\beta) \{ \underbrace{\cos((\omega_c - n\omega_m) t)} \,-\, \cos((\omega_c + n\omega_m) t) \} \\ - = - (-1)^k \cdot -\sum_{k=- \infty}^{-1} J_{2k+1}(\beta) \overbrace{\cos((\omega_c + (2k+1)\omega_m) t)} - \,-\, (-1)^k \cdot -\sum_{k=0}^\infty J_{2k+1}(\beta) \cos((\omega_c + (2k+1)\omega_m) t) + &= + \sum_{n=- \infty}^{0} J_{n}(\beta) \overbrace{\cos((\omega_c + n \omega_m) t)} + \,-\, \sum_{n=0}^\infty J_{-n}(\beta) \cos((\omega_c + n\omega_m) t) \end{align*} -dieser Term. -Zusätzlich dabei noch die letzte Besselindentität \eqref{fm:eq:besselid3} brauchen, ist bei allen ungeraden negativen \(n : J_{-n}(\beta) = -1\cdot J_n(\beta)\). -Somit wird neg.Teil zum Term -\[ - (-1)^k \cdot \sum_{k= \infty}^{1} -1 \cdot J_{2k+1}(\beta) \cos((\omega_c+(2k+1)\omega_m) t). -\] -TODO (jetzt habe ich zwei Summen die immer positiv sind? ) -Wenn dabei \(2k +1\) durch alle ungeraden Zahlen von \(-\infty \to \infty\) mit \(n\) substituiert vereinfacht sich die Summe zu +Auch hier wurde wieder eine zweite Summe \(\sum_{-\infty}^{-1}\) gebraucht um das Minus zu einem Plus zu wandeln. +Wenn \(n = 0 \) ist der Minuend gleich dem Subtrahend und somit dieser Teil \(=0\), das bedeutet \(n\) ended bei \(-1\) und started bei \(1\). +\begin{align*} + s(t) + &= + \sum_{n=- \infty}^{-1} J_{n}(\beta) \cos((\omega_c + n \omega_m) t) + \underbrace{\,-\, \sum_{n=1}^\infty J_{-n}(\beta)} \cos((\omega_c + n\omega_m) t) +\end{align*} +Um aus diesem Subtrahend eine Addition zu kreiernen, wird die Besselindentität \eqref{fm:eq:besselid3} gebraucht, +jedoch so \(-1 \cdot J_{-n}(\beta) = J_n(\beta)\) und daraus wird dann: +\begin{align*} + s(t) + &= + \sum_{n=- \infty}^{-1} J_{n}(\beta) \cos((\omega_c + n \omega_m) t) + \,+\, \sum_{n=1}^\infty J_{n}(\beta) \cos((\omega_c + n\omega_m) t) +\end{align*} +Da \(n\) immer ungerade ist und \(0\) nicht zu den ungeraden zahlen zählt, kann man dies so vereinfacht \[ - \sum_{n\, \text{ungerade}} -1 \cdot J_{n}(\beta) \cos((\omega_c + n\omega_m) t). - \label{fm:eq:ungerade} + s(t) + = + \sum_{n\, \text{ungerade}} -1 \cdot J_{n}(\beta) \cos((\omega_c + n\omega_m) t). + \label{fm:eq:ungerade} \] -Substituiert man nun noch \(n \text{mit} -n \) so fällt das \(-1\) weg. - +schreiben. +%------------------------------------------------------------------------------------------ \subsubsection{Summe Zusammenführen} Beide Teile \eqref{fm:eq:gerade} Gerade \[ @@ -151,10 +183,9 @@ ergeben zusammen \] Somit ist \eqref{fm:eq:proof} bewiesen. \newpage - -%---------------------------------------------------------------------------- +%----------------------------------------------------------------------------------------- \subsection{Bessel und Frequenzspektrum} -Um sich das ganze noch einwenig Bildlicher vorzustellenhier einmal die Besselfunktion \(J_{k}(\beta)\) in geplottet. +Um sich das ganze noch einwenig Bildlicher vorzustellenhier einmal die Bessel-Funktion \(J_{k}(\beta)\) in geplottet. \begin{figure} \centering \input{papers/fm/Python animation/bessel.pgf} @@ -168,7 +199,7 @@ Nun einmal das Modulierte FM signal im Frequenzspektrum mit den einzelen Summen TODO Hier wird beschrieben wie die Bessel Funktion der FM im Frequenzspektrum hilft, wieso diese gebrauch wird und ihre Vorteile. \begin{itemize} - \item Zuerest einmal die Herleitung von FM zu der Besselfunktion + \item Zuerest einmal die Herleitung von FM zu der Bessel-Funktion \item Im Frequenzspektrum darstellen mit Farben, ersichtlich machen. \item Parameter tuing der Trägerfrequenz, Modulierende frequenz und Beta. \end{itemize} -- cgit v1.2.1 From ded30e493c1b05f1f412f2e78636d7195ea054e0 Mon Sep 17 00:00:00 2001 From: Kuster Yanik Date: Thu, 4 Aug 2022 21:24:11 +0200 Subject: added new subsection wird das Ziel erreicht? --- buch/papers/lambertw/Bilder/Intuition.pdf | Bin 186972 -> 187016 bytes buch/papers/lambertw/Bilder/Strategie.py | 4 +- .../Bilder/lambertAbstandBauchgef\303\274hl.py" | 10 +- buch/papers/lambertw/teil0.tex | 5 +- buch/papers/lambertw/teil1.tex | 101 +++++++++++++++++---- 5 files changed, 93 insertions(+), 27 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/lambertw/Bilder/Intuition.pdf b/buch/papers/lambertw/Bilder/Intuition.pdf index 236212a..739b02b 100644 Binary files a/buch/papers/lambertw/Bilder/Intuition.pdf and b/buch/papers/lambertw/Bilder/Intuition.pdf differ diff --git a/buch/papers/lambertw/Bilder/Strategie.py b/buch/papers/lambertw/Bilder/Strategie.py index d7d06cb..975e248 100644 --- a/buch/papers/lambertw/Bilder/Strategie.py +++ b/buch/papers/lambertw/Bilder/Strategie.py @@ -34,8 +34,8 @@ ax.quiver(X, Y, U, W, angles='xy', scale_units='xy', scale=1, headwidth=5, headl ax.plot([V[0], (VZ+V)[0]], [V[1], (VZ+V)[1]], 'k--') ax.plot(np.vstack([V, Z])[:, 0], np.vstack([V, Z])[:,1], 'bo', markersize=10) -ax.set_xlabel("x") -ax.set_ylabel("y") +ax.set_xlabel("x", size=20) +ax.set_ylabel("y", size=20) ax.text(2.5, 4.5, "Visierlinie", size=20, rotation=10) diff --git "a/buch/papers/lambertw/Bilder/lambertAbstandBauchgef\303\274hl.py" "b/buch/papers/lambertw/Bilder/lambertAbstandBauchgef\303\274hl.py" index 9031bfc..3a90afa 100644 --- "a/buch/papers/lambertw/Bilder/lambertAbstandBauchgef\303\274hl.py" +++ "b/buch/papers/lambertw/Bilder/lambertAbstandBauchgef\303\274hl.py" @@ -39,8 +39,8 @@ plt.plot(0, ymin, 'bo', markersize=10) plt.plot([0, xmin], [ymin, ymin], 'k--') #plt.xlim(-0.1, 1) #plt.ylim(1, 2) -#plt.ylabel("y") -#plt.xlabel("x") +plt.ylabel("y") +plt.xlabel("x") plt.grid(True) plt.quiver(xmin, ymin, -0.2, 0, scale=1) @@ -53,6 +53,6 @@ plt.rcParams.update({ "font.serif": ["New Century Schoolbook"], }) -plt.text(xmin-0.11, ymin-0.12, r"$\dot{v}$", size=30) -plt.text(xmin-0.02, ymin+0.05, r"$V$", size=30, c='b') -plt.text(0.02, ymin+0.05, r"$Z$", size=30, c='b') \ No newline at end of file +plt.text(xmin-0.11, ymin-0.08, r"$\dot{v}$", size=20) +plt.text(xmin-0.02, ymin+0.05, r"$V$", size=20, c='b') +plt.text(0.02, ymin+0.05, r"$Z$", size=20, c='b') \ No newline at end of file diff --git a/buch/papers/lambertw/teil0.tex b/buch/papers/lambertw/teil0.tex index 088cb7b..6632eca 100644 --- a/buch/papers/lambertw/teil0.tex +++ b/buch/papers/lambertw/teil0.tex @@ -74,7 +74,7 @@ darstellen. Der Geschwindigkeitsvektor muss auf das Ziel zeigen, woraus folgt z-v \text{.} \end{equation} -Um den Richtungsvektor zu konstruieren kann der Einheitsvektor parallel zu $z-v$ um $\dot{v}$ gestreckt werden, was zu +Um den Richtungsvektor zu konstruieren kann der Einheitsvektor parallel zu $z-v$ um $|\dot{v}|$ gestreckt werden, was zu \begin{equation} \dot{v} = @@ -86,6 +86,7 @@ führt. Dies kann noch ausgeschrieben werden zu = |\dot{v}|\cdot\frac{z-v}{|z-v|} \text{.} + \label{lambertw:richtungsvektor} \end{equation} % Aus dem Verfolgungsproblem ist auch ersichtlich, dass die Punkte $V$ und $Z$ nicht am gleichen Ort starten und so eine Division durch Null ausgeschlossen ist. @@ -105,7 +106,7 @@ was algebraisch zu 1 \end{align} umgeformt werden kann. -Die Lösungen dieser Differentialgleichung sind die gesuchten Verfolgungskurven, insofern der Verfolger die Jagdstrategie verwendet. +Die Lösungen dieser Differentialgleichung sind die gesuchten Verfolgungskurven, sofern der Verfolger die Jagdstrategie verwendet. % \subsection{Ziel \label{lambertw:subsection:Ziel}} diff --git a/buch/papers/lambertw/teil1.tex b/buch/papers/lambertw/teil1.tex index 0fd0108..e8eca2c 100644 --- a/buch/papers/lambertw/teil1.tex +++ b/buch/papers/lambertw/teil1.tex @@ -30,7 +30,7 @@ Die Parametrisierung von $z(t)$ ist im Beispiel definiert als \left( \begin{array}{c} 0 \\ t \end{array} \right)\text{.} \end{equation} % -Die Parametrisierung von $v(t)$ ist von den Startbedingungen abhängig. Deshalb wird \eqref{bedingung_treffer} jeweils für die unterschiedlichen Startbedingungen separat analysiert. +Die Parametrisierung von $v(t)$ ist von den Startbedingungen abhängig. Deshalb wird die Bedingung \eqref{bedingung_treffer} jeweils für die unterschiedlichen Startbedingungen separat analysiert. % \subsection{Anfangsbedingung im ersten Quadranten} % @@ -41,7 +41,7 @@ Wenn der Verfolger im ersten Quadranten startet, dann kann $v(t)$ mit den Gleich x_0\cdot\sqrt{\frac{1}{\chi}W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right) \right)} \\ y(t) &= - \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(r_0-y_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right)\\ + \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right)\\ \chi &= \frac{r_0+y_0}{r_0-y_0}, \quad @@ -54,7 +54,7 @@ Wenn der Verfolger im ersten Quadranten startet, dann kann $v(t)$ mit den Gleich \text{.} \end{align} % -Der Folger ist durch +Der Verfolger ist durch \begin{equation} v(t) = @@ -76,31 +76,37 @@ Daher wird das Problem in zwei einzelne Teilprobleme zerlegt, wodurch die Beding &= y(t) = - \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(r_0-y_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right)\text{,} + \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right)\text{,} \end{align} % -welche Beide gleichzeitig erfüllt sein müssen, damit das Ziel erreicht wurde. +welche beide gleichzeitig erfüllt sein müssen, damit das Ziel erreicht wurde. Zuerst wird die Bedingung der $x$-Koordinate betrachtet. -Da $x_0 \neq 0$ und $\chi \neq 0$ mit +Da $x_0 \neq 0$ und $\chi \neq 0$ kann \begin{equation} 0 = x_0\sqrt{\frac{1}{\chi}W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right)\right)} \end{equation} -ist diese Bedingung genau dann erfüllt, wenn +algebraisch zu \begin{equation} 0 = W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right)\right) - \text{.} \end{equation} -% +umgeformt werden. Es ist zu beachten, dass $W(x)$ die Lambert W-Funktion ist, welche im Kapitel \eqref{buch:section:lambertw} behandelt wurde. -Diese Gleichung entspricht genau den Nullstellen der Lambert W-Funktion. Da die Lambert W-Funktion genau eine Nullstelle bei -\begin{equation} +Diese Gleichung entspricht genau den Nullstellen der Lambert W-Funktion. Mit der einzigen Nullstelle der Lambert W-Funktion bei +\begin{equation*} W(0)=0 + \text{,} +\end{equation*} +kann die Bedingung weiter vereinfacht werden zu +\begin{equation} + 0 + = + \chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right) + \text{.} \end{equation} -% Da $\chi\neq0$ und die Exponentialfunktion nie null sein kann, ist diese Bedingung unmöglich zu erfüllen. Beim Grenzwert für $t\rightarrow\infty$ geht die Exponentialfunktion gegen null. Dies nützt nicht viel, da unendlich viel Zeit vergehen müsste, damit ein Einholen möglich wäre. @@ -203,16 +209,18 @@ Durch quadrieren verschwindet die Wurzel des Betrages, womit % \begin{equation} |v-z|^2e_y\cdot v$. +Aus diesem Argument würde folgen, dass beim tiefsten Punkt der Verfolgungskurve im Beispiel den minimalen Abstand befindet. +% \begin{figure} \centering \includegraphics[scale=0.4]{./papers/lambertw/Bilder/Intuition.pdf} @@ -220,7 +228,8 @@ Es lässt sich vermuten, dass bei diesem Punkt der Abstand zum Ziel minimal sein \label{lambertw:grafic:intuition} \end{figure} % -Dies kann leicht überprüft werden, indem wir lokal alle relevanten benachbarten Punkte betrachten und das Vorzeichen der Änderung des Abstandes prüfen. + +Dieses Argument kann leicht überprüft werden, indem lokal alle relevanten benachbarten Punkte betrachtet und das Vorzeichen der Änderung des Abstandes überprüft wird. Dafür wird ein Ausdruck benötigt, der den Abstand und die benachbarten Punkte beschreibt. Der Richtungsvektor wird allgemein mit dem Winkel $\alpha \in[ 0, 2\pi)$ Die Ortsvektoren der Punkte können wiederum mit @@ -280,5 +289,61 @@ unterteilt werden. Von Interesse ist lediglich das Intervall $\alpha\in\left( \frac{\pi}{2}, \frac{3\pi}{2}\right)$, da der Verfolger sich stets in die negative $y$-Richtung bewegt. In diesem Intervall ist die Ableitung negativ, woraus folgt, dass jeglicher unmittelbar benachbarte Punkt, den der Verfolger als nächstes begehen könnte, stets näher am Ziel ist als zuvor. Dies bedeutet, dass der Scheitelpunkt der Verfolgungskurve nie ein lokales Minimum bezüglich des Abstandes sein kann. +% +\subsection{Wo ist der Abstand minimal?} +Damit der Verfolger das Ziel erreicht muss die Bedingung \eqref{lambertw:minimumAbstand} erfüllt sein. +Somit ist es ausreichend zu zeigen, dass +\begin{equation} + \operatorname{min}(|z-v|) Date: Fri, 5 Aug 2022 11:27:41 +0200 Subject: Resolved issue in main.tex --- buch/papers/sturmliouville/main.tex | 1 - 1 file changed, 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/main.tex b/buch/papers/sturmliouville/main.tex index 559a448..4b5b8af 100644 --- a/buch/papers/sturmliouville/main.tex +++ b/buch/papers/sturmliouville/main.tex @@ -9,7 +9,6 @@ \begin{refsection} \chapterauthor{Réda Haddouche und Erik Löffler} -<<<<<<< HEAD \input{papers/sturmliouville/einleitung.tex} %einleitung "was ist das sturm-liouville-problem" \input{papers/sturmliouville/eigenschaften.tex} -- cgit v1.2.1 From 6ec66a72b31ad7a47eb54d373d24f494318d35fb Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Fri, 5 Aug 2022 12:05:26 +0200 Subject: Added partial solution to X equation. --- .../sturmliouville/waermeleitung_beispiel.tex | 60 +++++++++++++++++++++- 1 file changed, 59 insertions(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index b25fc89..cc88f6a 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -130,7 +130,65 @@ Lösung \] führt. -Etwas aufwändiger wird es, die zweite Gleichung zu lösen. +Etwas aufwändiger wird es, die zweite Gleichung zu lösen. Aufgrund der Struktur +der Gleichung +\[ + X^{\prime \prime}(x) - \mu X(x) + = + 0 +\] +wird ein trigonometrischer Ansatz gewählt. Die Lösungen für $X(x)$ sind also +von der Form +\[ + X(x) + = + A \sin \left( \alpha x\right) + B \cos \left( \beta x\right). +\] + +Dieser Ansatz wird nun solange differenziert, bis alle in Gleichung (TODO: ref) +enthaltenen Ableitungen vorhanden sind. Man erhält also +\[ + X^{\prime}(x) + = + A \alpha \cos \left( \alpha x \right) - + B \beta \sin \left( \beta x \right) +\] +und +\[ + X^{\prime \prime}(x) + = + -A \alpha^{2} \sin \left( \alpha x \right) - + B \beta^{2} \cos \left( \beta x \right). +\] + +Eingesetzt in Gleichung (TDOD: ref) ergibt dies +\[ + -A\alpha^{2}\sin(\alpha x) - B\beta^{2}\cos(\beta x) - + \mu\left(A\sin(\alpha x) + B\cos(\beta x)\right) + = + 0 +\] +und durch umformen somit +\[ + \mu A\sin(\alpha x) + \mu B\cos(\beta x) + = + A\alpha^{2}\sin(\alpha x) + B\beta^{2}\cos(\beta x). +\] + +Durch Koeffizientenvergleich von +\[ + \mu A\sin(\alpha x) + = + A\alpha^{2}\sin(\alpha x) +\] +\[ + \mu B\cos(\beta x) + = + B\beta^{2}\cos(\beta x) +\] +ist schnell ersichtlich, dass $ \mu = \alpha^{2} = \beta^{2} $ gelten muss für +$ A \neq 0 $ und $ B \neq 0 $. Zur Berechnung von $ \mu $ bleiben also noch +$ \alpha $ und $ \beta $ zu bestimmen. % TODO: Rechenweg TODO: Rechenweg... Enden auf konstanter Temperatur: -- cgit v1.2.1 From 35be5a6ccc01f2fbd39b44134aa4f8bca6705901 Mon Sep 17 00:00:00 2001 From: daHugen Date: Fri, 5 Aug 2022 16:48:03 +0200 Subject: Update contains: syntax correction and changes in structure (added subsubsections). --- buch/papers/lambertw/teil4.tex | 37 ++++++++++++++++++++++--------------- 1 file changed, 22 insertions(+), 15 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/lambertw/teil4.tex b/buch/papers/lambertw/teil4.tex index 1053dd1..2de3663 100644 --- a/buch/papers/lambertw/teil4.tex +++ b/buch/papers/lambertw/teil4.tex @@ -6,11 +6,11 @@ \section{Beispiel einer Verfolgungskurve \label{lambertw:section:teil4}} \rhead{Beispiel einer Verfolgungskurve} -In diesem Abschnitt wird rechnerisch das Beispiel einer Verfolgungskurve mit der Verfolgungsstrategie ``Jagd'' beschreiben. Dafür werden zuerst Bewegungsraum, Anfangspositionen und Bewegungsverhalten definiert, in einem nächsten Schritt soll eine Differentialgleichung dafür aufgestellt und anschliessend gelöst werden. +In diesem Abschnitt wird rechnerisch das Beispiel einer Verfolgungskurve mit der Verfolgungsstrategie ``Jagd'' beschrieben. Dafür werden zuerst Bewegungsraum, Anfangspositionen und Bewegungsverhalten definiert, in einem nächsten Schritt soll eine Differentialgleichung dafür aufgestellt und anschliessend gelöst werden. \subsection{Anfangsbedingungen definieren und einsetzen \label{lambertw:subsection:Anfangsbedingungen}} -Das zu verfolgende Ziel \(Z\) bewegt sich entlang der \(y\)-Achse mit konstanter Geschwindigkeit \(|\dot{z}| = 1\), beginnend beim Ursprung des Kartesischen Koordinatensystems. Der Verfolger \(V\) startet auf einem beliebigen Punkt im ersten Quadranten und bewegt sich auch mit konstanter Geschwindigkeit \(|\dot{v}| = 1\) in Richtung Ziel. Diese Anfangspunkte oder Anfangsbedingungen können wie folgt formuliert werden: +Das zu verfolgende Ziel \(Z\) bewegt sich entlang der \(y\)-Achse mit konstanter Geschwindigkeit \(|\dot{z}| = 1\), beginnend beim Ursprung des kartesischen Koordinatensystems. Der Verfolger \(V\) startet auf einem beliebigen Punkt im ersten Quadranten und bewegt sich auch mit konstanter Geschwindigkeit \(|\dot{v}| = 1\) in Richtung Ziel. Diese Anfangspunkte oder Anfangsbedingungen können wie folgt formuliert werden: \begin{equation} Z = @@ -40,7 +40,7 @@ Diese DGL haben wir bereits in Kapitel \ref{lambertw:subsection:Verfolger} defin \subsection{Differentialgleichung vereinfachen \label{lambertw:subsection:DGLvereinfach}} -Nun haben wir eine Gleichung, es stellt sich aber die Frage, ob es überhaupt eine geschlossene Lösung dafür gibt. Eine Funktion welche die Beziehung \(y(x)\) beschreibt oder sogar \(x(t)\) und \(y(t)\) liefert. Zum jetzigen Zeitpunkt mag es nicht trivial scheinen, aber mit den gewählten Anfangsbedingungen \eqref{lambertw:Anfangsbed} ist es möglich eine geschlossene Lösung für die Gleichung \eqref{lambertw:eqMitAnfangsbed} zu finden. +Nun haben wir eine Gleichung, es stellt sich aber die Frage, ob es überhaupt eine geschlossene Lösung dafür gibt. Eine Funktion welche die Beziehung \(y(x)\) beschreibt oder sogar \(x(t)\) und \(y(t)\) liefert. Zum jetzigen Zeitpunkt mag es nicht trivial scheinen, aber mit den gewählten Anfangsbedingungen \eqref{lambertw:Anfangsbed} ist es möglich, eine geschlossene Lösung für die Gleichung \eqref{lambertw:eqMitAnfangsbed} zu finden. Auf dem Weg dahin muss die definierte DGL zuerst wesentlich vereinfacht werden, sei es mittels algebraischer Umformungen oder mit den Tools aus der Analysis. Da die nächsten Schritte sehr algebralastig sind und sie das Lesen dieses Papers träge machen würden, werden wir uns hier nur auf die wesentlichsten Schritte konzentrieren, welche notwendig sind, um den Lösungsweg nachvollziehen zu können. @@ -90,7 +90,7 @@ Versteckt im Ausdruck \eqref{lambertw:eqGeschwSubstituiert} befindet sich die er \label{lambertw:eqAlgVerinfacht} \end{equation} die faktorisierte Darstellung davon ist. -Da der linke Term gleich Null ist, muss auch der Inhalt des Quadrates gleich Null sein. Es ergibt sich eine weitere Vereinfachung, welche zu der im Vergleich zu \eqref{lambertw:eqOhneSkalarprod} wesentlich einfacheren DGL +Da der linke Term gleich Null ist, muss auch die Basis des Quadrates in \eqref{lambertw:eqAlgVerinfacht} gleich Null sein. Es ergibt sich eine weitere Vereinfachung, welche zu der im Vergleich zu \eqref{lambertw:eqOhneSkalarprod} wesentlich einfacheren DGL \begin{equation} x \dot{y} + (t-y) \dot{x} = 0 @@ -122,7 +122,7 @@ Der Grund dafür ist, dass \label{lambertw:eqQuotZeitAbleit} \end{equation} und somit kann der Quotient dieser zeitlichen Ableitungen in eine Ableitung nach \(x\) umgewandelt werden. -Nach dem die Eigenschaft \eqref{lambertw:eqQuotZeitAbleit} in \eqref{lambertw:eqVorKeineZeitAbleit} eingesetzt wird und vereinfacht wurde, entsteht die neue Gleichung +Nachdem die Eigenschaft \eqref{lambertw:eqQuotZeitAbleit} in \eqref{lambertw:eqVorKeineZeitAbleit} eingesetzt wurde, entsteht beim Vereinfachen die neue Gleichung \begin{equation} x y^{\prime} + t - y = 0. @@ -130,7 +130,7 @@ Nach dem die Eigenschaft \eqref{lambertw:eqQuotZeitAbleit} in \eqref{lambertw:eq \end{equation} \subsubsection{Variable \(t\) eliminieren - \label{lambertw:subsubsection:ZeitAbleit}} + \label{lambertw:subsubsection:VarTelimin}} Hier wäre es natürlich passend, wenn man die Abhängigkeit nach \(t\) komplett wegbringen könnte, aber wie? Wir wissen, dass sich der Verfolger mit Geschwindigkeit 1 bewegt, also legt er in der Zeit \(t\) die Strecke \(1\cdot t = t\) zurück. Längen und Strecken können auch mit der Bogenlänge repräsentiert werden, somit kann Zeit und zurückgelegte Strecke in der Gleichung \begin{equation} @@ -199,7 +199,7 @@ Wenn man in \eqref{lambertw:loesDGLmitU} die Substitution rückgängig macht, er \label{lambertw:loesDGLmitY} \end{equation} erster Ordnung, die bereits separiert ist. -Ersetzt man den \(\operatorname{sinh}\) durch seine exponentiellen Definition \(\operatorname{sinh}(x)=\frac{1}{2}(e^x-e^{-x})\), so resultiert auf sehr einfache Art die Lösung +Ersetzt man den \(\operatorname{sinh}\) durch seine exponentielle Definition \(\operatorname{sinh}(x)=\frac{1}{2}(e^x-e^{-x})\), so resultiert auf sehr einfache Art die Lösung \begin{equation} y = @@ -302,7 +302,8 @@ Wenn man die Koeffizienten \eqref{lambertw:eqKoeff1} und \eqref{lambertw:eqKoeff \begin{equation} y(x) = - \frac{1}{4}\left(\left(y_0+r_0\right)\eta+\left(y_0-r_0\right)\operatorname{ln}\left(\eta\right)-r_0+3y_0\right). + \frac{1}{4}\left(\left(y_0+r_0\right)\eta+\left(y_0-r_0\right) + \operatorname{ln}\left(\eta\right)-r_0+3y_0\right). \label{lambertw:eqAllgLoes} \end{equation} Damit die Funktion \eqref{lambertw:eqAllgLoes} trotzdem übersichtlich bleibt, wurden Anfangssteigung \(\eta\) und Anfangsentfernung \(r_0\) wie folgt definiert: @@ -321,7 +322,7 @@ Nun sind wir soweit, dass wir eine \(y(x)\)-Beziehung für beliebige Anfangswert \subsection{Funktion nach der Zeit \label{lambertw:subsection:FunkNachT}} -In diesem Abschnitt werden algebraischen Umformungen ein wenig detaillierter als zuvor beschrieben. Dies hat auch einen bestimmten Grund: Den Einsatz einer speziellen Funktion aufzeigen, sowie auch wann und wieso diese vorkommt. Welche spezielle Funktion? Fragst du dich wahrscheinlich in diesem Moment. Nun, um diese Frage kurz zu beantworten, es ist ``YouTube's favorite special function'' laut dem Mathematiker Michael Penn, die Lambert-\(W\)-Funktion \(W(x)\) welche im Kapitel \ref{buch:section:lambertw} bereits beschrieben wurde. +In diesem Abschnitt werden algebraischen Umformungen ein wenig detaillierter als zuvor beschrieben. Dies hat auch einen bestimmten Grund: Den Einsatz einer speziellen Funktion aufzeigen, sowie auch wann und wieso diese vorkommt. Welche spezielle Funktion? Fragt man sich wahrscheinlich in diesem Moment. Nun, um diese Frage kurz zu beantworten, es ist ``YouTube's favorite special function'' laut dem Mathematiker Michael Penn, die Lambert-\(W\)-Funktion \(W(x)\) welche im Kapitel \ref{buch:section:lambertw} bereits beschrieben wurde. \subsubsection{Zeitabhängigkeit wiederherstellen \label{lambertw:subsubsection:ZeitabhWiederherst}} @@ -342,7 +343,7 @@ Wie in \eqref{lambertw:eqDGLmitTnochmals} zu sehen ist, werden \(y\) und deren A \frac{1}{4}\left(\left(y_0+r_0\right)\eta+\left(y_0-r_0\right)\operatorname{ln}\left(\eta\right)-r_0+3y_0\right), \\ y^\prime &= - \frac{1}{2}\left(\left(y_0+r_0\right)\frac{x}{x_0^2}+\left(y_0-r_0\right)\frac{1}{x}\right). + \frac{1}{2}\biggl(\left(y_0+r_0\right)\frac{x}{x_0^2}+\left(y_0-r_0\right)\frac{1}{x}\biggr). \end{align} \end{subequations} @@ -372,16 +373,19 @@ und anschliessend \label{lambertw:eqMitExp} \end{equation} erhält. -Auf dem rechten Term von \eqref{lambertw:eqMitExp} beginnen wir langsam eine ähnliche Struktur wie \(\eta e^\eta\) zu erkennen, dies schreit nach der Struktur die benötigt wird um \(\eta\) mittels der Lambert-\(W\)-Funktion \(W(x)\) zu erhalten. Dies macht durchaus Sinn, wenn wir die Funktion \(x(t)\) finden wollen und \(W(x)\) die Umkehrfunktion von \(x e^x\) ist. +Auf dem rechten Term von \eqref{lambertw:eqMitExp} beginnen wir langsam eine ähnliche Struktur wie \(\eta e^\eta\) zu erkennen, dies schreit nach der Struktur, die benötigt wird, um \(\eta\) mittels der Lambert-\(W\)-Funktion \(W(x)\) zu erhalten. Dies macht durchaus Sinn, wenn wir die Funktion \(x(t)\) finden wollen und \(W(x)\) die Umkehrfunktion von \(x e^x\) ist. -Die erste Sache die uns in \eqref{lambertw:eqMitExp} stört ist, dass \(\eta\) als Potenz da steht. Dieses Problem können wir loswerden, indem wir beidseitig mit \(\:\displaystyle \frac{1}{r_0-y_0}\:\) potenzieren: +Die erste Sache, die uns in \eqref{lambertw:eqMitExp} stört ist, dass \(\eta\) als Potenz da steht. Dieses Problem können wir loswerden, indem wir beidseitig mit \(\:1 / (r_0-y_0)\:\) potenzieren: \begin{equation} \operatorname{exp}\left(\displaystyle \frac{-4t}{r_0-y_0}+\frac{y_0+r_0}{r_0-y_0}\right) = \eta\cdot \operatorname{exp}\left(\displaystyle \frac{y_0+r_0}{r_0-y_0}\eta\right). \label{lambertw:eqOhnePotenz} \end{equation} -Das nächste Problem auf welches wir in \eqref{lambertw:eqOhnePotenz} treffen ist, dass \(\eta\) nicht alleine im Exponent steht. Dies kann elegant mit der Substitution + +\subsubsection{Eine essenzielle Substitution + \label{lambertw:subsubsection:SubstChi}} +Das nächste Problem, auf welches wir in \eqref{lambertw:eqOhnePotenz} treffen, ist, dass \(\eta\) nicht alleine im Exponent steht. Dies kann elegant mit der Substitution \begin{equation} \chi = @@ -398,6 +402,9 @@ die auf dasselbe Ergebnis führen würden, aber \eqref{lambertw:eqChiSubst} lief \chi\eta\cdot e^{\displaystyle \chi\eta}. \label{lambertw:eqNachSubst} \end{equation} + +\subsubsection{Funktion nach der Zeit dank Lambert-\(W\) + \label{lambertw:subsubsection:LambertWundFvonT}} Nun sind wir endlich soweit, dass wir die angedeutete Lambert-\(W\)-Funktion \(W(x)\) einsetzen können. Wenn wir beidseitig \(W(x)\) anwenden, dann erhalten wir den Ausdruck \begin{equation} W\left(\chi\cdot \operatorname{exp}\left(\displaystyle \chi-\frac{4t}{r_0-y_0}\right)\right) @@ -417,14 +424,14 @@ Nach dem Auflösen nach \(x\) welches in \(\eta\) enthalten ist, erhalten wir di = y(t) &= - \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right). + \frac{1}{4}\biggl(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\biggl(\biggl(\frac{x(t)}{x_0}\biggr)^2\biggr)-r_0+3y_0\biggr). \end{align} \end{subequations} Nun haben wir unser letztes Ziel erreicht und sind in der Lage eine Verfolgung rechnerisch sowie graphisch zu repräsentieren. \subsubsection{Hinweise zur Lambert-\(W\)-Funktion \label{lambertw:subsubsection:HinwLambertW}} -Wir sind aber noch nicht ganz fertig, eine Frage muss noch beantwortet werden. Und zwar wieso, man schon bei der Gleichung \eqref{lambertw:eqFunkUndAbleitEingefuegt} weiss, dass die Lambert-\(W\)-Funktion zum Einsatz kommen wird. +Wir sind aber noch nicht ganz fertig, eine Frage muss noch beantwortet werden. Und zwar wieso man schon bei der Gleichung \eqref{lambertw:eqFunkUndAbleitEingefuegt} weiss, dass die Lambert-\(W\)-Funktion zum Einsatz kommen wird. Nun, der Grund dafür ist die Struktur \begin{equation} y -- cgit v1.2.1 From a5bde5f6142ac462d7c59f9d39485387c91242e2 Mon Sep 17 00:00:00 2001 From: tim30b Date: Fri, 5 Aug 2022 20:54:19 +0200 Subject: =?UTF-8?q?ImPro=20Buch=20als=20Referenz=20hinzugef=C3=BCgt?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- buch/papers/kreismembran/references.bib | 8 ++++++++ 1 file changed, 8 insertions(+) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/references.bib b/buch/papers/kreismembran/references.bib index acf8f90..3d9d0c1 100644 --- a/buch/papers/kreismembran/references.bib +++ b/buch/papers/kreismembran/references.bib @@ -52,6 +52,14 @@ url = {https://doi.org/10.1016/j.acha.2017.11.004} } +@book{kreismembran:Digital_Image_processing, + edition = {Fourth Edition}, + title = {Digital Image Processing}, + publisher = {Pearson}, + author = {Rafael C. Gozales and Richard E. Woods}, + date = {2018}, +} + @book{lokenath_debnath_integral_2015, edition = {Third Edition}, title = {Integral Tansforms and Their Applications}, -- cgit v1.2.1 From 28fd91bc332b07b3fbde985a92d71e0262a6e6fd Mon Sep 17 00:00:00 2001 From: tim30b Date: Fri, 5 Aug 2022 20:54:48 +0200 Subject: Simulation fast fertig --- buch/papers/kreismembran/teil4.tex | 123 ++++++++++++++++++++++++++++++++++++- 1 file changed, 120 insertions(+), 3 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil4.tex b/buch/papers/kreismembran/teil4.tex index 62a34c5..f660439 100644 --- a/buch/papers/kreismembran/teil4.tex +++ b/buch/papers/kreismembran/teil4.tex @@ -9,6 +9,7 @@ Um numerisch das Verhalten einer Membran zu ermitteln, muss eine numerische Darstellung definiert werden. Die Membran wird hier in Form der Matrix $ U $ digitalisiert. Jedes Element $ U_{ij} $ steh für die Auslenkung der Membran $ u(x,y,t) $ an der Stelle $ \{x,y\}=\{i,j\} $. +Zwischen benachbarten Elementen in der Matrix $ U $ liegt immer der Abstand $ dh $, eine Inkrementierung von $ i $ oder $ j $ entspricht somit einem Schritt in Richtung $ x $ oder $ y $ von Länge $ dh $ auf der Membran. Die zeitliche Dimension wird in Form des Array $ U[] $ aus $ z \times U $ Matrizen dargestellt, wobei $ z $ der Anzahl Zeitschritten entspricht. Das Element auf Zeile $ i $, Spalte $ j $ der $ w $-ten Matrix von $ U[] $ also $ U[w]_{ij} $ entspricht somit der Auslenkung $ u(i,j,w) $. Da die DGL von Zweiter Ordnung ist, reicht eine Zustandsvariabel pro Membran-Element nicht aus. @@ -20,8 +21,124 @@ Der Zustand einer Membran zum Zeitpunkt $ w $ wird mit $ X[w] $ beschrieben, was \subsection{Propagation} Um das Verhalten der Membran zu berechnen, muss aus einem gegebenen Zustand $ X[w] $ der Folgezustand $ X[w+1] $ gerechnet werden können, wobei dazwischen ein Zeitintervall $ dt $ vergeht. Die Berechnung von Folgezuständen kann anschliessend repetiert werden über das zu untersuchende Zeitfenster. -Da die Digitale Membran sich wie die analytisch untersuchte verhalten soll, muss auch sie +Die Folgeposition $ U[w+1] $ ergibt sich als +\begin{equation} + U[w+1] = U[w] + dt \cdot V[w], +\end{equation} +also die Ausgangslage $ + $ die Strecke welche während des Zeitintervall mit der Geschwindigkeit des Elementes zurückgelegt wurde. +Neben der Position muss auch die Geschwindigkeit aktualisiert werden. +Analog zur Folgeposition wird \begin{equation*} - \frac{1}{c^2}\frac{\partial^2u}{\partial t^2} = \Delta u + V[w+1] = V[w] + dt \cdot \frac{\partial^2u}{\partial t^2}. +\end{equation*} +Die Beschleunigung $ \frac{\partial^2u}{\partial t^2} $ eines Elementes ist durch die DGL \ref{kreismembran:Ausgang_DGL} gegeben als +\begin{equation*} + \frac{\partial^2u}{\partial t^2} = \Delta u \cdot c^2. +\end{equation*} +Die Geschwindigkeit des Folgezustandes kann somit mit +\begin{equation} + V[w+1] = V[w] + dt \cdot \Delta_h U \cdot c^2 +\end{equation} +berechnet werden. +Während $ c^2 $ lediglich eine Material spezifische Konstante ist, muss noch erläutert werden, wie der diskrete Laplace-Operator für $ \Delta_h u $ definiert ist. + +\subsection{Diskreter Laplace-Operator $\Delta_h$} +Die diskrete Ableitung zweiter Ordnung kann mit Hilfe der Taylor-Reihen-Entwicklung als +\begin{equation*} + \frac{\partial^2f}{\partial x^2} \approx \frac{f(x+dx)-2f(x)+f(x-dx)}{dx^2} +\end{equation*} +approximiert werden \cite{kreismembran:Digital_Image_processing}. +Dank der Linearität der Ableitung kann die Ableitung einer weiteren Dimension addiert werden. +Daraus folgt für den zweidimensionalen Fall +\begin{equation*} + \Delta_h u= \frac{u(x+dh,y,t)+u(x,y+dh,t)-4f(x)+u(x-dh,y,t)+u(x,y-dh,t)}{dh^2}. \end{equation*} -erfüllen. +Um $ \Delta_h $ auf eine Matrix anwenden zu können wird die Gleichung in Form einer Filtermaske + \begin{equation} + \Delta_h u= \frac{1}{dh^2} + \left[ {\begin{array}{ccc} + 0 & 1 & 0\\ + 1 & -4 & 1\\ + 0 & 1 & 0\\ + \end{array} } \right] + \end{equation} +formuliert. +Die Filtermaske kann dann auf jedes Element einzeln angewendet werden mit einer Matrizen-Faltung um $ \Delta_h U[] $ zu berechnen. + +\subsection{Simulation: Kreisförmige Membran} +Als Beispiel soll nun eine schwingende kreisförmige Membran simuliert werden. +\paragraph{Initialisierung} +Die Anzahl der simulierten Elementen soll $ m \times n $ was dementsprechend die Dimensionen von $ U $ und $ V $ vorgibt. +Als Anfangsbedingung wird eine Membran gewählt, welche bei $ t=0 $ mit einer Gauss-Kurve ausgelenkt wird. +Die Membran soll sich zu Beginn nicht bewegen, also wird $ V[0] $ mit Nullen initialisiert. +Die Auslenkung kann kompakt erreicht werden, wenn $ U[0] $ als Null-Matrix mit einer $ 1 $ in der Mitte initialisiert wird. +Diese Matrix wird anschliessend mit einer Filtermaske in Form einer Gauss-Glocke gefaltet. +Die Faltung mit einer Gauss-Glocke ist in Programmen wie Matlab eine Standartfunktion, da dies einm Tiefpassfilter in der Bildverarbeitung entspricht. + +\paragraph{Rand} +Bislang ist die definierte Matrix rechteckig. +Um eine kreisförmige Membran zu simulieren muss der Rand angepasst werden. +Da in den meisten Programme keine Möglichkeit besteht, mit runden Matrizen zu rechnen, wird der Rand in der Berechnung des Folgezustandes implementiert. +Der Rand bedeutet, das Membran-Elemente auf dem Rand sich nicht Bewegen können. +Die Position sowie die Geschwindigkeit aller Elemente welche nicht auf der definierten Membran sind müssen zu beliebiger Zeit $0$ entsprechen. +Hierzu wird eine Maske $M$ erstellt. +Diese Maske besteht aus einer binären Matrix von identischer Dimension wie $ U $ und $ V $. +Ist in der Matrix $M$ eine $1$ abgebildet so ist an jener stelle ein Element der Membran, ist es eine $0$ so befindet sich dieses Element auf dem Rand oder ausserhalb der Membran. +In dieser Anwendung ist $M$ eine Matrix mit einem Kreis voller $1$ umgeben von $0$ bis an den Rand der Matrix. +Die Maske wird angewendet indem das Resultat des nächsten Zustandes noch mit der Maske elementweise multipliziert wird. +Der Folgezustand kann also mit den Gleichungen +\begin{align} + \label{kreismembran:eq:folge_U} + U[w+1] &= (U[w] + dt \cdot V[w])*M\\ + \label{kreismembran:eq:folge_V} + V[w+1] &= (V[w] + dt \cdot \Delta_h u \cdot c^2)*M +\end{align} +berechnet werden. +\paragraph{Simulation} +Mit den gegebenen Gleichungen \ref{kreismembran:eq:folge_U} und \ref{kreismembran:eq:folge_V} das Verhalten der Membran mit einem Loop über das zu untersuchende Zeitintervall berechnet werden. +In der Abbildung \ref{kreismembran:im:simres_rund} sind Simulationsresultate zu sehen. +Die Erste Figur zeigt die Ausgangslage gefolgt von den Auslenkungen nach jeweils $ 50 $ weiteren Iterationsschritten. +Es ist zu erkennen, wie sich die Störung vom Zentrum an den Rand ausbreitet. +Erreicht die Störung den Rand wird sie reflektiert und nähert sich dem Zentrum. +\begin{figure} + \begin{center} + \label{kreismembran:im:simres_rund} + \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_1.png} + \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_2.png} + \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_3.png} + \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_4.png} + \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_5.png} + \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_6.png} + \caption{Simulations Resultate einer kreisförmigen Membran. Simuliert mit $ 200 \times 200 $ Elementen, dargestellt sind die Auslenkungen nach jeweils $ 50 $ Iterationsschritten.} + + \end{center} +\end{figure} +\subsection{Simulation: Unendliche Kreisförmige Membran} + +Um eine unendlich grosse Membran zu simulieren könnte der unpraktische weg gewählt werden die Matrix unendlich gross zu definieren, dies wird jedoch spätestens bei der numerischen Berechnung seine Probleme mit sich bringen. +Etwas geeigneter ist es die Matrix so gross wie möglich zu definieren wie es die Kapazitäten erlauben. +Wenn anschliessend nur das Verhalten im Zentrum, bei der Störung beobachtet wird, verhaltet sich die Membran wie eine unendliche. +Dies aber nur bis die Störung am Rand reflektiert wird und wieder das innere zu beobachtende Zentrum beeinflusst. +Soll erst gar keine Reflexion entstehen, muss ein Absorber modelliert werden welcher die Störung möglichst ohne Reflexion aufnimmt. + +\paragraph{Absorber} +Sehr knapp formuliert entstehen Reflexionen, wenn eine Welle von einem Material in ein anderes Material mit unterschiedlichen Eigenschaften eindringen möchte. +Je unterschiedlicher und abrupter der Übergang zwischen den Materialien umso ausgeprägter die Reflexion. +In diesem Fall sind die Eigenschaften vorgegeben. +Im Zentrum soll sich die Membran verhalten, wie von der DGL vorgegeben, am Rand jedoch muss sich jedes Membran-Element in der Ausgangslage befinden. +Der Spielraum welcher dem Absorber übrig bleibt ist die Art der Überganges. +Bei der endlichen kreisförmigen Membran hat die Maske $M$ ein binärer Übergang von Membran zu Rand bezweckt. +Anstelle dieses abrupten Wechsels wird nun eine Maske definiert, welche graduell von Membran $1$ zu Rand-Element $0$ wechselt. + + + + + + + + + + + + + -- cgit v1.2.1 From e37397bb3b3c9cf93dff1d1aaecb186ca10fc239 Mon Sep 17 00:00:00 2001 From: tim30b Date: Fri, 5 Aug 2022 23:03:46 +0200 Subject: =?UTF-8?q?minikorrekturen=20m=C3=BCller?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- buch/papers/kreismembran/teil0.tex | 46 +++++++++++++++++++++----------------- 1 file changed, 25 insertions(+), 21 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil0.tex b/buch/papers/kreismembran/teil0.tex index bb8188d..10cd476 100644 --- a/buch/papers/kreismembran/teil0.tex +++ b/buch/papers/kreismembran/teil0.tex @@ -5,32 +5,32 @@ % \section{Einleitung\label{kreismembran:section:teil0}} \rhead{Membran} -Eine Membran oder selten ein Schwingblatt ist laut Duden \cite{kreismembran:Duden:Membran} ein "dünnes Blättchen aus Metall, Papier o. Ä., das durch seine Schwingungsfähigkeit geeignet ist, Schallwellen zu übertragen ...". -Ein dünnes Blättchen aus Metall zeig jedoch nicht die selben dynamischen Eigenschaften auf, wie ein gespanntes Stück Papier. +Eine Membran oder selten ein Schwingblatt ist laut Duden \cite{kreismembran:Duden:Membran} ein ``dünnes Blättchen aus Metall, Papier o. Ä., das durch seine Schwingungsfähigkeit geeignet ist, Schallwellen zu übertragen ...''. +Ein dünnes Blättchen aus Metall zeig jedoch nicht die selben dynamischen Eigenschaften wie ein gespanntes Stück Papier. Beschreibt man das dynamische Verhalten, muss zwischen einer dünnen Platte und einer Membrane unterschieden werden \cite{kreismembran:membrane_vs_thin_plate}. Eine dünne Platte zum Beispiel aus Metall, wirkt selbst entgegen ihrer Deformation sobald sie gekrümmt wird. -Eine Membran auf der anderen Seite besteht aus einem Material welches sich ohne Kraftaufwand verbiegen lässt wie zum Beispiel Papier. -Bevor Papier als schwingende Membran betrachtet werden kann wird jedoch noch eine Spannung $ T $ benötigt welche das Material daran hindert aus der Ruhelage gebracht zu werden. +Eine Membran auf der anderen Seite besteht aus einem Material, welches sich ohne Kraftaufwand verbiegen lässt wie zum Beispiel Papier. +Bevor Papier als schwingende Membran betrachtet werden kann, wird jedoch noch eine Spannung $ T $ benötigt welche das Material daran hindert aus der Ruhelage gebracht zu werden. Ein geläufiges Beispiel einer Kreismembran ist eine runde Trommel. -Sie besteht herkömmlicher weise aus einem Leder (Fell), welches auf einen offenen Zylinder (Zargen) aufgespannt wird. +Sie besteht herkömmlicherweise aus einem Leder (Fell), welches auf einen offenen Zylinder (Zargen) aufgespannt wird. Das Leder alleine erzeugt nach einem Aufschlag keine hörbaren Schwingungen. -Sobald das Fell jedoch über den Zargen gespannt wird, kann das Fell auf verschiedensten weisen weiter schwingen, was für den Klang der Trommel verantwortlich ist. -Wie genau diese Schwingungen untersucht werden können wird in der Folgenden Arbeit Diskutiert. +Sobald das Fell jedoch über den Zargen gespannt wird, kann das Fell auf verschiedensten Weisen weiter schwingen, was für den Klang der Trommel verantwortlich ist. +Wie genau diese Schwingungen untersucht werden können wird in der folgenden Arbeit diskutiert. -\paragraph{Annahmen} +\subsection{Annahmen} Um die Wellengleichung herzuleiten \cite{kreismembran:wellengleichung_herleitung}, muss ein Modell einer Membran definiert werden. -Das untersuchte Modell einer Membrane Erfüllt folgende Eigenschaften: -\begin{enumerate}[i] +Das untersuchte Modell erfüllt folgende Eigenschaften: +\begin{enumerate}[i)] \item Die Membran ist homogen. Dies bedeutet, dass die Membran über die ganze Fläche die selbe Dichte $ \rho $ und Elastizität hat. Durch die konstante Elastizität ist die ganze Membran unter gleichmässiger Spannung $ T $. \item Die Membran ist perfekt flexibel. - Daraus folgt, dass die Membran ohne Kraftaufwand verbogen werden kann. - Die Membran ist dadurch nicht allein stehend schwing-fähig, hierzu muss sie gespannt werden mit einer Kraft $ T $. - \item Die Membran kann sich nur in Richtung ihrer Normalen in kleinem Ausmass Auslenken. - Auslenkungen in der ebene der Membran sind nicht möglich. + Damit ist gemeint, dass die Membran ohne Kraftaufwand verbogen werden kann. + Die Membran ist dadurch nicht allein stehend schwingfähig, hierzu muss sie gespannt werden mit einer Kraft $ T $. + \item Die Membran kann sich nur in Richtung ihrer Normalen in kleinem Ausmass auslenken. + Auslenkungen in der Ebene der Membran sind nicht möglich. \item Die Membran erfährt keine Art von Dämpfung. Die Membran wird also nicht durch ihr umliegendes Medium abgebremst noch erfährt sie Wärmeverluste durch Deformation. Die resultierende Schwingung wird daher nicht gedämpft sein. @@ -38,18 +38,18 @@ Das untersuchte Modell einer Membrane Erfüllt folgende Eigenschaften: \end{enumerate} \subsection{Wellengleichung} Um die Wellengleichung einer Membran herzuleiten wird vorerst eine schwingende Saite betrachtet. -Es lohnt sich das Verhalten einer Saite zu beschreiben da eine Saite das selbe Verhalten wie eine Membran aufweist mit dem Unterschied einer fehlenden Dimension. +Es lohnt sich das Verhalten einer Saite zu beschreiben, da eine Saite das selbe Verhalten wie eine Membran aufweist mit dem Unterschied einer fehlenden Dimension. Die Verbindung zwischen Membran und Saite ist intuitiv ersichtlich, stellt man sich einen Querschnitt einer Trommel vor. %Wie analog zur Membran kann eine Saite erst unter Spannung schwingen. Abbildung \ref{TODO} ist ein infinitesimales Stück einer Saite mit Länge $ dx $ skizziert. -Wie für die Membran ist die Annahme iii gültig, keine Bewegung in die Richtung $ \hat{x} $. +Wie für die Membran ist die Annahme iii) gültig, keine Bewegung in die Richtung $ \hat{x} $. Um dies zu erfüllen muss der Punkt $ P_1 $ gleich stark in Richtung $ -\hat{x} $ gezogen werden wie der Punkt $ P_2 $ in Richtung $ \hat{x} $ gezogen wird. Ist $ T_1 $ die Kraft welche mit Winkel $ \alpha $ auf Punkt $ P_1 $ wirkt sowie $ T_2 $ und $ \beta$ das analoge für Punkt $ P_2 $ ist, so können die Kräfte \begin{equation}\label{kreismembran:eq:no_translation} T_1 \cos \alpha = T_2 \cos \beta = T \end{equation} gleichgesetzt werden. -Das dynamische verhalten der senkrechten Auslenkung $ u(x,t) $ muss das newtonsche Gesetz +Das dynamische Verhalten der senkrechten Auslenkung $ u(x,t) $ muss das newtonsche Gesetz \begin{equation*} \sum F = m \cdot a \end{equation*} @@ -69,14 +69,18 @@ geschrieben werden. Der $ \tan \alpha $ entspricht der örtlichen Ableitung von $ u(x,t) $ an der Stelle $ x_0 $ und analog der $ \tan \beta $ für die Stelle $ x_0 + dx $. Die Gleichung wird dadurch zu \begin{equation*} - \frac{\partial u}{\partial x} \big\vert_{x_0 + dx} - \frac{\partial u}{\partial x} \big\vert_{x_0} = \frac{\rho dx}{T} \frac{\partial^2 u}{\partial t^2}. + \frac{\partial u}{\partial x} \bigg|_{x_0 + dx} - \frac{\partial u}{\partial x} \bigg|_{x_0} = \frac{\rho dx}{T} \frac{\partial^2 u}{\partial t^2}. \end{equation*} Durch die Division mit $ dx $ entsteht \begin{equation*} - \frac{1}{dx} \bigg[\frac{\partial u}{\partial x} \big\vert_{x_0 + dx} - \frac{\partial u}{\partial x} \big\vert_{x_0}\bigg] = \frac{\rho}{T}\frac{\partial^2 u}{\partial t^2}. + \frac{1}{dx} \bigg[\frac{\partial u}{\partial x} \bigg|_{x_0 + dx} - \frac{\partial u}{\partial x} \bigg|_{x_0}\bigg] = \frac{\rho}{T}\frac{\partial^2 u}{\partial t^2}. \end{equation*} -Auf der Linken Seite der Gleichung wird die Differenz der Steigungen durch die Intervall-Länge geteilt, in anderen Worten die zweite Ableitung von $ u(x,t) $ nach $ x $ berechnet. Der Term $ \frac{\rho}{T} $ wird mit $ c^2 $ ersetzt, da der Bruch für eine gegebene Membran eine positive Konstante sein muss. Somit resultiert die, in der Literatur gebräuchliche Form +Auf der linken Seite der Gleichung wird die Differenz der Steigungen durch die Intervalllänge geteilt, in anderen Worten die zweite Ableitung von $ u(x,t) $ nach $ x $ berechnet. +Der Term $ \frac{\rho}{T} $ wird durch $ c^2 $ ersetzt, da der Bruch für eine gegebene Membran eine positive Konstante sein muss. +Somit resultiert die in der Literatur gebräuchliche Form \begin{equation} + \label{kreismembran:Ausgang_DGL} \frac{1}{c^2}\frac{\partial^2u}{\partial t^2} = \Delta u. \end{equation} -In dieser Form ist die Gleichung auch gültig für eine Membran. Für den Fall einer Membran muss lediglich die Ableitung in zwei Dimensionen gerechnet werden. \ No newline at end of file +In dieser Form ist die Gleichung auch gültig für eine Membran. +Für den Fall einer Membran muss lediglich der Laplace-Operator $\Delta$ in zwei Dimensionen gerechnet werden. \ No newline at end of file -- cgit v1.2.1 From 40c2d617a90a81c57489a5c9e220ef577f6882a5 Mon Sep 17 00:00:00 2001 From: tim30b Date: Fri, 5 Aug 2022 23:34:04 +0200 Subject: simulation fertig --- buch/papers/kreismembran/teil4.tex | 61 ++++++++++++++++++++++++++++++++++---- 1 file changed, 55 insertions(+), 6 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil4.tex b/buch/papers/kreismembran/teil4.tex index f660439..b67e9e7 100644 --- a/buch/papers/kreismembran/teil4.tex +++ b/buch/papers/kreismembran/teil4.tex @@ -101,8 +101,9 @@ Die Erste Figur zeigt die Ausgangslage gefolgt von den Auslenkungen nach jeweils Es ist zu erkennen, wie sich die Störung vom Zentrum an den Rand ausbreitet. Erreicht die Störung den Rand wird sie reflektiert und nähert sich dem Zentrum. \begin{figure} + \begin{center} - \label{kreismembran:im:simres_rund} + \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_1.png} \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_2.png} \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_3.png} @@ -110,10 +111,11 @@ Erreicht die Störung den Rand wird sie reflektiert und nähert sich dem Zentrum \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_5.png} \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_6.png} \caption{Simulations Resultate einer kreisförmigen Membran. Simuliert mit $ 200 \times 200 $ Elementen, dargestellt sind die Auslenkungen nach jeweils $ 50 $ Iterationsschritten.} + \label{kreismembran:im:simres_rund} \end{center} \end{figure} -\subsection{Simulation: Unendliche Kreisförmige Membran} +\subsection{Simulation: Unendliche Membran} Um eine unendlich grosse Membran zu simulieren könnte der unpraktische weg gewählt werden die Matrix unendlich gross zu definieren, dies wird jedoch spätestens bei der numerischen Berechnung seine Probleme mit sich bringen. Etwas geeigneter ist es die Matrix so gross wie möglich zu definieren wie es die Kapazitäten erlauben. @@ -129,12 +131,59 @@ Im Zentrum soll sich die Membran verhalten, wie von der DGL vorgegeben, am Rand Der Spielraum welcher dem Absorber übrig bleibt ist die Art der Überganges. Bei der endlichen kreisförmigen Membran hat die Maske $M$ ein binärer Übergang von Membran zu Rand bezweckt. Anstelle dieses abrupten Wechsels wird nun eine Maske definiert, welche graduell von Membran $1$ zu Rand-Element $0$ wechselt. +Die Elemente werden auf Basis ihres Abstand $r$ zum Zentrum definiert. +Der Abstand entspricht +\begin{equation*} + r(i,j) = \sqrt{|i-\frac{m}{2}|^2+|j-\frac{n}{2}|^2}, +\end{equation*} +wobei $ m $ und $n$ den Dimensionen der Matrix entsprechen. +Für einen Stufenlosen Übergang werden die Elemente der Maske auf +\begin{align} + M_{ij} = \begin{cases} 1-e^{(r(i,j)-b)a} & \text{wenn $x > b$} \\ + 0 & \text{sonst} \end{cases} +\end{align} +gesetzt. +Der Parameter $a > 0$ bestimmt wie Steil der Übergang sein soll, $b$ bestimmt wie weit weg vom Zentrum sich der Übergang befindet. +In der Abbildung \ref{kreismembran:im:masks} ist der Unterschied der beiden Masken zu sehen. +\begin{figure} + + \begin{center} + + \includegraphics[width=0.45\textwidth]{papers/kreismembran/images/mask_disk.png} + \includegraphics[width=0.45\textwidth]{papers/kreismembran/images/mask_absorber.png} + \caption{Vergleich von Masken: Links Binär für eine endliche Membran, rechts mit Absorber für eine unendliche Membran} + \label{kreismembran:im:masks} + \end{center} +\end{figure} +\paragraph{Simulation} +Bis auf die Absorber-Maske kann nun identisch zur endlichen Membran simuliert werden. +Auch hier wurde eine Gauss-Glocke als Anfangsbedingung gewählt. +Die Simulationsresultate von Abbildung \ref{kreismembran:im:simres_unendlich} - - - - +\begin{figure} + + \begin{center} + + \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_2_1.png} + \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_2_2.png} + \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_2_3.png} + \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_2_4.png} + \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_2_5.png} + \includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_2_6.png} + \caption{Simulations Resultate einer unendlichen Membran. Simuliert mit $ 200 \times 200 $ Elementen, dargestellt sind die Auslenkungen nach jeweils $ 50 $ Iterationsschritten.} + \label{kreismembran:im:simres_unendlich} + + \end{center} +\end{figure} +zeigen deutlich wie die Störung vom Zentrum weg verläuft. +Nähert sich die Störung dem Rand, so wird sie immer stärker abgeschwächt. +Die Wirkung des Absorber ist an der letzten Figur zu erkennen, in welcher kaum noch Auslenkungen zu sehen sind. + +\section{Schlusswort} +Auch wenn ein Physikalisches Verhalten bereits durch Annahmen und Annäherungen deutlich vereinfacht wird, bestehen auch dann noch eine Vielzahl von Lösungsansätzen. +Lösungen einer unendlich grosse Membran scheinen fern der Realität zu sein, doch dies darf es im Sinne der Mathematik. +Und wer weis, für eine Ameise auf einem Trampolin ist eine unendliche Membran vielleicht eine ganz gute Annäherung. -- cgit v1.2.1 From 152b3d55898b7aebbd4fd0182a9c45914514a7d8 Mon Sep 17 00:00:00 2001 From: tim30b Date: Fri, 5 Aug 2022 23:34:51 +0200 Subject: beginn mit besseren referenzen auf annahmen --- buch/papers/kreismembran/teil0.tex | 18 +++++++++--------- 1 file changed, 9 insertions(+), 9 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil0.tex b/buch/papers/kreismembran/teil0.tex index 10cd476..ad41406 100644 --- a/buch/papers/kreismembran/teil0.tex +++ b/buch/papers/kreismembran/teil0.tex @@ -17,30 +17,30 @@ Sie besteht herkömmlicherweise aus einem Leder (Fell), welches auf einen offene Das Leder alleine erzeugt nach einem Aufschlag keine hörbaren Schwingungen. Sobald das Fell jedoch über den Zargen gespannt wird, kann das Fell auf verschiedensten Weisen weiter schwingen, was für den Klang der Trommel verantwortlich ist. Wie genau diese Schwingungen untersucht werden können wird in der folgenden Arbeit diskutiert. - + \subsection{Annahmen} Um die Wellengleichung herzuleiten \cite{kreismembran:wellengleichung_herleitung}, muss ein Modell einer Membran definiert werden. Das untersuchte Modell erfüllt folgende Eigenschaften: \begin{enumerate}[i)] \item Die Membran ist homogen. - Dies bedeutet, dass die Membran über die ganze Fläche die selbe Dichte $ \rho $ und Elastizität hat. - Durch die konstante Elastizität ist die ganze Membran unter gleichmässiger Spannung $ T $. + %Dies bedeutet, dass die Membran über die ganze Fläche die selbe Dichte $ \rho $ und Elastizität hat. + %Durch die konstante Elastizität ist die ganze Membran unter gleichmässiger Spannung $ T $. \item Die Membran ist perfekt flexibel. - Damit ist gemeint, dass die Membran ohne Kraftaufwand verbogen werden kann. - Die Membran ist dadurch nicht allein stehend schwingfähig, hierzu muss sie gespannt werden mit einer Kraft $ T $. + %Damit ist gemeint, dass die Membran ohne Kraftaufwand verbogen werden kann. + %Die Membran ist dadurch nicht allein stehend schwingfähig, hierzu muss sie gespannt werden mit einer Kraft $ T $. \item Die Membran kann sich nur in Richtung ihrer Normalen in kleinem Ausmass auslenken. - Auslenkungen in der Ebene der Membran sind nicht möglich. + %Auslenkungen in der Ebene der Membran sind nicht möglich. \item Die Membran erfährt keine Art von Dämpfung. - Die Membran wird also nicht durch ihr umliegendes Medium abgebremst noch erfährt sie Wärmeverluste durch Deformation. - Die resultierende Schwingung wird daher nicht gedämpft sein. + %Die Membran wird also nicht durch ihr umliegendes Medium abgebremst noch erfährt sie Wärmeverluste durch Deformation. + %Die resultierende Schwingung wird daher nicht gedämpft sein. \end{enumerate} \subsection{Wellengleichung} Um die Wellengleichung einer Membran herzuleiten wird vorerst eine schwingende Saite betrachtet. Es lohnt sich das Verhalten einer Saite zu beschreiben, da eine Saite das selbe Verhalten wie eine Membran aufweist mit dem Unterschied einer fehlenden Dimension. Die Verbindung zwischen Membran und Saite ist intuitiv ersichtlich, stellt man sich einen Querschnitt einer Trommel vor. -%Wie analog zur Membran kann eine Saite erst unter Spannung schwingen. + Abbildung \ref{TODO} ist ein infinitesimales Stück einer Saite mit Länge $ dx $ skizziert. Wie für die Membran ist die Annahme iii) gültig, keine Bewegung in die Richtung $ \hat{x} $. -- cgit v1.2.1 From 07467478fad0ab552c794e41442a36e18c296111 Mon Sep 17 00:00:00 2001 From: tim30b Date: Sat, 6 Aug 2022 14:58:37 +0200 Subject: add images --- buch/papers/kreismembran/images/Saite.pdf | Bin 0 -> 17845 bytes buch/papers/kreismembran/images/mask_absorber.png | Bin 0 -> 83443 bytes buch/papers/kreismembran/images/mask_disk.png | Bin 0 -> 15936 bytes buch/papers/kreismembran/images/sim_1_1.png | Bin 0 -> 28449 bytes buch/papers/kreismembran/images/sim_1_2.png | Bin 0 -> 40121 bytes buch/papers/kreismembran/images/sim_1_3.png | Bin 0 -> 47092 bytes buch/papers/kreismembran/images/sim_1_4.png | Bin 0 -> 50305 bytes buch/papers/kreismembran/images/sim_1_5.png | Bin 0 -> 54324 bytes buch/papers/kreismembran/images/sim_1_6.png | Bin 0 -> 49234 bytes buch/papers/kreismembran/images/sim_2_1.png | Bin 0 -> 28449 bytes buch/papers/kreismembran/images/sim_2_2.png | Bin 0 -> 36804 bytes 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buch/papers/kreismembran/teil1.tex | 11 +++++++---- buch/papers/kreismembran/teil3.tex | 3 ++- 2 files changed, 9 insertions(+), 5 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil1.tex b/buch/papers/kreismembran/teil1.tex index 377ba48..f0d478f 100644 --- a/buch/papers/kreismembran/teil1.tex +++ b/buch/papers/kreismembran/teil1.tex @@ -7,7 +7,7 @@ \section{Lösungsmethode 1: Separationsmethode  \label{kreismembran:section:teil1}} \rhead{Lösungsmethode 1: Separationsmethode} -An diesem Punkt bleibt also nur noch die Lösung der partiellen Differentialgleichung. In diesem Kapitel wird sie mit Hilfe der Separationsmethode gelöst. +An diesem Punkt bleibt also nur noch die Lösung der partiellen Differentialgleichung. In diesem Abschnitt wird sie mit Hilfe der Separationsmethode gelöst. \subsection{Aufgabestellung\label{sub:aufgabestellung}} Wie im vorherigen Abschnitt gezeigt, lautet die partielle Differentialgleichung, die die Schwingungen einer Membran beschreibt: @@ -30,7 +30,7 @@ Da es sich um eine Kreisscheibe handelt, werden Polarkoordinaten verwendet, so d ergibt. Es wird eine runde elastische Membran berücksichtigt, die das Gebiet $\Omega$ abdeckt und am Rand $\Gamma$ befestigt ist. -Es wirken keine äusseren Kräfte. Es handelt sich somit von einer kreisförmligen eingespannten homogenen schwingenden Membran. +Es wirken keine äusseren Kräfte. Es handelt sich somit von einer kreisförmligen eingespannten homogenen schwingenden Membran nach den Annahmen von \ref{kreimembran:annahmen}. Daher ist die Membranabweichung im Punkt $(r,\varphi)$ $\in$ $\overline{\rm \Omega}$ zum Zeitpunkt $t$: \begin{align*} @@ -56,7 +56,10 @@ Dank der Randbedingungen kann gefordert werden, dass $F(R)=0$ ist, und natürlic \begin{equation*} \frac{1}{c^2}\frac{T''(t)}{T(t)}=-\kappa^2=\frac{F''(r)}{F(r)}+\frac{1}{r}\frac{F'(r)}{F(r)}+\frac{1}{r^2}\frac{G''(\varphi)}{G(\varphi)}. \end{equation*} -Da die linke Seite nur von $t$ und die rechte Seite nur von $r$ und $\varphi$ abhängt, müssen sie gleich einer reellen Zahl sein. Aus physikalischen Gründen suchen wir nach Lösungen, die weder exponentiell in der Zeit wachsen noch exponentiell abklingen. Dies bedeutet, dass die Konstante negativ sein muss, also schreibt man $-\kappa^2$. Daraus ergeben sich die folgenden zwei Gleichungen: +Da die linke Seite nur von $t$ und die rechte Seite nur von $r$ und $\varphi$ abhängt, müssen sie gleich einer reellen Zahl sein. +Laut Annahme iv) in \ref{kreimembran:annahmen} erfährt die Membran keine Dämpfung. +Daher werden Lösungen gesucht, die weder exponentiell in der Zeit wachsen noch exponentiell abklingen. +Dies bedeutet, dass die Konstante negativ sein muss, also schreibt man $-\kappa^2$. Daraus ergeben sich die folgenden zwei Gleichungen: \begin{align*} T''(t) + c^2\kappa^2T(t) &= 0\\ r^2\frac{F''(r)}{F(r)} + r \frac{F'(r)}{F(r)} +\kappa^2 r^2 &= - \frac{G''(\varphi)}{G(\varphi)}. @@ -118,4 +121,4 @@ für die Anzahl der Knotenlinien steht. Es gibt bestimmte Bereiche auf der Membr \end{figure} -An diesem Punkt stellte sich die Frage, ob es möglich wäre, die partielle Differentialgleichung mit einer anderen Methode als der der Trennung der Variablen zu lösen. Nach einer kurzen Recherche wurde festgestellt, dass die beste Methode die Transformationsmethode ist, genauer gesagt die Anwendung der Hankel-Transformation. Im nächsten Kapitel wird daher diese Integraltransformation vorgestellt und entwickelt, und es wird erläutert, warum sie für diese Art von Problem geeignet ist. +An diesem Punkt stellte sich die Frage, ob es möglich wäre, die partielle Differentialgleichung mit einer anderen Methode als der der Trennung der Variablen zu lösen. Nach einer kurzen Recherche wurde festgestellt, dass eine weitere Methode die Transformationsmethode ist, genauer gesagt die Anwendung der Hankel-Transformation. Im nächsten Kapitel wird daher diese Integraltransformation vorgestellt und entwickelt, und es wird erläutert, warum sie für diese Art von Problem geeignet ist. diff --git a/buch/papers/kreismembran/teil3.tex b/buch/papers/kreismembran/teil3.tex index 014b6e6..276f911 100644 --- a/buch/papers/kreismembran/teil3.tex +++ b/buch/papers/kreismembran/teil3.tex @@ -78,7 +78,8 @@ kann man die Lösungsmethoden 1 und 2 vergleichen. \subsection{Vergleich der Analytischen Lösungen \label{kreismembran:vergleich}} -Bei der Analyse der Gleichungen \eqref{eq:lösung_endliche_generelle} und \eqref{eq:lösung_unendliche_generelle} fällt sofort auf, dass die Gleichung \eqref{eq:lösung_unendliche_generelle} nicht mehr von $m$ und $n$ abhängt, sondern nur noch von $n$ \cite{nishanth_p_vibrations_2018}. Das macht Sinn, denn $n$ beschreibt die Anzahl der Knotenlinien, und in einer unendlichen Membran gibt es keine, dato che abbiamo assunto che la soluzione è rotationssymmetrisch. Tatsächlich werden $a_{m0}$, $b_{m0}$ und $\kappa_{m0}$ in $a_m$, $b_m$ bzw. $\kappa_m$ umbenannt. Die beiden Termen $\cos(n\varphi)$ und $\sin(n\varphi)$ verschwinden ebenfalls, da für $n=0$ der $\cos(n\varphi)$ gleich 1 und der $\sin(n \varphi)$ gleich 0 ist. +Bei der Analyse der Gleichungen \eqref{eq:lösung_endliche_generelle} und \eqref{eq:lösung_unendliche_generelle} fällt sofort auf, dass die Gleichung \eqref{eq:lösung_unendliche_generelle} nicht mehr von $m$ und $n$ abhängt, sondern nur noch von $n$ \cite{nishanth_p_vibrations_2018}. +Das macht Sinn, denn $n$ beschreibt die Anzahl der Knotenlinien, welche unter der Annahme einer rotationssymmetrischen Lösung nicht vorhanden sein können. Tatsächlich werden $a_{m0}$, $b_{m0}$ und $\kappa_{m0}$ in $a_m$, $b_m$ bzw. $\kappa_m$ umbenannt. Die beiden Termen $\cos(n\varphi)$ und $\sin(n\varphi)$ verschwinden ebenfalls, da für $n=0$ der $\cos(n\varphi)$ gleich 1 und der $\sin(n \varphi)$ gleich 0 ist. Die Funktion hängt also nicht mehr von der Besselfunktionen $n$-ter Ordnung ab, sondern nur von der $0$-ter Ordnung. -- cgit v1.2.1 From 617271699ec4a2ad9a0b8ca9940cc19a21901382 Mon Sep 17 00:00:00 2001 From: tim30b Date: Sat, 6 Aug 2022 15:00:32 +0200 Subject: referenzen und equation fix --- buch/papers/kreismembran/teil0.tex | 28 +++++++++++++++++----------- buch/papers/kreismembran/teil4.tex | 2 ++ 2 files changed, 19 insertions(+), 11 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil0.tex b/buch/papers/kreismembran/teil0.tex index ad41406..6f55358 100644 --- a/buch/papers/kreismembran/teil0.tex +++ b/buch/papers/kreismembran/teil0.tex @@ -19,30 +19,36 @@ Sobald das Fell jedoch über den Zargen gespannt wird, kann das Fell auf verschi Wie genau diese Schwingungen untersucht werden können wird in der folgenden Arbeit diskutiert. -\subsection{Annahmen} +\subsection{Annahmen} \label{kreimembran:annahmen} Um die Wellengleichung herzuleiten \cite{kreismembran:wellengleichung_herleitung}, muss ein Modell einer Membran definiert werden. Das untersuchte Modell erfüllt folgende Eigenschaften: \begin{enumerate}[i)] \item Die Membran ist homogen. - %Dies bedeutet, dass die Membran über die ganze Fläche die selbe Dichte $ \rho $ und Elastizität hat. - %Durch die konstante Elastizität ist die ganze Membran unter gleichmässiger Spannung $ T $. + Dies bedeutet, dass die Membran über die ganze Fläche die selbe Dichte $ \rho $ und Elastizität hat. + Durch die konstante Elastizität ist die ganze Membran unter gleichmässiger Spannung $ T $. \item Die Membran ist perfekt flexibel. - %Damit ist gemeint, dass die Membran ohne Kraftaufwand verbogen werden kann. - %Die Membran ist dadurch nicht allein stehend schwingfähig, hierzu muss sie gespannt werden mit einer Kraft $ T $. + Damit ist gemeint, dass die Membran ohne Kraftaufwand verbogen werden kann. + Die Membran ist dadurch nicht allein stehend schwingfähig, hierzu muss sie gespannt werden mit einer Kraft $ T $. \item Die Membran kann sich nur in Richtung ihrer Normalen in kleinem Ausmass auslenken. - %Auslenkungen in der Ebene der Membran sind nicht möglich. + Auslenkungen in der Ebene der Membran sind nicht möglich. \item Die Membran erfährt keine Art von Dämpfung. - %Die Membran wird also nicht durch ihr umliegendes Medium abgebremst noch erfährt sie Wärmeverluste durch Deformation. - %Die resultierende Schwingung wird daher nicht gedämpft sein. + Die Membran wird also nicht durch ihr umliegendes Medium abgebremst noch erfährt sie Wärmeverluste durch Deformation. \end{enumerate} \subsection{Wellengleichung} Um die Wellengleichung einer Membran herzuleiten wird vorerst eine schwingende Saite betrachtet. Es lohnt sich das Verhalten einer Saite zu beschreiben, da eine Saite das selbe Verhalten wie eine Membran aufweist mit dem Unterschied einer fehlenden Dimension. Die Verbindung zwischen Membran und Saite ist intuitiv ersichtlich, stellt man sich einen Querschnitt einer Trommel vor. +\begin{figure} + + \begin{center} + \includegraphics[width=5cm,angle=-90]{papers/kreismembran/images/Saite.pdf} + \caption{Infinitesimales Stück einer Saite} + \label{kreismembran:im:Saite} + \end{center} +\end{figure} - -Abbildung \ref{TODO} ist ein infinitesimales Stück einer Saite mit Länge $ dx $ skizziert. +Abbildung \ref{kreismembran:im:Saite} ist ein infinitesimales Stück einer Saite mit Länge $ dx $ skizziert. Wie für die Membran ist die Annahme iii) gültig, keine Bewegung in die Richtung $ \hat{x} $. Um dies zu erfüllen muss der Punkt $ P_1 $ gleich stark in Richtung $ -\hat{x} $ gezogen werden wie der Punkt $ P_2 $ in Richtung $ \hat{x} $ gezogen wird. Ist $ T_1 $ die Kraft welche mit Winkel $ \alpha $ auf Punkt $ P_1 $ wirkt sowie $ T_2 $ und $ \beta$ das analoge für Punkt $ P_2 $ ist, so können die Kräfte \begin{equation}\label{kreismembran:eq:no_translation} @@ -73,7 +79,7 @@ Die Gleichung wird dadurch zu \end{equation*} Durch die Division mit $ dx $ entsteht \begin{equation*} - \frac{1}{dx} \bigg[\frac{\partial u}{\partial x} \bigg|_{x_0 + dx} - \frac{\partial u}{\partial x} \bigg|_{x_0}\bigg] = \frac{\rho}{T}\frac{\partial^2 u}{\partial t^2}. + \frac{1}{dx} \left[\frac{\partial u}{\partial x} \bigg|_{x_0 + dx} - \frac{\partial u}{\partial x} \bigg|_{x_0}\right] = \frac{\rho}{T}\frac{\partial^2 u}{\partial t^2}. \end{equation*} Auf der linken Seite der Gleichung wird die Differenz der Steigungen durch die Intervalllänge geteilt, in anderen Worten die zweite Ableitung von $ u(x,t) $ nach $ x $ berechnet. Der Term $ \frac{\rho}{T} $ wird durch $ c^2 $ ersetzt, da der Bruch für eine gegebene Membran eine positive Konstante sein muss. diff --git a/buch/papers/kreismembran/teil4.tex b/buch/papers/kreismembran/teil4.tex index b67e9e7..74bb87d 100644 --- a/buch/papers/kreismembran/teil4.tex +++ b/buch/papers/kreismembran/teil4.tex @@ -179,6 +179,8 @@ Die Simulationsresultate von Abbildung \ref{kreismembran:im:simres_unendlich} zeigen deutlich wie die Störung vom Zentrum weg verläuft. Nähert sich die Störung dem Rand, so wird sie immer stärker abgeschwächt. Die Wirkung des Absorber ist an der letzten Figur zu erkennen, in welcher kaum noch Auslenkungen zu sehen sind. +Dieses Verhalten spricht für den Absorber-Ansatz, es soll jedoch erwähnt sein, dass der Übergangsbereich eine sanft ansteigende Dämpfung in das System bringt. +Die DGL \ref{kreismembran:Ausgang_DGL} welche simuliert wird geht jedoch von der Annahme \ref{kreimembran:annahmen} iv) aus, dass die Membran keine Art von Dämpfung erfährt. \section{Schlusswort} Auch wenn ein Physikalisches Verhalten bereits durch Annahmen und Annäherungen deutlich vereinfacht wird, bestehen auch dann noch eine Vielzahl von Lösungsansätzen. -- cgit v1.2.1 From 77966f8e9049697adcebb519e87cc57115578f45 Mon Sep 17 00:00:00 2001 From: daHugen Date: Sat, 6 Aug 2022 15:13:34 +0200 Subject: Changed something in subsection \ref{lambertw:subsection:Anfangsbedingungen} --- buch/papers/lambertw/teil4.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/lambertw/teil4.tex b/buch/papers/lambertw/teil4.tex index 2de3663..ba32696 100644 --- a/buch/papers/lambertw/teil4.tex +++ b/buch/papers/lambertw/teil4.tex @@ -10,7 +10,7 @@ In diesem Abschnitt wird rechnerisch das Beispiel einer Verfolgungskurve mit der \subsection{Anfangsbedingungen definieren und einsetzen \label{lambertw:subsection:Anfangsbedingungen}} -Das zu verfolgende Ziel \(Z\) bewegt sich entlang der \(y\)-Achse mit konstanter Geschwindigkeit \(|\dot{z}| = 1\), beginnend beim Ursprung des kartesischen Koordinatensystems. Der Verfolger \(V\) startet auf einem beliebigen Punkt im ersten Quadranten und bewegt sich auch mit konstanter Geschwindigkeit \(|\dot{v}| = 1\) in Richtung Ziel. Diese Anfangspunkte oder Anfangsbedingungen können wie folgt formuliert werden: +Das zu verfolgende Ziel \(Z\) bewegt sich entlang der \(y\)-Achse mit konstanter Geschwindigkeit \(|\dot{z}| = 1\), beginnend beim Ursprung des kartesischen Koordinatensystems. Der Verfolger \(V\) startet auf einem beliebigen Punkt im ersten Quadranten und bewegt sich auch mit konstanter Geschwindigkeit \(|\dot{v}| = 1\) in Richtung Ziel. Aus diesen Bedingungen ergibt sich den ersten Quadranten als Bewegungsraum für \(V\). Diese Anfangspunkte oder Anfangsbedingungen können wie folgt formuliert werden: \begin{equation} Z = -- cgit v1.2.1 From 9b9ef5890fe375831f015e2a42ca719c102cadd9 Mon Sep 17 00:00:00 2001 From: Kuster Yanik Date: Sat, 6 Aug 2022 15:26:51 +0200 Subject: Removed unused pictures. Applied last changes to teil0 and teil1. --- buch/papers/lambertw/Bilder/Intuition.pdf | Bin 187016 -> 149406 bytes buch/papers/lambertw/Bilder/Strategie.pdf | Bin 151684 -> 148667 bytes buch/papers/lambertw/Bilder/Strategie.py | 11 +- buch/papers/lambertw/Bilder/Strategie.svg | 790 --------------------- .../Bilder/lambertAbstandBauchgef\303\274hl.py" | 10 +- buch/papers/lambertw/Bilder/pursuerDGL.ggb | Bin 36225 -> 0 bytes buch/papers/lambertw/Bilder/pursuerDGL.svg | 1 - buch/papers/lambertw/Bilder/pursuerDGL2.ggb | Bin 21894 -> 0 bytes buch/papers/lambertw/Bilder/pursuerDGL2.pdf | Bin 21894 -> 0 bytes buch/papers/lambertw/Bilder/pursuerDGL2.png | Bin 48606 -> 0 bytes buch/papers/lambertw/Bilder/pursuerDGL2.svg | 1 - buch/papers/lambertw/teil0.tex | 10 +- buch/papers/lambertw/teil1.tex | 48 +- 13 files changed, 40 insertions(+), 831 deletions(-) delete mode 100644 buch/papers/lambertw/Bilder/Strategie.svg delete mode 100644 buch/papers/lambertw/Bilder/pursuerDGL.ggb delete mode 100644 buch/papers/lambertw/Bilder/pursuerDGL.svg delete mode 100644 buch/papers/lambertw/Bilder/pursuerDGL2.ggb delete mode 100644 buch/papers/lambertw/Bilder/pursuerDGL2.pdf delete mode 100644 buch/papers/lambertw/Bilder/pursuerDGL2.png delete mode 100644 buch/papers/lambertw/Bilder/pursuerDGL2.svg (limited to 'buch/papers') diff --git a/buch/papers/lambertw/Bilder/Intuition.pdf b/buch/papers/lambertw/Bilder/Intuition.pdf index 739b02b..964b348 100644 Binary files a/buch/papers/lambertw/Bilder/Intuition.pdf and b/buch/papers/lambertw/Bilder/Intuition.pdf differ diff --git a/buch/papers/lambertw/Bilder/Strategie.pdf b/buch/papers/lambertw/Bilder/Strategie.pdf index b5428f5..42cae0d 100644 Binary files a/buch/papers/lambertw/Bilder/Strategie.pdf and b/buch/papers/lambertw/Bilder/Strategie.pdf differ diff --git a/buch/papers/lambertw/Bilder/Strategie.py b/buch/papers/lambertw/Bilder/Strategie.py index 975e248..f09edfb 100644 --- a/buch/papers/lambertw/Bilder/Strategie.py +++ b/buch/papers/lambertw/Bilder/Strategie.py @@ -9,6 +9,9 @@ import pylatex import numpy as np import matplotlib.pyplot as plt + + + N = np.array([0, 0]) V = np.array([1, 4]) Z = np.array([5, 5]) @@ -34,9 +37,10 @@ ax.quiver(X, Y, U, W, angles='xy', scale_units='xy', scale=1, headwidth=5, headl ax.plot([V[0], (VZ+V)[0]], [V[1], (VZ+V)[1]], 'k--') ax.plot(np.vstack([V, Z])[:, 0], np.vstack([V, Z])[:,1], 'bo', markersize=10) -ax.set_xlabel("x", size=20) -ax.set_ylabel("y", size=20) +ax.tick_params(labelsize=15) +plt.xticks(ticks=range(0, 7)) +plt.yticks(ticks=range(0, 7)) ax.text(2.5, 4.5, "Visierlinie", size=20, rotation=10) plt.rcParams.update({ @@ -48,6 +52,7 @@ plt.rcParams.update({ ax.text(1.6, 4.3, r"$\dot{v}$", size=20) ax.text(0.65, 3.9, r"$V$", size=20, c='b') ax.text(5.15, 4.85, r"$Z$", size=20, c='b') - +ax.set_xlabel(r"$x$", size=20) +ax.set_ylabel(r"$y$", size=20) diff --git a/buch/papers/lambertw/Bilder/Strategie.svg b/buch/papers/lambertw/Bilder/Strategie.svg deleted file mode 100644 index 30f9f22..0000000 --- a/buch/papers/lambertw/Bilder/Strategie.svg +++ /dev/null @@ -1,790 +0,0 @@ - - - - - - - - - 2022-07-29T16:52:06.315252 - image/svg+xml - - - Matplotlib v3.3.2, https://matplotlib.org/ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - diff --git "a/buch/papers/lambertw/Bilder/lambertAbstandBauchgef\303\274hl.py" "b/buch/papers/lambertw/Bilder/lambertAbstandBauchgef\303\274hl.py" index 3a90afa..73b322c 100644 --- "a/buch/papers/lambertw/Bilder/lambertAbstandBauchgef\303\274hl.py" +++ "b/buch/papers/lambertw/Bilder/lambertAbstandBauchgef\303\274hl.py" @@ -39,9 +39,11 @@ plt.plot(0, ymin, 'bo', markersize=10) plt.plot([0, xmin], [ymin, ymin], 'k--') #plt.xlim(-0.1, 1) #plt.ylim(1, 2) -plt.ylabel("y") -plt.xlabel("x") + plt.grid(True) +plt.tick_params(labelsize=15) +#plt.xticks(ticks=range(0, 7)) +#plt.yticks(ticks=range(0, 7)) plt.quiver(xmin, ymin, -0.2, 0, scale=1) plt.text(xmin+0.1, ymin-0.1, "Verfolgungskurve", size=20, rotation=20, color='r') @@ -55,4 +57,6 @@ plt.rcParams.update({ plt.text(xmin-0.11, ymin-0.08, r"$\dot{v}$", size=20) plt.text(xmin-0.02, ymin+0.05, r"$V$", size=20, c='b') -plt.text(0.02, ymin+0.05, r"$Z$", size=20, c='b') \ No newline at end of file +plt.text(0.02, ymin+0.05, r"$Z$", size=20, c='b') +plt.ylabel(r"$y$", size=20) +plt.xlabel(r"$x$", size=20) \ No newline at end of file diff --git a/buch/papers/lambertw/Bilder/pursuerDGL.ggb b/buch/papers/lambertw/Bilder/pursuerDGL.ggb deleted file mode 100644 index 3fb3a78..0000000 Binary files a/buch/papers/lambertw/Bilder/pursuerDGL.ggb and /dev/null differ diff --git a/buch/papers/lambertw/Bilder/pursuerDGL.svg b/buch/papers/lambertw/Bilder/pursuerDGL.svg deleted file mode 100644 index d91e5e1..0000000 --- a/buch/papers/lambertw/Bilder/pursuerDGL.svg +++ /dev/null @@ -1 +0,0 @@ -–0.7–0.7–0.7–0.6–0.6–0.6–0.5–0.5–0.5–0.4–0.4–0.4–0.3–0.3–0.3–0.2–0.2–0.2–0.1–0.1–0.10.10.10.10.20.20.20.30.30.30.40.40.40.50.50.50.60.60.60.70.70.70.80.80.80.90.90.91111.11.11.11.21.21.21.31.31.31.41.41.41.51.51.51.61.61.61.71.71.71.81.81.81.91.91.92222.12.12.12.22.22.22.32.32.32.42.42.42.52.52.52.62.62.62.72.72.72.82.82.82.92.92.93333.13.13.13.23.23.23.33.33.30.10.10.10.20.20.20.30.30.30.40.40.40.50.50.50.60.60.60.70.70.70.80.80.80.90.90.91111.11.11.11.21.21.21.31.31.31.41.41.41.51.51.51.61.61.61.71.71.71.81.81.81.91.91.92222.12.12.12.22.22.22.32.32.32.42.42.4000AOAOAOPOPOPOPAPAPAPPPAAA \ No newline at end of file diff --git a/buch/papers/lambertw/Bilder/pursuerDGL2.ggb b/buch/papers/lambertw/Bilder/pursuerDGL2.ggb deleted file mode 100644 index 3c4500b..0000000 Binary files a/buch/papers/lambertw/Bilder/pursuerDGL2.ggb and /dev/null differ diff --git a/buch/papers/lambertw/Bilder/pursuerDGL2.pdf b/buch/papers/lambertw/Bilder/pursuerDGL2.pdf deleted file mode 100644 index 932d9d9..0000000 Binary files a/buch/papers/lambertw/Bilder/pursuerDGL2.pdf and /dev/null differ diff --git a/buch/papers/lambertw/Bilder/pursuerDGL2.png b/buch/papers/lambertw/Bilder/pursuerDGL2.png deleted file mode 100644 index f41dffe..0000000 Binary files a/buch/papers/lambertw/Bilder/pursuerDGL2.png and /dev/null differ diff --git a/buch/papers/lambertw/Bilder/pursuerDGL2.svg b/buch/papers/lambertw/Bilder/pursuerDGL2.svg deleted file mode 100644 index 0c4a11d..0000000 --- a/buch/papers/lambertw/Bilder/pursuerDGL2.svg +++ /dev/null @@ -1 +0,0 @@ -–0.2–0.20.20.20.40.40.60.60.80.8111.21.21.41.41.61.61.81.8222.22.22.42.42.62.62.82.8333.23.2–0.2–0.20.20.20.40.40.60.60.80.8111.21.21.41.41.61.61.81.8222.22.22.42.42.62.62.82.83300Visierlinie \ No newline at end of file diff --git a/buch/papers/lambertw/teil0.tex b/buch/papers/lambertw/teil0.tex index 6632eca..baee9ea 100644 --- a/buch/papers/lambertw/teil0.tex +++ b/buch/papers/lambertw/teil0.tex @@ -78,17 +78,12 @@ Um den Richtungsvektor zu konstruieren kann der Einheitsvektor parallel zu $z-v$ \begin{equation} \dot{v} = - |\dot{v}|\cdot e_{z-v} -\end{equation} -führt. Dies kann noch ausgeschrieben werden zu -\begin{equation} - \dot{v} + |\dot{v}|\cdot (z-v)^\circ = |\dot{v}|\cdot\frac{z-v}{|z-v|} - \text{.} \label{lambertw:richtungsvektor} \end{equation} -% +führt. Aus dem Verfolgungsproblem ist auch ersichtlich, dass die Punkte $V$ und $Z$ nicht am gleichen Ort starten und so eine Division durch Null ausgeschlossen ist. Wenn die Punkte $V$ und $Z$ trotzdem am gleichen Ort starten, ist die Lösung trivial. @@ -97,6 +92,7 @@ Nun wird die Gleichung mit $\dot{v}$ skalar multipliziert, um das Gleichungssyst \frac{z-v}{|z-v|}\cdot|\dot{v}|\cdot\dot{v} &= |\dot{v}|^2 + \text{,} \end{align} was algebraisch zu \begin{align} diff --git a/buch/papers/lambertw/teil1.tex b/buch/papers/lambertw/teil1.tex index e8eca2c..8c30375 100644 --- a/buch/papers/lambertw/teil1.tex +++ b/buch/papers/lambertw/teil1.tex @@ -11,7 +11,7 @@ Sehr oft kommt es vor, dass bei Verfolgungsproblemen die Frage auftaucht, ob das Wenn zum Beispiel die Geschwindigkeit des Verfolgers kleiner ist als diejenige des Ziels, gibt es Anfangsbedingungen bei denen das Ziel nie erreicht wird. Im Anschluss dieser Frage stellt sich meist die nächste Frage, wie lange es dauert bis das Ziel erreicht wird. Diese beiden Fragen werden in diesem Kapitel behandelt und am Beispiel aus \ref{lambertw:section:teil4} betrachtet. -Das Beispiel wird bei dieser Betrachtung noch etwas erweitert indem alle Punkte auf der gesamtem $xy$-Ebene als Startwerte zugelassen werden. +Das Beispiel wird bei dieser Betrachtung noch etwas erweitert, indem alle Punkte auf der gesamtem $xy$-Ebene als Startwerte zugelassen werden. Nun gilt es zu definieren, wann das Ziel erreicht wird. Da sowohl Ziel und Verfolger als Punkte modelliert wurden, gilt das Ziel als erreicht, wenn die Koordinaten des Verfolgers mit denen des Ziels bei einem diskreten Zeitpunkt $t_1$ übereinstimmen. @@ -34,14 +34,14 @@ Die Parametrisierung von $v(t)$ ist von den Startbedingungen abhängig. Deshalb % \subsection{Anfangsbedingung im ersten Quadranten} % -Wenn der Verfolger im ersten Quadranten startet, dann kann $v(t)$ mit den Gleichungen aus \eqref{lambertw:eqFunkXNachT}, welche +Wenn der Verfolger im ersten Quadranten startet, dann kann $v(t)$ mit den Gleichungen aus \eqref{lambertw:eqFunkXNachT}, welche sind \begin{align} x\left(t\right) &= x_0\cdot\sqrt{\frac{1}{\chi}W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right) \right)} \\ y(t) &= - \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right)\\ + \frac{1}{4}\biggl(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\biggl(\left(\frac{x(t)}{x_0}\right)^2\biggr)-r_0+3y_0\biggr)\\ \chi &= \frac{r_0+y_0}{r_0-y_0}, \quad @@ -51,9 +51,10 @@ Wenn der Verfolger im ersten Quadranten startet, dann kann $v(t)$ mit den Gleich r_0 = \sqrt{x_0^2+y_0^2} - \text{.} + \text{,} \end{align} % +die Verfolgungskurve beschrieben werden. Der Verfolger ist durch \begin{equation} v(t) @@ -76,7 +77,8 @@ Daher wird das Problem in zwei einzelne Teilprobleme zerlegt, wodurch die Beding &= y(t) = - \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right)\text{,} + \frac{1}{4}\biggl(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\biggl(\left(\frac{x(t)}{x_0}\right)^2\biggr)-r_0+3y_0\biggr) + \text{,} \end{align} % welche beide gleichzeitig erfüllt sein müssen, damit das Ziel erreicht wurde. @@ -110,7 +112,7 @@ kann die Bedingung weiter vereinfacht werden zu Da $\chi\neq0$ und die Exponentialfunktion nie null sein kann, ist diese Bedingung unmöglich zu erfüllen. Beim Grenzwert für $t\rightarrow\infty$ geht die Exponentialfunktion gegen null. Dies nützt nicht viel, da unendlich viel Zeit vergehen müsste, damit ein Einholen möglich wäre. -Somit kann nach den gestellten Bedingungen das Ziel nie erreicht werden. +Somit kann unter den gestellten Bedingungen das Ziel nie erreicht werden. % % % @@ -155,7 +157,7 @@ Dies kann veranschaulicht werden anhand 1\text{.} \end{equation} % -Da der $y$-Anteil der Geschwindigkeit des Ziels grösser-gleich der des Verfolgers ist, können die $y$-Koordinaten nie übereinstimmen. +Da der $y$-Anteil der Geschwindigkeit des Ziels mindestens so gross wie die des Verfolgers ist, können die $y$-Koordinaten nie übereinstimmen. % \subsection{Anfangsbedingung auf positiven $y$-Achse} Wenn der Verfolger auf der positiven $y$-Achse startet, befindet er sich direkt auf der Fluchtgeraden des Ziels. @@ -194,8 +196,8 @@ Somit wird das Ziel immer erreicht bei $t_1$, wenn der Verfolger auf der positiv \subsection{Fazit} Durch die Symmetrie der Fluchtkurve an der $y$-Achse führen die Anfangsbedingungen im ersten und zweiten Quadranten zu den gleichen Ergebnissen. Nun ist klar, dass lediglich Anfangspunkte auf der positiven $y$-Achse oder direkt auf dem Ziel dazu führen, dass der Verfolger das Ziel bei $t_1$ einholt. Bei allen anderen Anfangspunkten wird der Verfolger das Ziel nie erreichen. -Dieses Resultat ist aber eher akademischer Natur, weil der Verfolger und das Ziel als Punkt betrachtet wurden. -Wobei aber in Realität nicht von Punkten sondern von Objekten mit einer räumlichen Ausdehnung gesprochen werden kann. + +Dieses Resultat ist aber eher akademischer Natur, weil der Verfolger und das Ziel als Punkt betrachtet wurden, während in der Realität nicht von Punkten sondern von Objekten mit einer räumlichen Ausdehnung gesprochen werden kann. Somit wird in einer nächsten Betrachtung untersucht, ob der Verfolger dem Ziel näher kommt als ein definierter Trefferradius. Falls dies stattfinden sollte, wird dies als Treffer interpretiert. Mathematisch kann dies mit @@ -205,7 +207,7 @@ Mathematisch kann dies mit \end{equation} % beschrieben werden, wobei $a_{\text{min}}$ dem Trefferradius entspricht. -Durch quadrieren verschwindet die Wurzel des Betrages, womit +Durch Quadrieren verschwindet die Wurzel des Betrages, womit % \begin{equation} |v-z|^2e_y\cdot v$. Aus diesem Argument würde folgen, dass beim tiefsten Punkt der Verfolgungskurve im Beispiel den minimalen Abstand befindet. % \begin{figure} \centering - \includegraphics[scale=0.4]{./papers/lambertw/Bilder/Intuition.pdf} + \includegraphics[scale=0.7]{./papers/lambertw/Bilder/Intuition.pdf} \caption{Intuition} \label{lambertw:grafic:intuition} \end{figure} % - Dieses Argument kann leicht überprüft werden, indem lokal alle relevanten benachbarten Punkte betrachtet und das Vorzeichen der Änderung des Abstandes überprüft wird. Dafür wird ein Ausdruck benötigt, der den Abstand und die benachbarten Punkte beschreibt. -Der Richtungsvektor wird allgemein mit dem Winkel $\alpha \in[ 0, 2\pi)$ + +$\dot{v}$ wird allgemein mit dem Winkel $\alpha \in[ 0, 2\pi)$ beschrieben, um alle unmittelbar benachbarten Punkte prüfen zu können. Die Ortsvektoren der Punkte können wiederum mit \begin{align} v @@ -242,7 +244,7 @@ Die Ortsvektoren der Punkte können wiederum mit &= \left(\begin{array}{c} 0 \\ t \end{array}\right) \end{align} -beschrieben werden. Der Verfolger wurde allgemein für jede Richtung $\alpha$ definiert, um alle unmittelbar benachbarten Punkte beschreiben zu können. +beschrieben werden. Da der Abstand \begin{equation} a @@ -251,7 +253,7 @@ Da der Abstand \geq 0 \end{equation} -ist, kann durch quadrieren ohne Informationsverlust die Rechnung vereinfacht werden zu +ist, kann durch Quadrieren ohne Informationsverlust die Rechnung vereinfacht werden zu \begin{equation} a^2 = @@ -331,18 +333,12 @@ Durch algebraische Umwandlung kann die Gleichung in die Form \dot{z}\dot{v}=|\dot{v}|^2 \end{equation} gebracht werden. -Da $|\dot{v}|=|\dot{z}|$ folgt +Wenn für den Winkel zwischen den Richtungsvektoren $\alpha$ und die Eigenschaft $|\dot{z}|=|\dot{v}|$ verwendet wird entsteht \begin{equation} \cos(\alpha)=1 - \text{,} -\end{equation} -wobei $\alpha$ der Winkel zwischen den Richtungsvektoren ist. -Mit $|\dot{z}|=|\dot{v}|=1$ entsteht -\begin{equation} - \cos(\alpha)=1 - \text{,} + \text{.} \end{equation} -woraus folgt, dass nur bei $\alpha=0$, wenn $\alpha \in [0,2\pi)$, ein lokales als auch globales Minimum vorhanden sein kann. +Jetzt ist klar, dass nur bei $\alpha=0$, wenn $\alpha \in [0,2\pi)$, ein lokales als auch globales Minimum vorhanden sein kann. $\alpha=0$ bedeutet, dass $\dot{v}=\dot{z}$ sein muss. Da die Richtungsvektoren bei $t\rightarrow\infty$ immer in die gleiche Richtung zeigen ist dort die Bedingung immer erfüllt. Dies entspricht gerade dem einen Rand von $t$, der andere Rand bei $t=0$ muss auch auf lokales bzw. globales Minimum untersucht werden. -- cgit v1.2.1 From 96ac18247b4b63c31f36971b7b4afeb189fafe85 Mon Sep 17 00:00:00 2001 From: runterer Date: Sat, 6 Aug 2022 16:02:56 +0200 Subject: simple corrections --- buch/papers/zeta/analytic_continuation.tex | 69 +++++++++++++++++++----------- buch/papers/zeta/euler_product.tex | 2 +- buch/papers/zeta/zeta_gamma.tex | 2 +- 3 files changed, 45 insertions(+), 28 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/zeta/analytic_continuation.tex b/buch/papers/zeta/analytic_continuation.tex index 0ccc116..8484b28 100644 --- a/buch/papers/zeta/analytic_continuation.tex +++ b/buch/papers/zeta/analytic_continuation.tex @@ -3,12 +3,12 @@ Die analytische Fortsetzung der Riemannschen Zetafunktion ist äusserst interessant. Sie ermöglicht die Berechnung von $\zeta(-1)$ und weiterer spannender Werte. -So liegen zum Beispiel unendlich viele Nullstellen der Zetafunktion bei $\Re(s) = 0.5$. +So liegen zum Beispiel unendlich viele Nullstellen der Zetafunktion bei $\Re(s) = \frac{1}{2}$. Diese sind relevant für die Primzahlverteilung und sind Gegenstand der Riemannschen Vermutung. Es werden zwei verschiedene Fortsetzungen benötigt. Die erste erweitert die Zetafunktion auf $\Re(s) > 0$. -Die zweite verwendet eine Spiegelung an der $\Re(s) = 0.5$ Linie und erschliesst damit die ganze komplexe Ebene. +Die zweite verwendet eine Spiegelung an der $\Re(s) = \frac{1}{2}$ Geraden und erschliesst damit die ganze komplexe Ebene. Eine grafische Darstellung dieses Plans ist in Abbildung \ref{zeta:fig:continuation_overview} zu sehen. \begin{figure} \centering @@ -23,7 +23,7 @@ Eine grafische Darstellung dieses Plans ist in Abbildung \ref{zeta:fig:continuat \end{figure} \subsection{Fortsetzung auf $\Re(s) > 0$} \label{zeta:subsection:auf_bereich_ge_0} -Zuerst definieren die Dirichletsche Etafunktion als +Zuerst definieren wir die Dirichletsche Etafunktion als \begin{equation}\label{zeta:equation:eta} \eta(s) = @@ -36,26 +36,40 @@ Diese Etafunktion konvergiert gemäss dem Leibnitz-Kriterium im Bereich $\Re(s) Wenn wir es nun schaffen, die sehr ähnliche Zetafunktion durch die Etafunktion auszudrücken, dann haben die gesuchte Fortsetzung. Zuerst wiederholen wir zweimal die Definition der Zetafunktion \eqref{zeta:equation1}, wobei wir sie einmal durch $2^{s-1}$ teilen \begin{align} - \zeta(s) + \color{red} + \zeta(s) &= \sum_{n=1}^{\infty} - \frac{1}{n^s} \label{zeta:align1} + \color{red} + \frac{1}{n^s} \label{zeta:align1} \\ - \frac{1}{2^{s-1}} - \zeta(s) + \color{blue} + \frac{1}{2^{s-1}} + \zeta(s) &= \sum_{n=1}^{\infty} - \frac{2}{(2n)^s}. \label{zeta:align2} + \color{blue} + \frac{2}{(2n)^s}. \label{zeta:align2} \end{align} Durch Subtraktion der beiden Gleichungen \eqref{zeta:align1} minus \eqref{zeta:align2}, ergibt sich \begin{align} - \left(1 - \frac{1}{2^{s-1}} \right) + \left({\color{red}1} - {\color{blue}\frac{1}{2^{s-1}}} \right) \zeta(s) &= - \frac{1}{1^s} - \underbrace{-\frac{2}{2^s} + \frac{1}{2^s}}_{-\frac{1}{2^s}} - + \frac{1}{3^s} - \underbrace{-\frac{2}{4^s} + \frac{1}{4^s}}_{-\frac{1}{4^s}} + {\color{red}\frac{1}{1^s}} + \underbrace{ + - + {\color{blue}\frac{2}{2^s}} + + + {\color{red}\frac{1}{2^s}} + }_{-\frac{1}{2^s}} + + + {\color{red}\frac{1}{3^s}} + \underbrace{- + {\color{blue}\frac{2}{4^s}} + + + {\color{red}\frac{1}{4^s}} + }_{-\frac{1}{4^s}} \ldots \\ &= \eta(s). @@ -87,14 +101,15 @@ Nun substituieren wir $t$ mit $t = \pi n^2 x$ und $dt=\pi n^2 dx$ und erhalten \end{equation} Analog zum Abschnitt \ref{zeta:section:zusammenhang_mit_gammafunktion} teilen wir durch $(\pi n^2)^{\frac{s}{2}}$ \begin{equation} - \frac{\Gamma \left( \frac{s}{2} \right)}{\pi^{\frac{s}{2}} n^s} + \frac{\Gamma \left( \frac{s}{2} \right)}{\pi^{\frac{s}{2}}} + \frac{1}{n^s} = \int_0^{\infty} x^{\frac{s}{2}-1} e^{-\pi n^2 x} \,dx, \end{equation} -und finden Zeta durch die Summenbildung $\sum_{n=1}^{\infty}$ +und finden $\zeta(s)$ durch die Summenbildung $\sum_{n=1}^{\infty}$ \begin{equation} \frac{\Gamma \left( \frac{s}{2} \right)}{\pi^{\frac{s}{2}}} \zeta(s) @@ -137,13 +152,13 @@ wobei wir die Summen so verändern müssen, dass sie bei $n=1$ beginnen und wir 1 &= \frac{1}{\sqrt{x}} - \left( + \Biggl( 2 \sum_{n=1}^{\infty} e^{\frac{-n^2 \pi}{x}} + 1 - \right) + \Biggr) \\ 2 \psi(x) @@ -189,7 +204,7 @@ Zunächst teilen wir nun das Integral aus \eqref{zeta:equation:integral1} auf al = I_1 + I_2, \end{equation} -wobei wir uns nun auf den ersten Teil $I_1$ konzentrieren werden. +wobei wir uns zunächst auf den ersten Teil $I_1$ konzentrieren werden. Dabei setzen wir die Definition von $\psi(x)$ aus \eqref{zeta:equation:psi} ein und erhalten \begin{align} I_1 @@ -201,11 +216,11 @@ Dabei setzen wir die Definition von $\psi(x)$ aus \eqref{zeta:equation:psi} ein &= \int_0^{1} x^{\frac{s}{2}-1} - \left( + \Biggl( - \frac{1}{2} + \frac{\psi\left(\frac{1}{x} \right)}{\sqrt{x}} + \frac{1}{2 \sqrt{x}} - \right) + \Biggr) \,dx \\ &= @@ -237,7 +252,7 @@ Dabei setzen wir die Definition von $\psi(x)$ aus \eqref{zeta:equation:psi} ein \,dx }_{I_4}. \label{zeta:equation:integral3} \end{align} -Dabei kann das zweite Integral $I_4$ gelöst werden als +Darin kann das zweite Integral $I_4$ gelöst werden als \begin{equation} I_4 = @@ -278,8 +293,8 @@ Dies ergibt \,dx, \end{align} wobei wir durch Multiplikation mit $(-1)$ die Integralgrenzen tauschen dürfen. -Es ist zu beachten das diese Grenzen nun identisch mit den Grenzen des zweiten Integrals von \eqref{zeta:equation:integral2} sind. -Wir setzen beide Lösungen ein in Gleichung \eqref{zeta:equation:integral3} und erhalten +Es ist zu beachten das diese Grenzen nun identisch mit den Grenzen des zweiten Integrals $I_2$ von \eqref{zeta:equation:integral2} sind. +Wir setzen beide Lösungen in Gleichung \eqref{zeta:equation:integral3} ein und erhalten \begin{equation} I_1 = @@ -356,17 +371,19 @@ Somit haben wir die analytische Fortsetzung gefunden als \zeta(s) = \frac{\Gamma \left( \frac{1-s}{2} \right)}{\pi^{\frac{1-s}{2}}} - \zeta(1-s). + \zeta(1-s), \end{equation} +was einer Spiegelung an der $\Re(s) = \frac{1}{2}$ Geraden entspricht. +Eine ganz ähnliche Spiegelungseigenschaft wurde bereits in Kapitel \ref{buch:funktionentheorie:subsection:gammareflektion} für die Gammafunktion gefunden. %TODO Definitionen und Gleichungen klarer unterscheiden \subsection{Poissonsche Summenformel} \label{zeta:subsec:poisson_summation} -Der Beweis für Gleichung \ref{zeta:equation:psi} folgt direkt durch die poissonsche Summenformel. +Der Beweis für Gleichung \eqref{zeta:equation:psi} folgt direkt durch die poissonsche Summenformel. Um diese zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta Funktion. \begin{lemma} - Die Fourierreihe der periodischen Dirac Delta Funktion $\sum \delta(x - 2\pi k)$ ist + Die Fourierreihe der periodischen Dirac $\delta$ Funktion $\sum \delta(x - 2\pi k)$ ist \begin{equation} \label{zeta:equation:fourier_dirac} \sum_{k=-\infty}^{\infty} \delta(x - 2\pi k) diff --git a/buch/papers/zeta/euler_product.tex b/buch/papers/zeta/euler_product.tex index a6ed512..5f4f5ca 100644 --- a/buch/papers/zeta/euler_product.tex +++ b/buch/papers/zeta/euler_product.tex @@ -64,7 +64,7 @@ Die Verteilung der Primzahlen ist Gegenstand der Riemannschen Vermutung, welche \begin{equation} n = \prod_i p_i^{k_i} \quad \forall \quad n \in \mathbb{N}. \end{equation} - Jeder Summand der Summen in \eqref{zeta:equation:eulerprodukt2} ist somit eine Zahl $n$. + Jeder Summand der Summen in \eqref{zeta:equation:eulerprodukt2} ist somit der Kehrwert genau einer natürlichen Zahl $n \in \mathbb{N}$. Da die Summen alle möglichen Kombinationen von Exponenten und Primzahlen in \eqref{zeta:equation:eulerprodukt2} enthält haben wir \begin{equation} \sum_{k_1=0}^{\infty} diff --git a/buch/papers/zeta/zeta_gamma.tex b/buch/papers/zeta/zeta_gamma.tex index db41676..1f10a33 100644 --- a/buch/papers/zeta/zeta_gamma.tex +++ b/buch/papers/zeta/zeta_gamma.tex @@ -19,7 +19,7 @@ Durch die Substitution von $t$ mit $t = nu$ und $dt = n\,du$ wird daraus &= \int_0^{\infty} n^s u^{s-1} e^{-nu} \,du. \end{align*} -Durch Division mit durch $n^s$ ergibt sich die Quotienten +Durch Division durch $n^s$ ergeben sich die Quotienten \begin{equation*} \frac{\Gamma(s)}{n^s} = -- cgit v1.2.1 From 79c0198f5082851ce28945e8278ab01b82496901 Mon Sep 17 00:00:00 2001 From: runterer Date: Sat, 6 Aug 2022 19:21:47 +0200 Subject: restructured 19.4.2 --- buch/papers/zeta/analytic_continuation.tex | 364 ++++++++++++++++------------- 1 file changed, 203 insertions(+), 161 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/zeta/analytic_continuation.tex b/buch/papers/zeta/analytic_continuation.tex index 8484b28..a45791e 100644 --- a/buch/papers/zeta/analytic_continuation.tex +++ b/buch/papers/zeta/analytic_continuation.tex @@ -110,78 +110,24 @@ Analog zum Abschnitt \ref{zeta:section:zusammenhang_mit_gammafunktion} teilen wi \,dx, \end{equation} und finden $\zeta(s)$ durch die Summenbildung $\sum_{n=1}^{\infty}$ -\begin{equation} +\begin{align} \frac{\Gamma \left( \frac{s}{2} \right)}{\pi^{\frac{s}{2}}} \zeta(s) - = + &= \int_0^{\infty} x^{\frac{s}{2}-1} \sum_{n=1}^{\infty} e^{-\pi n^2 x} - \,dx. \label{zeta:equation:integral1} -\end{equation} -Die Summe kürzen wir ab als $\psi(x) = \sum_{n=1}^{\infty} e^{-\pi n^2 x}$. -Im Abschnitt \ref{zeta:subsec:poisson_summation} wird die poissonsche Summenformel $\sum f(n) = \sum F(n)$ bewiesen. -In unserem Problem ist $f(n) = e^{-\pi n^2 x}$ und die zugehörige Fouriertransformation $F(n)$ ist -\begin{equation} - F(n) - = - \mathcal{F} - ( - e^{-\pi n^2 x} - ) - = - \frac{1}{\sqrt{x}} - e^{\frac{-n^2 \pi}{x}}. -\end{equation} -Dadurch ergibt sich -\begin{equation}\label{zeta:equation:psi} - \sum_{n=-\infty}^{\infty} - e^{-\pi n^2 x} - = - \frac{1}{\sqrt{x}} - \sum_{n=-\infty}^{\infty} - e^{\frac{-n^2 \pi}{x}}, -\end{equation} -wobei wir die Summen so verändern müssen, dass sie bei $n=1$ beginnen und wir $\psi(x)$ erhalten als -\begin{align} - 2 - \sum_{n=1}^{\infty} - e^{-\pi n^2 x} - + - 1 - &= - \frac{1}{\sqrt{x}} - \Biggl( - 2 - \sum_{n=1}^{\infty} - e^{\frac{-n^2 \pi}{x}} - + - 1 - \Biggr) + \,dx\label{zeta:equation:integral1} \\ - 2 - \psi(x) - + - 1 &= - \frac{1}{\sqrt{x}} - \left( - 2 - \psi\left(\frac{1}{x}\right) - + - 1 - \right) - \\ + \int_0^{\infty} + x^{\frac{s}{2}-1} \psi(x) - &= - - \frac{1}{2} - + \frac{\psi\left(\frac{1}{x} \right)}{\sqrt{x}} - + \frac{1}{2 \sqrt{x}}.\label{zeta:equation:psi} + \,dx, \end{align} -Diese Gleichung wird später wichtig werden. - -Zunächst teilen wir nun das Integral aus \eqref{zeta:equation:integral1} auf als +wobei die Summe $\sum_{n=1}^{\infty} e^{-\pi n^2 x}$ als $\psi(x)$ abgekürzt wird. +Zunächst teilen wir nun das Integral auf in zwei Teile \begin{equation}\label{zeta:equation:integral2} \int_0^{\infty} x^{\frac{s}{2}-1} @@ -202,100 +148,11 @@ Zunächst teilen wir nun das Integral aus \eqref{zeta:equation:integral1} auf al \,dx }_{I_2} = - I_1 + I_2, -\end{equation} -wobei wir uns zunächst auf den ersten Teil $I_1$ konzentrieren werden. -Dabei setzen wir die Definition von $\psi(x)$ aus \eqref{zeta:equation:psi} ein und erhalten -\begin{align} - I_1 - = - \int_0^{1} - x^{\frac{s}{2}-1} - \psi(x) - \,dx - &= - \int_0^{1} - x^{\frac{s}{2}-1} - \Biggl( - - \frac{1}{2} - + \frac{\psi\left(\frac{1}{x} \right)}{\sqrt{x}} - + \frac{1}{2 \sqrt{x}} - \Biggr) - \,dx - \\ - &= - \int_0^{1} - x^{\frac{s}{2}-\frac{3}{2}} - \psi \left( \frac{1}{x} \right) - + \frac{1}{2} - \biggl( - x^{\frac{s}{2}-\frac{3}{2}} - - - x^{\frac{s}{2}-1} - \biggl) - \,dx - \\ - &= - \underbrace{ - \int_0^{1} - x^{\frac{s}{2}-\frac{3}{2}} - \psi \left( \frac{1}{x} \right) - \,dx - }_{I_3} - + - \underbrace{ - \frac{1}{2} - \int_0^1 - x^{\frac{s}{2}-\frac{3}{2}} - - - x^{\frac{s}{2}-1} - \,dx - }_{I_4}. \label{zeta:equation:integral3} -\end{align} -Darin kann das zweite Integral $I_4$ gelöst werden als -\begin{equation} - I_4 - = - \frac{1}{2} - \int_0^1 - x^{\frac{s}{2}-\frac{3}{2}} - - - x^{\frac{s}{2}-1} - \,dx - = - \frac{1}{s(s-1)}. + I_1 + I_2. \end{equation} -Das erste Integral $I_3$ aus \eqref{zeta:equation:integral3} mit $\psi \left(\frac{1}{x} \right)$ ist nicht lösbar in dieser Form. -Deshalb substituieren wir $x = \frac{1}{u}$ und $dx = -\frac{1}{u^2}du$. -Die untere Integralgrenze wechselt ebenfalls zu $x_0 = 0 \rightarrow u_0 = \infty$. -Dies ergibt -\begin{align} - I_3 - = - \int_{\infty}^{1} - \left( - \frac{1}{u} - \right)^{\frac{s}{2}-\frac{3}{2}} - \psi(u) - \frac{-du}{u^2} - &= - \int_{1}^{\infty} - \left( - \frac{1}{u} - \right)^{\frac{s}{2}-\frac{3}{2}} - \psi(u) - \frac{du}{u^2} - \\ - &= - \int_{1}^{\infty} - x^{(-1) \left(\frac{s}{2}+\frac{1}{2}\right)} - \psi(x) - \,dx, -\end{align} -wobei wir durch Multiplikation mit $(-1)$ die Integralgrenzen tauschen dürfen. -Es ist zu beachten das diese Grenzen nun identisch mit den Grenzen des zweiten Integrals $I_2$ von \eqref{zeta:equation:integral2} sind. -Wir setzen beide Lösungen in Gleichung \eqref{zeta:equation:integral3} ein und erhalten -\begin{equation} +Abschnitt \ref{zeta:subsubsec:intcal} beschreibt wie das Integral $I_1$ umgestellt werden kann um ebenfalls die Integrationsgrenzen $1$ und $\infty$ zu bekommen. +Die Lösung, beschrieben in Gleichung \eqref{zeta:equation:intcal_res}, lautet +\begin{equation*} I_1 = \int_0^{1} @@ -309,8 +166,8 @@ Wir setzen beide Lösungen in Gleichung \eqref{zeta:equation:integral3} ein und \,dx + \frac{1}{s(s-1)}. -\end{equation} -Dieses Resultat setzen wir wiederum ein in \eqref{zeta:equation:integral2}, um schlussendlich +\end{equation*} +Dieses Resultat setzen wir nun ein in \eqref{zeta:equation:integral2}, um schlussendlich \begin{align} \frac{\Gamma \left( \frac{s}{2} \right)}{\pi^{\frac{s}{2}}} \zeta(s) @@ -375,12 +232,14 @@ Somit haben wir die analytische Fortsetzung gefunden als \end{equation} was einer Spiegelung an der $\Re(s) = \frac{1}{2}$ Geraden entspricht. Eine ganz ähnliche Spiegelungseigenschaft wurde bereits in Kapitel \ref{buch:funktionentheorie:subsection:gammareflektion} für die Gammafunktion gefunden. -%TODO Definitionen und Gleichungen klarer unterscheiden -\subsection{Poissonsche Summenformel} \label{zeta:subsec:poisson_summation} +\subsection{Berechnung des Integrals $I_1 = \int_0^{1} x^{\frac{s}{2}-1} \psi(x) \,dx$} \label{zeta:subsubsec:intcal} -Der Beweis für Gleichung \eqref{zeta:equation:psi} folgt direkt durch die poissonsche Summenformel. -Um diese zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta Funktion. +Ziel dieses Abschnittes ist es, zu zeigen wie das Integral $I_1$ aus Gleichung \eqref{zeta:equation:integral2} durch ein neues Integral mit den Integrationsgrenzen $1$ und $\infty$ ersetzt werden kann. +Da dieser Schritt ziemlich aufwendig ist, wird er hier in einem eigenen Abschnitt behandelt. +Zunächst wird die poissonsche Summenformel hergeleitet, da diese verwendet werden kann um $\psi(x)$ zu berechnen. + +Um die poissonsche Summenformel zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta Funktion. \begin{lemma} Die Fourierreihe der periodischen Dirac $\delta$ Funktion $\sum \delta(x - 2\pi k)$ ist @@ -492,3 +351,186 @@ Um diese zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta F f(k). \end{equation} \end{proof} + +Erinnern wir uns nochmals an unser Integral aus Gleichung \eqref{zeta:equation:integral2} +\begin{align*} + I_1 + &= + \int_0^{1} + x^{\frac{s}{2}-1} + \sum_{n=1}^{\infty} + e^{-\pi n^2 x} + \,dx + \\ + &= + \int_0^{1} + x^{\frac{s}{2}-1} + \psi(x) + \,dx + . +\end{align*} + +Wir wenden nun diese poissonsche Summenformel $\sum f(n) = \sum F(n)$ an auf $\psi(x)$. +In unserem Problem ist also $f(n) = e^{-\pi n^2 x}$ und die zugehörige Fouriertransformation $F(n)$ ist +\begin{equation} + F(n) + = + \mathcal{F} + ( + e^{-\pi n^2 x} + ) + = + \frac{1}{\sqrt{x}} + e^{\frac{-n^2 \pi}{x}}. +\end{equation} +Dadurch ergibt sich +\begin{equation}\label{zeta:equation:psi} + \sum_{n=-\infty}^{\infty} + e^{-\pi n^2 x} + = + \frac{1}{\sqrt{x}} + \sum_{n=-\infty}^{\infty} + e^{\frac{-n^2 \pi}{x}}, +\end{equation} +wobei wir die Summen so verändern müssen, dass sie bei $n=1$ beginnen und wir $\psi(x)$ erhalten als +\begin{align} + 2 + \sum_{n=1}^{\infty} + e^{-\pi n^2 x} + + + 1 + &= + \frac{1}{\sqrt{x}} + \Biggl( + 2 + \sum_{n=1}^{\infty} + e^{\frac{-n^2 \pi}{x}} + + + 1 + \Biggr) + \\ + 2 + \psi(x) + + + 1 + &= + \frac{1}{\sqrt{x}} + \left( + 2 + \psi\left(\frac{1}{x}\right) + + + 1 + \right) + \\ + \psi(x) + &= + - \frac{1}{2} + + \frac{\psi\left(\frac{1}{x} \right)}{\sqrt{x}} + + \frac{1}{2 \sqrt{x}}.\label{zeta:equation:psi} +\end{align} +Diese Form von $\psi(x)$ eingesetzt in $I_1$ ergibt +\begin{align} + I_1 + = + \int_0^{1} + x^{\frac{s}{2}-1} + \psi(x) + \,dx + &= + \int_0^{1} + x^{\frac{s}{2}-1} + \Biggl( + - \frac{1}{2} + + \frac{\psi\left(\frac{1}{x} \right)}{\sqrt{x}} + + \frac{1}{2 \sqrt{x}} + \Biggr) + \,dx + \\ + &= + \int_0^{1} + x^{\frac{s}{2}-\frac{3}{2}} + \psi \left( \frac{1}{x} \right) + + \frac{1}{2} + \biggl( + x^{\frac{s}{2}-\frac{3}{2}} + - + x^{\frac{s}{2}-1} + \biggl) + \,dx + \\ + &= + \underbrace{ + \int_0^{1} + x^{\frac{s}{2}-\frac{3}{2}} + \psi \left( \frac{1}{x} \right) + \,dx + }_{I_3} + + + \underbrace{ + \frac{1}{2} + \int_0^1 + x^{\frac{s}{2}-\frac{3}{2}} + - + x^{\frac{s}{2}-1} + \,dx + }_{I_4}. \label{zeta:equation:integral3} +\end{align} +Darin kann für das zweite Integral $I_4$ eine Lösung gefunden werden als +\begin{equation} + I_4 + = + \frac{1}{2} + \int_0^1 + x^{\frac{s}{2}-\frac{3}{2}} + - + x^{\frac{s}{2}-1} + \,dx + = + \frac{1}{s(s-1)}. +\end{equation} +Das erste Integral $I_3$ aus \eqref{zeta:equation:integral3} mit $\psi \left(\frac{1}{x} \right)$ ist hingegen nicht lösbar in dieser Form. +Deshalb substituieren wir $x = \frac{1}{u}$ und $dx = -\frac{1}{u^2}du$. +Die untere Integralgrenze wechselt ebenfalls zu $x_0 = 0 \rightarrow u_0 = \infty$. +Dies ergibt +\begin{align} + I_3 + = + \int_{\infty}^{1} + \left( + \frac{1}{u} + \right)^{\frac{s}{2}-\frac{3}{2}} + \psi(u) + \frac{-du}{u^2} + &= + \int_{1}^{\infty} + \left( + \frac{1}{u} + \right)^{\frac{s}{2}-\frac{3}{2}} + \psi(u) + \frac{du}{u^2} + \\ + &= + \int_{1}^{\infty} + x^{(-1) \left(\frac{s}{2}+\frac{1}{2}\right)} + \psi(x) + \,dx, +\end{align} +wobei wir durch Multiplikation mit $(-1)$ die Integralgrenzen tauschen dürfen. +Es ist zu beachten das diese Grenzen nun identisch mit den Grenzen des zweiten Integrals $I_2$ von \eqref{zeta:equation:integral2} sind. +Wir setzen beide Lösungen in Gleichung \eqref{zeta:equation:integral3} ein und erhalten +\begin{equation} + I_1 + = + \int_0^{1} + x^{\frac{s}{2}-1} + \psi(x) + \,dx + = + \int_{1}^{\infty} + x^{(-1) \left(\frac{s}{2}+\frac{1}{2}\right)} + \psi(x) + \,dx + + + \frac{1}{s(s-1)}. \label{zeta:equation:intcal_res} +\end{equation} +Diese Form des Integrals $I_1$ hat die gewünschten Integrationsgrenzen und ein essentieller Bestandteil des Beweises der Funktionalgleichung in Abschnitt \ref{zeta:subsection:auf_ganz}. -- cgit v1.2.1 From 77dfbc3727334b88dcf19c673d9ef9812df1806a Mon Sep 17 00:00:00 2001 From: runterer Date: Sun, 7 Aug 2022 17:30:30 +0200 Subject: wip conlcusion not finished --- buch/papers/zeta/analytic_continuation.tex | 11 +- buch/papers/zeta/continuation_overview.tikz.tex | 18 - buch/papers/zeta/einleitung.tex | 32 +- buch/papers/zeta/euler_product.tex | 11 +- buch/papers/zeta/fazit.tex | 28 + .../zeta/images/continuation_overview.tikz.tex | 18 + buch/papers/zeta/images/primzahlfunktion.pgf | 505 ++++++++ buch/papers/zeta/images/primzahlfunktion_paper.pgf | 505 ++++++++ buch/papers/zeta/images/youtube_screenshot.png | Bin 0 -> 378662 bytes buch/papers/zeta/images/zeta_re_-1_plot.pgf | 1147 ++++++++++++++++++ buch/papers/zeta/images/zeta_re_0.5_plot.pgf | 1206 +++++++++++++++++++ buch/papers/zeta/images/zeta_re_0_plot.pgf | 1242 ++++++++++++++++++++ buch/papers/zeta/main.tex | 2 +- buch/papers/zeta/presentation/presentation.tex | 12 +- .../zeta/presentation/youtube_screenshot.png | Bin 378662 -> 0 bytes buch/papers/zeta/primzahlfunktion.pgf | 505 -------- buch/papers/zeta/references.bib | 57 +- buch/papers/zeta/zeta_color_plot-img0.png | Bin 0 -> 37362 bytes buch/papers/zeta/zeta_color_plot.pgf | 402 +++++++ buch/papers/zeta/zeta_re_-1_plot.pgf | 1147 ------------------ buch/papers/zeta/zeta_re_0.5_plot.pgf | 1206 ------------------- buch/papers/zeta/zeta_re_0_plot.pgf | 1242 -------------------- 22 files changed, 5137 insertions(+), 4159 deletions(-) delete mode 100644 buch/papers/zeta/continuation_overview.tikz.tex create mode 100644 buch/papers/zeta/fazit.tex create mode 100644 buch/papers/zeta/images/continuation_overview.tikz.tex create mode 100644 buch/papers/zeta/images/primzahlfunktion.pgf create mode 100644 buch/papers/zeta/images/primzahlfunktion_paper.pgf create mode 100644 buch/papers/zeta/images/youtube_screenshot.png create mode 100644 buch/papers/zeta/images/zeta_re_-1_plot.pgf create mode 100644 buch/papers/zeta/images/zeta_re_0.5_plot.pgf create mode 100644 buch/papers/zeta/images/zeta_re_0_plot.pgf delete mode 100644 buch/papers/zeta/presentation/youtube_screenshot.png delete mode 100644 buch/papers/zeta/primzahlfunktion.pgf create mode 100644 buch/papers/zeta/zeta_color_plot-img0.png create mode 100644 buch/papers/zeta/zeta_color_plot.pgf delete mode 100644 buch/papers/zeta/zeta_re_-1_plot.pgf delete mode 100644 buch/papers/zeta/zeta_re_0.5_plot.pgf delete mode 100644 buch/papers/zeta/zeta_re_0_plot.pgf (limited to 'buch/papers') diff --git a/buch/papers/zeta/analytic_continuation.tex b/buch/papers/zeta/analytic_continuation.tex index a45791e..4046bb7 100644 --- a/buch/papers/zeta/analytic_continuation.tex +++ b/buch/papers/zeta/analytic_continuation.tex @@ -4,7 +4,7 @@ Die analytische Fortsetzung der Riemannschen Zetafunktion ist äusserst interessant. Sie ermöglicht die Berechnung von $\zeta(-1)$ und weiterer spannender Werte. So liegen zum Beispiel unendlich viele Nullstellen der Zetafunktion bei $\Re(s) = \frac{1}{2}$. -Diese sind relevant für die Primzahlverteilung und sind Gegenstand der Riemannschen Vermutung. +Wie bereits erwähnt sind diese Gegenstand der Riemannschen Vermutung. Es werden zwei verschiedene Fortsetzungen benötigt. Die erste erweitert die Zetafunktion auf $\Re(s) > 0$. @@ -12,7 +12,7 @@ Die zweite verwendet eine Spiegelung an der $\Re(s) = \frac{1}{2}$ Geraden und e Eine grafische Darstellung dieses Plans ist in Abbildung \ref{zeta:fig:continuation_overview} zu sehen. \begin{figure} \centering - \input{papers/zeta/continuation_overview.tikz.tex} + \input{papers/zeta/images/continuation_overview.tikz.tex} \caption{ Die verschiedenen Abschnitte der Riemannschen Zetafunktion. Die originale Definition von \eqref{zeta:equation1} ist im grünen Bereich gültig. @@ -237,7 +237,7 @@ Eine ganz ähnliche Spiegelungseigenschaft wurde bereits in Kapitel \ref{buch:fu Ziel dieses Abschnittes ist es, zu zeigen wie das Integral $I_1$ aus Gleichung \eqref{zeta:equation:integral2} durch ein neues Integral mit den Integrationsgrenzen $1$ und $\infty$ ersetzt werden kann. Da dieser Schritt ziemlich aufwendig ist, wird er hier in einem eigenen Abschnitt behandelt. -Zunächst wird die poissonsche Summenformel hergeleitet, da diese verwendet werden kann um $\psi(x)$ zu berechnen. +Zunächst wird die poissonsche Summenformel hergeleitet \cite{zeta:online:poisson}, da diese verwendet werden kann um $\psi(x)$ zu berechnen. Um die poissonsche Summenformel zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta Funktion. @@ -330,7 +330,7 @@ Um die poissonsche Summenformel zu beweisen, berechnen wir zunächst die Fourier \sum_{k=-\infty}^{\infty} \delta(x + k). \end{align} - Wenn wir dies einsetzen und erhalten wir den gesuchten Beweis für die poissonsche Summenformel + Wenn wir dies einsetzen und erhalten wir \begin{equation} \sum_{k=-\infty}^{\infty} F(k) @@ -348,8 +348,9 @@ Um die poissonsche Summenformel zu beweisen, berechnen wir zunächst die Fourier \, dx = \sum_{k=-\infty}^{\infty} - f(k). + f(k), \end{equation} + was der gesuchte Beweis für die poissonsche Summenformel ist. \end{proof} Erinnern wir uns nochmals an unser Integral aus Gleichung \eqref{zeta:equation:integral2} diff --git a/buch/papers/zeta/continuation_overview.tikz.tex b/buch/papers/zeta/continuation_overview.tikz.tex deleted file mode 100644 index 836ab1d..0000000 --- a/buch/papers/zeta/continuation_overview.tikz.tex +++ /dev/null @@ -1,18 +0,0 @@ -\begin{tikzpicture}[>=stealth', auto, node distance=0.9cm, scale=2, - dot/.style={fill, circle, inner sep=0, minimum size=0.1cm}] - - \draw[->] (-2,0) -- (-1,0) node[dot]{} node[anchor=north]{$-1$} -- (0,0) node[anchor=north west]{$0$} -- (0.5,0) node[anchor=north west]{$0.5$}-- (1,0) node[anchor=north west]{$1$} -- (2,0) node[anchor=west]{$\Re(s)$}; - - \draw[->] (0,-1.2) -- (0,1.2) node[anchor=south]{$\Im(s)$}; - \begin{scope}[yscale=0.1] - \draw[] (1,-1) -- (1,1); - \end{scope} - \draw[dotted] (0.5,-1) -- (0.5,1); - - \begin{scope}[] - \fill[opacity=0.2, red] (-1.8,1) rectangle (0, -1); - \fill[opacity=0.2, blue] (0,1) rectangle (1, -1); - \fill[opacity=0.2, green] (1,1) rectangle (1.8, -1); - \end{scope} - -\end{tikzpicture} diff --git a/buch/papers/zeta/einleitung.tex b/buch/papers/zeta/einleitung.tex index 3b70531..ad87fec 100644 --- a/buch/papers/zeta/einleitung.tex +++ b/buch/papers/zeta/einleitung.tex @@ -1,11 +1,41 @@ \section{Einleitung} \label{zeta:section:einleitung} \rhead{Einleitung} -Die Riemannsche Zetafunktion ist für alle komplexe $s$ mit $\Re(s) > 1$ definiert als +Die Riemannsche Zetafunktion $\zeta(s)$ ist für alle komplexe $s$ mit $\Re(s) > 1$ definiert als \begin{equation}\label{zeta:equation1} \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}. \end{equation} +Die Zetafunktion ist bekannt als Bestandteil der Riemannschen Vermutung, welche besagt das alle nichttrivialen Nullstellen der Zetafunktion einen Realteil von $\frac{1}{2}$ haben. +Mithilfe dieser Vermutung kann eine gute Annäherung an die Primzahlfunktion gefunden werden. +Die Primzahlfunktion steigt immer an, sobald eine Primzahl vorkommt. +Eine Darstellung davon ist in Abbildung \ref{fig:zeta:primzahlfunktion} zu finden. +Die Riemannsche Vermutung ist eines der ungelösten Millennium-Probleme der Mathematik, auf deren Lösung eine Belohnung von einer Million Doller ausgesetzt ist \cite{zeta:online:millennium}. +Auf eine genauere Beschreibung der Riemannschen Vermutung wird im Rahmen dieses Papers nicht eingegangen. +\begin{figure} + \centering + \input{papers/zeta/images/primzahlfunktion_paper.pgf} + \caption{Die Primzahlfunktion von $0$ bis $30$.} + \label{fig:zeta:primzahlfunktion} +\end{figure} +Der grundlegende Zusammenhang der Primzahlen und der Zetafunktion wird im ersten Abschnitt \ref{zeta:section:eulerprodukt} über das Eulerprodukt gezeigt. +Danach folgt die Verbindung zur bereits bekannten Gammafunktion in Abschnitt \ref{zeta:section:zusammenhang_mit_gammafunktion}. +Schlussendlich folgt die Beschreibung der analytischen Fortsetzung die gesamte komplexe Ebene in Abschnitt \ref{zeta:section:analytische_fortsetzung}. + +Diese analytische Fortsetzung wird für die Riemannsche Vermutung benötigt, ermöglicht aber auch andere interessante Aussagen. +So findet sich zum Beispiel immer wieder die aberwitzige Behauptung, das die Summe aller natürlichen Zahlen +\begin{equation*} + \sum{n=1}^{\infty} n + = + \sum_{n=1}^{\infty} + \frac{1}{n^{-1}} + = + -\frac{1}{12} +\end{equation*} +sei. +Obwohl diese Behauptung offensichtlich Falsch ist, hat sie doch ihre Berechtigung, wie durch die analytische Fortsetzung gezeigt werden wird. + +Die folgenden mathematischen Herleitungen sind, sofern nicht anders gekennzeichnet, eigene Darstellungen basierend auf den überaus umfangreichen Wikipedia-Artikeln auf Deutsch \cite{zeta:online:wiki_de} und Englisch \cite{zeta:online:wiki_en} sowie einer Video-Playlist \cite{zeta:online:mryoumath}. diff --git a/buch/papers/zeta/euler_product.tex b/buch/papers/zeta/euler_product.tex index 5f4f5ca..7915c84 100644 --- a/buch/papers/zeta/euler_product.tex +++ b/buch/papers/zeta/euler_product.tex @@ -1,9 +1,9 @@ \section{Eulerprodukt} \label{zeta:section:eulerprodukt} \rhead{Eulerprodukt} -Das Eulerprodukt stellt die Verbindung der Zetafunktion und der Primzahlen her. -Diese Verbindung ist sehr wichtig, da durch sie eine Aussage zur Primzahlverteilung gemacht werden kann. -Die Verteilung der Primzahlen ist Gegenstand der Riemannschen Vermutung, welche eines der grössten ungelösten Probleme der Mathematik ist. +Das Eulerprodukt stellt die gesuchte Verbindung der Zetafunktion und der Primzahlen her. +Wie der Name bereits sagt, wurde das Eulerprodukt bereits 1727 von Euler entdeckt. +Um daraus die Riemannsche Vermutung herzuleiten, wäre aber noch einiges mehr nötig. \begin{satz} Für alle Zahlen $s$ mit $\Re(s) > 1$ ist die Zetafunktion identisch mit dem unendlichen Eulerprodukt @@ -65,7 +65,7 @@ Die Verteilung der Primzahlen ist Gegenstand der Riemannschen Vermutung, welche n = \prod_i p_i^{k_i} \quad \forall \quad n \in \mathbb{N}. \end{equation} Jeder Summand der Summen in \eqref{zeta:equation:eulerprodukt2} ist somit der Kehrwert genau einer natürlichen Zahl $n \in \mathbb{N}$. - Da die Summen alle möglichen Kombinationen von Exponenten und Primzahlen in \eqref{zeta:equation:eulerprodukt2} enthält haben wir + Da die Summen alle möglichen Kombinationen von Exponenten und Primzahlen in \eqref{zeta:equation:eulerprodukt2} enthält, haben wir \begin{equation} \sum_{k_1=0}^{\infty} \sum_{k_2=0}^{\infty} @@ -79,7 +79,8 @@ Die Verteilung der Primzahlen ist Gegenstand der Riemannschen Vermutung, welche \sum_{n=1}^\infty \frac{1}{n^s} = - \zeta(s) + \zeta(s), \end{equation} + wodurch das Eulerprudukt bewiesen ist. \end{proof} diff --git a/buch/papers/zeta/fazit.tex b/buch/papers/zeta/fazit.tex new file mode 100644 index 0000000..f696f83 --- /dev/null +++ b/buch/papers/zeta/fazit.tex @@ -0,0 +1,28 @@ +\section{Fazit} \label{zeta:section:fazit} +\rhead{Fazit} + +Ganz zu Beginn dieses Papers wurde die Behauptung erwähnt, dass die Summe aller natürlichen Zahlen $-\frac{1}{12}$ sei. +Diese Summe ist nichts anderes als die Zetafunktion am Wert $s=-1$. +Da wir die analytische Fortsetzung mit der Funktionalgleichung \eqref{zeta:equation:functional} gefunden haben, können wir diese Behauptung prüfen. +Zunächst berechnen wir $\zeta(1-s) = \zeta(2) = \frac{\pi^2}{6}$, welches im konvergenten Bereich der Reihe liegt und auch bekannt ist als das Basler Problem. +Somit haben wir +\begin{align*} + \zeta(s) = \zeta(-1) + &= + \frac{\Gamma \left( \frac{1-s}{2} \right)}{\pi^{\frac{1-s}{2}}} + \zeta(1-s) + \frac{\pi^{\frac{s}{2}}}{\Gamma \left( \frac{s}{2} \right)} + \\ + &= + \frac{\Gamma(1)}{\pi} + \frac{\pi^2}{6} + \frac{\pi^{\frac{-1}{2}}}{\Gamma \left( \frac{-1}{2} \right)} + \\ + &= + \frac{1}{\pi} + \frac{\pi^2}{6} + \frac{1}{\sqrt{\pi} (-2\sqrt{\pi})} + &= + -\frac{1}{12}, +\end{align*} +wobei die Werte der Gammafunktion TODO berechnet werden. diff --git a/buch/papers/zeta/images/continuation_overview.tikz.tex b/buch/papers/zeta/images/continuation_overview.tikz.tex new file mode 100644 index 0000000..836ab1d --- /dev/null +++ b/buch/papers/zeta/images/continuation_overview.tikz.tex @@ -0,0 +1,18 @@ +\begin{tikzpicture}[>=stealth', auto, node distance=0.9cm, scale=2, + dot/.style={fill, circle, inner sep=0, minimum size=0.1cm}] + + \draw[->] (-2,0) -- (-1,0) node[dot]{} node[anchor=north]{$-1$} -- (0,0) node[anchor=north west]{$0$} -- (0.5,0) node[anchor=north west]{$0.5$}-- (1,0) node[anchor=north west]{$1$} -- (2,0) node[anchor=west]{$\Re(s)$}; + + \draw[->] (0,-1.2) -- (0,1.2) node[anchor=south]{$\Im(s)$}; + \begin{scope}[yscale=0.1] + \draw[] (1,-1) -- (1,1); + \end{scope} + \draw[dotted] (0.5,-1) -- (0.5,1); + + \begin{scope}[] + \fill[opacity=0.2, red] (-1.8,1) rectangle (0, -1); + \fill[opacity=0.2, blue] (0,1) rectangle (1, -1); + \fill[opacity=0.2, green] (1,1) rectangle (1.8, -1); + \end{scope} + +\end{tikzpicture} diff --git a/buch/papers/zeta/images/primzahlfunktion.pgf b/buch/papers/zeta/images/primzahlfunktion.pgf new file mode 100644 index 0000000..7d4f4fc --- /dev/null +++ b/buch/papers/zeta/images/primzahlfunktion.pgf @@ -0,0 +1,505 @@ +%% Creator: Matplotlib, PGF backend +%% +%% To include the figure in your LaTeX document, write +%% \input{.pgf} +%% +%% Make sure the required packages are loaded in your preamble +%% \usepackage{pgf} +%% +%% and, on pdftex +%% \usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-} +%% +%% or, on luatex and xetex +%% \usepackage{unicode-math} +%% +%% Figures using additional raster images can only be included by \input if +%% they are in the same directory as the main LaTeX file. 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+%% +%% To include the figure in your LaTeX document, write +%% \input{.pgf} +%% +%% Make sure the required packages are loaded in your preamble +%% \usepackage{pgf} +%% +%% and, on pdftex +%% \usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-} +%% +%% or, on luatex and xetex +%% \usepackage{unicode-math} +%% +%% Figures using additional raster images can only be included by \input if +%% they are in the same directory as the main LaTeX file. For loading figures +%% from other directories you can use the `import` package +%% \usepackage{import} +%% +%% and then include the figures with +%% \import{}{.pgf} +%% +%% Matplotlib used the following preamble +%% +\begingroup% +\makeatletter% +\begin{pgfpicture}% +\pgfpathrectangle{\pgfpointorigin}{\pgfqpoint{5.440000in}{3.480000in}}% +\pgfusepath{use as bounding box, clip}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetmiterjoin% +\definecolor{currentfill}{rgb}{1.000000,1.000000,1.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.000000pt}% +\definecolor{currentstroke}{rgb}{1.000000,1.000000,1.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{5.440000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{5.440000in}{3.480000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{3.480000in}}% +\pgfpathclose% +\pgfusepath{fill}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetmiterjoin% 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a/buch/papers/zeta/images/zeta_re_-1_plot.pgf b/buch/papers/zeta/images/zeta_re_-1_plot.pgf new file mode 100644 index 0000000..dd15ba1 --- /dev/null +++ b/buch/papers/zeta/images/zeta_re_-1_plot.pgf @@ -0,0 +1,1147 @@ +%% Creator: Matplotlib, PGF backend +%% +%% To include the figure in your LaTeX document, write +%% \input{.pgf} +%% +%% Make sure the required packages are loaded in your preamble +%% \usepackage{pgf} +%% +%% and, on pdftex +%% \usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-} +%% +%% or, on luatex and xetex +%% \usepackage{unicode-math} +%% +%% Figures using additional raster images can only be included by \input if +%% they are in the same directory as the main LaTeX file. For loading figures +%% from other directories you can use the `import` package +%% \usepackage{import} +%% +%% and then include the figures with +%% \import{}{.pgf} +%% +%% Matplotlib used the following preamble +%% +\begingroup% +\makeatletter% +\begin{pgfpicture}% +\pgfpathrectangle{\pgfpointorigin}{\pgfqpoint{6.400000in}{4.800000in}}% +\pgfusepath{use as bounding box, clip}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetmiterjoin% +\definecolor{currentfill}{rgb}{1.000000,1.000000,1.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.000000pt}% +\definecolor{currentstroke}{rgb}{1.000000,1.000000,1.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{6.400000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{6.400000in}{4.800000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{4.800000in}}% +\pgfpathclose% +\pgfusepath{fill}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetmiterjoin% 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+\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{0.991229in}{0.528000in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=0.991229in,y=0.430778in,,top]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle {-15}\)}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{1.678290in}{0.528000in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=1.678290in,y=0.430778in,,top]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle {-10}\)}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{2.365352in}{0.528000in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=2.365352in,y=0.430778in,,top]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle {-5}\)}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{3.052413in}{0.528000in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=3.052413in,y=0.430778in,,top]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle {0}\)}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{3.739474in}{0.528000in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=3.739474in,y=0.430778in,,top]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle {5}\)}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{4.426535in}{0.528000in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=4.426535in,y=0.430778in,,top]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle {10}\)}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{5.113597in}{0.528000in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=5.113597in,y=0.430778in,,top]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle {15}\)}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=3.280000in,y=0.276457in,,top]{\color{textcolor}\rmfamily\fontsize{8.000000}{9.600000}\selectfont \(\displaystyle \Re\)}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% 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b/buch/papers/zeta/images/zeta_re_0.5_plot.pgf new file mode 100644 index 0000000..3ac7df8 --- /dev/null +++ b/buch/papers/zeta/images/zeta_re_0.5_plot.pgf @@ -0,0 +1,1206 @@ +%% Creator: Matplotlib, PGF backend +%% +%% To include the figure in your LaTeX document, write +%% \input{.pgf} +%% +%% Make sure the required packages are loaded in your preamble +%% \usepackage{pgf} +%% +%% and, on pdftex +%% \usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-} +%% +%% or, on luatex and xetex +%% \usepackage{unicode-math} +%% +%% Figures using additional raster images can only be included by \input if +%% they are in the same directory as the main LaTeX file. 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+\pgfpathmoveto{\pgfqpoint{0.800000in}{4.224000in}}% +\pgfpathlineto{\pgfqpoint{5.760000in}{4.224000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\end{pgfpicture}% +\makeatother% +\endgroup% diff --git a/buch/papers/zeta/main.tex b/buch/papers/zeta/main.tex index caddace..de297a0 100644 --- a/buch/papers/zeta/main.tex +++ b/buch/papers/zeta/main.tex @@ -8,12 +8,12 @@ \begin{refsection} \chapterauthor{Raphael Unterer} -%TODO Einleitung \input{papers/zeta/einleitung.tex} \input{papers/zeta/euler_product.tex} \input{papers/zeta/zeta_gamma.tex} \input{papers/zeta/analytic_continuation.tex} +\input{papers/zeta/fazit} \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/zeta/presentation/presentation.tex b/buch/papers/zeta/presentation/presentation.tex index e106089..53fd305 100644 --- a/buch/papers/zeta/presentation/presentation.tex +++ b/buch/papers/zeta/presentation/presentation.tex @@ -129,7 +129,7 @@ \begin{frame} \frametitle{Summe aller Natürlichen Zahlen} \begin{center} - \includegraphics[width=0.7\textwidth]{youtube_screenshot.png} + \includegraphics[width=0.7\textwidth]{../images/youtube_screenshot.png} \end{center} \end{frame} \begin{frame} @@ -168,7 +168,7 @@ \begin{frame} \frametitle{Plan für die Analytische Fortsetzung von $\zeta(s)$} \begin{center} - \input{../continuation_overview.tikz.tex} + \input{../images/continuation_overview.tikz.tex} \end{center} \end{frame} \begin{frame} @@ -331,7 +331,7 @@ \begin{frame} \frametitle{Primzahlfunktion} \begin{center} - \scalebox{0.5}{\input{../primzahlfunktion.pgf}} + \scalebox{0.5}{\input{../images/primzahlfunktion.pgf}} \end{center} \end{frame} @@ -348,19 +348,19 @@ \begin{frame} \frametitle{Konstanter Realteil $\Re(s)=-1$ und $\Im(s)=0\ldots40$} \begin{center} - \scalebox{0.6}{\input{../zeta_re_-1_plot.pgf}} + \scalebox{0.6}{\input{../images/zeta_re_-1_plot.pgf}} \end{center} \end{frame} \begin{frame} \frametitle{Konstanter Realteil $\Re(s)=0$ und $\Im(s)=0\ldots40$} \begin{center} - \scalebox{0.6}{\input{../zeta_re_0_plot.pgf}} + \scalebox{0.6}{\input{../images/zeta_re_0_plot.pgf}} \end{center} \end{frame} \begin{frame} \frametitle{Konstanter Realteil $\Re(s)=0.5$ und $\Im(s)=0\ldots40$} \begin{center} - \scalebox{0.6}{\input{../zeta_re_0.5_plot.pgf}} + \scalebox{0.6}{\input{../images/zeta_re_0.5_plot.pgf}} \end{center} \end{frame} diff --git a/buch/papers/zeta/presentation/youtube_screenshot.png b/buch/papers/zeta/presentation/youtube_screenshot.png deleted file mode 100644 index 434041b..0000000 Binary files a/buch/papers/zeta/presentation/youtube_screenshot.png and /dev/null differ diff --git a/buch/papers/zeta/primzahlfunktion.pgf b/buch/papers/zeta/primzahlfunktion.pgf deleted file mode 100644 index 7d4f4fc..0000000 --- a/buch/papers/zeta/primzahlfunktion.pgf +++ /dev/null @@ -1,505 +0,0 @@ -%% Creator: Matplotlib, PGF backend -%% -%% To include the figure in your LaTeX document, write -%% \input{.pgf} -%% -%% Make sure the required packages are loaded in your preamble -%% \usepackage{pgf} -%% -%% and, on pdftex -%% \usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-} -%% -%% or, on luatex and xetex -%% \usepackage{unicode-math} -%% -%% Figures using additional raster images can only be included by \input if -%% they are in the same directory as the main LaTeX file. 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url = {https://de.wikipedia.org/wiki/BibTeX}, - date = {2020-02-06}, - year = {2020}, - month = {2}, - day = {6} +@online{zeta:online:millennium, + title = {The Millennium Prize Problems}, + url = {https://www.claymath.org/millennium-problems/millennium-prize-problems}, + year = {2022}, + month = {8}, + day = {4} } -@book{zeta:numerical-analysis, - title = {Numerical Analysis}, - author = {David Kincaid and Ward Cheney}, - publisher = {American Mathematical Society}, - year = {2002}, - isbn = {978-8-8218-4788-6}, - inseries = {Pure and applied undegraduate texts}, - volume = {2} +@online{zeta:online:wiki_en, + title = {Riemann zeta function}, + url = {https://en.wikipedia.org/wiki/Riemann_zeta_function}, + year = {2022}, + month = {8}, + day = {7} +} +@online{zeta:online:wiki_de, + title = {Riemannsche Zeta-Funktion}, + url = {https://de.wikipedia.org/wiki/Riemannsche_Zeta-Funktion}, + year = {2022}, + month = {8}, + day = {7} } -@article{zeta:mendezmueller, - author = { Tabea Méndez and Andreas Müller }, - title = { Noncommutative harmonic analysis and image registration }, - journal = { Appl. Comput. Harmon. Anal.}, - year = 2019, - volume = 47, - pages = {607--627}, - url = {https://doi.org/10.1016/j.acha.2017.11.004} +@online{zeta:online:poisson, + title = {Deriving the Poisson Summation Formula}, + url = {https://www.youtube.com/watch?v=4Bex-4BFYWo}, + author = {Physics and Math Lectures}, + year = {2022}, + month = {8}, + day = {7} } +@online{zeta:online:mryoumath, + title = {Riemann Zeta Function Playlist}, + url = {https://www.youtube.com/playlist?list=PL32446FDD4DA932C9}, + author = {MrYouMath}, + year = {2022}, + month = {8}, + day = {7} +} diff --git a/buch/papers/zeta/zeta_color_plot-img0.png b/buch/papers/zeta/zeta_color_plot-img0.png new file mode 100644 index 0000000..b8c7298 Binary files /dev/null and b/buch/papers/zeta/zeta_color_plot-img0.png differ diff --git a/buch/papers/zeta/zeta_color_plot.pgf b/buch/papers/zeta/zeta_color_plot.pgf new file mode 100644 index 0000000..0fd7cb8 --- /dev/null +++ b/buch/papers/zeta/zeta_color_plot.pgf @@ -0,0 +1,402 @@ +%% Creator: Matplotlib, PGF backend +%% +%% To include the figure in your LaTeX document, write +%% \input{.pgf} +%% +%% Make sure the required packages are loaded in your preamble +%% \usepackage{pgf} +%% +%% and, on pdftex +%% \usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-} +%% +%% or, on luatex and xetex +%% \usepackage{unicode-math} +%% +%% Figures using additional raster images can only be included by \input if +%% they are in the same directory as the main LaTeX file. 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-\pgfpathmoveto{\pgfqpoint{0.800000in}{4.224000in}}% -\pgfpathlineto{\pgfqpoint{5.760000in}{4.224000in}}% -\pgfusepath{stroke}% -\end{pgfscope}% -\end{pgfpicture}% -\makeatother% -\endgroup% -- cgit v1.2.1 From 676b8ffe21376e27f7f21c093bee2fd8692b7a4b Mon Sep 17 00:00:00 2001 From: runterer Date: Mon, 8 Aug 2022 00:01:33 +0200 Subject: finished fazit --- buch/papers/zeta/fazit.tex | 85 +- buch/papers/zeta/images/zeta_re_0.5_paper.pgf | 1137 +++++++++++++++++++++++++ buch/papers/zeta/references.bib | 15 + 3 files changed, 1228 insertions(+), 9 deletions(-) create mode 100644 buch/papers/zeta/images/zeta_re_0.5_paper.pgf (limited to 'buch/papers') diff --git a/buch/papers/zeta/fazit.tex b/buch/papers/zeta/fazit.tex index f696f83..fe2d35d 100644 --- a/buch/papers/zeta/fazit.tex +++ b/buch/papers/zeta/fazit.tex @@ -3,26 +3,93 @@ Ganz zu Beginn dieses Papers wurde die Behauptung erwähnt, dass die Summe aller natürlichen Zahlen $-\frac{1}{12}$ sei. Diese Summe ist nichts anderes als die Zetafunktion am Wert $s=-1$. -Da wir die analytische Fortsetzung mit der Funktionalgleichung \eqref{zeta:equation:functional} gefunden haben, können wir diese Behauptung prüfen. -Zunächst berechnen wir $\zeta(1-s) = \zeta(2) = \frac{\pi^2}{6}$, welches im konvergenten Bereich der Reihe liegt und auch bekannt ist als das Basler Problem. -Somit haben wir +Da wir die analytische Fortsetzung mit der Funktionalgleichung \eqref{zeta:equation:functional} gefunden haben, können wir den Wert $s=-1$ einsetzen und erhalten \begin{align*} - \zeta(s) = \zeta(-1) + \zeta(s) &= \frac{\Gamma \left( \frac{1-s}{2} \right)}{\pi^{\frac{1-s}{2}}} \zeta(1-s) \frac{\pi^{\frac{s}{2}}}{\Gamma \left( \frac{s}{2} \right)} \\ + \zeta(-1) &= \frac{\Gamma(1)}{\pi} - \frac{\pi^2}{6} - \frac{\pi^{\frac{-1}{2}}}{\Gamma \left( \frac{-1}{2} \right)} + \zeta(2) + \frac{\pi^{-\frac{1}{2}}}{\Gamma \left( -\frac{1}{2} \right)}. +\end{align*} +Also fehlen uns drei Werte, $\zeta(2)$, $\Gamma(1)$ und $\Gamma\left(-\frac{1}{2}\right)$. + +Zunächst konzentrieren wir uns auf $\zeta(2)$, welches im konvergenten Bereich der Reihe liegt und auch bekannt ist als das Basler Problem. +Wir lösen das Basler Problem \cite{zeta:online:basel} mithilfe der parsevalschen Gleichung \cite{zeta:online:pars} +\begin{align} + \int_{-\pi}^{\pi} |f(x)|^2 dx + &= + 2\pi \sum_{n=-\infty}^{\infty} |c_n|^2 \\ + c_n + &= + \frac{1}{2\pi} + \int_{-\pi}^{\pi}f(x)e^{-inx} dx, +\end{align} +welche besagt dass die Summe der quadrierten Fourierkoeffizienten einer Funktion identisch ist mit dem Integral der quadrierten Funktion. +Wenn wir dies für $f(x) = x$ auswerten erhalten wir +\begin{align} + c_n + &= + \begin{cases} + \frac{(-1)^n}{n} i, & \text{for } n\neq0, \\ + 0, & \text{for } n=0 + \end{cases} + \\ + \int_{-\pi}^{\pi} x^2 dx + &= + 2\pi \sum_{n=-\infty}^{\infty} |c_n|^2 + = + 4\pi \underbrace{\sum_{n=1}^{\infty} \frac{1}{n^2}}_{\zeta(2)}. +\end{align} +Durch einfaches Umstellen erhalten wir somit die Lösung des Basler Problems als +\begin{equation} + \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{1}{4\pi} + \int_{-\pi}^{\pi} x^2 dx + = \frac{\pi^2}{6}. +\end{equation} + +Als nächstes berechnen wir $\Gamma(1)$ und $\Gamma\left(-\frac{1}{2}\right)$ mithilfe der Integraldefinition der Gammafunktion \ref{buch:rekursion:def:gamma}. +Da das Integral für $\Gamma\left(-\frac{1}{2}\right)$ nicht konvergiert, wird die Reflektionsformel aus \ref{buch:funktionentheorie:subsection:gammareflektion} verwendet, welche das konvergierende Integral von $\Gamma\left(\frac{3}{2}\right)$ verwendet. +Es ergeben sich die Werte +\begin{align*} + \Gamma(1) + &= 1\\ + \Gamma\left(-\frac{1}{2}\right) + &= \frac{\pi}{\sin\left(-\frac{\pi}{2}\right) + \Gamma\left(\frac{3}{2}\right)} + = -\frac{\sqrt{\pi}}{2}. +\end{align*} + +Wenn wir diese Werte in die Funktionalgleichung einsetzen, erhalten wir das gewünschte Ergebnis +\begin{align*} + \zeta(-1) + &= + \frac{\Gamma(1)}{\pi} + \zeta(2) + \frac{\pi^{-\frac{1}{2}}}{\Gamma \left( -\frac{1}{2} \right)} \\ &= \frac{1}{\pi} \frac{\pi^2}{6} - \frac{1}{\sqrt{\pi} (-2\sqrt{\pi})} + \frac{\pi^{-\frac{1}{2}}}{ + -\frac{\sqrt{\pi}}{2}} + \\ &= - -\frac{1}{12}, + -\frac{1}{12}. \end{align*} -wobei die Werte der Gammafunktion TODO berechnet werden. + +Weiter wurde zu Beginn dieses Papers auf die Riemannsche Vermutung hingewiesen, wonach alle nichttrivialen Nullstellen der Zetafunktion auf der $\Re(s)=\frac{1}{2}$ Geraden liegen. +Abbildung \ref{zeta:fig:einzweitel} zeigt die Funktionswerte dieser Geraden. +%TODO colorplot does not work.. Ausserdem zeigt Abbildung \ref{zeta:fig:colorplot} die farbcodierte Zetafunktion für Werte der analytischen Fortsetzung und des originalen Definitionsbereichs. +\begin{figure} + \centering + \input{papers/zeta/images/zeta_re_0.5_paper.pgf} + \caption{Die komplexen Werte der Zetafunktion für die kritische Gerade $\Re(s)=\frac{1}{2}$ im Bereich $\Im(s) = 0\dots40$. + Klar sichtbar sind die immer wiederkehrenden Nullstellen, wie sie Gegenstand der Riemannschen Vermutung sind.} + \label{zeta:fig:einzweitel} +\end{figure} diff --git a/buch/papers/zeta/images/zeta_re_0.5_paper.pgf b/buch/papers/zeta/images/zeta_re_0.5_paper.pgf new file mode 100644 index 0000000..44fffce --- /dev/null +++ b/buch/papers/zeta/images/zeta_re_0.5_paper.pgf @@ -0,0 +1,1137 @@ +%% Creator: Matplotlib, PGF backend +%% +%% To include the figure in your LaTeX document, write +%% \input{.pgf} +%% +%% Make sure the required packages are loaded in your preamble +%% \usepackage{pgf} +%% +%% and, on pdftex +%% \usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-} +%% +%% or, on luatex and xetex +%% \usepackage{unicode-math} +%% +%% Figures using additional raster images can only be included by \input if +%% they are in the same directory as the main LaTeX file. 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+\pgfpathlineto{\pgfqpoint{3.330000in}{2.728000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\end{pgfpicture}% +\makeatother% +\endgroup% diff --git a/buch/papers/zeta/references.bib b/buch/papers/zeta/references.bib index e8d6b22..f2a2f31 100644 --- a/buch/papers/zeta/references.bib +++ b/buch/papers/zeta/references.bib @@ -44,3 +44,18 @@ month = {8}, day = {7} } + +@online{zeta:online:basel, + title = {Basel Problem}, + url = {https://en.wikipedia.org/wiki/Basel_problem}, + year = {2022}, + month = {8}, + day = {7} +} +@online{zeta:online:pars, + title = {Parseval's identity}, + url = {https://en.wikipedia.org/wiki/Parseval%27s_identity}, + year = {2022}, + month = {8}, + day = {7} +} -- cgit v1.2.1 From ebbf6e36246d36a2ec842b8c89a1f09a5dbec9de Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 8 Aug 2022 10:27:50 +0200 Subject: Corrected sign error in coefficient comparison. --- buch/papers/sturmliouville/waermeleitung_beispiel.tex | 14 +++++++------- 1 file changed, 7 insertions(+), 7 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index cc88f6a..7310186 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -170,23 +170,23 @@ Eingesetzt in Gleichung (TDOD: ref) ergibt dies \] und durch umformen somit \[ - \mu A\sin(\alpha x) + \mu B\cos(\beta x) + -A\alpha^{2}\sin(\alpha x) - B\beta^{2}\cos(\beta x) = - A\alpha^{2}\sin(\alpha x) + B\beta^{2}\cos(\beta x). + \mu A\sin(\alpha x) + \mu B\cos(\beta x). \] Durch Koeffizientenvergleich von \[ - \mu A\sin(\alpha x) + -A\alpha^{2}\sin(\alpha x) = - A\alpha^{2}\sin(\alpha x) + \mu A\sin(\alpha x) \] \[ - \mu B\cos(\beta x) + -B\beta^{2}\cos(\beta x) = - B\beta^{2}\cos(\beta x) + \mu B\cos(\beta x) \] -ist schnell ersichtlich, dass $ \mu = \alpha^{2} = \beta^{2} $ gelten muss für +ist schnell ersichtlich, dass $ \mu = -\alpha^{2} = -\beta^{2} $ gelten muss für $ A \neq 0 $ und $ B \neq 0 $. Zur Berechnung von $ \mu $ bleiben also noch $ \alpha $ und $ \beta $ zu bestimmen. -- cgit v1.2.1 From 2b1eb4b5979f4e0e7f2eee7414a8e0b3d9eae402 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 8 Aug 2022 13:04:13 +0200 Subject: Changed equation syntax to match rest of the Sturm-Liouville chapter. --- .../sturmliouville/waermeleitung_beispiel.tex | 106 ++++++++++----------- 1 file changed, 50 insertions(+), 56 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 7310186..0c9dd8e 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -1,6 +1,5 @@ % % waermeleitung_beispiel.tex -- Beispiel Wärmeleitung in homogenem Stab. -%%%%%%%%%%%%%%%%%%%%%%%%%%% Erster Entwurf %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % @@ -14,10 +13,10 @@ physikalischen Phänomenes auftritt. Zunächst wird ein eindimensionaler homogener Stab der Länge $l$ und Wärmeleitkoeffizient $\kappa$. Somit ergibt sich für das Wärmeleitungsproblem die partielle Differentialgleichung -\[ +\begin{equation} \frac{\partial u}{\partial t} = \kappa \frac{\partial^{2}u}{{\partial x}^{2}} -\] +\end{equation} wobei der Stab in diesem Fall auf der X-Achse im Intervall $[0,l]$ liegt. Da diese Differentialgleichung das Problem allgemein für einen homogenen @@ -32,13 +31,13 @@ Tempreatur gehalten werden. Die Enden des Stabes auf konstanter Temperatur zu halten bedeutet, dass die Lösungsfunktion $u(t,x)$ bei $x = 0$ und $x = l$ nur die vorgegebene Temperatur zurückgeben darf. Es folgen nun -\[ +\begin{equation} u(t,0) = u(t,l) = 0 -\] +\end{equation} als Randbedingungen. %%%%%%%%%%%%% Randbedingungen für Stab mit isolierten Enden %%%%%%%%%%%%%%%%%%% @@ -54,13 +53,13 @@ Temperatur fliesst. Um Wärmefluss zu unterdrücken, muss also dafür gesorgt werden, dass am Rand des Stabes keine Temperaturdifferenz existiert oder dass die partiellen Ableitungen von $u(t,x)$ nach $x$ bei $x = 0$ und $x = l$ verschwinden. Somit folgen -\[ +\begin{equation} \frac{\partial}{\partial x} u(t, 0) = \frac{\partial}{\partial x} u(t, l) = 0 -\] +\end{equation} als Randbedingungen. %%%%%%%%%%% Lösung der Differenzialgleichung %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% @@ -72,41 +71,40 @@ als Randbedingungen. Da die Lösungsfunktion von zwei Variablen abhängig ist, wird als Lösungsansatz die Separationsmethode verwendet. Dazu wird -\[ +\begin{equation} u(t,x) = T(t)X(x) -\] +\end{equation} in die ursprüngliche Differenzialgleichung eingesetzt. Daraus ergibt sich -\[ +\begin{equation} T^{\prime}(t)X(x) = \kappa T(t)X^{\prime \prime}(x) -\] +\end{equation} als neue Form. Nun können alle von $t$ abhängigen Ausdrücke auf die linke Seite, sowie alle von $x$ abhängigen Ausdrücke auf die rechte Seite gebracht werden und mittels der neuen Variablen $\mu$ gekoppelt werden: -\[ +\begin{equation} \frac{T^{\prime}(t)}{\kappa T(t)} = \frac{X^{\prime \prime}(x)}{X(x)} = \mu -\] +\end{equation} Durch die Einführung von $\mu$ kann das Problem nun in zwei separate Differenzialgleichungen aufgeteilt werden: -\[ +\begin{equation} T^{\prime}(t) - \kappa \mu T(t) - = + &= 0 -\] -\[ + \\ X^{\prime \prime}(x) - \mu X(x) - = + &= 0 -\] +\end{equation} Es ist an dieser Stelle zu bemerken, dass die Gleichung in $x$ in Sturm-Liouville-Form ist. Erfüllen die Randbedingungen des Stab-Problems auch @@ -116,108 +114,104 @@ werden. Widmen wir uns zunächst der ersten Gleichung. Diese Lösen wir über das charakteristische Polynom -\[ +\begin{equation} \lambda - \kappa \mu = 0. -\] +\end{equation} Es ist direkt ersichtlich, dass $\lambda = \kappa \mu$ gelten muss, was zur Lösung -\[ +\begin{equation} T(t) = e^{\kappa \mu t} -\] +\end{equation} führt. Etwas aufwändiger wird es, die zweite Gleichung zu lösen. Aufgrund der Struktur der Gleichung -\[ +\begin{equation} X^{\prime \prime}(x) - \mu X(x) = 0 -\] +\end{equation} wird ein trigonometrischer Ansatz gewählt. Die Lösungen für $X(x)$ sind also von der Form -\[ +\begin{equation} X(x) = A \sin \left( \alpha x\right) + B \cos \left( \beta x\right). -\] +\end{equation} Dieser Ansatz wird nun solange differenziert, bis alle in Gleichung (TODO: ref) enthaltenen Ableitungen vorhanden sind. Man erhält also -\[ +\begin{equation} X^{\prime}(x) = A \alpha \cos \left( \alpha x \right) - B \beta \sin \left( \beta x \right) -\] +\end{equation} und -\[ +\begin{equation} X^{\prime \prime}(x) = -A \alpha^{2} \sin \left( \alpha x \right) - B \beta^{2} \cos \left( \beta x \right). -\] +\end{equation} Eingesetzt in Gleichung (TDOD: ref) ergibt dies -\[ +\begin{equation} -A\alpha^{2}\sin(\alpha x) - B\beta^{2}\cos(\beta x) - \mu\left(A\sin(\alpha x) + B\cos(\beta x)\right) = 0 -\] +\end{equation} und durch umformen somit -\[ +\begin{equation} -A\alpha^{2}\sin(\alpha x) - B\beta^{2}\cos(\beta x) = \mu A\sin(\alpha x) + \mu B\cos(\beta x). -\] +\end{equation} Durch Koeffizientenvergleich von -\[ +\begin{equation} -A\alpha^{2}\sin(\alpha x) - = + &= \mu A\sin(\alpha x) -\] -\[ + \\ -B\beta^{2}\cos(\beta x) - = + &= \mu B\cos(\beta x) -\] +\end{equation} ist schnell ersichtlich, dass $ \mu = -\alpha^{2} = -\beta^{2} $ gelten muss für $ A \neq 0 $ und $ B \neq 0 $. Zur Berechnung von $ \mu $ bleiben also noch $ \alpha $ und $ \beta $ zu bestimmen. % TODO: Rechenweg TODO: Rechenweg... Enden auf konstanter Temperatur: -\[ +\begin{equation} u(t,x) - = + &= \sum_{n=1}^{\infty}a_{n}e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t} \sin\left(\frac{n\pi}{l}x\right) -\] -\[ + \\ a_{n} - = + &= \frac{2}{l}\int_{0}^{l}u(0,x)sin\left(\frac{n\pi}{l}x\right) dx -\] +\end{equation} TODO: Rechenweg... Enden isoliert: -\[ +\begin{equation} u(t,x) - = + &= a_{0} + \sum_{n=1}^{\infty}a_{n}e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t} \cos\left(\frac{n\pi}{l}x\right) -\] -\[ + \\ a_{0} - = + &= \frac{1}{l}\int_{0}^{l}u(0,x) dx -\] -\[ + \\ a_{n} - = + &= \frac{2}{l}\int_{0}^{l}u(0,x)sin\left(\frac{n\pi}{l}x\right) dx -\] +\end{equation} -- cgit v1.2.1 From 95ce389d41871e3e1a7dba350bf3dcdc1d67f80c Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 8 Aug 2022 13:12:23 +0200 Subject: Fixed alignment issue in fourier example. --- buch/papers/sturmliouville/waermeleitung_beispiel.tex | 8 ++++++++ 1 file changed, 8 insertions(+) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 0c9dd8e..27a7574 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -97,6 +97,7 @@ der neuen Variablen $\mu$ gekoppelt werden: Durch die Einführung von $\mu$ kann das Problem nun in zwei separate Differenzialgleichungen aufgeteilt werden: \begin{equation} +\begin{aligned} T^{\prime}(t) - \kappa \mu T(t) &= 0 @@ -104,6 +105,7 @@ Differenzialgleichungen aufgeteilt werden: X^{\prime \prime}(x) - \mu X(x) &= 0 +\end{aligned} \end{equation} Es ist an dieser Stelle zu bemerken, dass die Gleichung in $x$ in @@ -175,6 +177,7 @@ und durch umformen somit Durch Koeffizientenvergleich von \begin{equation} +\begin{aligned} -A\alpha^{2}\sin(\alpha x) &= \mu A\sin(\alpha x) @@ -182,6 +185,7 @@ Durch Koeffizientenvergleich von -B\beta^{2}\cos(\beta x) &= \mu B\cos(\beta x) +\end{aligned} \end{equation} ist schnell ersichtlich, dass $ \mu = -\alpha^{2} = -\beta^{2} $ gelten muss für $ A \neq 0 $ und $ B \neq 0 $. Zur Berechnung von $ \mu $ bleiben also noch @@ -190,6 +194,7 @@ $ \alpha $ und $ \beta $ zu bestimmen. % TODO: Rechenweg TODO: Rechenweg... Enden auf konstanter Temperatur: \begin{equation} +\begin{aligned} u(t,x) &= \sum_{n=1}^{\infty}a_{n}e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t} @@ -198,10 +203,12 @@ TODO: Rechenweg... Enden auf konstanter Temperatur: a_{n} &= \frac{2}{l}\int_{0}^{l}u(0,x)sin\left(\frac{n\pi}{l}x\right) dx +\end{aligned} \end{equation} TODO: Rechenweg... Enden isoliert: \begin{equation} +\begin{aligned} u(t,x) &= a_{0} + \sum_{n=1}^{\infty}a_{n}e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t} @@ -214,4 +221,5 @@ TODO: Rechenweg... Enden isoliert: a_{n} &= \frac{2}{l}\int_{0}^{l}u(0,x)sin\left(\frac{n\pi}{l}x\right) dx +\end{aligned} \end{equation} -- cgit v1.2.1 From 2cb7c0466bdaaa3eff6757382a913b3c955a0751 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 8 Aug 2022 16:57:13 +0200 Subject: Reordered fourier example. --- .../sturmliouville/waermeleitung_beispiel.tex | 160 ++++++++++++--------- 1 file changed, 90 insertions(+), 70 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 27a7574..da25b36 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -11,9 +11,11 @@ homogenen Stab und wie das Sturm-Liouville-Problem bei der Beschreibung dieses physikalischen Phänomenes auftritt. Zunächst wird ein eindimensionaler homogener Stab der Länge $l$ und -Wärmeleitkoeffizient $\kappa$. Somit ergibt sich für das Wärmeleitungsproblem +Wärmeleitkoeffizient $\kappa$. +Somit ergibt sich für das Wärmeleitungsproblem die partielle Differentialgleichung \begin{equation} + \label{eq:slp-example-fourier-heat-equation} \frac{\partial u}{\partial t} = \kappa \frac{\partial^{2}u}{{\partial x}^{2}} \end{equation} @@ -30,7 +32,8 @@ Tempreatur gehalten werden. Die Enden des Stabes auf konstanter Temperatur zu halten bedeutet, dass die Lösungsfunktion $u(t,x)$ bei $x = 0$ und $x = l$ nur die vorgegebene -Temperatur zurückgeben darf. Es folgen nun +Temperatur zurückgeben darf. +Es folgen nun \begin{equation} u(t,0) = @@ -52,7 +55,8 @@ Aus der Physik ist bekannt, dass Wärme immer von der höheren zur tieferen Temperatur fliesst. Um Wärmefluss zu unterdrücken, muss also dafür gesorgt werden, dass am Rand des Stabes keine Temperaturdifferenz existiert oder dass die partiellen Ableitungen von $u(t,x)$ nach $x$ bei $x = 0$ und $x = l$ -verschwinden. Somit folgen +verschwinden. +Somit folgen \begin{equation} \frac{\partial}{\partial x} u(t, 0) = @@ -70,18 +74,20 @@ als Randbedingungen. % TODO: Formeln sauber in Text einbinden. Da die Lösungsfunktion von zwei Variablen abhängig ist, wird als Lösungsansatz -die Separationsmethode verwendet. Dazu wird -\begin{equation} +die Separationsmethode verwendet. +Dazu wird +\[ u(t,x) = T(t)X(x) -\end{equation} -in die ursprüngliche Differenzialgleichung eingesetzt. Daraus ergibt sich -\begin{equation} +\] +in die ursprüngliche Differenzialgleichung eingesetzt. +Daraus ergibt sich +\[ T^{\prime}(t)X(x) = \kappa T(t)X^{\prime \prime}(x) -\end{equation} +\] als neue Form. Nun können alle von $t$ abhängigen Ausdrücke auf die linke Seite, sowie alle @@ -97,103 +103,117 @@ der neuen Variablen $\mu$ gekoppelt werden: Durch die Einführung von $\mu$ kann das Problem nun in zwei separate Differenzialgleichungen aufgeteilt werden: \begin{equation} -\begin{aligned} - T^{\prime}(t) - \kappa \mu T(t) - &= - 0 - \\ + \label{eq:slp-example-fourier-separated-x} X^{\prime \prime}(x) - \mu X(x) - &= - 0 -\end{aligned} -\end{equation} - -Es ist an dieser Stelle zu bemerken, dass die Gleichung in $x$ in -Sturm-Liouville-Form ist. Erfüllen die Randbedingungen des Stab-Problems auch -die Randbedingungen des Sturm-Liouville-Problems, kann bereits die Aussage -getroffen werden, dass alle Lösungen für die Gleichung in $x$ orthogonal sein -werden. - -Widmen wir uns zunächst der ersten Gleichung. Diese Lösen wir über das -charakteristische Polynom -\begin{equation} - \lambda - \kappa \mu = - 0. + 0 \end{equation} -Es ist direkt ersichtlich, dass $\lambda = \kappa \mu$ gelten muss, was zur -Lösung \begin{equation} - T(t) + \label{eq:slp-example-fourier-separated-t} + T^{\prime}(t) - \kappa \mu T(t) = - e^{\kappa \mu t} + 0 \end{equation} -führt. -Etwas aufwändiger wird es, die zweite Gleichung zu lösen. Aufgrund der Struktur -der Gleichung -\begin{equation} +Es ist an dieser Stelle zu bemerken, dass die Gleichung in $x$ in +Sturm-Liouville-Form ist. +Erfüllen die Randbedingungen des Stab-Problems auch die Randbedingungen des +Sturm-Liouville-Problems, kann bereits die Aussage getroffen werden, dass alle +Lösungen für die Gleichung in $x$ orthogonal sein werden. + +Widmen wir uns zunächst der ersten Gleichung. +Aufgrund der Struktur der Gleichung +\[ X^{\prime \prime}(x) - \mu X(x) = 0 -\end{equation} -wird ein trigonometrischer Ansatz gewählt. Die Lösungen für $X(x)$ sind also -von der Form -\begin{equation} +\] +wird ein trigonometrischer Ansatz gewählt. +Die Lösungen für $X(x)$ sind also von der Form +\[ X(x) = A \sin \left( \alpha x\right) + B \cos \left( \beta x\right). -\end{equation} +\] -Dieser Ansatz wird nun solange differenziert, bis alle in Gleichung (TODO: ref) -enthaltenen Ableitungen vorhanden sind. Man erhält also -\begin{equation} +Dieser Ansatz wird nun solange differenziert, bis alle in Gleichung +\eqref{eq:slp-example-fourier-separated-x} enthaltenen Ableitungen vorhanden +sind. +Man erhält also +\[ X^{\prime}(x) = - A \alpha \cos \left( \alpha x \right) - - B \beta \sin \left( \beta x \right) -\end{equation} + \alpha A \cos \left( \alpha x \right) - + \beta B \sin \left( \beta x \right) +\] und -\begin{equation} +\[ X^{\prime \prime}(x) = - -A \alpha^{2} \sin \left( \alpha x \right) - - B \beta^{2} \cos \left( \beta x \right). -\end{equation} + -\alpha^{2} A \sin \left( \alpha x \right) - + \beta^{2} B \cos \left( \beta x \right). +\] -Eingesetzt in Gleichung (TDOD: ref) ergibt dies -\begin{equation} - -A\alpha^{2}\sin(\alpha x) - B\beta^{2}\cos(\beta x) - +Eingesetzt in Gleichung \eqref{eq:slp-example-fourier-separated-x} ergibt dies +\[ + -\alpha^{2}A\sin(\alpha x) - \beta^{2}B\cos(\beta x) - \mu\left(A\sin(\alpha x) + B\cos(\beta x)\right) = 0 -\end{equation} +\] und durch umformen somit -\begin{equation} - -A\alpha^{2}\sin(\alpha x) - B\beta^{2}\cos(\beta x) +\[ + -\alpha^{2}A\sin(\alpha x) - \beta^{2}B\cos(\beta x) = \mu A\sin(\alpha x) + \mu B\cos(\beta x). -\end{equation} +\] Durch Koeffizientenvergleich von -\begin{equation} +\[ \begin{aligned} - -A\alpha^{2}\sin(\alpha x) + -\alpha^{2}A\sin(\alpha x) &= \mu A\sin(\alpha x) \\ - -B\beta^{2}\cos(\beta x) + -\beta^{2}B\cos(\beta x) &= \mu B\cos(\beta x) \end{aligned} -\end{equation} +\] ist schnell ersichtlich, dass $ \mu = -\alpha^{2} = -\beta^{2} $ gelten muss für -$ A \neq 0 $ und $ B \neq 0 $. Zur Berechnung von $ \mu $ bleiben also noch -$ \alpha $ und $ \beta $ zu bestimmen. +$ A \neq 0 $ und $ B \neq 0 $. +Zur Berechnung von $ \mu $ bleiben also noch $ \alpha $ und $ \beta $ zu +bestimmen. + +TODO: randbedingungen!!---- + +Betrachten wir nun die zweite Gleichung +\eqref{eq:slp-example-fourier-separated-t}. +Diese Lösen wir über das charakteristische Polynom +\[ + \lambda - \kappa \mu + = + 0. +\] +Es ist direkt ersichtlich, dass $\lambda = \kappa \mu$ gelten muss, was zur +Lösung +\[ + T(t) + = + e^{-\kappa \mu t} +\] +führt. +Und mit mit dem Resultat von zuvor die Lösung +\[ + T(t) + = + e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t} +\] +ergibt. % TODO: Rechenweg TODO: Rechenweg... Enden auf konstanter Temperatur: -\begin{equation} +\[ \begin{aligned} u(t,x) &= @@ -204,10 +224,10 @@ TODO: Rechenweg... Enden auf konstanter Temperatur: &= \frac{2}{l}\int_{0}^{l}u(0,x)sin\left(\frac{n\pi}{l}x\right) dx \end{aligned} -\end{equation} +\] TODO: Rechenweg... Enden isoliert: -\begin{equation} +\[ \begin{aligned} u(t,x) &= @@ -222,4 +242,4 @@ TODO: Rechenweg... Enden isoliert: &= \frac{2}{l}\int_{0}^{l}u(0,x)sin\left(\frac{n\pi}{l}x\right) dx \end{aligned} -\end{equation} +\] -- cgit v1.2.1 From a37eaf082bc34c696c40efe33cf868c41dd765a0 Mon Sep 17 00:00:00 2001 From: Andrea Mozzini Vellen Date: Mon, 8 Aug 2022 19:00:45 +0200 Subject: last commit --- buch/papers/kreismembran/teil1.tex | 22 +++++++++++----------- buch/papers/kreismembran/teil2.tex | 16 ++++++++-------- buch/papers/kreismembran/teil3.tex | 24 ++++++++++++++---------- 3 files changed, 33 insertions(+), 29 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil1.tex b/buch/papers/kreismembran/teil1.tex index f0d478f..a872ed1 100644 --- a/buch/papers/kreismembran/teil1.tex +++ b/buch/papers/kreismembran/teil1.tex @@ -23,7 +23,7 @@ Da es sich um eine Kreisscheibe handelt, werden Polarkoordinaten verwendet, so d \frac1r \frac{\partial}{\partial r} + - \frac{1}{r 2} + \frac{1}{r^2} \frac{\partial^2}{\partial\varphi^2} \label{buch:pde:kreis:laplace} \end{equation*} @@ -39,16 +39,16 @@ Daher ist die Membranabweichung im Punkt $(r,\varphi)$ $\in$ $\overline{\rm \Ome \end{align*} Um die Vergleichbarkeit der beiden nachfolgend vorgestellten Lösungsverfahren in Abschnitt \ref{kreismembran:vergleich} zu vereinfachen, werden keine Randbedingungen vorgegeben. -Um eine eindeutige Lösung bestimmen zu können, werden die folgenden Anfangsbedingungen festgelegt zur zeit $t = \text{0}$: +Um eine eindeutige Lösung bestimmen zu können, werden die folgenden Anfangsbedingungen festgelegt zur Zeit $t = \text{0}$: \begin{align*} u(r,\varphi, 0) &= f(r,\varphi)\\ u_t(r,\varphi, 0) &= g(r,\varphi). \end{align*} \subsection{Lösung\label{sub:lösung1}} -Nun wird das in Abschnitt \ref{sub:aufgabestellung} vorgestellte Problem mit Hilfe der varibalen Trennungsmethode gelöst. +Nun wird das in Abschnitt \ref{sub:aufgabestellung} vorgestellte Problem mit Hilfe der Separationsmethode gelöst. \subsubsection{Ansatz der Separation der Variablen\label{subsub:ansatz_separation}} -Bezug muss an dieser Stelle von einer Separation der Variablen ausgegangen werden: +Hierfür wird folgenden Ansatz gemacht: \begin{equation*} u(r,\varphi, t) = F(r)G(\varphi)T(t) \end{equation*} @@ -64,26 +64,26 @@ Dies bedeutet, dass die Konstante negativ sein muss, also schreibt man $-\kappa^ T''(t) + c^2\kappa^2T(t) &= 0\\ r^2\frac{F''(r)}{F(r)} + r \frac{F'(r)}{F(r)} +\kappa^2 r^2 &= - \frac{G''(\varphi)}{G(\varphi)}. \end{align*} -In der zweiten Gleichung hängt die linke Seite nur von $r$ ab, während die rechte Seite nur von $\varphi$ abhängt. Sie müssen also wiederum gleich einer reellen Zahl $\nu$ sein. Also das: +In der zweiten Gleichung hängt die linke Seite nur von $r$ ab, während die rechte Seite nur von $\varphi$ abhängt. Sie müssen also wiederum gleich einer reellen Zahl $\nu$ sein. Also: \begin{align*} - r^2F''(r) + rF'(r) + (\kappa^2 r^2 - \nu)F(r) &= 0 \\ - G''(\varphi) &= \nu G(\varphi). + r^2F''(r) + rF'(r) + (\kappa^2 r^2 - \nu)F(r) = 0 \quad \text{und} \quad + G''(\varphi) = \nu G(\varphi). \end{align*} \subsubsection{Lösung für $G(\varphi)$\label{subsub:lösung_G}} Da für die zweite Gleichung Lösungen von Schwingungen erwartet werden, für die $G''(\varphi)=-\omega^2 G(\varphi)$ gilt, schreibt man die gemeinsame Konstante als $\nu=-\omega^2$, was die Formeln später vereinfacht. Also: \begin{equation*} - G(\varphi) = C_n \cos(\varphi) + D_n \sin(\varphi) + G(\varphi) = C_n \cos(\nu\varphi) + D_n \sin(\nu\varphi) \label{eq:cos_sin_überlagerung} \end{equation*} \subsubsection{Lösung für $F(r)$\label{subsub:lösung_F}} -Die Gleichung für $F$ hat die Gestalt (verweis auf \ref{buch:differentialgleichungen:bessel-operator}) +Die Gleichung für $F$ hat die Gestalt (Verweis auf \label{buch:differentialgleichungen:bessel-operator} \begin{align} r^2F''(r) + rF'(r) + (\kappa^2 r^2 - n^2)F(r) = 0 \label{eq:2nd_degree_PDE} \end{align} -Wir bereits in Kapitel \ref{buch:differntialgleichungen:section:bessel} gezeigt, sind die Besselfunktionen +Wir bereits in Kapitel \ref{buch:differntialgleichungen:section:bessel} gezeigt, sind die Bessel-Funktionen \begin{equation*} J_{\nu}(x) = r^\nu \displaystyle\sum_{m=0}^{\infty} \frac{(-1)^m x^{2m}}{2^{2m+\nu}m! \Gamma (\nu + m+1)} \end{equation*} @@ -104,7 +104,7 @@ Durch Überlagerung aller Ergebnisse erhält man die Lösung \end{align} Dabei sind $m$ und $n$ ganze Zahlen, wobei $m$ für die Anzahl der Knotenkreise und $n$ -für die Anzahl der Knotenlinien steht. Es gibt bestimmte Bereiche auf der Membran, in denen es keine Bewegung oder Vibration gibt. Wenn der nicht schwingende Bereich ein Kreis ist, nennt man ihn einen Knotenkreis, und wenn er eine Linie ist, nennt man ihn ebenfalls eine Knotenlinie; siehe Abbildung \ref{buch:pde:kreis:fig:pauke}. $Jn(\kappa_{mn}r)$ ist die Besselfunktion $n$-ter Ordnung, wobei $\kappa mn$ die Wellenzahl und $r$ der Radius ist. $a_{mn}$ und $b_{mn}$ sind die zu bestimmenden Konstanten. +für die Anzahl der Knotenlinien steht. Es gibt bestimmte Bereiche auf der Membran, in denen es keine Bewegung oder Vibration gibt. Wenn der nicht schwingende Bereich ein Kreis ist, nennt man ihn einen Knotenkreis, und wenn er eine Linie ist, nennt man ihn ebenfalls eine Knotenlinie (siehe Abbildung \ref{buch:pde:kreis:fig:pauke}). $J_n(\kappa_{mn}r)$ ist die Besselfunktion $n$-ter Ordnung, wobei $\kappa mn$ die Wellenzahl und $r$ der Radius ist. $a_{mn}$ und $b_{mn}$ sind die zu bestimmenden Konstanten. \begin{figure} \centering diff --git a/buch/papers/kreismembran/teil2.tex b/buch/papers/kreismembran/teil2.tex index 4fb139c..133ee31 100644 --- a/buch/papers/kreismembran/teil2.tex +++ b/buch/papers/kreismembran/teil2.tex @@ -34,7 +34,7 @@ Unter Verwendung der Integraldarstellung J_n(\kappa r)=\frac{1}{2\pi}\int_{\phi_{0}}^{2\pi + \phi_{0}}e^{i(n\alpha-\kappa r \sin \alpha)} \; d\alpha \label{equation:bessel_n_ordnung} \end{equation*} - der Besselfunktion vom Ordnung $n$ \eqref{buch:fourier:eqn:bessel-integraldarstellung} wird \eqref{equation:F_ohne_bessel} zu: + der Bessel-Funktion vom Ordnung $n$ \eqref{buch:fourier:eqn:bessel-integraldarstellung} wird \eqref{equation:F_ohne_bessel} zu: \begin{align} F(k,\phi)&=e^{in(\phi-\frac{\pi}{2})}\int_{0}^{\infty}rJ_n(\kappa r) f(r) \; dr \nonumber \\ &=e^{in(\phi-\frac{\pi}{2})}\tilde{f}_n(\kappa), @@ -69,10 +69,10 @@ verwendet werden, um die Hankel-Transformation \eqref{equation:hankel} und ihre Insbesondere die Hankel-Transformation der nullten Ordnung ($n=0$) und der ersten Ordnung ($n=1$) sind häufig nützlich, um Lösungen für Probleme mit der Laplace Gleichung in einer achsensymmetrischen zylindrischen Geometrie zu finden. \subsection{Operatoreigenschaften der Hankel-Transformation \label{sub:op_properties_hankel}} -In diesem Kapitel werden die operativen Eigenschaften der Hankel-Transformation aufgeführt. Der Beweis für ihre Gültigkeit wird jedoch nicht analysiert. +In diesem Kapitel werden die operativen Eigenschaften der Hankel-Transformation aufgeführt. Die Beweise für ihre Gültigkeit werden jedoch nicht analysiert, dies ist in Buch \textit{Integral Tansforms and Their Applications} \cite{lokenath_debnath_integral_2015} zu finden. \begin{satz}{Skalierung:} - Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann: + Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann gilt: \begin{equation*} \mathscr{H}_n\{f(ar)\}=\frac{1}{a^{2}}\tilde{f}_n \left(\frac{\kappa}{a}\right), \quad a>0. @@ -80,7 +80,7 @@ In diesem Kapitel werden die operativen Eigenschaften der Hankel-Transformation \end{satz} \begin{satz}{Parsevalsche Relation:} -Wenn $\tilde{f}(\kappa)=\mathscr{H}_n\{f(r)\}$ und $\tilde{g}(\kappa)=\mathscr{H}_n\{g(r)\}$, dann: +Wenn $\tilde{f}(\kappa)=\mathscr{H}_n\{f(r)\}$ und $\tilde{g}(\kappa)=\mathscr{H}_n\{g(r)\}$, dann gilt: \begin{equation*} \int_{0}^{\infty}rf(r)g(r) \; dr = \int_{0}^{\infty}\kappa\tilde{f}(\kappa)\tilde{g}(\kappa) \; d\kappa. @@ -88,20 +88,20 @@ Wenn $\tilde{f}(\kappa)=\mathscr{H}_n\{f(r)\}$ und $\tilde{g}(\kappa)=\mathscr{H \end{satz} \begin{satz}{Hankel-Transformationen von Ableitungen:} -Wenn $\tilde{f}_n(\kappa)=\mathscr{H}_n\{f(r)\}$, dann: +Wenn $\tilde{f}_n(\kappa)=\mathscr{H}_n\{f(r)\}$, dann gilt: \begin{align*} &\mathscr{H}_n\{f'(r)\}=\frac{\kappa}{2n}\left[(n-1)\tilde{f}_{n+1}(\kappa)-(n+1)\tilde{f}_{n-1}(\kappa)\right], \quad n\geq1, \\ &\mathscr{H}_1\{f'(r)\}=-\kappa \tilde{f}_0(\kappa), \end{align*} -vorausgesetzt dass $[rf(r)]$ verschwindet wenn $r\to0$ und $r\to\infty$. +vorausgesetzt, dass $rf(r)$ verschwindet wenn $r\to0$ und $r\to\infty$. \end{satz} \begin{satz} -Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann: +Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann gilt: \begin{equation*} \mathscr{H}_n \left\{ \left( \nabla^2 - \frac{n^2}{r^2} f(r)\right)\right\}= \mathscr{H}_n\left\{\frac{1}{r}\frac{d}{dr}\left(r\frac{df}{dr}\right) - \frac{n^2}{r^2}f(r)\right\}=-\kappa^2\tilde{f}_{n}(\kappa), \end{equation*} -bereitgestellt dass $rf'(r)$ und $rf(r)$ verschwinden für $r\to0$ und $r\to\infty$. +bereitgestellt, dass $rf'(r)$ und $rf(r)$ verschwinden für $r\to0$ und $r\to\infty$. \end{satz} diff --git a/buch/papers/kreismembran/teil3.tex b/buch/papers/kreismembran/teil3.tex index 276f911..468ee24 100644 --- a/buch/papers/kreismembran/teil3.tex +++ b/buch/papers/kreismembran/teil3.tex @@ -6,25 +6,22 @@ \section{Lösungsmethode 2: Transformationsmethode \label{kreismembran:section:teil3}} \rhead{Lösungsmethode 2: Transformationsmethode} -Die Hankel-Transformation wird dann zur Lösung der Differentialgleichung verwendet. Es müssen jedoch einige Änderungen an dem Problem vorgenommen werden, damit es mit den Annahmen übereinstimmt, die für die Verwendung der Hankel-Transformation erforderlich sind. Das heisst, dass die Funktion $u$ nur von der Entfernung zum Ausgangspunkt abhängt. +Die Hankel-Transformation kann hier zur Lösung der Differentialgleichung verwendet werden. Es müssen jedoch einige Änderungen an dem Problem vorgenommen werden, damit es mit den Annahmen übereinstimmt, die für die Verwendung der Hankel-Transformation erforderlich sind. Das heisst, dass die Funktion $u$ nur von der Entfernung zum Ausgangspunkt abhängt. \subsubsection{Transformation und Reduktion auf eine algebraische Gleichung\label{subsub:transf_reduktion}} Führt man also das Konzept einer unendlichen und achsensymmetrischen Membran ein: -\begin{equation*} +\begin{align} \frac{\partial^2u}{\partial t^2} = c^2 \left(\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} - \frac{\partial u}{\partial r} \right), \quad 00 - \label{eq:PDE_inf_membane} -\end{equation*} - -\begin{align} - u(r,0)=f(r), \quad u_t(r,0) = g(r), \quad \text{für} \quad 00 \label{eq:PDE_inf_membane} \\ + u(r,0)=f(r), \quad u_t(r,0) = g(r), \quad \text{für} \quad 0 Date: Tue, 9 Aug 2022 07:26:30 +0200 Subject: =?UTF-8?q?Vorschlag=20f=C3=BCr=20Primzahlfunktion-Graph?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- buch/papers/zeta/images/Makefile | 5 ++ buch/papers/zeta/images/primzahlfunktion2.pdf | Bin 0 -> 17496 bytes buch/papers/zeta/images/primzahlfunktion2.tex | 63 ++++++++++++++++++++++++++ 3 files changed, 68 insertions(+) create mode 100644 buch/papers/zeta/images/Makefile create mode 100644 buch/papers/zeta/images/primzahlfunktion2.pdf create mode 100644 buch/papers/zeta/images/primzahlfunktion2.tex (limited to 'buch/papers') diff --git a/buch/papers/zeta/images/Makefile b/buch/papers/zeta/images/Makefile new file mode 100644 index 0000000..c8deeec --- /dev/null +++ b/buch/papers/zeta/images/Makefile @@ -0,0 +1,5 @@ +# +# Makefile to build images +# +primzahlfunktion2.pdf: primzahlfunktion2.tex + pdflatex primzahlfunktion2.tex diff --git a/buch/papers/zeta/images/primzahlfunktion2.pdf b/buch/papers/zeta/images/primzahlfunktion2.pdf new file mode 100644 index 0000000..8998fb8 Binary files /dev/null and b/buch/papers/zeta/images/primzahlfunktion2.pdf differ diff --git a/buch/papers/zeta/images/primzahlfunktion2.tex b/buch/papers/zeta/images/primzahlfunktion2.tex new file mode 100644 index 0000000..7425ce5 --- /dev/null +++ b/buch/papers/zeta/images/primzahlfunktion2.tex @@ -0,0 +1,63 @@ +% +% primzahlfunktion2.tex -- Primzahlfunktion, alternativer Vorschlag +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usetikzlibrary{arrows,intersections,math} +\begin{document} +\def\skala{1} +\begin{tikzpicture}[>=latex,thick,scale=\skala] + +\def\dx{0.38} +\def\dy{0.5} + +\foreach \x in {1,...,30}{ + \draw[color=gray!20] ({\x*\dx},0) -- ({\x*\dx},{10.5*\dy}); +} +\foreach \y in {1,...,10}{ + \draw[color=gray!20] (0,{\y*\dy}) -- ({30.5*\dx},{\y*\dy}); +} + +\draw[->] (-0.1,0) -- ({30.8*\dx},0) coordinate[label={$x$}]; +\draw[->] (0,-0.1) -- (0,{10.9*\dy}) coordinate[label={right:$\pi(x)$}]; + +\def\segment#1#2#3{ + %\draw[line width=0.1pt] ({#3*\dx},0) -- ({#3*\dx},{#2*\dy}); + \draw[color=blue,line width=1.4pt] + ({#1*\dx},{#2*\dy}) -- ({#3*\dx},{#2*\dy}); + \draw[color=blue,line width=0.3pt] + ({#3*\dx},{#2*\dy}) -- ({#3*\dx},{(#2+1)*\dy}); + \draw ({#3*\dx},-0.1) -- ({#3*\dx},0.1); + \node at ({(#3)*\dx},-0.1) [below] {$#3\mathstrut$}; +} + +\foreach \y in {2,4,...,10}{ + \draw (-0.1,{\y*\dy}) -- (0.1,{\y*\dy}); + \node at (-0.1,{\y*\dy}) [left] {$\y\mathstrut$}; +} + +\begin{scope} +\clip (0,-0.5) rectangle ({30*\dx},{10.1*\dy}); + +\segment{0}{0}{2} +\segment{2}{1}{3} +\segment{3}{2}{5} +\segment{5}{3}{7} +\segment{7}{4}{11} +\segment{11}{5}{13} +\segment{13}{6}{17} +\segment{17}{7}{19} +\segment{19}{8}{23} +\segment{23}{9}{29} +\segment{29}{10}{31} +\end{scope} + +\end{tikzpicture} +\end{document} + -- cgit v1.2.1 From 1a6b529c9f88bd92579714a43bfa2c9fa32e6a12 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Andreas=20M=C3=BCller?= Date: Tue, 9 Aug 2022 08:27:08 +0200 Subject: add zetaplot --- buch/papers/zeta/images/Makefile | 8 ++++++ buch/papers/zeta/images/zetaplot.m | 23 +++++++++++++++++ buch/papers/zeta/images/zetaplot.pdf | Bin 0 -> 41448 bytes buch/papers/zeta/images/zetaplot.tex | 47 +++++++++++++++++++++++++++++++++++ 4 files changed, 78 insertions(+) create mode 100644 buch/papers/zeta/images/zetaplot.m create mode 100644 buch/papers/zeta/images/zetaplot.pdf create mode 100644 buch/papers/zeta/images/zetaplot.tex (limited to 'buch/papers') diff --git a/buch/papers/zeta/images/Makefile b/buch/papers/zeta/images/Makefile index c8deeec..d9cc20d 100644 --- a/buch/papers/zeta/images/Makefile +++ b/buch/papers/zeta/images/Makefile @@ -1,5 +1,13 @@ # # Makefile to build images # +all: primzahlfunktion2.pdf zetaplot.pdf + primzahlfunktion2.pdf: primzahlfunktion2.tex pdflatex primzahlfunktion2.tex + +zetapath.tex: zetaplot.m zeta.m + octave zetaplot.m + +zetaplot.pdf: zetaplot.tex zetapath.tex + pdflatex zetaplot.tex diff --git a/buch/papers/zeta/images/zetaplot.m b/buch/papers/zeta/images/zetaplot.m new file mode 100644 index 0000000..984b645 --- /dev/null +++ b/buch/papers/zeta/images/zetaplot.m @@ -0,0 +1,23 @@ +% +% zetaplot.m +% +% (c) 2022 Prof Dr Andreas Müller +% +s = 1; +h = 0.02; +m = 40; + +fn = fopen("zetapath.tex", "w"); +fprintf(fn, "\\def\\zetapath{\n"); +counter = 0; +for y = (0:h:m) + if (counter > 0) + fprintf(fn, "\n\t--"); + end + z = zeta(0.5 + i*y); + fprintf(fn, " ({%.4f*\\dx},{%.4f*\\dy})", real(z), imag(z)); + counter = counter + 1; +end +fprintf(fn, "\n}\n"); +fclose(fn); + diff --git a/buch/papers/zeta/images/zetaplot.pdf b/buch/papers/zeta/images/zetaplot.pdf new file mode 100644 index 0000000..5a59ce6 Binary files /dev/null and b/buch/papers/zeta/images/zetaplot.pdf differ diff --git a/buch/papers/zeta/images/zetaplot.tex b/buch/papers/zeta/images/zetaplot.tex new file mode 100644 index 0000000..1cd3259 --- /dev/null +++ b/buch/papers/zeta/images/zetaplot.tex @@ -0,0 +1,47 @@ +% +% zetaplot.tex -- Abbildung der kritischen Geraden +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\usepackage{csvsimple} +\usetikzlibrary{arrows,intersections,math} +\begin{document} +\def\skala{1} +\begin{tikzpicture}[>=latex,thick,scale=\skala] + +\def\dx{2} +\def\dy{2} + +\draw[->] ({-1.6*\dx},0) -- ({3.4*\dx},0) + coordinate[label={$\Re\zeta(\frac12+it)$}]; +\draw[->] (0,{-2.1*\dx}) -- (0,{2.2*\dx}) + coordinate[label={left:$\Im\zeta(\frac12+it)$}]; + +\foreach \x in {-1,1,2,3}{ + \node at ({\x*\dx},-0.1) [below] {$\x$}; +} +\node at (-0.1,{1*\dy}) [above left] {$i$}; +\node at (-0.1,{2*\dy}) [left] {$2i$}; +\node at (-0.1,{-1*\dy}) [below left] {$-i$}; +\node at (-0.1,{-2*\dy}) [left] {$-2i$}; + +\foreach \x in {-1,1,2,3}{ + \draw ({\x*\dx},-0.1) -- ({\x*\dx},0.1); +} +\foreach \y in {1,2}{ + \draw (-0.1,{\y*\dy}) -- (0.1,{\y*\dy}); + \draw (-0.1,{-\y*\dy}) -- (0.1,{-\y*\dy}); +} + +\input{zetapath.tex} + +\draw[color=blue,line width=1pt] \zetapath; + +\end{tikzpicture} +\end{document} + -- cgit v1.2.1 From 5f0bd2f0f44f0111977a9eece9ac555b06208ca9 Mon Sep 17 00:00:00 2001 From: runterer Date: Tue, 9 Aug 2022 10:05:41 +0200 Subject: used matlab to calculate zetapath.tex --- buch/papers/zeta/images/Makefile | 3 - buch/papers/zeta/images/zetapath.tex | 2003 ++++++++++++++++++++++++++++++++++ buch/papers/zeta/images/zetaplot.pdf | Bin 41448 -> 37863 bytes 3 files changed, 2003 insertions(+), 3 deletions(-) create mode 100644 buch/papers/zeta/images/zetapath.tex (limited to 'buch/papers') diff --git a/buch/papers/zeta/images/Makefile b/buch/papers/zeta/images/Makefile index d9cc20d..611662d 100644 --- a/buch/papers/zeta/images/Makefile +++ b/buch/papers/zeta/images/Makefile @@ -6,8 +6,5 @@ all: primzahlfunktion2.pdf zetaplot.pdf primzahlfunktion2.pdf: primzahlfunktion2.tex pdflatex primzahlfunktion2.tex -zetapath.tex: zetaplot.m zeta.m - octave zetaplot.m - zetaplot.pdf: zetaplot.tex zetapath.tex pdflatex zetaplot.tex diff --git a/buch/papers/zeta/images/zetapath.tex b/buch/papers/zeta/images/zetapath.tex new file mode 100644 index 0000000..75e1522 --- /dev/null +++ b/buch/papers/zeta/images/zetapath.tex @@ -0,0 +1,2003 @@ +\def\zetapath{ + ({-1.4604*\dx},{0.0000*\dy}) + -- ({-1.4572*\dx},{-0.0783*\dy}) + -- ({-1.4476*\dx},{-0.1559*\dy}) + -- ({-1.4319*\dx},{-0.2320*\dy}) + -- ({-1.4104*\dx},{-0.3058*\dy}) + -- ({-1.3834*\dx},{-0.3769*\dy}) + -- ({-1.3514*\dx},{-0.4446*\dy}) + -- ({-1.3149*\dx},{-0.5085*\dy}) + -- ({-1.2745*\dx},{-0.5682*\dy}) + -- ({-1.2308*\dx},{-0.6235*\dy}) + -- ({-1.1843*\dx},{-0.6742*\dy}) + -- ({-1.1358*\dx},{-0.7202*\dy}) + -- ({-1.0856*\dx},{-0.7617*\dy}) + -- ({-1.0344*\dx},{-0.7985*\dy}) + -- ({-0.9826*\dx},{-0.8309*\dy}) + -- ({-0.9306*\dx},{-0.8591*\dy}) + -- ({-0.8788*\dx},{-0.8833*\dy}) + -- ({-0.8275*\dx},{-0.9037*\dy}) + -- ({-0.7770*\dx},{-0.9205*\dy}) + -- ({-0.7275*\dx},{-0.9341*\dy}) + -- ({-0.6792*\dx},{-0.9446*\dy}) + -- ({-0.6322*\dx},{-0.9525*\dy}) + -- ({-0.5867*\dx},{-0.9578*\dy}) + -- ({-0.5427*\dx},{-0.9609*\dy}) + -- ({-0.5002*\dx},{-0.9620*\dy}) + -- ({-0.4593*\dx},{-0.9613*\dy}) + -- ({-0.4200*\dx},{-0.9589*\dy}) + -- ({-0.3823*\dx},{-0.9552*\dy}) + -- ({-0.3462*\dx},{-0.9502*\dy}) + -- ({-0.3116*\dx},{-0.9441*\dy}) + -- ({-0.2785*\dx},{-0.9371*\dy}) + -- ({-0.2469*\dx},{-0.9292*\dy}) + -- ({-0.2167*\dx},{-0.9206*\dy}) + -- ({-0.1878*\dx},{-0.9114*\dy}) + -- ({-0.1603*\dx},{-0.9017*\dy}) + -- ({-0.1340*\dx},{-0.8916*\dy}) + -- ({-0.1089*\dx},{-0.8811*\dy}) + -- ({-0.0849*\dx},{-0.8703*\dy}) + -- ({-0.0621*\dx},{-0.8593*\dy}) + -- ({-0.0402*\dx},{-0.8481*\dy}) + -- ({-0.0194*\dx},{-0.8367*\dy}) + -- ({0.0004*\dx},{-0.8253*\dy}) + -- ({0.0194*\dx},{-0.8137*\dy}) + -- ({0.0375*\dx},{-0.8022*\dy}) + -- ({0.0549*\dx},{-0.7906*\dy}) + -- ({0.0714*\dx},{-0.7790*\dy}) + -- ({0.0872*\dx},{-0.7675*\dy}) + -- ({0.1024*\dx},{-0.7560*\dy}) + -- ({0.1168*\dx},{-0.7446*\dy}) + -- ({0.1307*\dx},{-0.7333*\dy}) + -- ({0.1439*\dx},{-0.7221*\dy}) + -- ({0.1566*\dx},{-0.7110*\dy}) + -- ({0.1688*\dx},{-0.7000*\dy}) + -- ({0.1805*\dx},{-0.6890*\dy}) + -- ({0.1917*\dx},{-0.6783*\dy}) + -- ({0.2024*\dx},{-0.6676*\dy}) + -- ({0.2127*\dx},{-0.6571*\dy}) + -- ({0.2226*\dx},{-0.6467*\dy}) + -- ({0.2321*\dx},{-0.6364*\dy}) + -- ({0.2412*\dx},{-0.6263*\dy}) + -- ({0.2500*\dx},{-0.6163*\dy}) + -- ({0.2585*\dx},{-0.6064*\dy}) + -- ({0.2666*\dx},{-0.5967*\dy}) + -- ({0.2745*\dx},{-0.5871*\dy}) + -- ({0.2820*\dx},{-0.5777*\dy}) + -- ({0.2893*\dx},{-0.5684*\dy}) + -- ({0.2963*\dx},{-0.5592*\dy}) + -- ({0.3031*\dx},{-0.5501*\dy}) + -- ({0.3096*\dx},{-0.5412*\dy}) + -- ({0.3159*\dx},{-0.5324*\dy}) + -- ({0.3220*\dx},{-0.5237*\dy}) + -- ({0.3279*\dx},{-0.5152*\dy}) + -- ({0.3336*\dx},{-0.5067*\dy}) + -- ({0.3391*\dx},{-0.4984*\dy}) + -- ({0.3445*\dx},{-0.4902*\dy}) + -- ({0.3496*\dx},{-0.4821*\dy}) + -- ({0.3546*\dx},{-0.4742*\dy}) + -- ({0.3595*\dx},{-0.4663*\dy}) + -- ({0.3642*\dx},{-0.4586*\dy}) + -- ({0.3687*\dx},{-0.4510*\dy}) + -- ({0.3732*\dx},{-0.4434*\dy}) + -- ({0.3774*\dx},{-0.4360*\dy}) + -- ({0.3816*\dx},{-0.4287*\dy}) + -- ({0.3857*\dx},{-0.4214*\dy}) + -- ({0.3896*\dx},{-0.4143*\dy}) + -- ({0.3934*\dx},{-0.4073*\dy}) + -- ({0.3972*\dx},{-0.4003*\dy}) + -- ({0.4008*\dx},{-0.3935*\dy}) + -- ({0.4043*\dx},{-0.3867*\dy}) + -- ({0.4078*\dx},{-0.3800*\dy}) + -- ({0.4111*\dx},{-0.3734*\dy}) + -- ({0.4144*\dx},{-0.3669*\dy}) + -- ({0.4176*\dx},{-0.3605*\dy}) + -- ({0.4207*\dx},{-0.3541*\dy}) + -- ({0.4237*\dx},{-0.3478*\dy}) + -- ({0.4267*\dx},{-0.3416*\dy}) + -- ({0.4296*\dx},{-0.3355*\dy}) + -- ({0.4324*\dx},{-0.3294*\dy}) + -- ({0.4352*\dx},{-0.3234*\dy}) + -- ({0.4379*\dx},{-0.3175*\dy}) + -- ({0.4405*\dx},{-0.3116*\dy}) + -- ({0.4431*\dx},{-0.3059*\dy}) + -- ({0.4457*\dx},{-0.3001*\dy}) + -- ({0.4482*\dx},{-0.2945*\dy}) + -- ({0.4506*\dx},{-0.2889*\dy}) + -- ({0.4530*\dx},{-0.2833*\dy}) + -- ({0.4554*\dx},{-0.2778*\dy}) + -- ({0.4577*\dx},{-0.2724*\dy}) + -- ({0.4599*\dx},{-0.2670*\dy}) + -- ({0.4622*\dx},{-0.2617*\dy}) + -- ({0.4643*\dx},{-0.2565*\dy}) + -- ({0.4665*\dx},{-0.2513*\dy}) + -- ({0.4686*\dx},{-0.2461*\dy}) + -- ({0.4707*\dx},{-0.2410*\dy}) + -- ({0.4727*\dx},{-0.2359*\dy}) + -- ({0.4747*\dx},{-0.2309*\dy}) + -- ({0.4767*\dx},{-0.2259*\dy}) + -- ({0.4787*\dx},{-0.2210*\dy}) + -- ({0.4806*\dx},{-0.2161*\dy}) + -- ({0.4825*\dx},{-0.2113*\dy}) + -- ({0.4844*\dx},{-0.2065*\dy}) + -- ({0.4862*\dx},{-0.2018*\dy}) + -- ({0.4880*\dx},{-0.1971*\dy}) + -- ({0.4898*\dx},{-0.1924*\dy}) + -- ({0.4916*\dx},{-0.1878*\dy}) + -- ({0.4934*\dx},{-0.1832*\dy}) + -- ({0.4951*\dx},{-0.1786*\dy}) + -- ({0.4968*\dx},{-0.1741*\dy}) + -- ({0.4985*\dx},{-0.1696*\dy}) + -- ({0.5002*\dx},{-0.1652*\dy}) + -- ({0.5019*\dx},{-0.1608*\dy}) + -- ({0.5035*\dx},{-0.1564*\dy}) + -- ({0.5052*\dx},{-0.1521*\dy}) + -- ({0.5068*\dx},{-0.1478*\dy}) + -- ({0.5084*\dx},{-0.1435*\dy}) + -- ({0.5100*\dx},{-0.1392*\dy}) + -- ({0.5116*\dx},{-0.1350*\dy}) + -- ({0.5132*\dx},{-0.1308*\dy}) + -- ({0.5147*\dx},{-0.1267*\dy}) + -- ({0.5163*\dx},{-0.1226*\dy}) + -- ({0.5178*\dx},{-0.1185*\dy}) + -- ({0.5193*\dx},{-0.1144*\dy}) + -- ({0.5208*\dx},{-0.1103*\dy}) + -- ({0.5224*\dx},{-0.1063*\dy}) + -- ({0.5239*\dx},{-0.1023*\dy}) + -- ({0.5254*\dx},{-0.0984*\dy}) + -- ({0.5268*\dx},{-0.0944*\dy}) + -- ({0.5283*\dx},{-0.0905*\dy}) + -- ({0.5298*\dx},{-0.0866*\dy}) + -- ({0.5313*\dx},{-0.0827*\dy}) + -- ({0.5327*\dx},{-0.0789*\dy}) + -- ({0.5342*\dx},{-0.0751*\dy}) + -- ({0.5357*\dx},{-0.0713*\dy}) + -- ({0.5371*\dx},{-0.0675*\dy}) + -- ({0.5386*\dx},{-0.0637*\dy}) + -- ({0.5400*\dx},{-0.0600*\dy}) + -- ({0.5414*\dx},{-0.0563*\dy}) + -- ({0.5429*\dx},{-0.0526*\dy}) + -- ({0.5443*\dx},{-0.0489*\dy}) + -- ({0.5458*\dx},{-0.0453*\dy}) + -- ({0.5472*\dx},{-0.0416*\dy}) + -- ({0.5486*\dx},{-0.0380*\dy}) + -- ({0.5501*\dx},{-0.0344*\dy}) + -- ({0.5515*\dx},{-0.0308*\dy}) + -- ({0.5529*\dx},{-0.0272*\dy}) + -- ({0.5544*\dx},{-0.0237*\dy}) + -- ({0.5558*\dx},{-0.0202*\dy}) + -- ({0.5572*\dx},{-0.0167*\dy}) + -- ({0.5587*\dx},{-0.0132*\dy}) + -- ({0.5601*\dx},{-0.0097*\dy}) + -- ({0.5615*\dx},{-0.0062*\dy}) + -- ({0.5630*\dx},{-0.0028*\dy}) + -- ({0.5644*\dx},{0.0006*\dy}) + -- ({0.5659*\dx},{0.0041*\dy}) + -- ({0.5673*\dx},{0.0075*\dy}) + -- ({0.5688*\dx},{0.0108*\dy}) + -- ({0.5702*\dx},{0.0142*\dy}) + -- ({0.5717*\dx},{0.0176*\dy}) + -- ({0.5731*\dx},{0.0209*\dy}) + -- ({0.5746*\dx},{0.0242*\dy}) + -- ({0.5761*\dx},{0.0275*\dy}) + -- ({0.5776*\dx},{0.0308*\dy}) + -- ({0.5790*\dx},{0.0341*\dy}) + -- ({0.5805*\dx},{0.0374*\dy}) + -- ({0.5820*\dx},{0.0407*\dy}) + -- ({0.5835*\dx},{0.0439*\dy}) + -- ({0.5850*\dx},{0.0471*\dy}) + -- ({0.5865*\dx},{0.0504*\dy}) + -- ({0.5880*\dx},{0.0536*\dy}) + -- ({0.5896*\dx},{0.0568*\dy}) + -- ({0.5911*\dx},{0.0599*\dy}) + -- ({0.5926*\dx},{0.0631*\dy}) + -- ({0.5942*\dx},{0.0663*\dy}) + -- ({0.5957*\dx},{0.0694*\dy}) + -- ({0.5973*\dx},{0.0725*\dy}) + -- ({0.5988*\dx},{0.0757*\dy}) + -- ({0.6004*\dx},{0.0788*\dy}) + -- ({0.6020*\dx},{0.0819*\dy}) + -- ({0.6036*\dx},{0.0850*\dy}) + -- ({0.6052*\dx},{0.0880*\dy}) + -- ({0.6068*\dx},{0.0911*\dy}) + -- ({0.6084*\dx},{0.0942*\dy}) + -- ({0.6100*\dx},{0.0972*\dy}) + -- ({0.6117*\dx},{0.1002*\dy}) + -- ({0.6133*\dx},{0.1033*\dy}) + -- ({0.6149*\dx},{0.1063*\dy}) + -- ({0.6166*\dx},{0.1093*\dy}) + -- ({0.6183*\dx},{0.1123*\dy}) + -- ({0.6200*\dx},{0.1152*\dy}) + -- ({0.6217*\dx},{0.1182*\dy}) + -- ({0.6234*\dx},{0.1212*\dy}) + -- ({0.6251*\dx},{0.1241*\dy}) + -- ({0.6268*\dx},{0.1271*\dy}) + -- ({0.6285*\dx},{0.1300*\dy}) + -- ({0.6303*\dx},{0.1329*\dy}) + -- ({0.6320*\dx},{0.1358*\dy}) + -- ({0.6338*\dx},{0.1387*\dy}) + -- ({0.6356*\dx},{0.1416*\dy}) + -- ({0.6374*\dx},{0.1445*\dy}) + -- ({0.6392*\dx},{0.1473*\dy}) + -- ({0.6410*\dx},{0.1502*\dy}) + -- ({0.6428*\dx},{0.1530*\dy}) + -- ({0.6446*\dx},{0.1559*\dy}) + -- ({0.6465*\dx},{0.1587*\dy}) + -- ({0.6484*\dx},{0.1615*\dy}) + -- ({0.6502*\dx},{0.1643*\dy}) + -- ({0.6521*\dx},{0.1671*\dy}) + -- ({0.6540*\dx},{0.1699*\dy}) + -- ({0.6560*\dx},{0.1727*\dy}) + -- ({0.6579*\dx},{0.1754*\dy}) + -- ({0.6598*\dx},{0.1782*\dy}) + -- ({0.6618*\dx},{0.1809*\dy}) + -- ({0.6638*\dx},{0.1837*\dy}) + -- ({0.6658*\dx},{0.1864*\dy}) + -- ({0.6678*\dx},{0.1891*\dy}) + -- ({0.6698*\dx},{0.1918*\dy}) + -- ({0.6718*\dx},{0.1945*\dy}) + -- ({0.6738*\dx},{0.1972*\dy}) + -- ({0.6759*\dx},{0.1998*\dy}) + -- ({0.6780*\dx},{0.2025*\dy}) + -- ({0.6801*\dx},{0.2052*\dy}) + -- ({0.6822*\dx},{0.2078*\dy}) + -- ({0.6843*\dx},{0.2104*\dy}) + -- ({0.6864*\dx},{0.2130*\dy}) + -- ({0.6886*\dx},{0.2156*\dy}) + -- ({0.6907*\dx},{0.2182*\dy}) + -- ({0.6929*\dx},{0.2208*\dy}) + -- ({0.6951*\dx},{0.2234*\dy}) + -- ({0.6973*\dx},{0.2260*\dy}) + -- ({0.6996*\dx},{0.2285*\dy}) + -- ({0.7018*\dx},{0.2310*\dy}) + -- ({0.7041*\dx},{0.2336*\dy}) + -- ({0.7064*\dx},{0.2361*\dy}) + -- ({0.7087*\dx},{0.2386*\dy}) + -- ({0.7110*\dx},{0.2411*\dy}) + -- ({0.7133*\dx},{0.2436*\dy}) + -- ({0.7156*\dx},{0.2460*\dy}) + -- ({0.7180*\dx},{0.2485*\dy}) + -- ({0.7204*\dx},{0.2509*\dy}) + -- ({0.7228*\dx},{0.2533*\dy}) + -- ({0.7252*\dx},{0.2558*\dy}) + -- ({0.7276*\dx},{0.2582*\dy}) + -- ({0.7301*\dx},{0.2606*\dy}) + -- ({0.7326*\dx},{0.2629*\dy}) + -- ({0.7350*\dx},{0.2653*\dy}) + -- ({0.7376*\dx},{0.2677*\dy}) + -- ({0.7401*\dx},{0.2700*\dy}) + -- ({0.7426*\dx},{0.2723*\dy}) + -- ({0.7452*\dx},{0.2746*\dy}) + -- ({0.7478*\dx},{0.2769*\dy}) + -- ({0.7504*\dx},{0.2792*\dy}) + -- ({0.7530*\dx},{0.2815*\dy}) + -- ({0.7556*\dx},{0.2837*\dy}) + -- ({0.7583*\dx},{0.2860*\dy}) + -- ({0.7609*\dx},{0.2882*\dy}) + -- ({0.7636*\dx},{0.2904*\dy}) + -- ({0.7663*\dx},{0.2926*\dy}) + -- ({0.7691*\dx},{0.2948*\dy}) + -- ({0.7718*\dx},{0.2969*\dy}) + -- ({0.7746*\dx},{0.2991*\dy}) + -- ({0.7774*\dx},{0.3012*\dy}) + -- ({0.7802*\dx},{0.3033*\dy}) + -- ({0.7830*\dx},{0.3054*\dy}) + -- ({0.7858*\dx},{0.3075*\dy}) + -- ({0.7887*\dx},{0.3096*\dy}) + -- ({0.7916*\dx},{0.3116*\dy}) + -- ({0.7945*\dx},{0.3137*\dy}) + -- ({0.7974*\dx},{0.3157*\dy}) + -- ({0.8004*\dx},{0.3177*\dy}) + -- ({0.8033*\dx},{0.3197*\dy}) + -- ({0.8063*\dx},{0.3216*\dy}) + -- ({0.8093*\dx},{0.3236*\dy}) + -- ({0.8123*\dx},{0.3255*\dy}) + -- ({0.8154*\dx},{0.3274*\dy}) + -- ({0.8184*\dx},{0.3293*\dy}) + -- ({0.8215*\dx},{0.3312*\dy}) + -- ({0.8246*\dx},{0.3330*\dy}) + -- ({0.8277*\dx},{0.3348*\dy}) + -- ({0.8309*\dx},{0.3367*\dy}) + -- ({0.8340*\dx},{0.3384*\dy}) + -- ({0.8372*\dx},{0.3402*\dy}) + -- ({0.8404*\dx},{0.3420*\dy}) + -- ({0.8437*\dx},{0.3437*\dy}) + -- ({0.8469*\dx},{0.3454*\dy}) + -- ({0.8502*\dx},{0.3471*\dy}) + -- ({0.8534*\dx},{0.3487*\dy}) + -- ({0.8567*\dx},{0.3504*\dy}) + -- ({0.8601*\dx},{0.3520*\dy}) + -- ({0.8634*\dx},{0.3536*\dy}) + -- ({0.8668*\dx},{0.3552*\dy}) + -- ({0.8702*\dx},{0.3567*\dy}) + -- ({0.8736*\dx},{0.3583*\dy}) + -- ({0.8770*\dx},{0.3598*\dy}) + -- ({0.8804*\dx},{0.3612*\dy}) + -- ({0.8839*\dx},{0.3627*\dy}) + -- ({0.8874*\dx},{0.3641*\dy}) + -- ({0.8909*\dx},{0.3655*\dy}) + -- ({0.8944*\dx},{0.3669*\dy}) + -- ({0.8980*\dx},{0.3683*\dy}) + -- ({0.9015*\dx},{0.3696*\dy}) + -- ({0.9051*\dx},{0.3709*\dy}) + -- ({0.9087*\dx},{0.3722*\dy}) + -- ({0.9124*\dx},{0.3734*\dy}) + -- ({0.9160*\dx},{0.3746*\dy}) + -- ({0.9197*\dx},{0.3758*\dy}) + -- ({0.9233*\dx},{0.3770*\dy}) + -- ({0.9271*\dx},{0.3781*\dy}) + -- ({0.9308*\dx},{0.3793*\dy}) + -- ({0.9345*\dx},{0.3803*\dy}) + -- ({0.9383*\dx},{0.3814*\dy}) + -- ({0.9421*\dx},{0.3824*\dy}) + -- ({0.9459*\dx},{0.3834*\dy}) + -- ({0.9497*\dx},{0.3844*\dy}) + -- ({0.9535*\dx},{0.3853*\dy}) + -- ({0.9574*\dx},{0.3862*\dy}) + -- ({0.9612*\dx},{0.3871*\dy}) + -- ({0.9651*\dx},{0.3879*\dy}) + -- ({0.9690*\dx},{0.3887*\dy}) + -- ({0.9730*\dx},{0.3895*\dy}) + -- ({0.9769*\dx},{0.3902*\dy}) + -- ({0.9809*\dx},{0.3910*\dy}) + -- ({0.9848*\dx},{0.3916*\dy}) + -- ({0.9888*\dx},{0.3923*\dy}) + -- ({0.9929*\dx},{0.3929*\dy}) + -- ({0.9969*\dx},{0.3935*\dy}) + -- ({1.0009*\dx},{0.3940*\dy}) + -- ({1.0050*\dx},{0.3945*\dy}) + -- ({1.0091*\dx},{0.3950*\dy}) + -- ({1.0132*\dx},{0.3954*\dy}) + -- ({1.0173*\dx},{0.3958*\dy}) + -- ({1.0214*\dx},{0.3962*\dy}) + -- ({1.0256*\dx},{0.3965*\dy}) + -- ({1.0297*\dx},{0.3968*\dy}) + -- ({1.0339*\dx},{0.3971*\dy}) + -- ({1.0381*\dx},{0.3973*\dy}) + -- ({1.0423*\dx},{0.3975*\dy}) + -- ({1.0465*\dx},{0.3976*\dy}) + -- ({1.0508*\dx},{0.3977*\dy}) + -- ({1.0550*\dx},{0.3977*\dy}) + -- ({1.0593*\dx},{0.3978*\dy}) + -- ({1.0636*\dx},{0.3977*\dy}) + -- ({1.0679*\dx},{0.3977*\dy}) + -- ({1.0722*\dx},{0.3976*\dy}) + -- ({1.0765*\dx},{0.3974*\dy}) + -- ({1.0808*\dx},{0.3973*\dy}) + -- ({1.0851*\dx},{0.3970*\dy}) + -- ({1.0895*\dx},{0.3968*\dy}) + -- ({1.0938*\dx},{0.3965*\dy}) + -- ({1.0982*\dx},{0.3961*\dy}) + -- ({1.1026*\dx},{0.3957*\dy}) + -- ({1.1070*\dx},{0.3953*\dy}) + -- ({1.1114*\dx},{0.3948*\dy}) + -- ({1.1158*\dx},{0.3943*\dy}) + -- ({1.1202*\dx},{0.3937*\dy}) + -- ({1.1247*\dx},{0.3931*\dy}) + -- ({1.1291*\dx},{0.3924*\dy}) + -- ({1.1336*\dx},{0.3917*\dy}) + -- ({1.1380*\dx},{0.3909*\dy}) + -- ({1.1425*\dx},{0.3901*\dy}) + -- ({1.1469*\dx},{0.3893*\dy}) + -- ({1.1514*\dx},{0.3884*\dy}) + -- ({1.1559*\dx},{0.3875*\dy}) + -- ({1.1604*\dx},{0.3865*\dy}) + -- ({1.1649*\dx},{0.3854*\dy}) + -- ({1.1694*\dx},{0.3843*\dy}) + -- ({1.1739*\dx},{0.3832*\dy}) + -- ({1.1784*\dx},{0.3820*\dy}) + -- ({1.1829*\dx},{0.3808*\dy}) + -- ({1.1874*\dx},{0.3795*\dy}) + -- ({1.1919*\dx},{0.3782*\dy}) + -- ({1.1965*\dx},{0.3768*\dy}) + -- ({1.2010*\dx},{0.3753*\dy}) + -- ({1.2055*\dx},{0.3738*\dy}) + -- ({1.2100*\dx},{0.3723*\dy}) + -- ({1.2145*\dx},{0.3707*\dy}) + -- ({1.2190*\dx},{0.3691*\dy}) + -- ({1.2236*\dx},{0.3674*\dy}) + -- ({1.2281*\dx},{0.3656*\dy}) + -- ({1.2326*\dx},{0.3638*\dy}) + -- ({1.2371*\dx},{0.3620*\dy}) + -- ({1.2416*\dx},{0.3600*\dy}) + -- ({1.2461*\dx},{0.3581*\dy}) + -- ({1.2506*\dx},{0.3561*\dy}) + -- ({1.2551*\dx},{0.3540*\dy}) + -- ({1.2596*\dx},{0.3519*\dy}) + -- ({1.2641*\dx},{0.3497*\dy}) + -- ({1.2686*\dx},{0.3475*\dy}) + -- ({1.2730*\dx},{0.3452*\dy}) + -- ({1.2775*\dx},{0.3428*\dy}) + -- ({1.2819*\dx},{0.3404*\dy}) + -- ({1.2864*\dx},{0.3379*\dy}) + -- ({1.2908*\dx},{0.3354*\dy}) + -- ({1.2952*\dx},{0.3329*\dy}) + -- ({1.2996*\dx},{0.3302*\dy}) + -- ({1.3040*\dx},{0.3275*\dy}) + -- ({1.3084*\dx},{0.3248*\dy}) + -- ({1.3128*\dx},{0.3220*\dy}) + -- ({1.3171*\dx},{0.3191*\dy}) + -- ({1.3215*\dx},{0.3162*\dy}) + -- ({1.3258*\dx},{0.3132*\dy}) + -- ({1.3301*\dx},{0.3102*\dy}) + -- ({1.3344*\dx},{0.3071*\dy}) + -- ({1.3386*\dx},{0.3040*\dy}) + -- ({1.3429*\dx},{0.3008*\dy}) + -- ({1.3471*\dx},{0.2975*\dy}) + -- ({1.3513*\dx},{0.2942*\dy}) + -- ({1.3555*\dx},{0.2908*\dy}) + -- ({1.3597*\dx},{0.2874*\dy}) + -- ({1.3638*\dx},{0.2839*\dy}) + -- ({1.3680*\dx},{0.2803*\dy}) + -- ({1.3721*\dx},{0.2767*\dy}) + -- ({1.3761*\dx},{0.2730*\dy}) + -- ({1.3802*\dx},{0.2693*\dy}) + -- ({1.3842*\dx},{0.2655*\dy}) + -- ({1.3882*\dx},{0.2616*\dy}) + -- ({1.3922*\dx},{0.2577*\dy}) + -- ({1.3961*\dx},{0.2537*\dy}) + -- ({1.4000*\dx},{0.2497*\dy}) + -- ({1.4039*\dx},{0.2456*\dy}) + -- ({1.4077*\dx},{0.2414*\dy}) + -- ({1.4115*\dx},{0.2372*\dy}) + -- ({1.4153*\dx},{0.2329*\dy}) + -- ({1.4191*\dx},{0.2286*\dy}) + -- ({1.4228*\dx},{0.2242*\dy}) + -- ({1.4264*\dx},{0.2197*\dy}) + -- ({1.4301*\dx},{0.2152*\dy}) + -- ({1.4337*\dx},{0.2107*\dy}) + -- ({1.4372*\dx},{0.2060*\dy}) + -- ({1.4407*\dx},{0.2014*\dy}) + -- ({1.4442*\dx},{0.1966*\dy}) + -- ({1.4476*\dx},{0.1918*\dy}) + -- ({1.4510*\dx},{0.1869*\dy}) + -- ({1.4544*\dx},{0.1820*\dy}) + -- ({1.4577*\dx},{0.1770*\dy}) + -- ({1.4609*\dx},{0.1720*\dy}) + -- ({1.4642*\dx},{0.1669*\dy}) + -- ({1.4673*\dx},{0.1618*\dy}) + -- ({1.4704*\dx},{0.1566*\dy}) + -- ({1.4735*\dx},{0.1513*\dy}) + -- ({1.4765*\dx},{0.1460*\dy}) + -- ({1.4795*\dx},{0.1406*\dy}) + -- ({1.4824*\dx},{0.1352*\dy}) + -- ({1.4853*\dx},{0.1297*\dy}) + -- ({1.4881*\dx},{0.1241*\dy}) + -- ({1.4908*\dx},{0.1185*\dy}) + -- ({1.4935*\dx},{0.1129*\dy}) + -- ({1.4961*\dx},{0.1072*\dy}) + -- ({1.4987*\dx},{0.1014*\dy}) + -- ({1.5012*\dx},{0.0956*\dy}) + -- ({1.5037*\dx},{0.0897*\dy}) + -- ({1.5061*\dx},{0.0838*\dy}) + -- ({1.5084*\dx},{0.0778*\dy}) + -- ({1.5107*\dx},{0.0718*\dy}) + -- ({1.5129*\dx},{0.0657*\dy}) + -- ({1.5151*\dx},{0.0596*\dy}) + -- ({1.5172*\dx},{0.0534*\dy}) + -- ({1.5192*\dx},{0.0472*\dy}) + -- ({1.5211*\dx},{0.0409*\dy}) + -- ({1.5230*\dx},{0.0346*\dy}) + -- ({1.5248*\dx},{0.0282*\dy}) + -- ({1.5265*\dx},{0.0218*\dy}) + -- ({1.5282*\dx},{0.0153*\dy}) + -- ({1.5298*\dx},{0.0088*\dy}) + -- ({1.5313*\dx},{0.0023*\dy}) + -- ({1.5328*\dx},{-0.0043*\dy}) + -- ({1.5341*\dx},{-0.0110*\dy}) + -- ({1.5354*\dx},{-0.0177*\dy}) + -- ({1.5366*\dx},{-0.0244*\dy}) + -- ({1.5378*\dx},{-0.0312*\dy}) + -- ({1.5388*\dx},{-0.0380*\dy}) + -- ({1.5398*\dx},{-0.0448*\dy}) + -- ({1.5407*\dx},{-0.0517*\dy}) + -- ({1.5415*\dx},{-0.0586*\dy}) + -- ({1.5422*\dx},{-0.0656*\dy}) + -- ({1.5429*\dx},{-0.0726*\dy}) + -- ({1.5434*\dx},{-0.0796*\dy}) + -- ({1.5439*\dx},{-0.0867*\dy}) + -- ({1.5443*\dx},{-0.0938*\dy}) + -- ({1.5446*\dx},{-0.1010*\dy}) + -- ({1.5448*\dx},{-0.1081*\dy}) + -- ({1.5449*\dx},{-0.1153*\dy}) + -- ({1.5449*\dx},{-0.1226*\dy}) + -- ({1.5449*\dx},{-0.1298*\dy}) + -- ({1.5447*\dx},{-0.1371*\dy}) + -- ({1.5444*\dx},{-0.1444*\dy}) + -- ({1.5441*\dx},{-0.1518*\dy}) + -- ({1.5437*\dx},{-0.1591*\dy}) + -- ({1.5431*\dx},{-0.1665*\dy}) + -- ({1.5425*\dx},{-0.1740*\dy}) + -- ({1.5418*\dx},{-0.1814*\dy}) + -- ({1.5409*\dx},{-0.1888*\dy}) + -- ({1.5400*\dx},{-0.1963*\dy}) + -- ({1.5390*\dx},{-0.2038*\dy}) + -- ({1.5378*\dx},{-0.2113*\dy}) + -- ({1.5366*\dx},{-0.2188*\dy}) + -- ({1.5353*\dx},{-0.2264*\dy}) + -- ({1.5339*\dx},{-0.2339*\dy}) + -- ({1.5323*\dx},{-0.2415*\dy}) + -- ({1.5307*\dx},{-0.2491*\dy}) + -- ({1.5289*\dx},{-0.2566*\dy}) + -- ({1.5271*\dx},{-0.2642*\dy}) + -- ({1.5251*\dx},{-0.2718*\dy}) + -- ({1.5230*\dx},{-0.2794*\dy}) + -- ({1.5209*\dx},{-0.2870*\dy}) + -- ({1.5186*\dx},{-0.2946*\dy}) + -- ({1.5162*\dx},{-0.3022*\dy}) + -- ({1.5137*\dx},{-0.3098*\dy}) + -- ({1.5111*\dx},{-0.3174*\dy}) + -- ({1.5084*\dx},{-0.3250*\dy}) + -- ({1.5055*\dx},{-0.3326*\dy}) + -- ({1.5026*\dx},{-0.3402*\dy}) + -- ({1.4995*\dx},{-0.3477*\dy}) + -- ({1.4963*\dx},{-0.3553*\dy}) + -- ({1.4931*\dx},{-0.3628*\dy}) + -- ({1.4897*\dx},{-0.3704*\dy}) + -- ({1.4862*\dx},{-0.3779*\dy}) + -- ({1.4825*\dx},{-0.3854*\dy}) + -- ({1.4788*\dx},{-0.3929*\dy}) + -- ({1.4749*\dx},{-0.4003*\dy}) + -- ({1.4710*\dx},{-0.4078*\dy}) + -- ({1.4669*\dx},{-0.4152*\dy}) + -- ({1.4627*\dx},{-0.4226*\dy}) + -- ({1.4584*\dx},{-0.4300*\dy}) + -- ({1.4539*\dx},{-0.4373*\dy}) + -- ({1.4494*\dx},{-0.4446*\dy}) + -- ({1.4447*\dx},{-0.4519*\dy}) + -- ({1.4399*\dx},{-0.4591*\dy}) + -- ({1.4350*\dx},{-0.4663*\dy}) + -- ({1.4300*\dx},{-0.4735*\dy}) + -- ({1.4249*\dx},{-0.4806*\dy}) + -- ({1.4197*\dx},{-0.4877*\dy}) + -- ({1.4143*\dx},{-0.4947*\dy}) + -- ({1.4088*\dx},{-0.5017*\dy}) + -- ({1.4032*\dx},{-0.5086*\dy}) + -- ({1.3975*\dx},{-0.5155*\dy}) + -- ({1.3916*\dx},{-0.5224*\dy}) + -- ({1.3857*\dx},{-0.5292*\dy}) + -- ({1.3796*\dx},{-0.5359*\dy}) + -- ({1.3734*\dx},{-0.5426*\dy}) + -- ({1.3671*\dx},{-0.5492*\dy}) + -- ({1.3607*\dx},{-0.5558*\dy}) + -- ({1.3542*\dx},{-0.5623*\dy}) + -- ({1.3475*\dx},{-0.5687*\dy}) + -- ({1.3408*\dx},{-0.5751*\dy}) + -- ({1.3339*\dx},{-0.5814*\dy}) + -- ({1.3269*\dx},{-0.5876*\dy}) + -- ({1.3198*\dx},{-0.5937*\dy}) + -- ({1.3126*\dx},{-0.5998*\dy}) + -- ({1.3052*\dx},{-0.6058*\dy}) + -- ({1.2978*\dx},{-0.6117*\dy}) + -- ({1.2902*\dx},{-0.6176*\dy}) + -- ({1.2826*\dx},{-0.6233*\dy}) + -- ({1.2748*\dx},{-0.6290*\dy}) + -- ({1.2669*\dx},{-0.6346*\dy}) + -- ({1.2589*\dx},{-0.6401*\dy}) + -- ({1.2508*\dx},{-0.6455*\dy}) + -- ({1.2426*\dx},{-0.6508*\dy}) + -- ({1.2343*\dx},{-0.6560*\dy}) + -- ({1.2259*\dx},{-0.6611*\dy}) + -- ({1.2173*\dx},{-0.6662*\dy}) + -- ({1.2087*\dx},{-0.6711*\dy}) + -- ({1.2000*\dx},{-0.6759*\dy}) + -- ({1.1911*\dx},{-0.6806*\dy}) + -- ({1.1822*\dx},{-0.6852*\dy}) + -- ({1.1731*\dx},{-0.6897*\dy}) + -- ({1.1640*\dx},{-0.6941*\dy}) + -- ({1.1548*\dx},{-0.6984*\dy}) + -- ({1.1454*\dx},{-0.7025*\dy}) + -- ({1.1360*\dx},{-0.7065*\dy}) + -- ({1.1265*\dx},{-0.7105*\dy}) + -- ({1.1169*\dx},{-0.7143*\dy}) + -- ({1.1072*\dx},{-0.7179*\dy}) + -- ({1.0974*\dx},{-0.7215*\dy}) + -- ({1.0875*\dx},{-0.7249*\dy}) + -- ({1.0775*\dx},{-0.7282*\dy}) + -- ({1.0674*\dx},{-0.7313*\dy}) + -- ({1.0573*\dx},{-0.7344*\dy}) + -- ({1.0471*\dx},{-0.7373*\dy}) + -- ({1.0368*\dx},{-0.7400*\dy}) + -- ({1.0264*\dx},{-0.7426*\dy}) + -- ({1.0159*\dx},{-0.7451*\dy}) + -- ({1.0054*\dx},{-0.7474*\dy}) + -- ({0.9948*\dx},{-0.7496*\dy}) + -- ({0.9841*\dx},{-0.7517*\dy}) + -- ({0.9734*\dx},{-0.7536*\dy}) + -- ({0.9625*\dx},{-0.7553*\dy}) + -- ({0.9516*\dx},{-0.7569*\dy}) + -- ({0.9407*\dx},{-0.7584*\dy}) + -- ({0.9297*\dx},{-0.7596*\dy}) + -- ({0.9186*\dx},{-0.7608*\dy}) + -- ({0.9075*\dx},{-0.7617*\dy}) + -- ({0.8963*\dx},{-0.7626*\dy}) + -- ({0.8850*\dx},{-0.7632*\dy}) + -- ({0.8738*\dx},{-0.7637*\dy}) + -- ({0.8624*\dx},{-0.7640*\dy}) + -- ({0.8510*\dx},{-0.7642*\dy}) + -- ({0.8396*\dx},{-0.7642*\dy}) + -- ({0.8281*\dx},{-0.7640*\dy}) + -- ({0.8166*\dx},{-0.7636*\dy}) + -- ({0.8050*\dx},{-0.7631*\dy}) + -- ({0.7934*\dx},{-0.7624*\dy}) + -- ({0.7818*\dx},{-0.7615*\dy}) + -- ({0.7702*\dx},{-0.7605*\dy}) + -- ({0.7585*\dx},{-0.7593*\dy}) + -- ({0.7468*\dx},{-0.7579*\dy}) + -- ({0.7350*\dx},{-0.7563*\dy}) + -- ({0.7233*\dx},{-0.7545*\dy}) + -- ({0.7115*\dx},{-0.7526*\dy}) + -- ({0.6998*\dx},{-0.7504*\dy}) + -- ({0.6880*\dx},{-0.7481*\dy}) + -- ({0.6762*\dx},{-0.7456*\dy}) + -- ({0.6644*\dx},{-0.7429*\dy}) + -- ({0.6526*\dx},{-0.7401*\dy}) + -- ({0.6407*\dx},{-0.7370*\dy}) + -- ({0.6289*\dx},{-0.7337*\dy}) + -- ({0.6171*\dx},{-0.7303*\dy}) + -- ({0.6054*\dx},{-0.7267*\dy}) + -- ({0.5936*\dx},{-0.7229*\dy}) + -- ({0.5818*\dx},{-0.7188*\dy}) + -- ({0.5701*\dx},{-0.7146*\dy}) + -- ({0.5583*\dx},{-0.7102*\dy}) + -- ({0.5466*\dx},{-0.7056*\dy}) + -- ({0.5350*\dx},{-0.7008*\dy}) + -- ({0.5233*\dx},{-0.6959*\dy}) + -- ({0.5117*\dx},{-0.6907*\dy}) + -- ({0.5002*\dx},{-0.6853*\dy}) + -- ({0.4886*\dx},{-0.6797*\dy}) + -- ({0.4772*\dx},{-0.6740*\dy}) + -- ({0.4657*\dx},{-0.6680*\dy}) + -- ({0.4543*\dx},{-0.6618*\dy}) + -- ({0.4430*\dx},{-0.6555*\dy}) + -- ({0.4317*\dx},{-0.6489*\dy}) + -- ({0.4205*\dx},{-0.6422*\dy}) + -- ({0.4094*\dx},{-0.6352*\dy}) + -- ({0.3983*\dx},{-0.6281*\dy}) + -- ({0.3873*\dx},{-0.6208*\dy}) + -- ({0.3763*\dx},{-0.6132*\dy}) + -- ({0.3655*\dx},{-0.6055*\dy}) + -- ({0.3547*\dx},{-0.5976*\dy}) + -- ({0.3440*\dx},{-0.5894*\dy}) + -- ({0.3334*\dx},{-0.5811*\dy}) + -- ({0.3229*\dx},{-0.5726*\dy}) + -- ({0.3125*\dx},{-0.5639*\dy}) + -- ({0.3022*\dx},{-0.5550*\dy}) + -- ({0.2920*\dx},{-0.5459*\dy}) + -- ({0.2818*\dx},{-0.5367*\dy}) + -- ({0.2718*\dx},{-0.5272*\dy}) + -- ({0.2620*\dx},{-0.5175*\dy}) + -- ({0.2522*\dx},{-0.5077*\dy}) + -- ({0.2425*\dx},{-0.4977*\dy}) + -- ({0.2330*\dx},{-0.4875*\dy}) + -- ({0.2236*\dx},{-0.4771*\dy}) + -- ({0.2144*\dx},{-0.4665*\dy}) + -- ({0.2052*\dx},{-0.4557*\dy}) + -- ({0.1962*\dx},{-0.4448*\dy}) + -- ({0.1874*\dx},{-0.4337*\dy}) + -- ({0.1787*\dx},{-0.4224*\dy}) + -- ({0.1701*\dx},{-0.4109*\dy}) + -- ({0.1617*\dx},{-0.3992*\dy}) + -- ({0.1535*\dx},{-0.3874*\dy}) + -- ({0.1454*\dx},{-0.3754*\dy}) + -- ({0.1375*\dx},{-0.3633*\dy}) + -- ({0.1297*\dx},{-0.3509*\dy}) + -- ({0.1221*\dx},{-0.3384*\dy}) + -- ({0.1147*\dx},{-0.3258*\dy}) + -- ({0.1074*\dx},{-0.3130*\dy}) + -- ({0.1004*\dx},{-0.3000*\dy}) + -- ({0.0935*\dx},{-0.2869*\dy}) + -- ({0.0868*\dx},{-0.2736*\dy}) + -- ({0.0803*\dx},{-0.2602*\dy}) + -- ({0.0740*\dx},{-0.2466*\dy}) + -- ({0.0679*\dx},{-0.2329*\dy}) + -- ({0.0620*\dx},{-0.2190*\dy}) + -- ({0.0563*\dx},{-0.2050*\dy}) + -- ({0.0508*\dx},{-0.1908*\dy}) + -- ({0.0455*\dx},{-0.1766*\dy}) + -- ({0.0404*\dx},{-0.1622*\dy}) + -- ({0.0355*\dx},{-0.1476*\dy}) + -- ({0.0309*\dx},{-0.1330*\dy}) + -- ({0.0264*\dx},{-0.1182*\dy}) + -- ({0.0222*\dx},{-0.1033*\dy}) + -- ({0.0183*\dx},{-0.0882*\dy}) + -- ({0.0145*\dx},{-0.0731*\dy}) + -- ({0.0110*\dx},{-0.0579*\dy}) + -- ({0.0077*\dx},{-0.0425*\dy}) + -- ({0.0047*\dx},{-0.0271*\dy}) + -- ({0.0019*\dx},{-0.0115*\dy}) + -- ({-0.0006*\dx},{0.0041*\dy}) + -- ({-0.0030*\dx},{0.0199*\dy}) + -- ({-0.0050*\dx},{0.0357*\dy}) + -- ({-0.0068*\dx},{0.0516*\dy}) + -- ({-0.0084*\dx},{0.0676*\dy}) + -- ({-0.0097*\dx},{0.0836*\dy}) + -- ({-0.0107*\dx},{0.0998*\dy}) + -- ({-0.0115*\dx},{0.1160*\dy}) + -- ({-0.0120*\dx},{0.1322*\dy}) + -- ({-0.0122*\dx},{0.1486*\dy}) + -- ({-0.0122*\dx},{0.1649*\dy}) + -- ({-0.0119*\dx},{0.1813*\dy}) + -- ({-0.0114*\dx},{0.1978*\dy}) + -- ({-0.0105*\dx},{0.2143*\dy}) + -- ({-0.0094*\dx},{0.2309*\dy}) + -- ({-0.0080*\dx},{0.2475*\dy}) + -- ({-0.0064*\dx},{0.2641*\dy}) + -- ({-0.0044*\dx},{0.2807*\dy}) + -- ({-0.0022*\dx},{0.2973*\dy}) + -- ({0.0003*\dx},{0.3140*\dy}) + -- ({0.0031*\dx},{0.3307*\dy}) + -- ({0.0061*\dx},{0.3473*\dy}) + -- ({0.0095*\dx},{0.3640*\dy}) + -- ({0.0131*\dx},{0.3807*\dy}) + -- ({0.0171*\dx},{0.3973*\dy}) + -- ({0.0213*\dx},{0.4140*\dy}) + -- ({0.0258*\dx},{0.4306*\dy}) + -- ({0.0306*\dx},{0.4472*\dy}) + -- ({0.0357*\dx},{0.4637*\dy}) + -- ({0.0411*\dx},{0.4803*\dy}) + -- ({0.0467*\dx},{0.4967*\dy}) + -- ({0.0527*\dx},{0.5132*\dy}) + -- ({0.0590*\dx},{0.5296*\dy}) + -- ({0.0655*\dx},{0.5459*\dy}) + -- ({0.0724*\dx},{0.5621*\dy}) + -- ({0.0795*\dx},{0.5783*\dy}) + -- ({0.0869*\dx},{0.5945*\dy}) + -- ({0.0947*\dx},{0.6105*\dy}) + -- ({0.1027*\dx},{0.6264*\dy}) + -- ({0.1110*\dx},{0.6423*\dy}) + -- ({0.1196*\dx},{0.6581*\dy}) + -- ({0.1285*\dx},{0.6737*\dy}) + -- ({0.1376*\dx},{0.6893*\dy}) + -- ({0.1471*\dx},{0.7048*\dy}) + -- ({0.1569*\dx},{0.7201*\dy}) + -- ({0.1669*\dx},{0.7353*\dy}) + -- ({0.1772*\dx},{0.7504*\dy}) + -- ({0.1878*\dx},{0.7653*\dy}) + -- ({0.1987*\dx},{0.7801*\dy}) + -- ({0.2099*\dx},{0.7948*\dy}) + -- ({0.2214*\dx},{0.8093*\dy}) + -- ({0.2331*\dx},{0.8236*\dy}) + -- ({0.2451*\dx},{0.8378*\dy}) + -- ({0.2574*\dx},{0.8519*\dy}) + -- ({0.2700*\dx},{0.8657*\dy}) + -- ({0.2828*\dx},{0.8794*\dy}) + -- ({0.2959*\dx},{0.8929*\dy}) + -- ({0.3093*\dx},{0.9062*\dy}) + -- ({0.3229*\dx},{0.9193*\dy}) + -- ({0.3368*\dx},{0.9322*\dy}) + -- ({0.3509*\dx},{0.9449*\dy}) + -- ({0.3653*\dx},{0.9574*\dy}) + -- ({0.3800*\dx},{0.9697*\dy}) + -- ({0.3949*\dx},{0.9817*\dy}) + -- ({0.4100*\dx},{0.9936*\dy}) + -- ({0.4254*\dx},{1.0052*\dy}) + -- ({0.4411*\dx},{1.0165*\dy}) + -- ({0.4569*\dx},{1.0276*\dy}) + -- ({0.4730*\dx},{1.0385*\dy}) + -- ({0.4894*\dx},{1.0491*\dy}) + -- ({0.5059*\dx},{1.0595*\dy}) + -- ({0.5227*\dx},{1.0696*\dy}) + -- ({0.5397*\dx},{1.0795*\dy}) + -- ({0.5569*\dx},{1.0890*\dy}) + -- ({0.5743*\dx},{1.0983*\dy}) + -- ({0.5919*\dx},{1.1073*\dy}) + -- ({0.6098*\dx},{1.1160*\dy}) + -- ({0.6278*\dx},{1.1245*\dy}) + -- ({0.6460*\dx},{1.1326*\dy}) + -- ({0.6644*\dx},{1.1405*\dy}) + -- ({0.6830*\dx},{1.1480*\dy}) + -- ({0.7017*\dx},{1.1552*\dy}) + -- ({0.7206*\dx},{1.1622*\dy}) + -- ({0.7397*\dx},{1.1688*\dy}) + -- ({0.7590*\dx},{1.1750*\dy}) + -- ({0.7784*\dx},{1.1810*\dy}) + -- ({0.7979*\dx},{1.1866*\dy}) + -- ({0.8176*\dx},{1.1919*\dy}) + -- ({0.8375*\dx},{1.1969*\dy}) + -- ({0.8574*\dx},{1.2015*\dy}) + -- ({0.8775*\dx},{1.2058*\dy}) + -- ({0.8978*\dx},{1.2098*\dy}) + -- ({0.9181*\dx},{1.2133*\dy}) + -- ({0.9385*\dx},{1.2166*\dy}) + -- ({0.9591*\dx},{1.2195*\dy}) + -- ({0.9797*\dx},{1.2220*\dy}) + -- ({1.0005*\dx},{1.2242*\dy}) + -- ({1.0213*\dx},{1.2260*\dy}) + -- ({1.0422*\dx},{1.2274*\dy}) + -- ({1.0632*\dx},{1.2285*\dy}) + -- ({1.0842*\dx},{1.2292*\dy}) + -- ({1.1053*\dx},{1.2295*\dy}) + -- ({1.1264*\dx},{1.2294*\dy}) + -- ({1.1476*\dx},{1.2290*\dy}) + -- ({1.1688*\dx},{1.2282*\dy}) + -- ({1.1901*\dx},{1.2270*\dy}) + -- ({1.2114*\dx},{1.2254*\dy}) + -- ({1.2327*\dx},{1.2234*\dy}) + -- ({1.2540*\dx},{1.2211*\dy}) + -- ({1.2753*\dx},{1.2184*\dy}) + -- ({1.2965*\dx},{1.2152*\dy}) + -- ({1.3178*\dx},{1.2117*\dy}) + -- ({1.3391*\dx},{1.2078*\dy}) + -- ({1.3603*\dx},{1.2035*\dy}) + -- ({1.3815*\dx},{1.1988*\dy}) + -- ({1.4027*\dx},{1.1938*\dy}) + -- ({1.4238*\dx},{1.1883*\dy}) + -- ({1.4448*\dx},{1.1824*\dy}) + -- ({1.4658*\dx},{1.1762*\dy}) + -- ({1.4867*\dx},{1.1695*\dy}) + -- ({1.5076*\dx},{1.1625*\dy}) + -- ({1.5283*\dx},{1.1550*\dy}) + -- ({1.5489*\dx},{1.1472*\dy}) + -- ({1.5695*\dx},{1.1390*\dy}) + -- ({1.5899*\dx},{1.1304*\dy}) + -- ({1.6102*\dx},{1.1214*\dy}) + -- ({1.6304*\dx},{1.1120*\dy}) + -- ({1.6504*\dx},{1.1023*\dy}) + -- ({1.6703*\dx},{1.0921*\dy}) + -- ({1.6901*\dx},{1.0816*\dy}) + -- ({1.7097*\dx},{1.0707*\dy}) + -- ({1.7291*\dx},{1.0594*\dy}) + -- ({1.7484*\dx},{1.0477*\dy}) + -- ({1.7675*\dx},{1.0357*\dy}) + -- ({1.7864*\dx},{1.0233*\dy}) + -- ({1.8051*\dx},{1.0105*\dy}) + -- ({1.8235*\dx},{0.9974*\dy}) + -- ({1.8418*\dx},{0.9839*\dy}) + -- ({1.8599*\dx},{0.9700*\dy}) + -- ({1.8777*\dx},{0.9558*\dy}) + -- ({1.8953*\dx},{0.9412*\dy}) + -- ({1.9127*\dx},{0.9263*\dy}) + -- ({1.9298*\dx},{0.9110*\dy}) + -- ({1.9467*\dx},{0.8954*\dy}) + -- ({1.9632*\dx},{0.8795*\dy}) + -- ({1.9796*\dx},{0.8632*\dy}) + -- ({1.9956*\dx},{0.8466*\dy}) + -- ({2.0114*\dx},{0.8296*\dy}) + -- ({2.0269*\dx},{0.8124*\dy}) + -- ({2.0420*\dx},{0.7948*\dy}) + -- ({2.0569*\dx},{0.7770*\dy}) + -- ({2.0715*\dx},{0.7588*\dy}) + -- ({2.0857*\dx},{0.7403*\dy}) + -- ({2.0996*\dx},{0.7215*\dy}) + -- ({2.1132*\dx},{0.7025*\dy}) + -- ({2.1264*\dx},{0.6831*\dy}) + -- ({2.1393*\dx},{0.6635*\dy}) + -- ({2.1519*\dx},{0.6437*\dy}) + -- ({2.1641*\dx},{0.6235*\dy}) + -- ({2.1759*\dx},{0.6031*\dy}) + -- ({2.1874*\dx},{0.5825*\dy}) + -- ({2.1985*\dx},{0.5616*\dy}) + -- ({2.2092*\dx},{0.5404*\dy}) + -- ({2.2195*\dx},{0.5191*\dy}) + -- ({2.2295*\dx},{0.4975*\dy}) + -- ({2.2390*\dx},{0.4757*\dy}) + -- ({2.2481*\dx},{0.4537*\dy}) + -- ({2.2569*\dx},{0.4315*\dy}) + -- ({2.2652*\dx},{0.4091*\dy}) + -- ({2.2731*\dx},{0.3865*\dy}) + -- ({2.2806*\dx},{0.3638*\dy}) + -- ({2.2877*\dx},{0.3408*\dy}) + -- ({2.2943*\dx},{0.3177*\dy}) + -- ({2.3005*\dx},{0.2945*\dy}) + -- ({2.3063*\dx},{0.2711*\dy}) + -- ({2.3116*\dx},{0.2476*\dy}) + -- ({2.3165*\dx},{0.2239*\dy}) + -- ({2.3210*\dx},{0.2002*\dy}) + -- ({2.3249*\dx},{0.1763*\dy}) + -- ({2.3285*\dx},{0.1523*\dy}) + -- ({2.3316*\dx},{0.1283*\dy}) + -- ({2.3342*\dx},{0.1041*\dy}) + -- ({2.3364*\dx},{0.0799*\dy}) + -- ({2.3381*\dx},{0.0556*\dy}) + -- ({2.3393*\dx},{0.0312*\dy}) + -- ({2.3401*\dx},{0.0068*\dy}) + -- ({2.3403*\dx},{-0.0176*\dy}) + -- ({2.3402*\dx},{-0.0421*\dy}) + -- ({2.3395*\dx},{-0.0665*\dy}) + -- ({2.3384*\dx},{-0.0910*\dy}) + -- ({2.3368*\dx},{-0.1155*\dy}) + -- ({2.3347*\dx},{-0.1400*\dy}) + -- ({2.3322*\dx},{-0.1644*\dy}) + -- ({2.3292*\dx},{-0.1889*\dy}) + -- ({2.3257*\dx},{-0.2133*\dy}) + -- ({2.3217*\dx},{-0.2376*\dy}) + -- ({2.3172*\dx},{-0.2619*\dy}) + -- ({2.3123*\dx},{-0.2861*\dy}) + -- ({2.3069*\dx},{-0.3102*\dy}) + -- ({2.3010*\dx},{-0.3342*\dy}) + -- ({2.2946*\dx},{-0.3582*\dy}) + -- ({2.2878*\dx},{-0.3820*\dy}) + -- ({2.2805*\dx},{-0.4057*\dy}) + -- ({2.2727*\dx},{-0.4293*\dy}) + -- ({2.2645*\dx},{-0.4528*\dy}) + -- ({2.2558*\dx},{-0.4761*\dy}) + -- ({2.2466*\dx},{-0.4992*\dy}) + -- ({2.2370*\dx},{-0.5222*\dy}) + -- ({2.2269*\dx},{-0.5450*\dy}) + -- ({2.2164*\dx},{-0.5676*\dy}) + -- ({2.2054*\dx},{-0.5900*\dy}) + -- ({2.1939*\dx},{-0.6122*\dy}) + -- ({2.1820*\dx},{-0.6342*\dy}) + -- ({2.1697*\dx},{-0.6560*\dy}) + -- ({2.1569*\dx},{-0.6775*\dy}) + -- ({2.1437*\dx},{-0.6988*\dy}) + -- ({2.1301*\dx},{-0.7198*\dy}) + -- ({2.1161*\dx},{-0.7406*\dy}) + -- ({2.1016*\dx},{-0.7611*\dy}) + -- ({2.0867*\dx},{-0.7813*\dy}) + -- ({2.0714*\dx},{-0.8012*\dy}) + -- ({2.0558*\dx},{-0.8208*\dy}) + -- ({2.0397*\dx},{-0.8401*\dy}) + -- ({2.0232*\dx},{-0.8591*\dy}) + -- ({2.0063*\dx},{-0.8778*\dy}) + -- ({1.9891*\dx},{-0.8961*\dy}) + -- ({1.9715*\dx},{-0.9141*\dy}) + -- ({1.9535*\dx},{-0.9318*\dy}) + -- ({1.9352*\dx},{-0.9491*\dy}) + -- ({1.9165*\dx},{-0.9660*\dy}) + -- ({1.8974*\dx},{-0.9825*\dy}) + -- ({1.8781*\dx},{-0.9987*\dy}) + -- ({1.8584*\dx},{-1.0144*\dy}) + -- ({1.8384*\dx},{-1.0298*\dy}) + -- ({1.8181*\dx},{-1.0448*\dy}) + -- ({1.7974*\dx},{-1.0593*\dy}) + -- ({1.7765*\dx},{-1.0735*\dy}) + -- ({1.7553*\dx},{-1.0872*\dy}) + -- ({1.7338*\dx},{-1.1004*\dy}) + -- ({1.7120*\dx},{-1.1133*\dy}) + -- ({1.6900*\dx},{-1.1257*\dy}) + -- ({1.6677*\dx},{-1.1376*\dy}) + -- ({1.6452*\dx},{-1.1491*\dy}) + -- ({1.6225*\dx},{-1.1601*\dy}) + -- ({1.5995*\dx},{-1.1707*\dy}) + -- ({1.5764*\dx},{-1.1808*\dy}) + -- ({1.5530*\dx},{-1.1904*\dy}) + -- ({1.5294*\dx},{-1.1995*\dy}) + -- ({1.5057*\dx},{-1.2081*\dy}) + -- ({1.4817*\dx},{-1.2162*\dy}) + -- ({1.4576*\dx},{-1.2239*\dy}) + -- ({1.4334*\dx},{-1.2310*\dy}) + -- ({1.4090*\dx},{-1.2377*\dy}) + -- ({1.3845*\dx},{-1.2438*\dy}) + -- ({1.3599*\dx},{-1.2494*\dy}) + -- ({1.3352*\dx},{-1.2545*\dy}) + -- ({1.3103*\dx},{-1.2591*\dy}) + -- ({1.2854*\dx},{-1.2632*\dy}) + -- ({1.2605*\dx},{-1.2667*\dy}) + -- ({1.2354*\dx},{-1.2698*\dy}) + -- ({1.2103*\dx},{-1.2723*\dy}) + -- ({1.1852*\dx},{-1.2742*\dy}) + -- ({1.1600*\dx},{-1.2757*\dy}) + -- ({1.1348*\dx},{-1.2766*\dy}) + -- ({1.1097*\dx},{-1.2770*\dy}) + -- ({1.0845*\dx},{-1.2768*\dy}) + -- ({1.0593*\dx},{-1.2762*\dy}) + -- ({1.0342*\dx},{-1.2750*\dy}) + -- ({1.0091*\dx},{-1.2732*\dy}) + -- ({0.9841*\dx},{-1.2710*\dy}) + -- ({0.9591*\dx},{-1.2682*\dy}) + -- ({0.9342*\dx},{-1.2649*\dy}) + -- ({0.9094*\dx},{-1.2610*\dy}) + -- ({0.8848*\dx},{-1.2567*\dy}) + -- ({0.8602*\dx},{-1.2518*\dy}) + -- ({0.8357*\dx},{-1.2464*\dy}) + -- ({0.8114*\dx},{-1.2404*\dy}) + -- ({0.7872*\dx},{-1.2340*\dy}) + -- ({0.7632*\dx},{-1.2271*\dy}) + -- ({0.7394*\dx},{-1.2196*\dy}) + -- ({0.7157*\dx},{-1.2116*\dy}) + -- ({0.6923*\dx},{-1.2032*\dy}) + -- ({0.6690*\dx},{-1.1942*\dy}) + -- ({0.6460*\dx},{-1.1847*\dy}) + -- ({0.6231*\dx},{-1.1748*\dy}) + -- ({0.6006*\dx},{-1.1644*\dy}) + -- ({0.5783*\dx},{-1.1535*\dy}) + -- ({0.5562*\dx},{-1.1421*\dy}) + -- ({0.5344*\dx},{-1.1303*\dy}) + -- ({0.5129*\dx},{-1.1180*\dy}) + -- ({0.4917*\dx},{-1.1052*\dy}) + -- ({0.4708*\dx},{-1.0920*\dy}) + -- ({0.4502*\dx},{-1.0784*\dy}) + -- ({0.4299*\dx},{-1.0643*\dy}) + -- ({0.4100*\dx},{-1.0498*\dy}) + -- ({0.3904*\dx},{-1.0349*\dy}) + -- ({0.3711*\dx},{-1.0195*\dy}) + -- ({0.3523*\dx},{-1.0038*\dy}) + -- ({0.3338*\dx},{-0.9877*\dy}) + -- ({0.3157*\dx},{-0.9712*\dy}) + -- ({0.2979*\dx},{-0.9543*\dy}) + -- ({0.2806*\dx},{-0.9370*\dy}) + -- ({0.2637*\dx},{-0.9194*\dy}) + -- ({0.2472*\dx},{-0.9014*\dy}) + -- ({0.2311*\dx},{-0.8831*\dy}) + -- ({0.2154*\dx},{-0.8645*\dy}) + -- ({0.2002*\dx},{-0.8455*\dy}) + -- ({0.1855*\dx},{-0.8262*\dy}) + -- ({0.1712*\dx},{-0.8066*\dy}) + -- ({0.1573*\dx},{-0.7868*\dy}) + -- ({0.1440*\dx},{-0.7666*\dy}) + -- ({0.1311*\dx},{-0.7462*\dy}) + -- ({0.1187*\dx},{-0.7256*\dy}) + -- ({0.1068*\dx},{-0.7047*\dy}) + -- ({0.0953*\dx},{-0.6835*\dy}) + -- ({0.0844*\dx},{-0.6621*\dy}) + -- ({0.0740*\dx},{-0.6406*\dy}) + -- ({0.0641*\dx},{-0.6188*\dy}) + -- ({0.0547*\dx},{-0.5968*\dy}) + -- ({0.0458*\dx},{-0.5747*\dy}) + -- ({0.0375*\dx},{-0.5524*\dy}) + -- ({0.0296*\dx},{-0.5299*\dy}) + -- ({0.0223*\dx},{-0.5074*\dy}) + -- ({0.0156*\dx},{-0.4847*\dy}) + -- ({0.0093*\dx},{-0.4618*\dy}) + -- ({0.0037*\dx},{-0.4389*\dy}) + -- ({-0.0015*\dx},{-0.4159*\dy}) + -- ({-0.0061*\dx},{-0.3928*\dy}) + -- ({-0.0101*\dx},{-0.3697*\dy}) + -- ({-0.0136*\dx},{-0.3465*\dy}) + -- ({-0.0166*\dx},{-0.3233*\dy}) + -- ({-0.0190*\dx},{-0.3001*\dy}) + -- ({-0.0208*\dx},{-0.2768*\dy}) + -- ({-0.0221*\dx},{-0.2536*\dy}) + -- ({-0.0229*\dx},{-0.2304*\dy}) + -- ({-0.0231*\dx},{-0.2072*\dy}) + -- ({-0.0228*\dx},{-0.1841*\dy}) + -- ({-0.0219*\dx},{-0.1610*\dy}) + -- ({-0.0204*\dx},{-0.1380*\dy}) + -- ({-0.0185*\dx},{-0.1151*\dy}) + -- ({-0.0159*\dx},{-0.0923*\dy}) + -- ({-0.0129*\dx},{-0.0696*\dy}) + -- ({-0.0093*\dx},{-0.0470*\dy}) + -- ({-0.0052*\dx},{-0.0245*\dy}) + -- ({-0.0005*\dx},{-0.0023*\dy}) + -- ({0.0047*\dx},{0.0199*\dy}) + -- ({0.0104*\dx},{0.0418*\dy}) + -- ({0.0166*\dx},{0.0635*\dy}) + -- ({0.0233*\dx},{0.0851*\dy}) + -- ({0.0306*\dx},{0.1064*\dy}) + -- ({0.0383*\dx},{0.1275*\dy}) + -- ({0.0466*\dx},{0.1484*\dy}) + -- ({0.0553*\dx},{0.1690*\dy}) + -- ({0.0645*\dx},{0.1893*\dy}) + -- ({0.0742*\dx},{0.2094*\dy}) + -- ({0.0843*\dx},{0.2291*\dy}) + -- ({0.0950*\dx},{0.2486*\dy}) + -- ({0.1060*\dx},{0.2677*\dy}) + -- ({0.1176*\dx},{0.2866*\dy}) + -- ({0.1295*\dx},{0.3050*\dy}) + -- ({0.1419*\dx},{0.3232*\dy}) + -- ({0.1547*\dx},{0.3410*\dy}) + -- ({0.1679*\dx},{0.3584*\dy}) + -- ({0.1816*\dx},{0.3754*\dy}) + -- ({0.1956*\dx},{0.3921*\dy}) + -- ({0.2100*\dx},{0.4083*\dy}) + -- ({0.2248*\dx},{0.4242*\dy}) + -- ({0.2399*\dx},{0.4396*\dy}) + -- ({0.2554*\dx},{0.4546*\dy}) + -- ({0.2712*\dx},{0.4691*\dy}) + -- ({0.2874*\dx},{0.4832*\dy}) + -- ({0.3038*\dx},{0.4969*\dy}) + -- ({0.3206*\dx},{0.5101*\dy}) + -- ({0.3377*\dx},{0.5228*\dy}) + -- ({0.3550*\dx},{0.5351*\dy}) + -- ({0.3727*\dx},{0.5468*\dy}) + -- ({0.3906*\dx},{0.5581*\dy}) + -- ({0.4087*\dx},{0.5689*\dy}) + -- ({0.4270*\dx},{0.5792*\dy}) + -- ({0.4456*\dx},{0.5889*\dy}) + -- ({0.4644*\dx},{0.5982*\dy}) + -- ({0.4834*\dx},{0.6069*\dy}) + -- ({0.5025*\dx},{0.6151*\dy}) + -- ({0.5218*\dx},{0.6228*\dy}) + -- ({0.5413*\dx},{0.6299*\dy}) + -- ({0.5609*\dx},{0.6365*\dy}) + -- ({0.5806*\dx},{0.6426*\dy}) + -- ({0.6005*\dx},{0.6481*\dy}) + -- ({0.6204*\dx},{0.6531*\dy}) + -- ({0.6404*\dx},{0.6575*\dy}) + -- ({0.6605*\dx},{0.6614*\dy}) + -- ({0.6806*\dx},{0.6647*\dy}) + -- ({0.7007*\dx},{0.6674*\dy}) + -- ({0.7209*\dx},{0.6696*\dy}) + -- ({0.7411*\dx},{0.6713*\dy}) + -- ({0.7613*\dx},{0.6723*\dy}) + -- ({0.7814*\dx},{0.6729*\dy}) + -- ({0.8015*\dx},{0.6728*\dy}) + -- ({0.8216*\dx},{0.6722*\dy}) + -- ({0.8416*\dx},{0.6711*\dy}) + -- ({0.8615*\dx},{0.6694*\dy}) + -- ({0.8813*\dx},{0.6672*\dy}) + -- ({0.9010*\dx},{0.6644*\dy}) + -- ({0.9205*\dx},{0.6610*\dy}) + -- ({0.9400*\dx},{0.6571*\dy}) + -- ({0.9592*\dx},{0.6527*\dy}) + -- ({0.9783*\dx},{0.6478*\dy}) + -- ({0.9972*\dx},{0.6423*\dy}) + -- ({1.0159*\dx},{0.6363*\dy}) + -- ({1.0344*\dx},{0.6298*\dy}) + -- ({1.0527*\dx},{0.6227*\dy}) + -- ({1.0707*\dx},{0.6152*\dy}) + -- ({1.0885*\dx},{0.6071*\dy}) + -- ({1.1060*\dx},{0.5986*\dy}) + -- ({1.1232*\dx},{0.5896*\dy}) + -- ({1.1401*\dx},{0.5801*\dy}) + -- ({1.1567*\dx},{0.5701*\dy}) + -- ({1.1730*\dx},{0.5597*\dy}) + -- ({1.1889*\dx},{0.5488*\dy}) + -- ({1.2045*\dx},{0.5374*\dy}) + -- ({1.2197*\dx},{0.5257*\dy}) + -- ({1.2346*\dx},{0.5135*\dy}) + -- ({1.2491*\dx},{0.5008*\dy}) + -- ({1.2631*\dx},{0.4878*\dy}) + -- ({1.2768*\dx},{0.4744*\dy}) + -- ({1.2900*\dx},{0.4606*\dy}) + -- ({1.3029*\dx},{0.4464*\dy}) + -- ({1.3152*\dx},{0.4319*\dy}) + -- ({1.3271*\dx},{0.4170*\dy}) + -- ({1.3386*\dx},{0.4018*\dy}) + -- ({1.3496*\dx},{0.3862*\dy}) + -- ({1.3601*\dx},{0.3704*\dy}) + -- ({1.3701*\dx},{0.3542*\dy}) + -- ({1.3796*\dx},{0.3378*\dy}) + -- ({1.3886*\dx},{0.3211*\dy}) + -- ({1.3971*\dx},{0.3041*\dy}) + -- ({1.4051*\dx},{0.2869*\dy}) + -- ({1.4125*\dx},{0.2694*\dy}) + -- ({1.4195*\dx},{0.2518*\dy}) + -- ({1.4258*\dx},{0.2339*\dy}) + -- ({1.4316*\dx},{0.2159*\dy}) + -- ({1.4369*\dx},{0.1977*\dy}) + -- ({1.4416*\dx},{0.1793*\dy}) + -- ({1.4457*\dx},{0.1608*\dy}) + -- ({1.4493*\dx},{0.1422*\dy}) + -- ({1.4523*\dx},{0.1235*\dy}) + -- ({1.4547*\dx},{0.1047*\dy}) + -- ({1.4565*\dx},{0.0858*\dy}) + -- ({1.4577*\dx},{0.0668*\dy}) + -- ({1.4584*\dx},{0.0478*\dy}) + -- ({1.4584*\dx},{0.0288*\dy}) + -- ({1.4579*\dx},{0.0098*\dy}) + -- ({1.4568*\dx},{-0.0092*\dy}) + -- ({1.4551*\dx},{-0.0282*\dy}) + -- ({1.4528*\dx},{-0.0472*\dy}) + -- ({1.4499*\dx},{-0.0661*\dy}) + -- ({1.4464*\dx},{-0.0849*\dy}) + -- ({1.4424*\dx},{-0.1036*\dy}) + -- ({1.4377*\dx},{-0.1222*\dy}) + -- ({1.4325*\dx},{-0.1407*\dy}) + -- ({1.4267*\dx},{-0.1591*\dy}) + -- ({1.4202*\dx},{-0.1773*\dy}) + -- ({1.4133*\dx},{-0.1953*\dy}) + -- ({1.4057*\dx},{-0.2131*\dy}) + -- ({1.3976*\dx},{-0.2307*\dy}) + -- ({1.3889*\dx},{-0.2481*\dy}) + -- ({1.3797*\dx},{-0.2652*\dy}) + -- ({1.3699*\dx},{-0.2821*\dy}) + -- ({1.3595*\dx},{-0.2987*\dy}) + -- ({1.3487*\dx},{-0.3151*\dy}) + -- ({1.3373*\dx},{-0.3311*\dy}) + -- ({1.3253*\dx},{-0.3468*\dy}) + -- ({1.3129*\dx},{-0.3621*\dy}) + -- ({1.2999*\dx},{-0.3771*\dy}) + -- ({1.2865*\dx},{-0.3917*\dy}) + -- ({1.2725*\dx},{-0.4060*\dy}) + -- ({1.2581*\dx},{-0.4198*\dy}) + -- ({1.2432*\dx},{-0.4333*\dy}) + -- ({1.2279*\dx},{-0.4463*\dy}) + -- ({1.2121*\dx},{-0.4589*\dy}) + -- ({1.1959*\dx},{-0.4710*\dy}) + -- ({1.1792*\dx},{-0.4826*\dy}) + -- ({1.1622*\dx},{-0.4938*\dy}) + -- ({1.1447*\dx},{-0.5045*\dy}) + -- ({1.1269*\dx},{-0.5147*\dy}) + -- ({1.1087*\dx},{-0.5243*\dy}) + -- ({1.0901*\dx},{-0.5335*\dy}) + -- ({1.0712*\dx},{-0.5420*\dy}) + -- ({1.0520*\dx},{-0.5501*\dy}) + -- ({1.0324*\dx},{-0.5575*\dy}) + -- ({1.0126*\dx},{-0.5644*\dy}) + -- ({0.9924*\dx},{-0.5708*\dy}) + -- ({0.9721*\dx},{-0.5765*\dy}) + -- ({0.9514*\dx},{-0.5816*\dy}) + -- ({0.9305*\dx},{-0.5861*\dy}) + -- ({0.9094*\dx},{-0.5900*\dy}) + -- ({0.8881*\dx},{-0.5933*\dy}) + -- ({0.8666*\dx},{-0.5959*\dy}) + -- ({0.8450*\dx},{-0.5979*\dy}) + -- ({0.8232*\dx},{-0.5993*\dy}) + -- ({0.8013*\dx},{-0.6000*\dy}) + -- ({0.7792*\dx},{-0.6000*\dy}) + -- ({0.7571*\dx},{-0.5994*\dy}) + -- ({0.7349*\dx},{-0.5981*\dy}) + -- ({0.7126*\dx},{-0.5961*\dy}) + -- ({0.6903*\dx},{-0.5935*\dy}) + -- ({0.6680*\dx},{-0.5902*\dy}) + -- ({0.6457*\dx},{-0.5862*\dy}) + -- ({0.6234*\dx},{-0.5815*\dy}) + -- ({0.6012*\dx},{-0.5762*\dy}) + -- ({0.5790*\dx},{-0.5701*\dy}) + -- ({0.5568*\dx},{-0.5634*\dy}) + -- ({0.5348*\dx},{-0.5560*\dy}) + -- ({0.5129*\dx},{-0.5479*\dy}) + -- ({0.4911*\dx},{-0.5391*\dy}) + -- ({0.4695*\dx},{-0.5297*\dy}) + -- ({0.4481*\dx},{-0.5196*\dy}) + -- ({0.4269*\dx},{-0.5087*\dy}) + -- ({0.4058*\dx},{-0.4973*\dy}) + -- ({0.3850*\dx},{-0.4851*\dy}) + -- ({0.3645*\dx},{-0.4723*\dy}) + -- ({0.3442*\dx},{-0.4589*\dy}) + -- ({0.3243*\dx},{-0.4448*\dy}) + -- ({0.3046*\dx},{-0.4300*\dy}) + -- ({0.2853*\dx},{-0.4146*\dy}) + -- ({0.2663*\dx},{-0.3986*\dy}) + -- ({0.2477*\dx},{-0.3820*\dy}) + -- ({0.2295*\dx},{-0.3647*\dy}) + -- ({0.2117*\dx},{-0.3468*\dy}) + -- ({0.1943*\dx},{-0.3284*\dy}) + -- ({0.1774*\dx},{-0.3093*\dy}) + -- ({0.1609*\dx},{-0.2897*\dy}) + -- ({0.1449*\dx},{-0.2695*\dy}) + -- ({0.1294*\dx},{-0.2488*\dy}) + -- ({0.1144*\dx},{-0.2275*\dy}) + -- ({0.0999*\dx},{-0.2057*\dy}) + -- ({0.0860*\dx},{-0.1834*\dy}) + -- ({0.0726*\dx},{-0.1606*\dy}) + -- ({0.0598*\dx},{-0.1373*\dy}) + -- ({0.0476*\dx},{-0.1135*\dy}) + -- ({0.0360*\dx},{-0.0893*\dy}) + -- ({0.0250*\dx},{-0.0646*\dy}) + -- ({0.0147*\dx},{-0.0395*\dy}) + -- ({0.0050*\dx},{-0.0140*\dy}) + -- ({-0.0040*\dx},{0.0119*\dy}) + -- ({-0.0124*\dx},{0.0382*\dy}) + -- ({-0.0201*\dx},{0.0648*\dy}) + -- ({-0.0270*\dx},{0.0918*\dy}) + -- ({-0.0333*\dx},{0.1191*\dy}) + -- ({-0.0388*\dx},{0.1466*\dy}) + -- ({-0.0436*\dx},{0.1745*\dy}) + -- ({-0.0477*\dx},{0.2027*\dy}) + -- ({-0.0510*\dx},{0.2311*\dy}) + -- ({-0.0535*\dx},{0.2597*\dy}) + -- ({-0.0553*\dx},{0.2885*\dy}) + -- ({-0.0563*\dx},{0.3175*\dy}) + -- ({-0.0565*\dx},{0.3466*\dy}) + -- ({-0.0559*\dx},{0.3759*\dy}) + -- ({-0.0545*\dx},{0.4054*\dy}) + -- ({-0.0523*\dx},{0.4349*\dy}) + -- ({-0.0494*\dx},{0.4645*\dy}) + -- ({-0.0455*\dx},{0.4941*\dy}) + -- ({-0.0409*\dx},{0.5238*\dy}) + -- ({-0.0355*\dx},{0.5535*\dy}) + -- ({-0.0292*\dx},{0.5832*\dy}) + -- ({-0.0221*\dx},{0.6128*\dy}) + -- ({-0.0142*\dx},{0.6424*\dy}) + -- ({-0.0054*\dx},{0.6719*\dy}) + -- ({0.0042*\dx},{0.7013*\dy}) + -- ({0.0146*\dx},{0.7305*\dy}) + -- ({0.0258*\dx},{0.7597*\dy}) + -- ({0.0379*\dx},{0.7886*\dy}) + -- ({0.0508*\dx},{0.8174*\dy}) + -- ({0.0645*\dx},{0.8459*\dy}) + -- ({0.0790*\dx},{0.8742*\dy}) + -- ({0.0943*\dx},{0.9022*\dy}) + -- ({0.1105*\dx},{0.9299*\dy}) + -- ({0.1274*\dx},{0.9573*\dy}) + -- ({0.1452*\dx},{0.9844*\dy}) + -- ({0.1637*\dx},{1.0111*\dy}) + -- ({0.1830*\dx},{1.0375*\dy}) + -- ({0.2031*\dx},{1.0634*\dy}) + -- ({0.2239*\dx},{1.0889*\dy}) + -- ({0.2455*\dx},{1.1140*\dy}) + -- ({0.2679*\dx},{1.1386*\dy}) + -- ({0.2910*\dx},{1.1627*\dy}) + -- ({0.3148*\dx},{1.1863*\dy}) + -- ({0.3393*\dx},{1.2094*\dy}) + -- ({0.3645*\dx},{1.2319*\dy}) + -- ({0.3904*\dx},{1.2539*\dy}) + -- ({0.4169*\dx},{1.2752*\dy}) + -- ({0.4442*\dx},{1.2960*\dy}) + -- ({0.4720*\dx},{1.3161*\dy}) + -- ({0.5005*\dx},{1.3355*\dy}) + -- ({0.5296*\dx},{1.3543*\dy}) + -- ({0.5593*\dx},{1.3724*\dy}) + -- ({0.5896*\dx},{1.3898*\dy}) + -- ({0.6204*\dx},{1.4065*\dy}) + -- ({0.6518*\dx},{1.4224*\dy}) + -- ({0.6837*\dx},{1.4376*\dy}) + -- ({0.7161*\dx},{1.4520*\dy}) + -- ({0.7489*\dx},{1.4656*\dy}) + -- ({0.7823*\dx},{1.4784*\dy}) + -- ({0.8160*\dx},{1.4903*\dy}) + -- ({0.8503*\dx},{1.5015*\dy}) + -- ({0.8849*\dx},{1.5118*\dy}) + -- ({0.9198*\dx},{1.5213*\dy}) + -- ({0.9552*\dx},{1.5298*\dy}) + -- ({0.9908*\dx},{1.5375*\dy}) + -- ({1.0268*\dx},{1.5443*\dy}) + -- ({1.0631*\dx},{1.5502*\dy}) + -- ({1.0996*\dx},{1.5552*\dy}) + -- ({1.1364*\dx},{1.5593*\dy}) + -- ({1.1733*\dx},{1.5624*\dy}) + -- ({1.2105*\dx},{1.5646*\dy}) + -- ({1.2478*\dx},{1.5658*\dy}) + -- ({1.2853*\dx},{1.5661*\dy}) + -- ({1.3229*\dx},{1.5655*\dy}) + -- ({1.3606*\dx},{1.5638*\dy}) + -- ({1.3983*\dx},{1.5613*\dy}) + -- ({1.4361*\dx},{1.5577*\dy}) + -- ({1.4738*\dx},{1.5531*\dy}) + -- ({1.5116*\dx},{1.5476*\dy}) + -- ({1.5493*\dx},{1.5411*\dy}) + -- ({1.5870*\dx},{1.5337*\dy}) + -- ({1.6245*\dx},{1.5252*\dy}) + -- ({1.6620*\dx},{1.5158*\dy}) + -- ({1.6993*\dx},{1.5054*\dy}) + -- ({1.7364*\dx},{1.4940*\dy}) + -- ({1.7733*\dx},{1.4816*\dy}) + -- ({1.8100*\dx},{1.4683*\dy}) + -- ({1.8464*\dx},{1.4540*\dy}) + -- ({1.8826*\dx},{1.4388*\dy}) + -- ({1.9184*\dx},{1.4226*\dy}) + -- ({1.9539*\dx},{1.4055*\dy}) + -- ({1.9891*\dx},{1.3874*\dy}) + -- ({2.0238*\dx},{1.3684*\dy}) + -- ({2.0582*\dx},{1.3484*\dy}) + -- ({2.0921*\dx},{1.3276*\dy}) + -- ({2.1255*\dx},{1.3059*\dy}) + -- ({2.1585*\dx},{1.2832*\dy}) + -- ({2.1910*\dx},{1.2597*\dy}) + -- ({2.2229*\dx},{1.2353*\dy}) + -- ({2.2543*\dx},{1.2101*\dy}) + -- ({2.2850*\dx},{1.1840*\dy}) + -- ({2.3152*\dx},{1.1571*\dy}) + -- ({2.3448*\dx},{1.1294*\dy}) + -- ({2.3736*\dx},{1.1009*\dy}) + -- ({2.4019*\dx},{1.0716*\dy}) + -- ({2.4294*\dx},{1.0416*\dy}) + -- ({2.4562*\dx},{1.0108*\dy}) + -- ({2.4822*\dx},{0.9793*\dy}) + -- ({2.5075*\dx},{0.9471*\dy}) + -- ({2.5321*\dx},{0.9142*\dy}) + -- ({2.5558*\dx},{0.8807*\dy}) + -- ({2.5787*\dx},{0.8465*\dy}) + -- ({2.6008*\dx},{0.8117*\dy}) + -- ({2.6220*\dx},{0.7762*\dy}) + -- ({2.6423*\dx},{0.7402*\dy}) + -- ({2.6618*\dx},{0.7037*\dy}) + -- ({2.6803*\dx},{0.6666*\dy}) + -- ({2.6980*\dx},{0.6290*\dy}) + -- ({2.7147*\dx},{0.5909*\dy}) + -- ({2.7305*\dx},{0.5524*\dy}) + -- ({2.7453*\dx},{0.5134*\dy}) + -- ({2.7591*\dx},{0.4740*\dy}) + -- ({2.7719*\dx},{0.4342*\dy}) + -- ({2.7838*\dx},{0.3941*\dy}) + -- ({2.7946*\dx},{0.3537*\dy}) + -- ({2.8045*\dx},{0.3129*\dy}) + -- ({2.8133*\dx},{0.2719*\dy}) + -- ({2.8210*\dx},{0.2306*\dy}) + -- ({2.8278*\dx},{0.1891*\dy}) + -- ({2.8335*\dx},{0.1474*\dy}) + -- ({2.8381*\dx},{0.1055*\dy}) + -- ({2.8417*\dx},{0.0636*\dy}) + -- ({2.8442*\dx},{0.0215*\dy}) + -- ({2.8457*\dx},{-0.0207*\dy}) + -- ({2.8461*\dx},{-0.0629*\dy}) + -- ({2.8454*\dx},{-0.1052*\dy}) + -- ({2.8436*\dx},{-0.1474*\dy}) + -- ({2.8408*\dx},{-0.1896*\dy}) + -- ({2.8369*\dx},{-0.2317*\dy}) + -- ({2.8320*\dx},{-0.2738*\dy}) + -- ({2.8260*\dx},{-0.3157*\dy}) + -- ({2.8189*\dx},{-0.3574*\dy}) + -- ({2.8108*\dx},{-0.3990*\dy}) + -- ({2.8016*\dx},{-0.4404*\dy}) + -- ({2.7914*\dx},{-0.4815*\dy}) + -- ({2.7801*\dx},{-0.5224*\dy}) + -- ({2.7678*\dx},{-0.5629*\dy}) + -- ({2.7545*\dx},{-0.6032*\dy}) + -- ({2.7401*\dx},{-0.6430*\dy}) + -- ({2.7248*\dx},{-0.6825*\dy}) + -- ({2.7084*\dx},{-0.7216*\dy}) + -- ({2.6911*\dx},{-0.7603*\dy}) + -- ({2.6728*\dx},{-0.7985*\dy}) + -- ({2.6535*\dx},{-0.8362*\dy}) + -- ({2.6333*\dx},{-0.8734*\dy}) + -- ({2.6122*\dx},{-0.9100*\dy}) + -- ({2.5901*\dx},{-0.9461*\dy}) + -- ({2.5671*\dx},{-0.9816*\dy}) + -- ({2.5433*\dx},{-1.0165*\dy}) + -- ({2.5186*\dx},{-1.0507*\dy}) + -- ({2.4930*\dx},{-1.0843*\dy}) + -- ({2.4667*\dx},{-1.1172*\dy}) + -- ({2.4395*\dx},{-1.1494*\dy}) + -- ({2.4115*\dx},{-1.1808*\dy}) + -- ({2.3827*\dx},{-1.2115*\dy}) + -- ({2.3532*\dx},{-1.2414*\dy}) + -- ({2.3230*\dx},{-1.2706*\dy}) + -- ({2.2921*\dx},{-1.2989*\dy}) + -- ({2.2605*\dx},{-1.3264*\dy}) + -- ({2.2282*\dx},{-1.3530*\dy}) + -- ({2.1953*\dx},{-1.3788*\dy}) + -- ({2.1618*\dx},{-1.4037*\dy}) + -- ({2.1277*\dx},{-1.4277*\dy}) + -- ({2.0930*\dx},{-1.4508*\dy}) + -- ({2.0578*\dx},{-1.4729*\dy}) + -- ({2.0221*\dx},{-1.4941*\dy}) + -- ({1.9859*\dx},{-1.5144*\dy}) + -- ({1.9493*\dx},{-1.5337*\dy}) + -- ({1.9123*\dx},{-1.5520*\dy}) + -- ({1.8748*\dx},{-1.5693*\dy}) + -- ({1.8370*\dx},{-1.5856*\dy}) + -- ({1.7988*\dx},{-1.6009*\dy}) + -- ({1.7604*\dx},{-1.6152*\dy}) + -- ({1.7216*\dx},{-1.6285*\dy}) + -- ({1.6826*\dx},{-1.6407*\dy}) + -- ({1.6434*\dx},{-1.6519*\dy}) + -- ({1.6040*\dx},{-1.6620*\dy}) + -- ({1.5644*\dx},{-1.6711*\dy}) + -- ({1.5247*\dx},{-1.6792*\dy}) + -- ({1.4848*\dx},{-1.6862*\dy}) + -- ({1.4449*\dx},{-1.6921*\dy}) + -- ({1.4050*\dx},{-1.6970*\dy}) + -- ({1.3650*\dx},{-1.7009*\dy}) + -- ({1.3250*\dx},{-1.7037*\dy}) + -- ({1.2851*\dx},{-1.7054*\dy}) + -- ({1.2453*\dx},{-1.7061*\dy}) + -- ({1.2056*\dx},{-1.7058*\dy}) + -- ({1.1660*\dx},{-1.7044*\dy}) + -- ({1.1265*\dx},{-1.7020*\dy}) + -- ({1.0873*\dx},{-1.6986*\dy}) + -- ({1.0482*\dx},{-1.6941*\dy}) + -- ({1.0095*\dx},{-1.6887*\dy}) + -- ({0.9709*\dx},{-1.6822*\dy}) + -- ({0.9327*\dx},{-1.6747*\dy}) + -- ({0.8949*\dx},{-1.6663*\dy}) + -- ({0.8574*\dx},{-1.6569*\dy}) + -- ({0.8202*\dx},{-1.6466*\dy}) + -- ({0.7835*\dx},{-1.6353*\dy}) + -- ({0.7473*\dx},{-1.6231*\dy}) + -- ({0.7115*\dx},{-1.6099*\dy}) + -- ({0.6761*\dx},{-1.5959*\dy}) + -- ({0.6413*\dx},{-1.5810*\dy}) + -- ({0.6071*\dx},{-1.5652*\dy}) + -- ({0.5734*\dx},{-1.5486*\dy}) + -- ({0.5403*\dx},{-1.5312*\dy}) + -- ({0.5078*\dx},{-1.5129*\dy}) + -- ({0.4759*\dx},{-1.4939*\dy}) + -- ({0.4447*\dx},{-1.4741*\dy}) + -- ({0.4141*\dx},{-1.4535*\dy}) + -- ({0.3843*\dx},{-1.4322*\dy}) + -- ({0.3552*\dx},{-1.4103*\dy}) + -- ({0.3268*\dx},{-1.3876*\dy}) + -- ({0.2991*\dx},{-1.3643*\dy}) + -- ({0.2722*\dx},{-1.3403*\dy}) + -- ({0.2461*\dx},{-1.3158*\dy}) + -- ({0.2208*\dx},{-1.2907*\dy}) + -- ({0.1963*\dx},{-1.2650*\dy}) + -- ({0.1726*\dx},{-1.2387*\dy}) + -- ({0.1498*\dx},{-1.2120*\dy}) + -- ({0.1279*\dx},{-1.1848*\dy}) + -- ({0.1068*\dx},{-1.1572*\dy}) + -- ({0.0865*\dx},{-1.1291*\dy}) + -- ({0.0672*\dx},{-1.1006*\dy}) + -- ({0.0487*\dx},{-1.0718*\dy}) + -- ({0.0312*\dx},{-1.0426*\dy}) + -- ({0.0146*\dx},{-1.0131*\dy}) + -- ({-0.0011*\dx},{-0.9833*\dy}) + -- ({-0.0159*\dx},{-0.9533*\dy}) + -- ({-0.0298*\dx},{-0.9230*\dy}) + -- ({-0.0427*\dx},{-0.8926*\dy}) + -- ({-0.0547*\dx},{-0.8620*\dy}) + -- ({-0.0658*\dx},{-0.8312*\dy}) + -- ({-0.0759*\dx},{-0.8003*\dy}) + -- ({-0.0851*\dx},{-0.7694*\dy}) + -- ({-0.0933*\dx},{-0.7384*\dy}) + -- ({-0.1007*\dx},{-0.7074*\dy}) + -- ({-0.1070*\dx},{-0.6764*\dy}) + -- ({-0.1125*\dx},{-0.6454*\dy}) + -- ({-0.1170*\dx},{-0.6145*\dy}) + -- ({-0.1206*\dx},{-0.5837*\dy}) + -- ({-0.1233*\dx},{-0.5530*\dy}) + -- ({-0.1251*\dx},{-0.5225*\dy}) + -- ({-0.1260*\dx},{-0.4921*\dy}) + -- ({-0.1261*\dx},{-0.4619*\dy}) + -- ({-0.1252*\dx},{-0.4320*\dy}) + -- ({-0.1235*\dx},{-0.4024*\dy}) + -- ({-0.1209*\dx},{-0.3730*\dy}) + -- ({-0.1175*\dx},{-0.3439*\dy}) + -- ({-0.1132*\dx},{-0.3152*\dy}) + -- ({-0.1081*\dx},{-0.2868*\dy}) + -- ({-0.1022*\dx},{-0.2589*\dy}) + -- ({-0.0956*\dx},{-0.2313*\dy}) + -- ({-0.0882*\dx},{-0.2042*\dy}) + -- ({-0.0800*\dx},{-0.1775*\dy}) + -- ({-0.0711*\dx},{-0.1513*\dy}) + -- ({-0.0614*\dx},{-0.1256*\dy}) + -- ({-0.0511*\dx},{-0.1005*\dy}) + -- ({-0.0401*\dx},{-0.0759*\dy}) + -- ({-0.0284*\dx},{-0.0518*\dy}) + -- ({-0.0161*\dx},{-0.0283*\dy}) + -- ({-0.0032*\dx},{-0.0055*\dy}) + -- ({0.0103*\dx},{0.0168*\dy}) + -- ({0.0243*\dx},{0.0384*\dy}) + -- ({0.0389*\dx},{0.0593*\dy}) + -- ({0.0541*\dx},{0.0796*\dy}) + -- ({0.0697*\dx},{0.0992*\dy}) + -- ({0.0858*\dx},{0.1181*\dy}) + -- ({0.1023*\dx},{0.1363*\dy}) + -- ({0.1193*\dx},{0.1537*\dy}) + -- ({0.1367*\dx},{0.1705*\dy}) + -- ({0.1544*\dx},{0.1864*\dy}) + -- ({0.1725*\dx},{0.2017*\dy}) + -- ({0.1908*\dx},{0.2161*\dy}) + -- ({0.2095*\dx},{0.2298*\dy}) + -- ({0.2284*\dx},{0.2427*\dy}) + -- ({0.2476*\dx},{0.2549*\dy}) + -- ({0.2669*\dx},{0.2662*\dy}) + -- ({0.2865*\dx},{0.2768*\dy}) + -- ({0.3062*\dx},{0.2865*\dy}) + -- ({0.3260*\dx},{0.2955*\dy}) + -- ({0.3459*\dx},{0.3036*\dy}) + -- ({0.3658*\dx},{0.3110*\dy}) + -- ({0.3858*\dx},{0.3175*\dy}) + -- ({0.4058*\dx},{0.3233*\dy}) + -- ({0.4258*\dx},{0.3283*\dy}) + -- ({0.4457*\dx},{0.3324*\dy}) + -- ({0.4655*\dx},{0.3358*\dy}) + -- ({0.4853*\dx},{0.3384*\dy}) + -- ({0.5049*\dx},{0.3403*\dy}) + -- ({0.5243*\dx},{0.3413*\dy}) + -- ({0.5436*\dx},{0.3416*\dy}) + -- ({0.5626*\dx},{0.3412*\dy}) + -- ({0.5814*\dx},{0.3400*\dy}) + -- ({0.5999*\dx},{0.3381*\dy}) + -- ({0.6182*\dx},{0.3355*\dy}) + -- ({0.6361*\dx},{0.3321*\dy}) + -- ({0.6537*\dx},{0.3281*\dy}) + -- ({0.6709*\dx},{0.3234*\dy}) + -- ({0.6877*\dx},{0.3181*\dy}) + -- ({0.7041*\dx},{0.3120*\dy}) + -- ({0.7201*\dx},{0.3054*\dy}) + -- ({0.7356*\dx},{0.2982*\dy}) + -- ({0.7506*\dx},{0.2903*\dy}) + -- ({0.7651*\dx},{0.2819*\dy}) + -- ({0.7791*\dx},{0.2730*\dy}) + -- ({0.7926*\dx},{0.2635*\dy}) + -- ({0.8055*\dx},{0.2535*\dy}) + -- ({0.8178*\dx},{0.2430*\dy}) + -- ({0.8296*\dx},{0.2320*\dy}) + -- ({0.8407*\dx},{0.2206*\dy}) + -- ({0.8512*\dx},{0.2088*\dy}) + -- ({0.8610*\dx},{0.1965*\dy}) + -- ({0.8702*\dx},{0.1839*\dy}) + -- ({0.8788*\dx},{0.1710*\dy}) + -- ({0.8866*\dx},{0.1577*\dy}) + -- ({0.8937*\dx},{0.1442*\dy}) + -- ({0.9002*\dx},{0.1304*\dy}) + -- ({0.9059*\dx},{0.1163*\dy}) + -- ({0.9109*\dx},{0.1020*\dy}) + -- ({0.9152*\dx},{0.0876*\dy}) + -- ({0.9187*\dx},{0.0730*\dy}) + -- ({0.9215*\dx},{0.0582*\dy}) + -- ({0.9236*\dx},{0.0434*\dy}) + -- ({0.9249*\dx},{0.0284*\dy}) + -- ({0.9254*\dx},{0.0135*\dy}) + -- ({0.9252*\dx},{-0.0015*\dy}) + -- ({0.9242*\dx},{-0.0165*\dy}) + -- ({0.9225*\dx},{-0.0314*\dy}) + -- ({0.9200*\dx},{-0.0463*\dy}) + -- ({0.9168*\dx},{-0.0610*\dy}) + -- ({0.9128*\dx},{-0.0756*\dy}) + -- ({0.9081*\dx},{-0.0901*\dy}) + -- ({0.9026*\dx},{-0.1044*\dy}) + -- ({0.8964*\dx},{-0.1184*\dy}) + -- ({0.8895*\dx},{-0.1322*\dy}) + -- ({0.8818*\dx},{-0.1458*\dy}) + -- ({0.8734*\dx},{-0.1590*\dy}) + -- ({0.8644*\dx},{-0.1720*\dy}) + -- ({0.8546*\dx},{-0.1845*\dy}) + -- ({0.8442*\dx},{-0.1967*\dy}) + -- ({0.8331*\dx},{-0.2085*\dy}) + -- ({0.8214*\dx},{-0.2198*\dy}) + -- ({0.8090*\dx},{-0.2307*\dy}) + -- ({0.7960*\dx},{-0.2411*\dy}) + -- ({0.7824*\dx},{-0.2510*\dy}) + -- ({0.7682*\dx},{-0.2603*\dy}) + -- ({0.7535*\dx},{-0.2691*\dy}) + -- ({0.7382*\dx},{-0.2773*\dy}) + -- ({0.7224*\dx},{-0.2849*\dy}) + -- ({0.7061*\dx},{-0.2919*\dy}) + -- ({0.6893*\dx},{-0.2982*\dy}) + -- ({0.6721*\dx},{-0.3039*\dy}) + -- ({0.6544*\dx},{-0.3089*\dy}) + -- ({0.6363*\dx},{-0.3132*\dy}) + -- ({0.6178*\dx},{-0.3168*\dy}) + -- ({0.5989*\dx},{-0.3196*\dy}) + -- ({0.5798*\dx},{-0.3217*\dy}) + -- ({0.5603*\dx},{-0.3230*\dy}) + -- ({0.5405*\dx},{-0.3235*\dy}) + -- ({0.5205*\dx},{-0.3232*\dy}) + -- ({0.5003*\dx},{-0.3222*\dy}) + -- ({0.4798*\dx},{-0.3203*\dy}) + -- ({0.4592*\dx},{-0.3175*\dy}) + -- ({0.4385*\dx},{-0.3140*\dy}) + -- ({0.4176*\dx},{-0.3095*\dy}) + -- ({0.3967*\dx},{-0.3043*\dy}) + -- ({0.3758*\dx},{-0.2981*\dy}) + -- ({0.3548*\dx},{-0.2911*\dy}) + -- ({0.3338*\dx},{-0.2832*\dy}) + -- ({0.3129*\dx},{-0.2745*\dy}) + -- ({0.2921*\dx},{-0.2649*\dy}) + -- ({0.2714*\dx},{-0.2544*\dy}) + -- ({0.2508*\dx},{-0.2430*\dy}) + -- ({0.2305*\dx},{-0.2307*\dy}) + -- ({0.2103*\dx},{-0.2176*\dy}) + -- ({0.1904*\dx},{-0.2036*\dy}) + -- ({0.1707*\dx},{-0.1887*\dy}) + -- ({0.1514*\dx},{-0.1730*\dy}) + -- ({0.1324*\dx},{-0.1564*\dy}) + -- ({0.1137*\dx},{-0.1390*\dy}) + -- ({0.0955*\dx},{-0.1207*\dy}) + -- ({0.0777*\dx},{-0.1016*\dy}) + -- ({0.0604*\dx},{-0.0817*\dy}) + -- ({0.0435*\dx},{-0.0610*\dy}) + -- ({0.0272*\dx},{-0.0395*\dy}) + -- ({0.0115*\dx},{-0.0173*\dy}) + -- ({-0.0037*\dx},{0.0058*\dy}) + -- ({-0.0182*\dx},{0.0295*\dy}) + -- ({-0.0321*\dx},{0.0540*\dy}) + -- ({-0.0454*\dx},{0.0792*\dy}) + -- ({-0.0579*\dx},{0.1051*\dy}) + -- ({-0.0697*\dx},{0.1316*\dy}) + -- ({-0.0807*\dx},{0.1588*\dy}) + -- ({-0.0910*\dx},{0.1866*\dy}) + -- ({-0.1005*\dx},{0.2150*\dy}) + -- ({-0.1091*\dx},{0.2439*\dy}) + -- ({-0.1169*\dx},{0.2734*\dy}) + -- ({-0.1238*\dx},{0.3035*\dy}) + -- ({-0.1298*\dx},{0.3340*\dy}) + -- ({-0.1349*\dx},{0.3649*\dy}) + -- ({-0.1391*\dx},{0.3964*\dy}) + -- ({-0.1423*\dx},{0.4282*\dy}) + -- ({-0.1445*\dx},{0.4604*\dy}) + -- ({-0.1457*\dx},{0.4929*\dy}) + -- ({-0.1459*\dx},{0.5257*\dy}) + -- ({-0.1451*\dx},{0.5589*\dy}) + -- ({-0.1433*\dx},{0.5922*\dy}) + -- ({-0.1404*\dx},{0.6258*\dy}) + -- ({-0.1365*\dx},{0.6595*\dy}) + -- ({-0.1314*\dx},{0.6934*\dy}) + -- ({-0.1253*\dx},{0.7274*\dy}) + -- ({-0.1181*\dx},{0.7615*\dy}) + -- ({-0.1098*\dx},{0.7956*\dy}) + -- ({-0.1004*\dx},{0.8298*\dy}) + -- ({-0.0899*\dx},{0.8638*\dy}) + -- ({-0.0783*\dx},{0.8978*\dy}) + -- ({-0.0655*\dx},{0.9317*\dy}) + -- ({-0.0517*\dx},{0.9655*\dy}) + -- ({-0.0367*\dx},{0.9990*\dy}) + -- ({-0.0206*\dx},{1.0324*\dy}) + -- ({-0.0034*\dx},{1.0654*\dy}) + -- ({0.0149*\dx},{1.0982*\dy}) + -- ({0.0344*\dx},{1.1307*\dy}) + -- ({0.0549*\dx},{1.1627*\dy}) + -- ({0.0765*\dx},{1.1944*\dy}) + -- ({0.0992*\dx},{1.2256*\dy}) + -- ({0.1230*\dx},{1.2563*\dy}) + -- ({0.1478*\dx},{1.2865*\dy}) + -- ({0.1737*\dx},{1.3162*\dy}) + -- ({0.2006*\dx},{1.3452*\dy}) + -- ({0.2285*\dx},{1.3737*\dy}) + -- ({0.2574*\dx},{1.4014*\dy}) + -- ({0.2874*\dx},{1.4285*\dy}) + -- ({0.3182*\dx},{1.4548*\dy}) + -- ({0.3501*\dx},{1.4804*\dy}) + -- ({0.3828*\dx},{1.5051*\dy}) + -- ({0.4165*\dx},{1.5291*\dy}) + -- ({0.4510*\dx},{1.5521*\dy}) + -- ({0.4864*\dx},{1.5743*\dy}) + -- ({0.5226*\dx},{1.5955*\dy}) + -- ({0.5597*\dx},{1.6158*\dy}) + -- ({0.5975*\dx},{1.6350*\dy}) + -- ({0.6360*\dx},{1.6533*\dy}) + -- ({0.6753*\dx},{1.6705*\dy}) + -- ({0.7152*\dx},{1.6867*\dy}) + -- ({0.7558*\dx},{1.7017*\dy}) + -- ({0.7970*\dx},{1.7157*\dy}) + -- ({0.8388*\dx},{1.7285*\dy}) + -- ({0.8812*\dx},{1.7401*\dy}) + -- ({0.9241*\dx},{1.7505*\dy}) + -- ({0.9674*\dx},{1.7598*\dy}) + -- ({1.0112*\dx},{1.7678*\dy}) + -- ({1.0554*\dx},{1.7746*\dy}) + -- ({1.1000*\dx},{1.7801*\dy}) + -- ({1.1449*\dx},{1.7843*\dy}) + -- ({1.1901*\dx},{1.7872*\dy}) + -- ({1.2355*\dx},{1.7889*\dy}) + -- ({1.2812*\dx},{1.7892*\dy}) + -- ({1.3270*\dx},{1.7882*\dy}) + -- ({1.3730*\dx},{1.7859*\dy}) + -- ({1.4190*\dx},{1.7822*\dy}) + -- ({1.4651*\dx},{1.7772*\dy}) + -- ({1.5111*\dx},{1.7708*\dy}) + -- ({1.5572*\dx},{1.7631*\dy}) + -- ({1.6031*\dx},{1.7540*\dy}) + -- ({1.6490*\dx},{1.7435*\dy}) + -- ({1.6946*\dx},{1.7317*\dy}) + -- ({1.7400*\dx},{1.7185*\dy}) + -- ({1.7852*\dx},{1.7040*\dy}) + -- ({1.8301*\dx},{1.6882*\dy}) + -- ({1.8746*\dx},{1.6709*\dy}) + -- ({1.9188*\dx},{1.6524*\dy}) + -- ({1.9625*\dx},{1.6325*\dy}) + -- ({2.0058*\dx},{1.6114*\dy}) + -- ({2.0485*\dx},{1.5889*\dy}) + -- ({2.0907*\dx},{1.5651*\dy}) + -- ({2.1322*\dx},{1.5401*\dy}) + -- ({2.1732*\dx},{1.5138*\dy}) + -- ({2.2134*\dx},{1.4862*\dy}) + -- ({2.2530*\dx},{1.4574*\dy}) + -- ({2.2918*\dx},{1.4275*\dy}) + -- ({2.3297*\dx},{1.3963*\dy}) + -- ({2.3669*\dx},{1.3640*\dy}) + -- ({2.4031*\dx},{1.3306*\dy}) + -- ({2.4385*\dx},{1.2961*\dy}) + -- ({2.4729*\dx},{1.2605*\dy}) + -- ({2.5063*\dx},{1.2238*\dy}) + -- ({2.5387*\dx},{1.1861*\dy}) + -- ({2.5701*\dx},{1.1474*\dy}) + -- ({2.6003*\dx},{1.1078*\dy}) + -- ({2.6295*\dx},{1.0672*\dy}) + -- ({2.6575*\dx},{1.0258*\dy}) + -- ({2.6843*\dx},{0.9835*\dy}) + -- ({2.7099*\dx},{0.9403*\dy}) + -- ({2.7343*\dx},{0.8964*\dy}) + -- ({2.7574*\dx},{0.8517*\dy}) + -- ({2.7793*\dx},{0.8063*\dy}) + -- ({2.7998*\dx},{0.7603*\dy}) + -- ({2.8190*\dx},{0.7136*\dy}) + -- ({2.8369*\dx},{0.6663*\dy}) + -- ({2.8534*\dx},{0.6185*\dy}) + -- ({2.8684*\dx},{0.5702*\dy}) + -- ({2.8821*\dx},{0.5214*\dy}) + -- ({2.8944*\dx},{0.4722*\dy}) + -- ({2.9052*\dx},{0.4227*\dy}) + -- ({2.9146*\dx},{0.3728*\dy}) + -- ({2.9225*\dx},{0.3226*\dy}) + -- ({2.9289*\dx},{0.2722*\dy}) + -- ({2.9339*\dx},{0.2216*\dy}) + -- ({2.9374*\dx},{0.1709*\dy}) + -- ({2.9394*\dx},{0.1200*\dy}) + -- ({2.9399*\dx},{0.0692*\dy}) + -- ({2.9389*\dx},{0.0183*\dy}) + -- ({2.9363*\dx},{-0.0326*\dy}) + -- ({2.9323*\dx},{-0.0833*\dy}) + -- ({2.9268*\dx},{-0.1339*\dy}) + -- ({2.9198*\dx},{-0.1843*\dy}) + -- ({2.9114*\dx},{-0.2345*\dy}) + -- ({2.9014*\dx},{-0.2844*\dy}) + -- ({2.8900*\dx},{-0.3340*\dy}) + -- ({2.8771*\dx},{-0.3832*\dy}) + -- ({2.8627*\dx},{-0.4319*\dy}) + -- ({2.8470*\dx},{-0.4802*\dy}) + -- ({2.8298*\dx},{-0.5280*\dy}) + -- ({2.8112*\dx},{-0.5753*\dy}) + -- ({2.7912*\dx},{-0.6219*\dy}) + -- ({2.7698*\dx},{-0.6679*\dy}) + -- ({2.7471*\dx},{-0.7131*\dy}) + -- ({2.7231*\dx},{-0.7577*\dy}) + -- ({2.6978*\dx},{-0.8015*\dy}) + -- ({2.6712*\dx},{-0.8444*\dy}) + -- ({2.6433*\dx},{-0.8865*\dy}) + -- ({2.6143*\dx},{-0.9277*\dy}) + -- ({2.5840*\dx},{-0.9680*\dy}) + -- ({2.5526*\dx},{-1.0073*\dy}) + -- ({2.5200*\dx},{-1.0455*\dy}) + -- ({2.4863*\dx},{-1.0828*\dy}) + -- ({2.4516*\dx},{-1.1189*\dy}) + -- ({2.4159*\dx},{-1.1539*\dy}) + -- ({2.3791*\dx},{-1.1878*\dy}) + -- ({2.3414*\dx},{-1.2205*\dy}) + -- ({2.3028*\dx},{-1.2520*\dy}) + -- ({2.2633*\dx},{-1.2823*\dy}) + -- ({2.2230*\dx},{-1.3113*\dy}) + -- ({2.1819*\dx},{-1.3390*\dy}) + -- ({2.1400*\dx},{-1.3654*\dy}) + -- ({2.0974*\dx},{-1.3904*\dy}) + -- ({2.0542*\dx},{-1.4141*\dy}) + -- ({2.0103*\dx},{-1.4364*\dy}) + -- ({1.9659*\dx},{-1.4573*\dy}) + -- ({1.9209*\dx},{-1.4768*\dy}) + -- ({1.8755*\dx},{-1.4948*\dy}) + -- ({1.8296*\dx},{-1.5114*\dy}) + -- ({1.7833*\dx},{-1.5266*\dy}) + -- ({1.7367*\dx},{-1.5402*\dy}) + -- ({1.6897*\dx},{-1.5524*\dy}) + -- ({1.6426*\dx},{-1.5631*\dy}) + -- ({1.5952*\dx},{-1.5723*\dy}) + -- ({1.5477*\dx},{-1.5800*\dy}) + -- ({1.5001*\dx},{-1.5862*\dy}) + -- ({1.4524*\dx},{-1.5909*\dy}) + -- ({1.4048*\dx},{-1.5941*\dy}) + -- ({1.3572*\dx},{-1.5958*\dy}) + -- ({1.3097*\dx},{-1.5960*\dy}) + -- ({1.2623*\dx},{-1.5947*\dy}) + -- ({1.2152*\dx},{-1.5919*\dy}) + -- ({1.1683*\dx},{-1.5876*\dy}) + -- ({1.1217*\dx},{-1.5818*\dy}) + -- ({1.0754*\dx},{-1.5747*\dy}) + -- ({1.0295*\dx},{-1.5660*\dy}) + -- ({0.9841*\dx},{-1.5559*\dy}) + -- ({0.9391*\dx},{-1.5445*\dy}) + -- ({0.8947*\dx},{-1.5316*\dy}) + -- ({0.8508*\dx},{-1.5173*\dy}) + -- ({0.8076*\dx},{-1.5017*\dy}) + -- ({0.7650*\dx},{-1.4848*\dy}) + -- ({0.7231*\dx},{-1.4666*\dy}) + -- ({0.6820*\dx},{-1.4471*\dy}) + -- ({0.6417*\dx},{-1.4263*\dy}) + -- ({0.6022*\dx},{-1.4043*\dy}) + -- ({0.5635*\dx},{-1.3811*\dy}) + -- ({0.5258*\dx},{-1.3568*\dy}) + -- ({0.4890*\dx},{-1.3313*\dy}) + -- ({0.4532*\dx},{-1.3048*\dy}) + -- ({0.4185*\dx},{-1.2771*\dy}) + -- ({0.3847*\dx},{-1.2485*\dy}) + -- ({0.3521*\dx},{-1.2188*\dy}) + -- ({0.3205*\dx},{-1.1882*\dy}) + -- ({0.2902*\dx},{-1.1567*\dy}) + -- ({0.2609*\dx},{-1.1243*\dy}) + -- ({0.2329*\dx},{-1.0911*\dy}) + -- ({0.2061*\dx},{-1.0571*\dy}) + -- ({0.1806*\dx},{-1.0224*\dy}) + -- ({0.1563*\dx},{-0.9870*\dy}) + -- ({0.1333*\dx},{-0.9509*\dy}) + -- ({0.1117*\dx},{-0.9142*\dy}) + -- ({0.0914*\dx},{-0.8769*\dy}) + -- ({0.0724*\dx},{-0.8391*\dy}) + -- ({0.0547*\dx},{-0.8008*\dy}) + -- ({0.0385*\dx},{-0.7622*\dy}) + -- ({0.0236*\dx},{-0.7231*\dy}) + -- ({0.0102*\dx},{-0.6837*\dy}) + -- ({-0.0019*\dx},{-0.6441*\dy}) + -- ({-0.0125*\dx},{-0.6042*\dy}) + -- ({-0.0217*\dx},{-0.5641*\dy}) + -- ({-0.0295*\dx},{-0.5239*\dy}) + -- ({-0.0359*\dx},{-0.4836*\dy}) + -- ({-0.0409*\dx},{-0.4433*\dy}) + -- ({-0.0444*\dx},{-0.4030*\dy}) + -- ({-0.0466*\dx},{-0.3628*\dy}) + -- ({-0.0473*\dx},{-0.3226*\dy}) + -- ({-0.0466*\dx},{-0.2827*\dy}) + -- ({-0.0445*\dx},{-0.2429*\dy}) + -- ({-0.0410*\dx},{-0.2034*\dy}) + -- ({-0.0362*\dx},{-0.1642*\dy}) + -- ({-0.0300*\dx},{-0.1254*\dy}) + -- ({-0.0224*\dx},{-0.0870*\dy}) + -- ({-0.0136*\dx},{-0.0490*\dy}) + -- ({-0.0034*\dx},{-0.0115*\dy}) + -- ({0.0081*\dx},{0.0255*\dy}) + -- ({0.0208*\dx},{0.0619*\dy}) + -- ({0.0348*\dx},{0.0976*\dy}) + -- ({0.0500*\dx},{0.1327*\dy}) + -- ({0.0663*\dx},{0.1671*\dy}) + -- ({0.0839*\dx},{0.2008*\dy}) + -- ({0.1026*\dx},{0.2336*\dy}) + -- ({0.1224*\dx},{0.2656*\dy}) + -- ({0.1432*\dx},{0.2968*\dy}) + -- ({0.1651*\dx},{0.3271*\dy}) + -- ({0.1880*\dx},{0.3564*\dy}) + -- ({0.2119*\dx},{0.3848*\dy}) + -- ({0.2367*\dx},{0.4122*\dy}) + -- ({0.2625*\dx},{0.4385*\dy}) + -- ({0.2890*\dx},{0.4638*\dy}) + -- ({0.3164*\dx},{0.4880*\dy}) + -- ({0.3446*\dx},{0.5111*\dy}) + -- ({0.3735*\dx},{0.5331*\dy}) + -- ({0.4031*\dx},{0.5539*\dy}) + -- ({0.4334*\dx},{0.5735*\dy}) + -- ({0.4643*\dx},{0.5919*\dy}) + -- ({0.4957*\dx},{0.6090*\dy}) + -- ({0.5276*\dx},{0.6250*\dy}) + -- ({0.5601*\dx},{0.6396*\dy}) + -- ({0.5929*\dx},{0.6530*\dy}) + -- ({0.6262*\dx},{0.6652*\dy}) + -- ({0.6597*\dx},{0.6760*\dy}) + -- ({0.6936*\dx},{0.6855*\dy}) + -- ({0.7277*\dx},{0.6937*\dy}) + -- ({0.7620*\dx},{0.7006*\dy}) + -- ({0.7964*\dx},{0.7062*\dy}) + -- ({0.8309*\dx},{0.7104*\dy}) + -- ({0.8655*\dx},{0.7133*\dy}) + -- ({0.9001*\dx},{0.7149*\dy}) + -- ({0.9346*\dx},{0.7152*\dy}) + -- ({0.9690*\dx},{0.7141*\dy}) + -- ({1.0033*\dx},{0.7117*\dy}) + -- ({1.0374*\dx},{0.7081*\dy}) + -- ({1.0712*\dx},{0.7031*\dy}) + -- ({1.1048*\dx},{0.6969*\dy}) + -- ({1.1380*\dx},{0.6894*\dy}) + -- ({1.1709*\dx},{0.6806*\dy}) + -- ({1.2033*\dx},{0.6706*\dy}) + -- ({1.2353*\dx},{0.6594*\dy}) + -- ({1.2667*\dx},{0.6470*\dy}) + -- ({1.2976*\dx},{0.6334*\dy}) + -- ({1.3279*\dx},{0.6186*\dy}) + -- ({1.3576*\dx},{0.6027*\dy}) + -- ({1.3866*\dx},{0.5857*\dy}) + -- ({1.4149*\dx},{0.5676*\dy}) + -- ({1.4424*\dx},{0.5485*\dy}) + -- ({1.4692*\dx},{0.5284*\dy}) + -- ({1.4951*\dx},{0.5072*\dy}) + -- ({1.5201*\dx},{0.4851*\dy}) + -- ({1.5443*\dx},{0.4621*\dy}) + -- ({1.5675*\dx},{0.4381*\dy}) + -- ({1.5898*\dx},{0.4133*\dy}) + -- ({1.6111*\dx},{0.3877*\dy}) + -- ({1.6314*\dx},{0.3614*\dy}) + -- ({1.6506*\dx},{0.3342*\dy}) + -- ({1.6687*\dx},{0.3064*\dy}) + -- ({1.6858*\dx},{0.2779*\dy}) + -- ({1.7017*\dx},{0.2488*\dy}) + -- ({1.7165*\dx},{0.2191*\dy}) + -- ({1.7302*\dx},{0.1889*\dy}) + -- ({1.7427*\dx},{0.1581*\dy}) + -- ({1.7540*\dx},{0.1270*\dy}) + -- ({1.7640*\dx},{0.0954*\dy}) + -- ({1.7729*\dx},{0.0635*\dy}) + -- ({1.7806*\dx},{0.0312*\dy}) + -- ({1.7870*\dx},{-0.0013*\dy}) + -- ({1.7921*\dx},{-0.0340*\dy}) + -- ({1.7960*\dx},{-0.0669*\dy}) + -- ({1.7987*\dx},{-0.1000*\dy}) + -- ({1.8000*\dx},{-0.1331*\dy}) + -- ({1.8002*\dx},{-0.1662*\dy}) + -- ({1.7990*\dx},{-0.1993*\dy}) + -- ({1.7967*\dx},{-0.2324*\dy}) + -- ({1.7930*\dx},{-0.2654*\dy}) + -- ({1.7881*\dx},{-0.2982*\dy}) + -- ({1.7820*\dx},{-0.3308*\dy}) + -- ({1.7747*\dx},{-0.3632*\dy}) + -- ({1.7661*\dx},{-0.3952*\dy}) + -- ({1.7563*\dx},{-0.4270*\dy}) + -- ({1.7454*\dx},{-0.4584*\dy}) + -- ({1.7332*\dx},{-0.4893*\dy}) + -- ({1.7199*\dx},{-0.5198*\dy}) + -- ({1.7055*\dx},{-0.5497*\dy}) + -- ({1.6900*\dx},{-0.5792*\dy}) + -- ({1.6733*\dx},{-0.6080*\dy}) + -- ({1.6556*\dx},{-0.6362*\dy}) + -- ({1.6368*\dx},{-0.6637*\dy}) + -- ({1.6171*\dx},{-0.6906*\dy}) + -- ({1.5963*\dx},{-0.7166*\dy}) + -- ({1.5746*\dx},{-0.7419*\dy}) + -- ({1.5519*\dx},{-0.7664*\dy}) + -- ({1.5283*\dx},{-0.7901*\dy}) + -- ({1.5039*\dx},{-0.8128*\dy}) + -- ({1.4787*\dx},{-0.8347*\dy}) + -- ({1.4526*\dx},{-0.8556*\dy}) + -- ({1.4258*\dx},{-0.8756*\dy}) + -- ({1.3983*\dx},{-0.8945*\dy}) + -- ({1.3700*\dx},{-0.9124*\dy}) + -- ({1.3412*\dx},{-0.9293*\dy}) + -- ({1.3117*\dx},{-0.9452*\dy}) + -- ({1.2817*\dx},{-0.9599*\dy}) + -- ({1.2511*\dx},{-0.9735*\dy}) + -- ({1.2201*\dx},{-0.9860*\dy}) + -- ({1.1886*\dx},{-0.9974*\dy}) + -- ({1.1567*\dx},{-1.0076*\dy}) + -- ({1.1245*\dx},{-1.0166*\dy}) + -- ({1.0920*\dx},{-1.0245*\dy}) + -- ({1.0592*\dx},{-1.0312*\dy}) + -- ({1.0262*\dx},{-1.0367*\dy}) + -- ({0.9931*\dx},{-1.0409*\dy}) + -- ({0.9598*\dx},{-1.0440*\dy}) + -- ({0.9264*\dx},{-1.0459*\dy}) + -- ({0.8930*\dx},{-1.0465*\dy}) + -- ({0.8596*\dx},{-1.0460*\dy}) + -- ({0.8263*\dx},{-1.0442*\dy}) + -- ({0.7930*\dx},{-1.0413*\dy}) +} diff --git a/buch/papers/zeta/images/zetaplot.pdf b/buch/papers/zeta/images/zetaplot.pdf index 5a59ce6..c6d3693 100644 Binary files a/buch/papers/zeta/images/zetaplot.pdf and b/buch/papers/zeta/images/zetaplot.pdf differ -- cgit v1.2.1 From 4fadfb233a3b7fdc3de486dd85d64fa62408b2a4 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Tue, 9 Aug 2022 18:34:54 +0200 Subject: Added some text, corrected a few errors and added two file extensions to gitignore. --- buch/papers/sturmliouville/.gitignore | 2 ++ .../papers/sturmliouville/waermeleitung_beispiel.tex | 20 +++++++++++++++++--- 2 files changed, 19 insertions(+), 3 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/.gitignore b/buch/papers/sturmliouville/.gitignore index a136337..47f7228 100644 --- a/buch/papers/sturmliouville/.gitignore +++ b/buch/papers/sturmliouville/.gitignore @@ -1 +1,3 @@ *.pdf +*.fls +*.fdb_latexmk \ No newline at end of file diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index da25b36..4885694 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -35,6 +35,7 @@ Lösungsfunktion $u(t,x)$ bei $x = 0$ und $x = l$ nur die vorgegebene Temperatur zurückgeben darf. Es folgen nun \begin{equation} + \label{eq:slp-example-fourier-boundary-condition-ends-constant} u(t,0) = u(t,l) @@ -58,6 +59,7 @@ dass die partiellen Ableitungen von $u(t,x)$ nach $x$ bei $x = 0$ und $x = l$ verschwinden. Somit folgen \begin{equation} + \label{eq:slp-example-fourier-boundary-condition-ends-isolated} \frac{\partial}{\partial x} u(t, 0) = \frac{\partial}{\partial x} u(t, l) @@ -120,6 +122,7 @@ Sturm-Liouville-Form ist. Erfüllen die Randbedingungen des Stab-Problems auch die Randbedingungen des Sturm-Liouville-Problems, kann bereits die Aussage getroffen werden, dass alle Lösungen für die Gleichung in $x$ orthogonal sein werden. +Mehr dazu später. Widmen wir uns zunächst der ersten Gleichung. Aufgrund der Struktur der Gleichung @@ -181,11 +184,22 @@ Durch Koeffizientenvergleich von \end{aligned} \] ist schnell ersichtlich, dass $ \mu = -\alpha^{2} = -\beta^{2} $ gelten muss für -$ A \neq 0 $ und $ B \neq 0 $. +$ A \neq 0 $ oder $ B \neq 0 $. Zur Berechnung von $ \mu $ bleiben also noch $ \alpha $ und $ \beta $ zu bestimmen. +Dazu werden die Randbedingungen +\eqref{eq:slp-example-fourier-boundary-condition-ends-constant} und +\eqref{eq:slp-example-fourier-boundary-condition-ends-isolated} benötigt. +Zu bemerken ist, dass die Randbedingungen nur Anforderungen in $x$ stellen und +somit direkt für $X(x)$ übernomen werden können. -TODO: randbedingungen!!---- +Daraus ergibt sich für einen Stab mit Enden auf konstanter Temperatur + +\begin{equation} + \mu + = + -\frac{n^{2}\pi^{2}}{l^{2}} +\end{equation} Betrachten wir nun die zweite Gleichung \eqref{eq:slp-example-fourier-separated-t}. @@ -203,7 +217,7 @@ Lösung e^{-\kappa \mu t} \] führt. -Und mit mit dem Resultat von zuvor die Lösung +Und mit dem Resultat (TODO) die Lösung \[ T(t) = -- cgit v1.2.1 From 2e6fd0152fc9c135ced14ea186ac7e2fc1b15f7e Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Tue, 9 Aug 2022 18:38:54 +0200 Subject: Removed file extensions from gitignore. --- buch/papers/sturmliouville/.gitignore | 4 +--- 1 file changed, 1 insertion(+), 3 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/.gitignore b/buch/papers/sturmliouville/.gitignore index 47f7228..f08278d 100644 --- a/buch/papers/sturmliouville/.gitignore +++ b/buch/papers/sturmliouville/.gitignore @@ -1,3 +1 @@ -*.pdf -*.fls -*.fdb_latexmk \ No newline at end of file +*.pdf \ No newline at end of file -- cgit v1.2.1 From 67dc3b04c3926f0c7beb5cd6781cc58a4c38e667 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Tue, 9 Aug 2022 21:12:25 +0200 Subject: Added section to show orthogonality with boundary conditions to fourier example. --- .../sturmliouville/waermeleitung_beispiel.tex | 54 ++++++++++++++++++++-- 1 file changed, 50 insertions(+), 4 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 4885694..92ecc49 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -122,7 +122,52 @@ Sturm-Liouville-Form ist. Erfüllen die Randbedingungen des Stab-Problems auch die Randbedingungen des Sturm-Liouville-Problems, kann bereits die Aussage getroffen werden, dass alle Lösungen für die Gleichung in $x$ orthogonal sein werden. -Mehr dazu später. + +Damit die Lösungen von $X$ orthogonal sind, müssen also die Gleichungen +\begin{equation} +\begin{aligned} + \label{eq:slp-example-fourier-randbedingungen} + k_a y(a) + h_a p(a) y'(a) &= 0 \\ + k_b y(b) + h_b p(b) y'(b) &= 0 +\end{aligned} +\end{equation} +erfüllt sein und es muss ausserdem +\begin{equation} +\begin{aligned} + \label{eq:slp-example-fourier-coefficient-constraints} + |k_a|^2 + |h_a|^2 &\neq 0\\ + |k_b|^2 + |h_b|^2 &\neq 0\\ +\end{aligned} +\end{equation} +gelten. + +Um zu verifizieren, ob die Randbedingungen erfüllt sind, benötigen wir zunächst +$p(x)$. +Dazu wird die Gleichung \eqref{eq:slp-example-fourier-separated-x} mit der +Sturm-Liouville-Form \eqref{eq:sturm-liouville-equation} verglichen, was zu +$p(x) = 1$ führt. + +Werden nun $p(x)$ und die Randbedingungen +\eqref{eq:slp-example-fourier-boundary-condition-ends-constant} in +\eqref{eq:slp-example-fourier-randbedingungen} eigesetzt, erhält man +\[ +\begin{aligned} + k_a y(0) + h_a y'(0) &= h_a y'(0) = 0 \\ + k_b y(l) + h_b y'(l) &= h_b y'(l) = 0. +\end{aligned} +\] +Damit die Gleichungen erfüllt sind, müssen $h_a = 0$ und $h_b = 0$ sein. +Zusätzlich müssen aber die Bedingungen +\eqref{eq:slp-example-fourier-coefficient-constraints} erfüllt sein und +da $y(0) = 0$ und $y(l) = 0$ sind, können belibige $k_a \neq 0$ und $k_b \neq 0$ +gewählt werden. + +Somit ist gezeigt, dass die Randbedingungen des Stab-Problems für Enden auf +konstanter Temperatur auch die Sturm-Liouville-Randbedingungen erfüllen und +alle daraus reultierenden Lösungen orthogonal sind. +Analog dazu kann gezeit werden, dass die Randbedingungen für einen Stab mit +isolierten Enden ebenfalls die Sturm-Liouville-Randbedingungen erfüllen und +somit auch zu orthogonalen Lösungen führen. Widmen wir uns zunächst der ersten Gleichung. Aufgrund der Struktur der Gleichung @@ -187,7 +232,7 @@ ist schnell ersichtlich, dass $ \mu = -\alpha^{2} = -\beta^{2} $ gelten muss fü $ A \neq 0 $ oder $ B \neq 0 $. Zur Berechnung von $ \mu $ bleiben also noch $ \alpha $ und $ \beta $ zu bestimmen. -Dazu werden die Randbedingungen +Dazu werden nochmals die Randbedingungen \eqref{eq:slp-example-fourier-boundary-condition-ends-constant} und \eqref{eq:slp-example-fourier-boundary-condition-ends-isolated} benötigt. Zu bemerken ist, dass die Randbedingungen nur Anforderungen in $x$ stellen und @@ -196,12 +241,13 @@ somit direkt für $X(x)$ übernomen werden können. Daraus ergibt sich für einen Stab mit Enden auf konstanter Temperatur \begin{equation} + \label{eq:slp-example-fourier-mu-solution} \mu = -\frac{n^{2}\pi^{2}}{l^{2}} \end{equation} -Betrachten wir nun die zweite Gleichung +Betrachten wir nun die zweite Gleichung der Separation \eqref{eq:slp-example-fourier-separated-t}. Diese Lösen wir über das charakteristische Polynom \[ @@ -217,7 +263,7 @@ Lösung e^{-\kappa \mu t} \] führt. -Und mit dem Resultat (TODO) die Lösung +Und mit dem Resultat \eqref{eq:slp-example-fourier-mu-solution} die Lösung \[ T(t) = -- cgit v1.2.1 From 970e6a8a2b2371834e8a4ff42123da59e3990fe4 Mon Sep 17 00:00:00 2001 From: runterer Date: Tue, 9 Aug 2022 21:59:31 +0200 Subject: Finished --- buch/papers/zeta/analytic_continuation.tex | 26 +++++++++++++------------- buch/papers/zeta/einleitung.tex | 8 ++++---- buch/papers/zeta/euler_product.tex | 14 +++++++------- buch/papers/zeta/fazit.tex | 17 ++++++++--------- buch/papers/zeta/images/zetaplot.tex | 2 +- buch/papers/zeta/zeta_gamma.tex | 4 ++-- 6 files changed, 35 insertions(+), 36 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/zeta/analytic_continuation.tex b/buch/papers/zeta/analytic_continuation.tex index 4046bb7..ed07e04 100644 --- a/buch/papers/zeta/analytic_continuation.tex +++ b/buch/papers/zeta/analytic_continuation.tex @@ -62,14 +62,14 @@ Durch Subtraktion der beiden Gleichungen \eqref{zeta:align1} minus \eqref{zeta:a {\color{blue}\frac{2}{2^s}} + {\color{red}\frac{1}{2^s}} - }_{-\frac{1}{2^s}} + }_{\displaystyle{-\frac{1}{2^s}}} + {\color{red}\frac{1}{3^s}} \underbrace{- {\color{blue}\frac{2}{4^s}} + {\color{red}\frac{1}{4^s}} - }_{-\frac{1}{4^s}} + }_{\displaystyle{-\frac{1}{4^s}}} \ldots \\ &= \eta(s). @@ -89,7 +89,7 @@ Wir beginnen damit, die Gammafunktion für den halben Funktionswert zu berechnen = \int_0^{\infty} t^{\frac{s}{2}-1} e^{-t} dt. \end{equation} -Nun substituieren wir $t$ mit $t = \pi n^2 x$ und $dt=\pi n^2 dx$ und erhalten +Nun substituieren wir $t = \pi n^2 x$ und $dt=\pi n^2 dx$ und erhalten \begin{equation} \Gamma \left( \frac{s}{2} \right) = @@ -109,7 +109,7 @@ Analog zum Abschnitt \ref{zeta:section:zusammenhang_mit_gammafunktion} teilen wi e^{-\pi n^2 x} \,dx, \end{equation} -und finden $\zeta(s)$ durch die Summenbildung $\sum_{n=1}^{\infty}$ +und finden $\zeta(s)$ durch die Summenbildung über alle $n$ \begin{align} \frac{\Gamma \left( \frac{s}{2} \right)}{\pi^{\frac{s}{2}}} \zeta(s) @@ -139,14 +139,14 @@ Zunächst teilen wir nun das Integral auf in zwei Teile x^{\frac{s}{2}-1} \psi(x) \,dx - }_{I_1} + }_{\displaystyle{I_1}} + \underbrace{ \int_1^{\infty} x^{\frac{s}{2}-1} \psi(x) \,dx - }_{I_2} + }_{\displaystyle{I_2}} = I_1 + I_2. \end{equation} @@ -231,11 +231,11 @@ Somit haben wir die analytische Fortsetzung gefunden als \zeta(1-s), \end{equation} was einer Spiegelung an der $\Re(s) = \frac{1}{2}$ Geraden entspricht. -Eine ganz ähnliche Spiegelungseigenschaft wurde bereits in Kapitel \ref{buch:funktionentheorie:subsection:gammareflektion} für die Gammafunktion gefunden. +Eine ganz ähnliche Spiegelungseigenschaft wurde bereits in Abschnitt \ref{buch:funktionentheorie:subsection:gammareflektion} für die Gammafunktion gefunden. \subsection{Berechnung des Integrals $I_1 = \int_0^{1} x^{\frac{s}{2}-1} \psi(x) \,dx$} \label{zeta:subsubsec:intcal} -Ziel dieses Abschnittes ist es, zu zeigen wie das Integral $I_1$ aus Gleichung \eqref{zeta:equation:integral2} durch ein neues Integral mit den Integrationsgrenzen $1$ und $\infty$ ersetzt werden kann. +Ziel dieses Abschnittes ist, zu zeigen wie das Integral $I_1$ aus Gleichung \eqref{zeta:equation:integral2} durch ein neues Integral mit den Integrationsgrenzen $1$ und $\infty$ ersetzt werden kann. Da dieser Schritt ziemlich aufwendig ist, wird er hier in einem eigenen Abschnitt behandelt. Zunächst wird die poissonsche Summenformel hergeleitet \cite{zeta:online:poisson}, da diese verwendet werden kann um $\psi(x)$ zu berechnen. @@ -313,8 +313,8 @@ Um die poissonsche Summenformel zu beweisen, berechnen wir zunächst die Fourier \underbrace{ \sum_{k=-\infty}^{\infty} e^{-i 2\pi x k} - }_{\text{\eqref{zeta:equation:fourier_dirac}}} - \, dx, + }_{\displaystyle{\text{\eqref{zeta:equation:fourier_dirac}}}} + \, dx, \label{zeta:equation:1934} \end{align} und verwenden die Fouriertransformation der Dirac Funktion aus \eqref{zeta:equation:fourier_dirac} \begin{align} @@ -330,7 +330,7 @@ Um die poissonsche Summenformel zu beweisen, berechnen wir zunächst die Fourier \sum_{k=-\infty}^{\infty} \delta(x + k). \end{align} - Wenn wir dies einsetzen und erhalten wir + Wenn wir dies einsetzen in \eqref{zeta:equation:1934} erhalten wir \begin{equation} \sum_{k=-\infty}^{\infty} F(k) @@ -465,7 +465,7 @@ Diese Form von $\psi(x)$ eingesetzt in $I_1$ ergibt x^{\frac{s}{2}-\frac{3}{2}} \psi \left( \frac{1}{x} \right) \,dx - }_{I_3} + }_{\displaystyle{I_3}} + \underbrace{ \frac{1}{2} @@ -474,7 +474,7 @@ Diese Form von $\psi(x)$ eingesetzt in $I_1$ ergibt - x^{\frac{s}{2}-1} \,dx - }_{I_4}. \label{zeta:equation:integral3} + }_{\displaystyle{I_4}}. \label{zeta:equation:integral3} \end{align} Darin kann für das zweite Integral $I_4$ eine Lösung gefunden werden als \begin{equation} diff --git a/buch/papers/zeta/einleitung.tex b/buch/papers/zeta/einleitung.tex index ad87fec..828678d 100644 --- a/buch/papers/zeta/einleitung.tex +++ b/buch/papers/zeta/einleitung.tex @@ -12,11 +12,11 @@ Die Zetafunktion ist bekannt als Bestandteil der Riemannschen Vermutung, welche Mithilfe dieser Vermutung kann eine gute Annäherung an die Primzahlfunktion gefunden werden. Die Primzahlfunktion steigt immer an, sobald eine Primzahl vorkommt. Eine Darstellung davon ist in Abbildung \ref{fig:zeta:primzahlfunktion} zu finden. -Die Riemannsche Vermutung ist eines der ungelösten Millennium-Probleme der Mathematik, auf deren Lösung eine Belohnung von einer Million Doller ausgesetzt ist \cite{zeta:online:millennium}. +Die Riemannsche Vermutung ist eines der ungelösten Millennium-Probleme der Mathematik, auf deren Lösung eine Belohnung von einer Million Dollar ausgesetzt ist \cite{zeta:online:millennium}. Auf eine genauere Beschreibung der Riemannschen Vermutung wird im Rahmen dieses Papers nicht eingegangen. \begin{figure} \centering - \input{papers/zeta/images/primzahlfunktion_paper.pgf} + \input{papers/zeta/images/primzahlfunktion2.tex} \caption{Die Primzahlfunktion von $0$ bis $30$.} \label{fig:zeta:primzahlfunktion} \end{figure} @@ -28,7 +28,7 @@ Schlussendlich folgt die Beschreibung der analytischen Fortsetzung die gesamte k Diese analytische Fortsetzung wird für die Riemannsche Vermutung benötigt, ermöglicht aber auch andere interessante Aussagen. So findet sich zum Beispiel immer wieder die aberwitzige Behauptung, das die Summe aller natürlichen Zahlen \begin{equation*} - \sum{n=1}^{\infty} n + \sum_{n=1}^{\infty} n = \sum_{n=1}^{\infty} \frac{1}{n^{-1}} @@ -36,6 +36,6 @@ So findet sich zum Beispiel immer wieder die aberwitzige Behauptung, das die Sum -\frac{1}{12} \end{equation*} sei. -Obwohl diese Behauptung offensichtlich Falsch ist, hat sie doch ihre Berechtigung, wie durch die analytische Fortsetzung gezeigt werden wird. +Obwohl diese Behauptung offensichtlich falsch ist, hat sie doch ihre Berechtigung, wie durch die analytische Fortsetzung gezeigt werden wird. Die folgenden mathematischen Herleitungen sind, sofern nicht anders gekennzeichnet, eigene Darstellungen basierend auf den überaus umfangreichen Wikipedia-Artikeln auf Deutsch \cite{zeta:online:wiki_de} und Englisch \cite{zeta:online:wiki_en} sowie einer Video-Playlist \cite{zeta:online:mryoumath}. diff --git a/buch/papers/zeta/euler_product.tex b/buch/papers/zeta/euler_product.tex index 7915c84..9c08dd2 100644 --- a/buch/papers/zeta/euler_product.tex +++ b/buch/papers/zeta/euler_product.tex @@ -28,9 +28,9 @@ Um daraus die Riemannsche Vermutung herzuleiten, wäre aber noch einiges mehr n = \prod_{p \in P} \sum_{k_i=0}^{\infty} - \left( + \biggl( \frac{1}{p_i^s} - \right)^{k_i} + \biggr)^{k_i} = \prod_{p \in P} \sum_{k_i=0}^{\infty} @@ -53,11 +53,11 @@ Um daraus die Riemannsche Vermutung herzuleiten, wäre aber noch einiges mehr n \sum_{k_1=0}^{\infty} \sum_{k_2=0}^{\infty} \ldots - \left( + \biggl( \frac{1}{p_1^{k_1}} \frac{1}{p_2^{k_2}} \ldots - \right)^s. + \biggr)^s. \label{zeta:equation:eulerprodukt2} \end{align} Der Fundamentalsatz der Arithmetik (Primfaktorzerlegung) besagt, dass jede beliebige Zahl $n \in \mathbb{N}$ durch eine eindeutige Primfaktorzerlegung beschrieben werden kann @@ -70,17 +70,17 @@ Um daraus die Riemannsche Vermutung herzuleiten, wäre aber noch einiges mehr n \sum_{k_1=0}^{\infty} \sum_{k_2=0}^{\infty} \ldots - \left( + \biggl( \frac{1}{p_1^{k_1}} \frac{1}{p_2^{k_2}} \ldots - \right)^s + \biggr)^s = \sum_{n=1}^\infty \frac{1}{n^s} = \zeta(s), \end{equation} - wodurch das Eulerprudukt bewiesen ist. + wodurch das Eulerprodukt bewiesen ist. \end{proof} diff --git a/buch/papers/zeta/fazit.tex b/buch/papers/zeta/fazit.tex index fe2d35d..e33083a 100644 --- a/buch/papers/zeta/fazit.tex +++ b/buch/papers/zeta/fazit.tex @@ -1,5 +1,5 @@ -\section{Fazit} \label{zeta:section:fazit} -\rhead{Fazit} +\section{Der Wert $\zeta(-1)$} \label{zeta:section:fazit} +\rhead{Der Wert $\zeta(-1)$} Ganz zu Beginn dieses Papers wurde die Behauptung erwähnt, dass die Summe aller natürlichen Zahlen $-\frac{1}{12}$ sei. Diese Summe ist nichts anderes als die Zetafunktion am Wert $s=-1$. @@ -17,7 +17,7 @@ Da wir die analytische Fortsetzung mit der Funktionalgleichung \eqref{zeta:equat \zeta(2) \frac{\pi^{-\frac{1}{2}}}{\Gamma \left( -\frac{1}{2} \right)}. \end{align*} -Also fehlen uns drei Werte, $\zeta(2)$, $\Gamma(1)$ und $\Gamma\left(-\frac{1}{2}\right)$. +Also fehlen uns drei Werte, $\zeta(2)$, $\Gamma(1)$ und $\Gamma(-\frac{1}{2})$. Zunächst konzentrieren wir uns auf $\zeta(2)$, welches im konvergenten Bereich der Reihe liegt und auch bekannt ist als das Basler Problem. Wir lösen das Basler Problem \cite{zeta:online:basel} mithilfe der parsevalschen Gleichung \cite{zeta:online:pars} @@ -44,7 +44,7 @@ Wenn wir dies für $f(x) = x$ auswerten erhalten wir &= 2\pi \sum_{n=-\infty}^{\infty} |c_n|^2 = - 4\pi \underbrace{\sum_{n=1}^{\infty} \frac{1}{n^2}}_{\zeta(2)}. + 4\pi \underbrace{\sum_{n=1}^{\infty} \frac{1}{n^2}}_{\displaystyle{\zeta(2)}}. \end{align} Durch einfaches Umstellen erhalten wir somit die Lösung des Basler Problems als \begin{equation} @@ -53,13 +53,13 @@ Durch einfaches Umstellen erhalten wir somit die Lösung des Basler Problems als = \frac{\pi^2}{6}. \end{equation} -Als nächstes berechnen wir $\Gamma(1)$ und $\Gamma\left(-\frac{1}{2}\right)$ mithilfe der Integraldefinition der Gammafunktion \ref{buch:rekursion:def:gamma}. -Da das Integral für $\Gamma\left(-\frac{1}{2}\right)$ nicht konvergiert, wird die Reflektionsformel aus \ref{buch:funktionentheorie:subsection:gammareflektion} verwendet, welche das konvergierende Integral von $\Gamma\left(\frac{3}{2}\right)$ verwendet. +Als nächstes berechnen wir $\Gamma(1)$ und $\Gamma(-\frac{1}{2})$ mithilfe der Integraldefinition der Gammafunktion (Definition \ref{buch:rekursion:def:gamma}). +Da das Integral für $\Gamma(-\frac{1}{2})$ nicht konvergiert, wird die Reflektionsformel aus \ref{buch:funktionentheorie:subsection:gammareflektion} verwendet, welche das konvergierende Integral von $\Gamma\left(\frac{3}{2}\right)$ verwendet. Es ergeben sich die Werte \begin{align*} \Gamma(1) &= 1\\ - \Gamma\left(-\frac{1}{2}\right) + \Gamma\biggl(-\frac{1}{2}\biggr) &= \frac{\pi}{\sin\left(-\frac{\pi}{2}\right) \Gamma\left(\frac{3}{2}\right)} = -\frac{\sqrt{\pi}}{2}. @@ -85,10 +85,9 @@ Wenn wir diese Werte in die Funktionalgleichung einsetzen, erhalten wir das gew Weiter wurde zu Beginn dieses Papers auf die Riemannsche Vermutung hingewiesen, wonach alle nichttrivialen Nullstellen der Zetafunktion auf der $\Re(s)=\frac{1}{2}$ Geraden liegen. Abbildung \ref{zeta:fig:einzweitel} zeigt die Funktionswerte dieser Geraden. -%TODO colorplot does not work.. Ausserdem zeigt Abbildung \ref{zeta:fig:colorplot} die farbcodierte Zetafunktion für Werte der analytischen Fortsetzung und des originalen Definitionsbereichs. \begin{figure} \centering - \input{papers/zeta/images/zeta_re_0.5_paper.pgf} + \input{papers/zeta/images/zetaplot.tex} \caption{Die komplexen Werte der Zetafunktion für die kritische Gerade $\Re(s)=\frac{1}{2}$ im Bereich $\Im(s) = 0\dots40$. Klar sichtbar sind die immer wiederkehrenden Nullstellen, wie sie Gegenstand der Riemannschen Vermutung sind.} \label{zeta:fig:einzweitel} diff --git a/buch/papers/zeta/images/zetaplot.tex b/buch/papers/zeta/images/zetaplot.tex index 1cd3259..521bb1a 100644 --- a/buch/papers/zeta/images/zetaplot.tex +++ b/buch/papers/zeta/images/zetaplot.tex @@ -38,7 +38,7 @@ \draw (-0.1,{-\y*\dy}) -- (0.1,{-\y*\dy}); } -\input{zetapath.tex} +\input{papers/zeta/images/zetapath.tex} \draw[color=blue,line width=1pt] \zetapath; diff --git a/buch/papers/zeta/zeta_gamma.tex b/buch/papers/zeta/zeta_gamma.tex index 1f10a33..dd422e3 100644 --- a/buch/papers/zeta/zeta_gamma.tex +++ b/buch/papers/zeta/zeta_gamma.tex @@ -11,7 +11,7 @@ Wir erinnern uns an die Definition der Gammafunktion in \eqref{buch:rekursion:ga \int_0^{\infty} t^{s-1} e^{-t} \,dt, \end{equation*} wobei die Notation an die Zetafunktion angepasst ist. -Durch die Substitution von $t$ mit $t = nu$ und $dt = n\,du$ wird daraus +Durch die Substitution $t = nu$ und $dt = n\,du$ wird daraus \begin{align*} \Gamma(s) &= @@ -57,5 +57,5 @@ Wenn wir dieses Resultat einsetzen in \eqref{zeta:equation:zeta_gamma1} und durc \frac{1}{\Gamma(s)} \int_0^{\infty} \frac{u^{s-1}}{e^u -1} - du \qed + du. \end{equation} -- cgit v1.2.1 From 7d2e4ff7b1b50b382af659fcfbbc38adb6dd7ace Mon Sep 17 00:00:00 2001 From: runterer Date: Tue, 9 Aug 2022 22:05:19 +0200 Subject: minor changes --- buch/papers/zeta/analytic_continuation.tex | 2 +- buch/papers/zeta/fazit.tex | 2 +- 2 files changed, 2 insertions(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/zeta/analytic_continuation.tex b/buch/papers/zeta/analytic_continuation.tex index ed07e04..d45a6ae 100644 --- a/buch/papers/zeta/analytic_continuation.tex +++ b/buch/papers/zeta/analytic_continuation.tex @@ -330,7 +330,7 @@ Um die poissonsche Summenformel zu beweisen, berechnen wir zunächst die Fourier \sum_{k=-\infty}^{\infty} \delta(x + k). \end{align} - Wenn wir dies einsetzen in \eqref{zeta:equation:1934} erhalten wir + Wenn wir dies einsetzen in Gleichung \eqref{zeta:equation:1934} erhalten wir \begin{equation} \sum_{k=-\infty}^{\infty} F(k) diff --git a/buch/papers/zeta/fazit.tex b/buch/papers/zeta/fazit.tex index e33083a..027f324 100644 --- a/buch/papers/zeta/fazit.tex +++ b/buch/papers/zeta/fazit.tex @@ -54,7 +54,7 @@ Durch einfaches Umstellen erhalten wir somit die Lösung des Basler Problems als \end{equation} Als nächstes berechnen wir $\Gamma(1)$ und $\Gamma(-\frac{1}{2})$ mithilfe der Integraldefinition der Gammafunktion (Definition \ref{buch:rekursion:def:gamma}). -Da das Integral für $\Gamma(-\frac{1}{2})$ nicht konvergiert, wird die Reflektionsformel aus \ref{buch:funktionentheorie:subsection:gammareflektion} verwendet, welche das konvergierende Integral von $\Gamma\left(\frac{3}{2}\right)$ verwendet. +Da das Integral für $\Gamma(-\frac{1}{2})$ nicht konvergiert, wird die Reflektionsformel aus \ref{buch:funktionentheorie:subsection:gammareflektion} verwendet, welche das konvergierende Integral von $\Gamma(\frac{3}{2})$ verwendet. Es ergeben sich die Werte \begin{align*} \Gamma(1) -- cgit v1.2.1 From d4e52d5bd83bed95d7712c34e14ccde3ff72810e Mon Sep 17 00:00:00 2001 From: Nicolas Tobler Date: Tue, 9 Aug 2022 23:54:32 +0200 Subject: Improved plot color choices --- buch/papers/ellfilter/einleitung.tex | 2 +- buch/papers/ellfilter/elliptic.tex | 53 +- buch/papers/ellfilter/jacobi.tex | 79 +- buch/papers/ellfilter/python/F_N_elliptic.pgf | 834 ----------------- buch/papers/ellfilter/python/elliptic.pgf | 1207 ++++++++++++++++++++----- buch/papers/ellfilter/python/elliptic.py | 6 +- buch/papers/ellfilter/python/elliptic2.py | 55 +- buch/papers/ellfilter/python/k.pgf | 86 -- buch/papers/ellfilter/tikz/arccos.tikz.tex | 28 +- buch/papers/ellfilter/tikz/arccos2.tikz.tex | 41 +- buch/papers/ellfilter/tikz/cd.tikz.tex | 53 +- buch/papers/ellfilter/tikz/sn.tikz.tex | 54 +- buch/papers/ellfilter/tschebyscheff.tex | 51 +- 13 files changed, 1183 insertions(+), 1366 deletions(-) delete mode 100644 buch/papers/ellfilter/python/F_N_elliptic.pgf (limited to 'buch/papers') diff --git a/buch/papers/ellfilter/einleitung.tex b/buch/papers/ellfilter/einleitung.tex index 18913fb..5bc2ead 100644 --- a/buch/papers/ellfilter/einleitung.tex +++ b/buch/papers/ellfilter/einleitung.tex @@ -56,7 +56,7 @@ Das Tschebyscheff-1 Filter ist maximal steil für eine definierte Welligkeit im Es scheint so als sind gewisse Eigenschaften dieser speziellen Funktionen verantwortlich für die Optimalität dieser Filter. Dieses Paper betrachtet die Theorie hinter dem elliptischen Filter, dem wohl exotischsten dieser Auswahl. -Es weist sich aus durch den Steilsten Übergangsbereich für eine gegebene Filterdesignspezifikation. +Es weist sich aus durch den steilsten Übergangsbereich für eine gegebene Filterdesignspezifikation. Des weiteren kann es als Verallgemeinerung des Tschebyscheff-Filters angesehen werden. % wenn $F_N(w)$ eine rationale Funktion ist, ist auch $H(\Omega)$ eine rationale Funktion und daher ein lineares Filter. %proof? diff --git a/buch/papers/ellfilter/elliptic.tex b/buch/papers/ellfilter/elliptic.tex index 861600b..8c60e46 100644 --- a/buch/papers/ellfilter/elliptic.tex +++ b/buch/papers/ellfilter/elliptic.tex @@ -6,47 +6,26 @@ Kommen wir nun zum eigentlichen Teil dieses Papers, den elliptischen rationalen &= \cd \left(N~\frac{K_1}{K}~\cd^{-1}(w, k), k_1)\right) , \quad k= 1/\xi, k_1 = 1/f(\xi) \\ &= \cd \left(N~K_1~z , k_1 \right), \quad w= \cd(z K, k) \end{align} - - -sieht ähnlich aus wie die trigonometrische Darstellung der Tschebyschef-Polynome \eqref{ellfilter:eq:chebychef_polynomials} +Beim Betrachten dieser Definition, fällt die Ähnlichkeit zur trigonometrische Darstellung der Tschebyschef-Polynome \eqref{ellfilter:eq:chebychef_polynomials} auf. Anstelle vom Kosinus kommt hier die $\cd$-Funktion zum Einsatz. Die Ordnungszahl $N$ kommt auch als Faktor for. Zusätzlich werden noch zwei verschiedene elliptische Module $k$ und $k_1$ gebraucht. +Bei $k = k_1 = 0$ wird der $\cd$ zum Kosinus und wir erhalten in diesem Spezialfall die Tschebyschef-Polynome. - - -Sinus entspricht $\sn$ - -Damit die Nullstellen an ähnlichen Positionen zu liegen kommen wie bei den Tschebyscheff-Polynomen, muss die $\cd$-Funktion gewählt werden. - +Durch das Konzept vom fundamentalen Rechteck, siehe Abbildung \ref{buch:elliptisch:fig:ellall} können für alle inversen Jacobi elliptischen Funktionen die Positionen der Null- und Polstellen anhand eines Diagramms ermittelt werden. Die $\cd^{-1}(w, k)$-Funktion ist um $K$ verschoben zur $\sn^{-1}(w, k)$-Funktion, wie ersichtlich in Abbildung \ref{ellfilter:fig:cd}. \begin{figure} \centering \input{papers/ellfilter/tikz/cd.tikz.tex} \caption{ - $z$-Ebene der Funktion $z = \sn^{-1}(w, k)$. + $z$-Ebene der Funktion $z = \cd^{-1}(w, k)$. Die Funktion ist in der realen Achse $4K$-periodisch und in der imaginären Achse $2jK^\prime$-periodisch. } \label{ellfilter:fig:cd} \end{figure} -Auffallend ist, dass sich alle Nullstellen und Polstellen um $K$ verschoben haben. - -Durch das Konzept vom fundamentalen Rechteck, siehe Abbildung \ref{ellfilter:fig:fundamental_rectangle} können für alle inversen Jacobi elliptischen Funktionen die Positionen der Null- und Polstellen anhand eines Diagramms ermittelt werden. -Der erste Buchstabe bestimmt die Position der Nullstelle und der zweite Buchstabe die Polstelle. -\begin{figure} - \centering - \input{papers/ellfilter/tikz/fundamental_rectangle.tikz.tex} - \caption{ - Fundamentales Rechteck der inversen Jacobi elliptischen Funktionen. - } - \label{ellfilter:fig:fundamental_rectangle} -\end{figure} - -Auffallend an der $w = \sn(z, k)$-Funktion ist, dass sich $w$ auf der reellen Achse wie der Kosinus immer zwischen $-1$ und $1$ bewegt, während bei $\mathrm{Im(z) = K^\prime}$ die Werte zwischen $\pm 1/k$ und $\pm \infty$ verlaufen. -Die Funktion hat also Equirippel-Verhalten um $w=0$ und um $w=\pm \infty$. -Falls es möglich ist diese Werte abzufahren im Sti der Tschebyscheff-Polynome, kann ein Filter gebaut werden, dass Equirippel-Verhalten im Durchlass- und Sperrbereich aufweist. - - +Auffallend an der $w = \cd(z, k)$-Funktion ist, dass sich $w$ auf der reellen Achse wie der Kosinus immer zwischen $-1$ und $1$ bewegt, während bei $\mathrm{Im(z) = K^\prime}$ die Werte zwischen $\pm 1/k$ und $\pm \infty$ verlaufen. +Die Funktion hat also Equirippel-Verhalten um $w=0$ und um $w=\pm \infty$. %TODO Check +Falls es möglich ist diese Werte abzufahren im Stil der Tschebyscheff-Polynome, kann ein Filter gebaut werden, dass Equirippel-Verhalten im Durchlass- und Sperrbereich aufweist. Analog zu Abbildung \ref{ellfilter:fig:arccos2} können wir auch bei den elliptisch rationalen Funktionen die komplexe $z$-Ebene betrachten, wie ersichtlich in Abbildung \ref{ellfilter:fig:cd2}, um die besser zu verstehen. \begin{figure} @@ -60,20 +39,10 @@ Analog zu Abbildung \ref{ellfilter:fig:arccos2} können wir auch bei den ellipti \end{figure} % Da die $\cd^{-1}$-Funktion - - -\begin{figure} - \centering - \input{papers/ellfilter/python/F_N_elliptic.pgf} - \caption{$F_N$ für ein elliptischs filter.} - \label{ellfilter:fig:elliptic} -\end{figure} - - \begin{figure} \centering \input{papers/ellfilter/python/elliptic.pgf} - \caption{Die resultierende frequenzantwort eines elliptischs filter.} + \caption{$F_N$ und die resultierende Frequenzantwort eines elliptischen Filters.} \label{ellfilter:fig:elliptic_freq} \end{figure} @@ -90,6 +59,10 @@ Dies trifft ein wenn die Gradengleichung erfüllt ist. Leider ist das lösen dieser Gleichung nicht trivial. Die Rechnung wird in \ref{ellfilter:bib:orfanidis} im Detail angeschaut. +$K$ und $K^\prime$ sind voneinender abhängig. + +Das Problem lässt sich grafisch darstellen. + \begin{figure} \centering \input{papers/ellfilter/python/k.pgf} @@ -108,8 +81,6 @@ Die Rechnung wird in \ref{ellfilter:bib:orfanidis} im Detail angeschaut. \caption{Die Gradgleichung als geometrisches Problem.} \end{figure} - - \subsection{Polynome?} Bei den Tschebyscheff-Polynomen haben wir gesehen, dass die Trigonometrische Formel zu einfachen Polynomen umgewandelt werden kann. diff --git a/buch/papers/ellfilter/jacobi.tex b/buch/papers/ellfilter/jacobi.tex index 6a208fa..3940171 100644 --- a/buch/papers/ellfilter/jacobi.tex +++ b/buch/papers/ellfilter/jacobi.tex @@ -2,14 +2,16 @@ %TODO $z$ or $u$ for parameter? -Für das elliptische Filter wird statt der, für das Tschebyscheff-Filter benutzen Kreis-Trigonometrie die elliptischen Funktionen gebraucht. +Für das elliptische Filter werden, wie es der Name bereits deutet, elliptische Funktionen gebraucht. +Wie die trigonometrischen Funktionen Zusammenhänge eines Kreises darlegen, beschreiben die elliptischen Funktionen Ellipsen. +Es ist daher naheliegend, dass Kosinus des Tschebyscheff-Filters mit einem elliptischen Pendant ausgetauscht werden könnte. Der Begriff elliptische Funktion wird für sehr viele Funktionen gebraucht, daher ist es hier wichtig zu erwähnen, dass es ausschliesslich um die Jacobischen elliptischen Funktionen geht. +Die Jacobi elliptischen Funktionen werden ausführlich im Kapitel \ref{buch:elliptisch:section:jacobi} behandelt. Im Wesentlichen erweitern die Jacobi elliptischen Funktionen die trigonometrische Funktionen für Ellipsen. Zum Beispiel gibt es analog zum Sinus den elliptischen $\sn(z, k)$. Im Gegensatz zum den trigonometrischen Funktionen haben die elliptischen Funktionen zwei parameter. -Zum einen gibt es den \textit{elliptische Modul} $k$, der die Exzentrizität der Ellipse parametrisiert. -Zum andern das Winkelargument $z$. +Den \textit{elliptische Modul} $k$, der die Exzentrizität der Ellipse parametrisiert und das Winkelargument $z$. Im Kreis ist der Radius für alle Winkel konstant, bei Ellipsen ändert sich das. Dies hat zur Folge, dass bei einer Ellipse die Kreisbodenstrecke nicht linear zum Winkel verläuft. Darum kann hier nicht der gewohnte Winkel verwendet werden. @@ -27,17 +29,17 @@ Das Winkelargument $z$ kann durch das elliptische Integral erster Art 1-k^2 \sin^2 \theta } } - = - \int_{0}^{\phi} - \frac{ - dt - }{ - \sqrt{ - (1-t^2)(1-k^2 t^2) - } - } %TODO which is right? are both functions from phi? + % = + % \int_{0}^{\phi} + % \frac{ + % dt + % }{ + % \sqrt{ + % (1-t^2)(1-k^2 t^2) + % } + % } %TODO which is right? are both functions from phi? \end{equation} -mit dem Winkel $\phi$ in Verbindung liegt. +mit dem Winkel $\phi$ in Verbindung gebracht werden. Dabei wird das vollständige und unvollständige Elliptische integral unterschieden. Beim vollständigen Integral @@ -53,9 +55,9 @@ Beim vollständigen Integral } } \end{equation} -wird über ein viertel Ellipsenbogen integriert also bis $\phi=\pi/2$ und liefert das Winkelargument für eine Vierteldrehung. +wird über ein viertel Ellipsenbogen integriert, also bis $\phi=\pi/2$ und liefert das Winkelargument für eine Vierteldrehung. Die Zahl wird oft auch abgekürzt mit $K = K(k)$ und ist für das elliptische Filter sehr relevant. -Alle elliptishen Funktionen sind somit $4K$-periodisch. +Alle elliptischen Funktionen sind somit $4K$-periodisch. Neben dem $\sn$ gibt es zwei weitere basis-elliptische Funktionen $\cn$ und $\dn$. Dazu kommen noch weitere abgeleitete Funktionen, die durch Quotienten und Kehrwerte dieser Funktionen zustande kommen. @@ -96,7 +98,7 @@ Mithilfe von $F^{-1}$ kann zum Beispiel $sn^{-1}$ mit dem Elliptischen integral w \end{equation} -\begin{equation} +\begin{equation} %TODO remove unnecessary equations \phi = F^{-1}(z, k) @@ -153,31 +155,9 @@ Beim $\cos^{-1}(x)$ haben wir gesehen, dass die analytische Fortsetzung bei $x < Wenn man das gleiche mit $\sn^{-1}(w, k)$ macht, erkennt man zwei interessante Stellen. Die erste ist die gleiche wie beim $\cos^{-1}(x)$ nämlich bei $t = \pm 1$. Der erste Term unter der Wurzel wird dann negativ, während der zweite noch positiv ist, da $k \leq 1$. -\begin{equation} - \frac{ - 1 - }{ - \sqrt{ - (1-t^2)(1-k^2 t^2) - } - } - \in \mathbb{R} - \quad \forall \quad - -1 \leq t \leq 1 -\end{equation} -Die zweite stelle passiert wenn beide Faktoren unter der Wurzel negativ werden, was bei $t = 1/k$ der Fall ist. - - - - -Funktion in relle und komplexe Richtung periodisch - -In der reellen Richtung ist sie $4K(k)$-periodisch und in der imaginären Richtung $4K^\prime(k)$-periodisch. - - - -%TODO sn^{-1} grafik - +Ab diesem Punkt verläuft knickt die Funktion in die imaginäre Richtung ab. +Bei $t = 1/k$ ist auch der zweite Term negativ und die Funktion verläuft in die negative reelle Richtung. +Abbildung \label{ellfilter:fig:sn} zeigt den Verlauf der Funktion in der komplexen Ebene. \begin{figure} \centering \input{papers/ellfilter/tikz/sn.tikz.tex} @@ -185,5 +165,20 @@ In der reellen Richtung ist sie $4K(k)$-periodisch und in der imaginären Richtu $z$-Ebene der Funktion $z = \sn^{-1}(w, k)$. Die Funktion ist in der realen Achse $4K$-periodisch und in der imaginären Achse $2jK^\prime$-periodisch. } - % \label{ellfilter:fig:cd2} + \label{ellfilter:fig:sn} \end{figure} +In der reellen Richtung ist sie $4K(k)$-periodisch und in der imaginären Richtung $4K^\prime(k)$-periodisch, wobei $K^\prime$ das komplemenäre vollständige Elliptische Integral ist: +\begin{equation} + K^\prime(k) + = + \int_{0}^{\pi / 2} + \frac{ + d\theta + }{ + \sqrt{ + 1-{k^\prime}^2 \sin^2 \theta + } + }, + \quad + k^\prime = \sqrt{1-k^2}. +\end{equation} diff --git a/buch/papers/ellfilter/python/F_N_elliptic.pgf b/buch/papers/ellfilter/python/F_N_elliptic.pgf deleted file mode 100644 index 50faaaa..0000000 --- a/buch/papers/ellfilter/python/F_N_elliptic.pgf +++ /dev/null @@ -1,834 +0,0 @@ -%% Creator: Matplotlib, PGF backend -%% -%% To include the figure in your LaTeX document, write -%% \input{.pgf} -%% -%% Make sure the required packages are loaded in your preamble -%% \usepackage{pgf} -%% -%% Also ensure that all the required font packages are loaded; for instance, -%% the lmodern package is sometimes necessary when using math font. -%% \usepackage{lmodern} -%% -%% Figures using additional raster images can only be included by \input if -%% they are in the same directory as the main LaTeX file. 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-\begin{pgfscope}% -\pgfsys@transformshift{2.247564in}{0.548769in}% -\pgfsys@useobject{currentmarker}{}% -\end{pgfscope}% -\end{pgfscope}% -\begin{pgfscope}% -\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{textcolor}% -\pgfsetfillcolor{textcolor}% -\pgftext[x=2.247564in,y=0.451547in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {1.0}\)}% -\end{pgfscope}% -\begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.733531in}{0.548769in}}{\pgfqpoint{3.028066in}{1.753186in}}% -\pgfusepath{clip}% -\pgfsetrectcap% -\pgfsetroundjoin% -\pgfsetlinewidth{0.803000pt}% -\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% -\pgfsetstrokecolor{currentstroke}% -\pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{3.004580in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{3.004580in}{2.301955in}}% -\pgfusepath{stroke}% -\end{pgfscope}% -\begin{pgfscope}% -\pgfsetbuttcap% -\pgfsetroundjoin% -\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetfillcolor{currentfill}% -\pgfsetlinewidth{0.803000pt}% -\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{currentstroke}% -\pgfsetdash{}{0pt}% -\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% -\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% -\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% -\pgfusepath{stroke,fill}% -}% -\begin{pgfscope}% -\pgfsys@transformshift{3.004580in}{0.548769in}% -\pgfsys@useobject{currentmarker}{}% -\end{pgfscope}% -\end{pgfscope}% -\begin{pgfscope}% -\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{textcolor}% -\pgfsetfillcolor{textcolor}% -\pgftext[x=3.004580in,y=0.451547in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {1.5}\)}% -\end{pgfscope}% -\begin{pgfscope}% 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-\begin{pgfscope}% -\pgfsys@transformshift{3.761597in}{0.548769in}% -\pgfsys@useobject{currentmarker}{}% -\end{pgfscope}% -\end{pgfscope}% -\begin{pgfscope}% -\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{textcolor}% -\pgfsetfillcolor{textcolor}% -\pgftext[x=3.761597in,y=0.451547in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {2.0}\)}% -\end{pgfscope}% -\begin{pgfscope}% -\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{textcolor}% -\pgfsetfillcolor{textcolor}% -\pgftext[x=2.247564in,y=0.272534in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle w\)}% -\end{pgfscope}% -\begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.733531in}{0.548769in}}{\pgfqpoint{3.028066in}{1.753186in}}% -\pgfusepath{clip}% -\pgfsetrectcap% -\pgfsetroundjoin% -\pgfsetlinewidth{0.803000pt}% -\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% -\pgfsetstrokecolor{currentstroke}% -\pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{0.733531in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{0.548769in}}% -\pgfusepath{stroke}% -\end{pgfscope}% -\begin{pgfscope}% -\pgfsetbuttcap% -\pgfsetroundjoin% -\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetfillcolor{currentfill}% -\pgfsetlinewidth{0.803000pt}% -\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{currentstroke}% -\pgfsetdash{}{0pt}% -\pgfsys@defobject{currentmarker}{\pgfqpoint{-0.048611in}{0.000000in}}{\pgfqpoint{-0.000000in}{0.000000in}}{% -\pgfpathmoveto{\pgfqpoint{-0.000000in}{0.000000in}}% -\pgfpathlineto{\pgfqpoint{-0.048611in}{0.000000in}}% -\pgfusepath{stroke,fill}% -}% -\begin{pgfscope}% -\pgfsys@transformshift{0.733531in}{0.548769in}% -\pgfsys@useobject{currentmarker}{}% -\end{pgfscope}% -\end{pgfscope}% -\begin{pgfscope}% -\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{textcolor}% 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-\pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{0.733531in}{1.250043in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{1.250043in}}% -\pgfusepath{stroke}% -\end{pgfscope}% -\begin{pgfscope}% -\pgfsetbuttcap% -\pgfsetroundjoin% -\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetfillcolor{currentfill}% -\pgfsetlinewidth{0.803000pt}% -\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{currentstroke}% -\pgfsetdash{}{0pt}% -\pgfsys@defobject{currentmarker}{\pgfqpoint{-0.048611in}{0.000000in}}{\pgfqpoint{-0.000000in}{0.000000in}}{% -\pgfpathmoveto{\pgfqpoint{-0.000000in}{0.000000in}}% -\pgfpathlineto{\pgfqpoint{-0.048611in}{0.000000in}}% -\pgfusepath{stroke,fill}% -}% -\begin{pgfscope}% -\pgfsys@transformshift{0.733531in}{1.250043in}% -\pgfsys@useobject{currentmarker}{}% -\end{pgfscope}% -\end{pgfscope}% -\begin{pgfscope}% -\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{textcolor}% -\pgfsetfillcolor{textcolor}% 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-\pgfpathlineto{\pgfqpoint{3.761597in}{2.301955in}}% -\pgfusepath{stroke}% -\end{pgfscope}% -\end{pgfpicture}% -\makeatother% -\endgroup% diff --git a/buch/papers/ellfilter/python/elliptic.pgf b/buch/papers/ellfilter/python/elliptic.pgf index 89ffb60..32485c1 100644 --- a/buch/papers/ellfilter/python/elliptic.pgf +++ b/buch/papers/ellfilter/python/elliptic.pgf @@ -23,7 +23,7 @@ \begingroup% \makeatletter% \begin{pgfpicture}% -\pgfpathrectangle{\pgfpointorigin}{\pgfqpoint{4.000000in}{2.500000in}}% +\pgfpathrectangle{\pgfpointorigin}{\pgfqpoint{5.000000in}{3.000000in}}% \pgfusepath{use as bounding box, clip}% \begin{pgfscope}% \pgfsetbuttcap% @@ -34,9 +34,9 @@ \pgfsetstrokeopacity{0.000000}% \pgfsetdash{}{0pt}% \pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% -\pgfpathlineto{\pgfqpoint{4.000000in}{0.000000in}}% -\pgfpathlineto{\pgfqpoint{4.000000in}{2.500000in}}% -\pgfpathlineto{\pgfqpoint{0.000000in}{2.500000in}}% +\pgfpathlineto{\pgfqpoint{5.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{5.000000in}{3.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{3.000000in}}% \pgfpathlineto{\pgfqpoint{0.000000in}{0.000000in}}% \pgfpathclose% \pgfusepath{}% @@ -51,16 +51,16 @@ \pgfsetstrokecolor{currentstroke}% \pgfsetstrokeopacity{0.000000}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{0.617954in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{2.301955in}}% -\pgfpathlineto{\pgfqpoint{0.617954in}{2.301955in}}% -\pgfpathlineto{\pgfqpoint{0.617954in}{0.548769in}}% +\pgfpathmoveto{\pgfqpoint{0.733531in}{1.746607in}}% +\pgfpathlineto{\pgfqpoint{4.727004in}{1.746607in}}% +\pgfpathlineto{\pgfqpoint{4.727004in}{2.850000in}}% +\pgfpathlineto{\pgfqpoint{0.733531in}{2.850000in}}% +\pgfpathlineto{\pgfqpoint{0.733531in}{1.746607in}}% \pgfpathclose% \pgfusepath{fill}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetbuttcap% \pgfsetmiterjoin% @@ -72,16 +72,16 @@ \pgfsetstrokecolor{currentstroke}% \pgfsetstrokeopacity{0.200000}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{0.617954in}{1.788459in}}% -\pgfpathlineto{\pgfqpoint{2.189776in}{1.788459in}}% -\pgfpathlineto{\pgfqpoint{2.189776in}{3.541645in}}% -\pgfpathlineto{\pgfqpoint{0.617954in}{3.541645in}}% -\pgfpathlineto{\pgfqpoint{0.617954in}{1.788459in}}% +\pgfpathmoveto{\pgfqpoint{0.733531in}{-108.151374in}}% +\pgfpathlineto{\pgfqpoint{2.730268in}{-108.151374in}}% +\pgfpathlineto{\pgfqpoint{2.730268in}{2.187964in}}% +\pgfpathlineto{\pgfqpoint{0.733531in}{2.187964in}}% +\pgfpathlineto{\pgfqpoint{0.733531in}{-108.151374in}}% \pgfpathclose% \pgfusepath{fill}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetbuttcap% \pgfsetmiterjoin% @@ -93,16 +93,16 @@ \pgfsetstrokecolor{currentstroke}% \pgfsetstrokeopacity{0.200000}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{2.189776in}{0.724087in}}% -\pgfpathlineto{\pgfqpoint{2.205368in}{0.724087in}}% -\pgfpathlineto{\pgfqpoint{2.205368in}{1.788459in}}% -\pgfpathlineto{\pgfqpoint{2.189776in}{1.788459in}}% -\pgfpathlineto{\pgfqpoint{2.189776in}{0.724087in}}% +\pgfpathmoveto{\pgfqpoint{2.730268in}{2.187964in}}% +\pgfpathlineto{\pgfqpoint{2.750075in}{2.187964in}}% +\pgfpathlineto{\pgfqpoint{2.750075in}{2.408643in}}% +\pgfpathlineto{\pgfqpoint{2.730268in}{2.408643in}}% +\pgfpathlineto{\pgfqpoint{2.730268in}{2.187964in}}% \pgfpathclose% \pgfusepath{fill}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetbuttcap% \pgfsetmiterjoin% @@ -114,16 +114,16 @@ \pgfsetstrokecolor{currentstroke}% \pgfsetstrokeopacity{0.200000}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{2.205368in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{3.777189in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{3.777189in}{0.724087in}}% -\pgfpathlineto{\pgfqpoint{2.205368in}{0.724087in}}% -\pgfpathlineto{\pgfqpoint{2.205368in}{0.548769in}}% +\pgfpathmoveto{\pgfqpoint{2.750075in}{2.408643in}}% +\pgfpathlineto{\pgfqpoint{4.746812in}{2.408643in}}% +\pgfpathlineto{\pgfqpoint{4.746812in}{2.850005in}}% +\pgfpathlineto{\pgfqpoint{2.750075in}{2.850005in}}% +\pgfpathlineto{\pgfqpoint{2.750075in}{2.408643in}}% \pgfpathclose% \pgfusepath{fill}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% @@ -131,8 +131,8 @@ \definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{0.617954in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{0.617954in}{2.301955in}}% +\pgfpathmoveto{\pgfqpoint{0.733531in}{1.746607in}}% +\pgfpathlineto{\pgfqpoint{0.733531in}{2.850000in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -150,7 +150,770 @@ \pgfusepath{stroke,fill}% }% \begin{pgfscope}% -\pgfsys@transformshift{0.617954in}{0.548769in}% +\pgfsys@transformshift{0.733531in}{1.746607in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{1.232715in}{1.746607in}}% +\pgfpathlineto{\pgfqpoint{1.232715in}{2.850000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{1.232715in}{1.746607in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{1.731899in}{1.746607in}}% +\pgfpathlineto{\pgfqpoint{1.731899in}{2.850000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{1.731899in}{1.746607in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{2.231083in}{1.746607in}}% +\pgfpathlineto{\pgfqpoint{2.231083in}{2.850000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{2.231083in}{1.746607in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{2.730268in}{1.746607in}}% +\pgfpathlineto{\pgfqpoint{2.730268in}{2.850000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{2.730268in}{1.746607in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{3.229452in}{1.746607in}}% +\pgfpathlineto{\pgfqpoint{3.229452in}{2.850000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{3.229452in}{1.746607in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{3.728636in}{1.746607in}}% +\pgfpathlineto{\pgfqpoint{3.728636in}{2.850000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{3.728636in}{1.746607in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{4.227820in}{1.746607in}}% +\pgfpathlineto{\pgfqpoint{4.227820in}{2.850000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{4.227820in}{1.746607in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{4.727004in}{1.746607in}}% +\pgfpathlineto{\pgfqpoint{4.727004in}{2.850000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{4.727004in}{1.746607in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.733531in}{1.746607in}}% +\pgfpathlineto{\pgfqpoint{4.727004in}{1.746607in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{-0.048611in}{0.000000in}}{\pgfqpoint{-0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{-0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{-0.048611in}{0.000000in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{0.733531in}{1.746607in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=0.348306in, y=1.698381in, left, base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {10^{-4}}\)}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.733531in}{2.187964in}}% +\pgfpathlineto{\pgfqpoint{4.727004in}{2.187964in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{-0.048611in}{0.000000in}}{\pgfqpoint{-0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{-0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{-0.048611in}{0.000000in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{0.733531in}{2.187964in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=0.435112in, y=2.139739in, left, base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {10^{0}}\)}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{1.746607in}}{\pgfqpoint{3.993473in}{1.103393in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.733531in}{2.629321in}}% +\pgfpathlineto{\pgfqpoint{4.727004in}{2.629321in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{-0.048611in}{0.000000in}}{\pgfqpoint{-0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{-0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{-0.048611in}{0.000000in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{0.733531in}{2.629321in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=0.435112in, y=2.581096in, left, base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {10^{4}}\)}% +\end{pgfscope}% +\begin{pgfscope}% +\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{textcolor}% +\pgfsetfillcolor{textcolor}% +\pgftext[x=0.292751in,y=2.298303in,,bottom,rotate=90.000000]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle F^2_N(w)\)}% 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+\pgfpathlineto{\pgfqpoint{4.746812in}{0.480557in}}% +\pgfpathlineto{\pgfqpoint{2.750075in}{0.480557in}}% +\pgfpathlineto{\pgfqpoint{2.750075in}{0.370218in}}% +\pgfpathclose% +\pgfusepath{fill}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.733531in}{0.370218in}}% +\pgfpathlineto{\pgfqpoint{0.733531in}{1.473611in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{0.733531in}{0.370218in}% \pgfsys@useobject{currentmarker}{}% \end{pgfscope}% \end{pgfscope}% @@ -158,10 +921,10 @@ \definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{textcolor}% \pgfsetfillcolor{textcolor}% -\pgftext[x=0.617954in,y=0.451547in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.0}\)}% +\pgftext[x=0.733531in,y=0.272996in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.00}\)}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% @@ -169,8 +932,8 @@ \definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{1.403865in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{1.403865in}{2.301955in}}% +\pgfpathmoveto{\pgfqpoint{1.232715in}{0.370218in}}% +\pgfpathlineto{\pgfqpoint{1.232715in}{1.473611in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -188,7 +951,7 @@ \pgfusepath{stroke,fill}% }% \begin{pgfscope}% -\pgfsys@transformshift{1.403865in}{0.548769in}% +\pgfsys@transformshift{1.232715in}{0.370218in}% \pgfsys@useobject{currentmarker}{}% \end{pgfscope}% \end{pgfscope}% @@ -196,10 +959,10 @@ \definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{textcolor}% \pgfsetfillcolor{textcolor}% -\pgftext[x=1.403865in,y=0.451547in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.5}\)}% +\pgftext[x=1.232715in,y=0.272996in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.25}\)}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% @@ -207,8 +970,8 @@ \definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{2.189776in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{2.189776in}{2.301955in}}% +\pgfpathmoveto{\pgfqpoint{1.731899in}{0.370218in}}% +\pgfpathlineto{\pgfqpoint{1.731899in}{1.473611in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -226,7 +989,7 @@ \pgfusepath{stroke,fill}% }% \begin{pgfscope}% -\pgfsys@transformshift{2.189776in}{0.548769in}% +\pgfsys@transformshift{1.731899in}{0.370218in}% \pgfsys@useobject{currentmarker}{}% \end{pgfscope}% \end{pgfscope}% @@ -234,10 +997,10 @@ \definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{textcolor}% \pgfsetfillcolor{textcolor}% -\pgftext[x=2.189776in,y=0.451547in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {1.0}\)}% +\pgftext[x=1.731899in,y=0.272996in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.50}\)}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% @@ -245,8 +1008,8 @@ \definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{2.975686in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{2.975686in}{2.301955in}}% +\pgfpathmoveto{\pgfqpoint{2.231083in}{0.370218in}}% +\pgfpathlineto{\pgfqpoint{2.231083in}{1.473611in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -264,7 +1027,7 @@ \pgfusepath{stroke,fill}% }% \begin{pgfscope}% -\pgfsys@transformshift{2.975686in}{0.548769in}% +\pgfsys@transformshift{2.231083in}{0.370218in}% \pgfsys@useobject{currentmarker}{}% \end{pgfscope}% \end{pgfscope}% @@ -272,10 +1035,10 @@ \definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{textcolor}% \pgfsetfillcolor{textcolor}% -\pgftext[x=2.975686in,y=0.451547in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {1.5}\)}% +\pgftext[x=2.231083in,y=0.272996in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.75}\)}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% @@ -283,8 +1046,8 @@ \definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{3.761597in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{2.301955in}}% +\pgfpathmoveto{\pgfqpoint{2.730268in}{0.370218in}}% +\pgfpathlineto{\pgfqpoint{2.730268in}{1.473611in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -302,7 +1065,7 @@ \pgfusepath{stroke,fill}% }% \begin{pgfscope}% -\pgfsys@transformshift{3.761597in}{0.548769in}% +\pgfsys@transformshift{2.730268in}{0.370218in}% \pgfsys@useobject{currentmarker}{}% \end{pgfscope}% \end{pgfscope}% @@ -310,16 +1073,48 @@ \definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{textcolor}% \pgfsetfillcolor{textcolor}% -\pgftext[x=3.761597in,y=0.451547in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {2.0}\)}% +\pgftext[x=2.730268in,y=0.272996in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {1.00}\)}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}% +\pgfusepath{clip}% +\pgfsetrectcap% +\pgfsetroundjoin% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{3.229452in}{0.370218in}}% +\pgfpathlineto{\pgfqpoint{3.229452in}{1.473611in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetroundjoin% +\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetfillcolor{currentfill}% +\pgfsetlinewidth{0.803000pt}% +\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% +\pgfusepath{stroke,fill}% +}% +\begin{pgfscope}% +\pgfsys@transformshift{3.229452in}{0.370218in}% +\pgfsys@useobject{currentmarker}{}% +\end{pgfscope}% \end{pgfscope}% \begin{pgfscope}% \definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{textcolor}% \pgfsetfillcolor{textcolor}% -\pgftext[x=2.189776in,y=0.272534in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle w\)}% +\pgftext[x=3.229452in,y=0.272996in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {1.25}\)}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% @@ -327,8 +1122,8 @@ \definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{0.617954in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{0.548769in}}% +\pgfpathmoveto{\pgfqpoint{3.728636in}{0.370218in}}% +\pgfpathlineto{\pgfqpoint{3.728636in}{1.473611in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -340,13 +1135,13 @@ \definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfsys@defobject{currentmarker}{\pgfqpoint{-0.048611in}{0.000000in}}{\pgfqpoint{-0.000000in}{0.000000in}}{% -\pgfpathmoveto{\pgfqpoint{-0.000000in}{0.000000in}}% -\pgfpathlineto{\pgfqpoint{-0.048611in}{0.000000in}}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% \pgfusepath{stroke,fill}% }% \begin{pgfscope}% -\pgfsys@transformshift{0.617954in}{0.548769in}% +\pgfsys@transformshift{3.728636in}{0.370218in}% \pgfsys@useobject{currentmarker}{}% \end{pgfscope}% \end{pgfscope}% @@ -354,10 +1149,10 @@ \definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{textcolor}% \pgfsetfillcolor{textcolor}% -\pgftext[x=0.343262in, y=0.500544in, left, base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.0}\)}% +\pgftext[x=3.728636in,y=0.272996in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {1.50}\)}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% @@ -365,8 +1160,8 @@ \definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{0.617954in}{0.899406in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{0.899406in}}% +\pgfpathmoveto{\pgfqpoint{4.227820in}{0.370218in}}% +\pgfpathlineto{\pgfqpoint{4.227820in}{1.473611in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -378,13 +1173,13 @@ \definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfsys@defobject{currentmarker}{\pgfqpoint{-0.048611in}{0.000000in}}{\pgfqpoint{-0.000000in}{0.000000in}}{% -\pgfpathmoveto{\pgfqpoint{-0.000000in}{0.000000in}}% -\pgfpathlineto{\pgfqpoint{-0.048611in}{0.000000in}}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% \pgfusepath{stroke,fill}% }% \begin{pgfscope}% -\pgfsys@transformshift{0.617954in}{0.899406in}% +\pgfsys@transformshift{4.227820in}{0.370218in}% \pgfsys@useobject{currentmarker}{}% \end{pgfscope}% \end{pgfscope}% @@ -392,10 +1187,10 @@ \definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{textcolor}% \pgfsetfillcolor{textcolor}% -\pgftext[x=0.343262in, y=0.851181in, left, base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.2}\)}% +\pgftext[x=4.227820in,y=0.272996in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {1.75}\)}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% @@ -403,8 +1198,8 @@ \definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{0.617954in}{1.250043in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{1.250043in}}% +\pgfpathmoveto{\pgfqpoint{4.727004in}{0.370218in}}% +\pgfpathlineto{\pgfqpoint{4.727004in}{1.473611in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -416,13 +1211,13 @@ \definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfsys@defobject{currentmarker}{\pgfqpoint{-0.048611in}{0.000000in}}{\pgfqpoint{-0.000000in}{0.000000in}}{% -\pgfpathmoveto{\pgfqpoint{-0.000000in}{0.000000in}}% -\pgfpathlineto{\pgfqpoint{-0.048611in}{0.000000in}}% +\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{% +\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}% +\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}% \pgfusepath{stroke,fill}% }% \begin{pgfscope}% -\pgfsys@transformshift{0.617954in}{1.250043in}% +\pgfsys@transformshift{4.727004in}{0.370218in}% \pgfsys@useobject{currentmarker}{}% \end{pgfscope}% \end{pgfscope}% @@ -430,10 +1225,10 @@ \definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{textcolor}% \pgfsetfillcolor{textcolor}% -\pgftext[x=0.343262in, y=1.201818in, left, base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.4}\)}% +\pgftext[x=4.727004in,y=0.272996in,,top]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {2.00}\)}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% @@ -441,8 +1236,8 @@ \definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{0.617954in}{1.600680in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{1.600680in}}% +\pgfpathmoveto{\pgfqpoint{0.733531in}{0.370218in}}% +\pgfpathlineto{\pgfqpoint{4.727004in}{0.370218in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -460,7 +1255,7 @@ \pgfusepath{stroke,fill}% }% \begin{pgfscope}% -\pgfsys@transformshift{0.617954in}{1.600680in}% +\pgfsys@transformshift{0.733531in}{0.370218in}% \pgfsys@useobject{currentmarker}{}% \end{pgfscope}% \end{pgfscope}% @@ -468,10 +1263,10 @@ \definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{textcolor}% \pgfsetfillcolor{textcolor}% -\pgftext[x=0.343262in, y=1.552455in, left, base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.6}\)}% +\pgftext[x=0.458839in, y=0.321992in, left, base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.0}\)}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% @@ -479,8 +1274,8 @@ \definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{0.617954in}{1.951318in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{1.951318in}}% +\pgfpathmoveto{\pgfqpoint{0.733531in}{0.921914in}}% +\pgfpathlineto{\pgfqpoint{4.727004in}{0.921914in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -498,7 +1293,7 @@ \pgfusepath{stroke,fill}% }% \begin{pgfscope}% -\pgfsys@transformshift{0.617954in}{1.951318in}% +\pgfsys@transformshift{0.733531in}{0.921914in}% \pgfsys@useobject{currentmarker}{}% \end{pgfscope}% \end{pgfscope}% @@ -506,10 +1301,10 @@ \definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{textcolor}% \pgfsetfillcolor{textcolor}% -\pgftext[x=0.343262in, y=1.903092in, left, base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.8}\)}% +\pgftext[x=0.458839in, y=0.873689in, left, base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {0.5}\)}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% @@ -517,8 +1312,8 @@ \definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{0.617954in}{2.301955in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{2.301955in}}% +\pgfpathmoveto{\pgfqpoint{0.733531in}{1.473611in}}% +\pgfpathlineto{\pgfqpoint{4.727004in}{1.473611in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -536,7 +1331,7 @@ \pgfusepath{stroke,fill}% }% \begin{pgfscope}% -\pgfsys@transformshift{0.617954in}{2.301955in}% +\pgfsys@transformshift{0.733531in}{1.473611in}% \pgfsys@useobject{currentmarker}{}% \end{pgfscope}% \end{pgfscope}% @@ -544,16 +1339,16 @@ \definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{textcolor}% \pgfsetfillcolor{textcolor}% -\pgftext[x=0.343262in, y=2.253730in, left, base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {1.0}\)}% +\pgftext[x=0.458839in, y=1.425386in, left, base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle {1.0}\)}% \end{pgfscope}% \begin{pgfscope}% 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-636,15 +1426,15 @@ \definecolor{currentstroke}{rgb}{1.000000,0.647059,0.000000}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{2.189776in}{1.789929in}}% -\pgfpathlineto{\pgfqpoint{2.198207in}{1.037413in}}% -\pgfpathlineto{\pgfqpoint{2.204331in}{0.756839in}}% -\pgfpathlineto{\pgfqpoint{2.205368in}{0.723819in}}% -\pgfpathlineto{\pgfqpoint{2.205368in}{0.723819in}}% +\pgfpathmoveto{\pgfqpoint{2.730268in}{1.151360in}}% +\pgfpathlineto{\pgfqpoint{2.739978in}{0.709346in}}% +\pgfpathlineto{\pgfqpoint{2.746805in}{0.536003in}}% +\pgfpathlineto{\pgfqpoint{2.750075in}{0.480388in}}% +\pgfpathlineto{\pgfqpoint{2.750075in}{0.480388in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{0.617954in}{0.548769in}}{\pgfqpoint{3.143642in}{1.753186in}}% +\pgfpathrectangle{\pgfqpoint{0.733531in}{0.370218in}}{\pgfqpoint{3.993473in}{1.103393in}}% \pgfusepath{clip}% \pgfsetrectcap% \pgfsetroundjoin% @@ -652,40 +1442,37 @@ \definecolor{currentstroke}{rgb}{1.000000,0.000000,0.000000}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{2.205368in}{0.723819in}}% -\pgfpathlineto{\pgfqpoint{2.211596in}{0.579693in}}% -\pgfpathlineto{\pgfqpoint{2.213153in}{0.554315in}}% -\pgfpathlineto{\pgfqpoint{2.213932in}{0.554729in}}% -\pgfpathlineto{\pgfqpoint{2.220938in}{0.631412in}}% -\pgfpathlineto{\pgfqpoint{2.227945in}{0.675812in}}% -\pgfpathlineto{\pgfqpoint{2.234951in}{0.701496in}}% -\pgfpathlineto{\pgfqpoint{2.241958in}{0.715641in}}% -\pgfpathlineto{\pgfqpoint{2.248186in}{0.721878in}}% -\pgfpathlineto{\pgfqpoint{2.254414in}{0.724034in}}% -\pgfpathlineto{\pgfqpoint{2.261420in}{0.723044in}}% -\pgfpathlineto{\pgfqpoint{2.269984in}{0.718482in}}% -\pgfpathlineto{\pgfqpoint{2.281661in}{0.708528in}}% -\pgfpathlineto{\pgfqpoint{2.300345in}{0.688017in}}% -\pgfpathlineto{\pgfqpoint{2.384424in}{0.591070in}}% -\pgfpathlineto{\pgfqpoint{2.417121in}{0.560087in}}% 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+\pgfpathmoveto{\pgfqpoint{0.733531in}{0.370218in}}% +\pgfpathlineto{\pgfqpoint{0.733531in}{1.473611in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -706,8 +1493,8 @@ \definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{3.761597in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{2.301955in}}% +\pgfpathmoveto{\pgfqpoint{4.727004in}{0.370218in}}% +\pgfpathlineto{\pgfqpoint{4.727004in}{1.473611in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -717,8 +1504,8 @@ \definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{0.617954in}{0.548769in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{0.548769in}}% +\pgfpathmoveto{\pgfqpoint{0.733531in}{0.370218in}}% +\pgfpathlineto{\pgfqpoint{4.727004in}{0.370218in}}% \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% @@ -728,8 +1515,8 @@ \definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% \pgfsetstrokecolor{currentstroke}% \pgfsetdash{}{0pt}% -\pgfpathmoveto{\pgfqpoint{0.617954in}{2.301955in}}% -\pgfpathlineto{\pgfqpoint{3.761597in}{2.301955in}}% +\pgfpathmoveto{\pgfqpoint{0.733531in}{1.473611in}}% +\pgfpathlineto{\pgfqpoint{4.727004in}{1.473611in}}% \pgfusepath{stroke}% \end{pgfscope}% \end{pgfpicture}% diff --git a/buch/papers/ellfilter/python/elliptic.py b/buch/papers/ellfilter/python/elliptic.py index c9cf5bd..6e0fd12 100644 --- a/buch/papers/ellfilter/python/elliptic.py +++ b/buch/papers/ellfilter/python/elliptic.py @@ -342,9 +342,9 @@ k = np.array([0.1,0.2,0.4,0.6,0.9,0.99]) K = ell_int(k) K_prime = ell_int(np.sqrt(1-k**2)) -axs[1].plot(K, K_prime, '.', color=last_color(), markersize=2) -for x, y, n in zip(K, K_prime, k): - axs[1].text(x+0.1, y+0.1, f"$k={n:.2f}$", rotation_mode="anchor") +# axs[1].plot(K, K_prime, '.', color=last_color(), markersize=2) +# for x, y, n in zip(K, K_prime, k): +# axs[1].text(x+0.1, y+0.1, f"$k={n:.2f}$", rotation_mode="anchor") axs[1].set_ylabel("$K^\prime$") axs[1].set_xlabel("$K$") axs[1].set_xlim([0,6]) diff --git a/buch/papers/ellfilter/python/elliptic2.py b/buch/papers/ellfilter/python/elliptic2.py index cfa16ea..20a7428 100644 --- a/buch/papers/ellfilter/python/elliptic2.py +++ b/buch/papers/ellfilter/python/elliptic2.py @@ -50,20 +50,20 @@ def ellip_filter(N, mode=-1): return w/omega_c, FN2 / epsilon2, mag, a, b -plt.figure(figsize=(4,2.5)) +f, axs = plt.subplots(2, 1, figsize=(5,3), sharex=True) for mode, c in enumerate(["green", "orange", "red"]): w, FN2, mag, a, b = ellip_filter(N, mode=mode) - plt.semilogy(w, FN2, label=f"$N={N}, k=0.1$", linewidth=1, color=c) + axs[0].semilogy(w, FN2, label=f"$N={N}, k=0.1$", linewidth=1, color=c) -plt.gca().add_patch(Rectangle( +axs[0].add_patch(Rectangle( (0, 0), 1, 1, fc ='green', alpha=0.2, lw = 10, )) -plt.gca().add_patch(Rectangle( +axs[0].add_patch(Rectangle( (1, 1), 0.00992, 1e2-1, fc ='orange', @@ -71,7 +71,7 @@ plt.gca().add_patch(Rectangle( lw = 10, )) -plt.gca().add_patch(Rectangle( +axs[0].add_patch(Rectangle( (1.00992, 100), 1, 1e6, fc ='red', @@ -83,54 +83,41 @@ zeros = [0,0.87,0.995] poles = [1.01,1.155] import matplotlib.transforms -plt.plot( # mark errors as vertical bars +axs[0].plot( # mark errors as vertical bars zeros, np.zeros_like(zeros), "o", mfc='none', color='black', transform=matplotlib.transforms.blended_transform_factory( - plt.gca().transData, - plt.gca().transAxes, + axs[0].transData, + axs[0].transAxes, ), ) -plt.plot( # mark errors as vertical bars +axs[0].plot( # mark errors as vertical bars poles, np.ones_like(poles), "x", mfc='none', color='black', transform=matplotlib.transforms.blended_transform_factory( - plt.gca().transData, - plt.gca().transAxes, + axs[0].transData, + axs[0].transAxes, ), ) -plt.xlim([0,2]) -plt.ylim([1e-4,1e6]) -plt.grid() -plt.xlabel("$w$") -plt.ylabel("$F^2_N(w)$") -# plt.legend() -plt.tight_layout() -plt.savefig("F_N_elliptic.pgf") -plt.show() - - - -plt.figure(figsize=(4,2.5)) for mode, c in enumerate(["green", "orange", "red"]): w, FN2, mag, a, b = ellip_filter(N, mode=mode) - plt.plot(w, mag, linewidth=1, color=c) + axs[1].plot(w, mag, linewidth=1, color=c) -plt.gca().add_patch(Rectangle( +axs[1].add_patch(Rectangle( (0, np.sqrt(2)/2), 1, 1, fc ='green', alpha=0.2, lw = 10, )) -plt.gca().add_patch(Rectangle( +axs[1].add_patch(Rectangle( (1, 0.1), 0.00992, np.sqrt(2)/2 - 0.1, fc ='orange', @@ -138,7 +125,7 @@ plt.gca().add_patch(Rectangle( lw = 10, )) -plt.gca().add_patch(Rectangle( +axs[1].add_patch(Rectangle( (1.00992, 0), 1, 0.1, fc ='red', @@ -146,11 +133,13 @@ plt.gca().add_patch(Rectangle( lw = 10, )) -plt.grid() -plt.xlim([0,2]) -plt.ylim([0,1]) -plt.xlabel("$w$") -plt.ylabel("$|H(w)|$") +axs[0].set_xlim([0,2]) +axs[0].set_ylim([1e-4,1e6]) +axs[0].grid() +axs[0].set_ylabel("$F^2_N(w)$") +axs[1].grid() +axs[1].set_ylim([0,1]) +axs[1].set_ylabel("$|H(w)|$") plt.tight_layout() plt.savefig("elliptic.pgf") plt.show() diff --git a/buch/papers/ellfilter/python/k.pgf b/buch/papers/ellfilter/python/k.pgf index 52dd705..bbb823a 100644 --- a/buch/papers/ellfilter/python/k.pgf +++ b/buch/papers/ellfilter/python/k.pgf @@ -1011,56 +1011,6 @@ \pgfusepath{stroke}% \end{pgfscope}% \begin{pgfscope}% -\pgfpathrectangle{\pgfqpoint{2.874885in}{0.548769in}}{\pgfqpoint{1.940523in}{1.753186in}}% -\pgfusepath{clip}% -\pgfsetbuttcap% -\pgfsetroundjoin% -\definecolor{currentfill}{rgb}{0.121569,0.466667,0.705882}% -\pgfsetfillcolor{currentfill}% -\pgfsetlinewidth{1.003750pt}% -\definecolor{currentstroke}{rgb}{0.121569,0.466667,0.705882}% -\pgfsetstrokecolor{currentstroke}% -\pgfsetdash{}{0pt}% -\pgfsys@defobject{currentmarker}{\pgfqpoint{-0.006944in}{-0.006944in}}{\pgfqpoint{0.006944in}{0.006944in}}{% -\pgfpathmoveto{\pgfqpoint{0.000000in}{-0.006944in}}% -\pgfpathcurveto{\pgfqpoint{0.001842in}{-0.006944in}}{\pgfqpoint{0.003608in}{-0.006213in}}{\pgfqpoint{0.004910in}{-0.004910in}}% -\pgfpathcurveto{\pgfqpoint{0.006213in}{-0.003608in}}{\pgfqpoint{0.006944in}{-0.001842in}}{\pgfqpoint{0.006944in}{0.000000in}}% -\pgfpathcurveto{\pgfqpoint{0.006944in}{0.001842in}}{\pgfqpoint{0.006213in}{0.003608in}}{\pgfqpoint{0.004910in}{0.004910in}}% -\pgfpathcurveto{\pgfqpoint{0.003608in}{0.006213in}}{\pgfqpoint{0.001842in}{0.006944in}}{\pgfqpoint{0.000000in}{0.006944in}}% -\pgfpathcurveto{\pgfqpoint{-0.001842in}{0.006944in}}{\pgfqpoint{-0.003608in}{0.006213in}}{\pgfqpoint{-0.004910in}{0.004910in}}% -\pgfpathcurveto{\pgfqpoint{-0.006213in}{0.003608in}}{\pgfqpoint{-0.006944in}{0.001842in}}{\pgfqpoint{-0.006944in}{0.000000in}}% -\pgfpathcurveto{\pgfqpoint{-0.006944in}{-0.001842in}}{\pgfqpoint{-0.006213in}{-0.003608in}}{\pgfqpoint{-0.004910in}{-0.004910in}}% -\pgfpathcurveto{\pgfqpoint{-0.003608in}{-0.006213in}}{\pgfqpoint{-0.001842in}{-0.006944in}}{\pgfqpoint{0.000000in}{-0.006944in}}% -\pgfpathlineto{\pgfqpoint{0.000000in}{-0.006944in}}% -\pgfpathclose% -\pgfusepath{stroke,fill}% -}% -\begin{pgfscope}% -\pgfsys@transformshift{3.384190in}{1.844597in}% -\pgfsys@useobject{currentmarker}{}% -\end{pgfscope}% -\begin{pgfscope}% -\pgfsys@transformshift{3.388110in}{1.606330in}% -\pgfsys@useobject{currentmarker}{}% -\end{pgfscope}% -\begin{pgfscope}% -\pgfsys@transformshift{3.405294in}{1.376014in}% -\pgfsys@useobject{currentmarker}{}% -\end{pgfscope}% -\begin{pgfscope}% -\pgfsys@transformshift{3.441114in}{1.248396in}% -\pgfsys@useobject{currentmarker}{}% -\end{pgfscope}% -\begin{pgfscope}% -\pgfsys@transformshift{3.612461in}{1.128939in}% -\pgfsys@useobject{currentmarker}{}% -\end{pgfscope}% -\begin{pgfscope}% -\pgfsys@transformshift{3.960478in}{1.102320in}% -\pgfsys@useobject{currentmarker}{}% -\end{pgfscope}% -\end{pgfscope}% -\begin{pgfscope}% \pgfsetrectcap% \pgfsetmiterjoin% \pgfsetlinewidth{0.803000pt}% @@ -1116,42 +1066,6 @@ \pgfsetfillcolor{textcolor}% \pgftext[x=3.415254in,y=0.583833in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle \pi/2\)}% \end{pgfscope}% -\begin{pgfscope}% -\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{textcolor}% -\pgfsetfillcolor{textcolor}% -\pgftext[x=3.416532in,y=1.879661in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle k=0.10\)}% -\end{pgfscope}% -\begin{pgfscope}% -\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{textcolor}% -\pgfsetfillcolor{textcolor}% -\pgftext[x=3.420452in,y=1.641394in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle k=0.20\)}% -\end{pgfscope}% -\begin{pgfscope}% -\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{textcolor}% -\pgfsetfillcolor{textcolor}% -\pgftext[x=3.437636in,y=1.411078in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle k=0.40\)}% -\end{pgfscope}% -\begin{pgfscope}% -\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{textcolor}% -\pgfsetfillcolor{textcolor}% -\pgftext[x=3.473456in,y=1.283460in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle k=0.60\)}% -\end{pgfscope}% -\begin{pgfscope}% -\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{textcolor}% -\pgfsetfillcolor{textcolor}% -\pgftext[x=3.644803in,y=1.164003in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle k=0.90\)}% -\end{pgfscope}% -\begin{pgfscope}% -\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}% -\pgfsetstrokecolor{textcolor}% -\pgfsetfillcolor{textcolor}% -\pgftext[x=3.992820in,y=1.137383in,left,base]{\color{textcolor}\rmfamily\fontsize{10.000000}{12.000000}\selectfont \(\displaystyle k=0.99\)}% -\end{pgfscope}% \end{pgfpicture}% \makeatother% \endgroup% diff --git a/buch/papers/ellfilter/tikz/arccos.tikz.tex b/buch/papers/ellfilter/tikz/arccos.tikz.tex index 4211053..a139fc4 100644 --- a/buch/papers/ellfilter/tikz/arccos.tikz.tex +++ b/buch/papers/ellfilter/tikz/arccos.tikz.tex @@ -23,26 +23,26 @@ \clip(-7.5,-2) rectangle (7.5,2); % \pause - \draw[ultra thick, ->, orange] (1, 0) -- (0,0); + \draw[ultra thick, ->, darkgreen] (1, 0) -- (0,0); % \pause - \draw[ultra thick, ->, darkgreen] (0, 0) -- (0,1.5); + \draw[ultra thick, ->, orange] (0, 0) -- (0,1.5); % \pause - \draw[ultra thick, ->, red] (2, 0) -- (1,0); + \draw[ultra thick, ->, cyan] (2, 0) -- (1,0); \draw[ultra thick, ->, blue] (2,1.5) -- (2, 0); % \pause \foreach \i in {-2,...,1} { \begin{scope}[xshift=\i*4cm] - \begin{scope}[opacity=0.5] - \draw[->, orange] (-1, 0) -- (0,0); - \draw[->, darkgreen] (0, 0) -- (0,1.5); - \draw[->, darkgreen] (0, 0) -- (0,-1.5); - \draw[->, orange] (1, 0) -- (0,0); - \draw[->, red] (2, 0) -- (1,0); + \begin{scope}[] + \draw[->, darkgreen] (-1, 0) -- (0,0); + \draw[->, orange] (0, 0) -- (0,1.5); + \draw[->, orange] (0, 0) -- (0,-1.5); + \draw[->, darkgreen] (1, 0) -- (0,0); + \draw[->, cyan] (2, 0) -- (1,0); \draw[->, blue] (2,1.5) -- (2, 0); \draw[->, blue] (2,-1.5) -- (2, 0); - \draw[->, red] (2, 0) -- (3,0); + \draw[->, cyan] (2, 0) -- (3,0); \end{scope} \node[zero] at (1,0) {}; \node[zero] at (3,0) {}; @@ -58,10 +58,10 @@ \draw[gray, ->] (-5,0) -- (5,0) node[anchor=west]{$w$}; - \draw[thick, ->, blue] (-4, 0) -- (-2, 0); - \draw[thick, ->, red] (-2, 0) -- (0, 0); - \draw[thick, ->, orange] (0, 0) -- (2, 0); - \draw[thick, ->, darkgreen] (2, 0) -- (4, 0); + \draw[ultra thick, ->, blue] (-4, 0) -- (-2, 0); + \draw[ultra thick, ->, cyan] (-2, 0) -- (0, 0); + \draw[ultra thick, ->, darkgreen] (0, 0) -- (2, 0); + \draw[ultra thick, ->, orange] (2, 0) -- (4, 0); \node[anchor=south] at (-4,0) {$-\infty$}; \node[anchor=south] at (-2,0) {$-1$}; diff --git a/buch/papers/ellfilter/tikz/arccos2.tikz.tex b/buch/papers/ellfilter/tikz/arccos2.tikz.tex index 755e8a0..c3f11bb 100644 --- a/buch/papers/ellfilter/tikz/arccos2.tikz.tex +++ b/buch/papers/ellfilter/tikz/arccos2.tikz.tex @@ -2,21 +2,34 @@ \tikzstyle{zero} = [draw, circle, inner sep =0, minimum height=0.15cm] \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}} + \tikzstyle{dot} = [fill, circle, inner sep =0, minimum height=0.1cm] - \begin{scope}[xscale=0.5] - \draw[gray, ->] (0,-2) -- (0,2) node[anchor=south]{$\mathrm{Im}~z_1$}; - \draw[gray, ->] (-10,0) -- (10,0) node[anchor=west]{$\mathrm{Re}~z_1$}; + \begin{scope}[xscale=0.75] + + \draw[gray, ->] (0,-1) -- (0,2) node[anchor=south]{$\mathrm{Im}~z_1$}; + \draw[gray, ->] (-2,0) -- (9,0) node[anchor=west]{$\mathrm{Re}~z_1$}; \begin{scope} - \draw[>->, line width=0.05, thick, blue] (2, 1.5) -- (2,0.05) -- node[anchor=south, pos=0.5]{$N=1$} (0.1,0.05) -- (0.1,1.5); - \draw[>->, line width=0.05, thick, orange] (4, 1.5) -- (4,0) -- node[anchor=south, pos=0.25]{$N=2$} (0,0) -- (0,1.5); - \draw[>->, line width=0.05, thick, red] (6, 1.5) node[anchor=north west]{$-\infty$} -- (6,-0.05) node[anchor=west]{$-1$} -- node[anchor=north]{$0$} node[anchor=south, pos=0.1666]{$N=3$} (-0.1,-0.05) node[anchor=east]{$1$} -- (-0.1,1.5) node[anchor=north east]{$\infty$}; + \draw[->, ultra thick, blue] (8, 1.5) -- node[align=center]{Sperrbereich} (8,0); + \draw[->, ultra thick, cyan] (8, 0) -- node[yshift=-0.5cm]{Durchlassbereich}(4,0); + \draw[->, ultra thick, darkgreen] (4, 0) -- node[yshift=-0.5cm]{Durchlassbereich} (0,0); + \draw[->, ultra thick, orange] (0, 0) -- node[align=center]{Sperrbereich} (0,1.5); + + \node[anchor=north east] at (8, 1.5) {$-\infty$}; + \draw (8, 0) node[dot]{} node[anchor=south east] {$1$}; + \draw (6, 0) node[dot]{} node[anchor=south] {$-1$}; + \draw (4, 0) node[dot]{} node[anchor=south] {$1$}; + \draw (2, 0) node[dot]{} node[anchor=south] {$-1$}; + \draw (0, 0) node[dot]{} node[anchor=south west] {$1$}; + \node[anchor=north west] at (0, 1.5){$\infty$}; + + \node at(4,1) {$N = 4$}; - \node[zero] at (-7,0) {}; - \node[zero] at (-5,0) {}; - \node[zero] at (-3,0) {}; + % \node[zero] at (-7,0) {}; + % \node[zero] at (-5,0) {}; + % \node[zero] at (-3,0) {}; \node[zero] at (-1,0) {}; \node[zero] at (1,0) {}; \node[zero] at (3,0) {}; @@ -25,9 +38,9 @@ \end{scope} - \node[gray, anchor=north] at (-8,0) {$-4\pi$}; - \node[gray, anchor=north] at (-6,0) {$-3\pi$}; - \node[gray, anchor=north] at (-4,0) {$-2\pi$}; + % \node[gray, anchor=north] at (-8,0) {$-4\pi$}; + % \node[gray, anchor=north] at (-6,0) {$-3\pi$}; + % \node[gray, anchor=north] at (-4,0) {$-2\pi$}; \node[gray, anchor=north] at (-2,0) {$-\pi$}; \node[gray, anchor=north] at (2,0) {$\pi$}; \node[gray, anchor=north] at (4,0) {$2\pi$}; @@ -35,12 +48,12 @@ \node[gray, anchor=north] at (8,0) {$4\pi$}; - \node[gray, anchor=east] at (0,-1.5) {$-\infty$}; + % \node[gray, anchor=east] at (0,-1.5) {$-\infty$}; \node[gray, anchor=east] at (0, 1.5) {$\infty$}; \end{scope} - \node[zero] at (4,2) (n) {}; + \node[zero] at (6.5,2) (n) {}; \node[anchor=west] at (n.east) {Zero}; \end{tikzpicture} \ No newline at end of file diff --git a/buch/papers/ellfilter/tikz/cd.tikz.tex b/buch/papers/ellfilter/tikz/cd.tikz.tex index b2b0090..cc5852c 100644 --- a/buch/papers/ellfilter/tikz/cd.tikz.tex +++ b/buch/papers/ellfilter/tikz/cd.tikz.tex @@ -22,32 +22,35 @@ \fill[yellow!30] (0,0) rectangle (1, 0.5); + \foreach \i in {-2,...,1} { + \foreach \j in {-2,...,1} { + \begin{scope}[xshift=\i*4cm, yshift=\j*1cm] + \draw[->, orange!50] (0, 0) -- (0,0.5); + \draw[->, darkgreen!50] (1, 0) -- (0,0); + \draw[->, cyan!50] (2, 0) -- (1,0); + \draw[->, blue!50] (2,0.5) -- (2, 0); + \draw[->, purple!50] (1, 0.5) -- (2,0.5); + \draw[->, red!50] (0, 0.5) -- (1,0.5); + \draw[->, orange!50] (0,1) -- (0,0.5); + \draw[->, blue!50] (2,0.5) -- (2, 1); + \draw[->, purple!50] (3, 0.5) -- (2,0.5); + \draw[->, red!50] (4, 0.5) -- (3,0.5); + \draw[->, cyan!50] (2, 0) -- (3,0); + \draw[->, darkgreen!50] (3, 0) -- (4,0); + \end{scope} + } + } - \draw[ultra thick, ->, darkgreen] (0, 0) -- (0,0.5); - \draw[ultra thick, ->, orange] (1, 0) -- (0,0); - \draw[ultra thick, ->, red] (2, 0) -- (1,0); + \draw[ultra thick, ->, orange] (0, 0) -- (0,0.5); + \draw[ultra thick, ->, darkgreen] (1, 0) -- (0,0); + \draw[ultra thick, ->, cyan] (2, 0) -- (1,0); \draw[ultra thick, ->, blue] (2,0.5) -- (2, 0); \draw[ultra thick, ->, purple] (1, 0.5) -- (2,0.5); - \draw[ultra thick, ->, cyan] (0, 0.5) -- (1,0.5); - - + \draw[ultra thick, ->, red] (0, 0.5) -- (1,0.5); \foreach \i in {-2,...,1} { \foreach \j in {-2,...,1} { \begin{scope}[xshift=\i*4cm, yshift=\j*1cm] - \draw[opacity=0.5, ->, darkgreen] (0, 0) -- (0,0.5); - \draw[opacity=0.5, ->, orange] (1, 0) -- (0,0); - \draw[opacity=0.5, ->, red] (2, 0) -- (1,0); - \draw[opacity=0.5, ->, blue] (2,0.5) -- (2, 0); - \draw[opacity=0.5, ->, purple] (1, 0.5) -- (2,0.5); - \draw[opacity=0.5, ->, cyan] (0, 0.5) -- (1,0.5); - \draw[opacity=0.5, ->, darkgreen] (0,1) -- (0,0.5); - \draw[opacity=0.5, ->, blue] (2,0.5) -- (2, 1); - \draw[opacity=0.5, ->, purple] (3, 0.5) -- (2,0.5); - \draw[opacity=0.5, ->, cyan] (4, 0.5) -- (3,0.5); - \draw[opacity=0.5, ->, red] (2, 0) -- (3,0); - \draw[opacity=0.5, ->, orange] (3, 0) -- (4,0); - \node[zero] at ( 1, 0) {}; \node[zero] at ( 3, 0) {}; \node[pole] at ( 1,0.5) {}; @@ -72,12 +75,12 @@ \draw[gray, ->] (-6,0) -- (6,0) node[anchor=west]{$w$}; - \draw[thick, ->, purple] (-5, 0) -- (-3, 0); - \draw[thick, ->, blue] (-3, 0) -- (-2, 0); - \draw[thick, ->, red] (-2, 0) -- (0, 0); - \draw[thick, ->, orange] (0, 0) -- (2, 0); - \draw[thick, ->, darkgreen] (2, 0) -- (3, 0); - \draw[thick, ->, cyan] (3, 0) -- (5, 0); + \draw[ultra thick, ->, purple] (-5, 0) -- (-3, 0); + \draw[ultra thick, ->, blue] (-3, 0) -- (-2, 0); + \draw[ultra thick, ->, cyan] (-2, 0) -- (0, 0); + \draw[ultra thick, ->, darkgreen] (0, 0) -- (2, 0); + \draw[ultra thick, ->, orange] (2, 0) -- (3, 0); + \draw[ultra thick, ->, red] (3, 0) -- (5, 0); \node[anchor=south] at (-5,0) {$-\infty$}; \node[anchor=south] at (-3,0) {$-1/k$}; diff --git a/buch/papers/ellfilter/tikz/sn.tikz.tex b/buch/papers/ellfilter/tikz/sn.tikz.tex index 8e4d223..c3df8d1 100644 --- a/buch/papers/ellfilter/tikz/sn.tikz.tex +++ b/buch/papers/ellfilter/tikz/sn.tikz.tex @@ -17,39 +17,45 @@ \begin{scope}[xshift=-1cm] + \foreach \i in {-2,...,2} { + \foreach \j in {-2,...,1} { + \begin{scope}[xshift=\i*4cm, yshift=\j*1cm] + \draw[<-, blue!50] (0, 0) -- (0,0.5); + \draw[<-, cyan!50] (1, 0) -- (0,0); + \draw[<-, darkgreen!50] (2, 0) -- (1,0); + \draw[<-, orange!50] (2,0.5) -- (2, 0); + \draw[<-, red!50] (1, 0.5) -- (2,0.5); + \draw[<-, purple!50] (0, 0.5) -- (1,0.5); + \draw[<-, blue!50] (0,1) -- (0,0.5); + \draw[<-, orange!50] (2,0.5) -- (2, 1); + \draw[<-, red!50] (3, 0.5) -- (2,0.5); + \draw[<-, purple!50] (4, 0.5) -- (3,0.5); + \draw[<-, darkgreen!50] (2, 0) -- (3,0); + \draw[<-, cyan!50] (3, 0) -- (4,0); + \end{scope} + } + } + % \pause - \draw[ultra thick, <-, orange] (2, 0) -- (1,0); + \draw[ultra thick, <-, darkgreen] (2, 0) -- (1,0); % \pause - \draw[ultra thick, <-, darkgreen] (2,0.5) -- (2, 0); + \draw[ultra thick, <-, orange] (2,0.5) -- (2, 0); % \pause - \draw[ultra thick, <-, cyan] (1, 0.5) -- (2,0.5); + \draw[ultra thick, <-, red] (1, 0.5) -- (2,0.5); % \pause \draw[ultra thick, <-, blue] (0, 0) -- (0,0.5); \draw[ultra thick, <-, purple] (0, 0.5) -- (1,0.5); - \draw[ultra thick, <-, red] (1, 0) -- (0,0); + \draw[ultra thick, <-, cyan] (1, 0) -- (0,0); % \pause + \foreach \i in {-2,...,2} { \foreach \j in {-2,...,1} { \begin{scope}[xshift=\i*4cm, yshift=\j*1cm] - \draw[opacity=0.5, <-, blue] (0, 0) -- (0,0.5); - \draw[opacity=0.5, <-, red] (1, 0) -- (0,0); - \draw[opacity=0.5, <-, orange] (2, 0) -- (1,0); - \draw[opacity=0.5, <-, darkgreen] (2,0.5) -- (2, 0); - \draw[opacity=0.5, <-, cyan] (1, 0.5) -- (2,0.5); - \draw[opacity=0.5, <-, purple] (0, 0.5) -- (1,0.5); - \draw[opacity=0.5, <-, blue] (0,1) -- (0,0.5); - \draw[opacity=0.5, <-, darkgreen] (2,0.5) -- (2, 1); - \draw[opacity=0.5, <-, cyan] (3, 0.5) -- (2,0.5); - \draw[opacity=0.5, <-, purple] (4, 0.5) -- (3,0.5); - \draw[opacity=0.5, <-, orange] (2, 0) -- (3,0); - \draw[opacity=0.5, <-, red] (3, 0) -- (4,0); - \node[zero] at ( 1, 0) {}; \node[zero] at ( 3, 0) {}; \node[pole] at ( 1,0.5) {}; \node[pole] at ( 3,0.5) {}; - \end{scope} } } @@ -72,12 +78,12 @@ \draw[gray, ->] (-6,0) -- (6,0) node[anchor=west]{$w$}; - \draw[thick, ->, purple] (-5, 0) -- (-3, 0); - \draw[thick, ->, blue] (-3, 0) -- (-2, 0); - \draw[thick, ->, red] (-2, 0) -- (0, 0); - \draw[thick, ->, orange] (0, 0) -- (2, 0); - \draw[thick, ->, darkgreen] (2, 0) -- (3, 0); - \draw[thick, ->, cyan] (3, 0) -- (5, 0); + \draw[ultra thick, ->, purple] (-5, 0) -- (-3, 0); + \draw[ultra thick, ->, blue] (-3, 0) -- (-2, 0); + \draw[ultra thick, ->, cyan] (-2, 0) -- (0, 0); + \draw[ultra thick, ->, darkgreen] (0, 0) -- (2, 0); + \draw[ultra thick, ->, orange] (2, 0) -- (3, 0); + \draw[ultra thick, ->, red] (3, 0) -- (5, 0); \node[anchor=south] at (-5,0) {$-\infty$}; \node[anchor=south] at (-3,0) {$-1/k$}; diff --git a/buch/papers/ellfilter/tschebyscheff.tex b/buch/papers/ellfilter/tschebyscheff.tex index 7d426b6..8a82c5f 100644 --- a/buch/papers/ellfilter/tschebyscheff.tex +++ b/buch/papers/ellfilter/tschebyscheff.tex @@ -1,8 +1,7 @@ \section{Tschebyscheff-Filter} -Als Einstieg betrachent Wir das Tschebyscheff-Filter, welches sehr verwand ist mit dem elliptischen Filter. +Als Einstieg betrachten wir das Tschebyscheff-Filter, welches sehr verwand ist mit dem elliptischen Filter. Genauer ausgedrückt sind die Tschebyscheff-1 und -2 Filter Spezialfälle davon. - Der Name des Filters deutet schon an, dass die Tschebyscheff-Polynome $T_N$ für das Filter relevant sind: \begin{align} T_{0}(x)&=1\\ @@ -16,7 +15,7 @@ Bemerkenswert ist, dass die Polynome im Intervall $[-1, 1]$ mit der trigonometri T_N(w) &= \cos \left( N \cos^{-1}(w) \right) \\ &= \cos \left(N~z \right), \quad w= \cos(z) \end{align} -übereinstimmt. +übereinstimmen. Der Zusammenhang lässt sich mit den Doppel- und Mehrfachwinkelfunktionen der trigonometrischen Funktionen erklären. Abbildung \ref{ellfilter:fig:chebychef_polynomials} zeigt einige Tschebyscheff-Polynome. \begin{figure} @@ -36,12 +35,11 @@ Wenn wir die Tschebyscheff-Polynome quadrieren, passen sie perfekt in die Voraus \label{ellfiter:fig:chebychef} \end{figure} - Die analytische Fortsetzung von \eqref{ellfilter:eq:chebychef_polynomials} über das Intervall $[-1,1]$ hinaus stimmt mit den Polynomen überein, wie es zu erwarten ist. -Die genauere Betrachtung wird uns dann helfen die elliptischen Filter besser zu verstehen. +Die genauere Betrachtung wird uns helfen die elliptischen Filter besser zu verstehen. -Starten wir mit der Funktion, die als erstes auf $w$ angewendet wird, dem Arcuscosinus. -Die invertierte Funktion des Kosinus kann als definites Integral dargestellt werden: +Starten wir mit der Funktion, die in \eqref{ellfilter:eq:chebychef_polynomials} als erstes auf $w$ angewendet wird, dem Arcuscosinus. +Die invertierte Funktion des Kosinus kann als bestimmtes Integral dargestellt werden: \begin{align} \cos^{-1}(x) &= @@ -88,46 +86,21 @@ Abbildung \ref{ellfilter:fig:arccos} zeigt den $\arccos$ in der komplexen Ebene. \caption{Die Funktion $z = \cos^{-1}(w)$ dargestellt in der komplexen ebene.} \label{ellfilter:fig:arccos} \end{figure} -Wegen der Periodizität des Kosinus ist auch der Arcuscosinus $2\pi$-periodisch und es entstehen periodische Nullstellen. -% \begin{equation} -% \frac{ -% 1 -% }{ -% \sqrt{ -% 1-z^2 -% } -% } -% \in \mathbb{R} -% \quad -% \forall -% \quad -% -1 \leq z \leq 1 -% \end{equation} -% \begin{equation} -% \frac{ -% 1 -% }{ -% \sqrt{ -% 1-z^2 -% } -% } -% = i \xi \quad | \quad \xi \in \mathbb{R} -% \quad -% \forall -% \quad -% z \leq -1 \cup z \geq 1 -% \end{equation} +Wegen der Periodizität des Kosinus ist auch der Arcuscosinus $2\pi$-periodisch. +Das Einzeichnen von Pol- und Nullstellen ist hilfreich für die Betrachtung der Funktion. + -Die Tschebyscheff-Polynome skalieren diese Nullstellen mit dem Ordnungsfaktor $N$, wie dargestellt in Abbildung \ref{ellfilter:fig:arccos2}. +In \eqref{ellfilter:eq:chebychef_polynomials} wird $z$ mit dem Ordnungsfaktor $N$ multipliziert und durch die Kosinusfunktion zurück transformiert. +Die Skalierung hat zur folge, dass bei der Rücktransformation durch den Kosinus mehrere Nullstellen durchlaufen werden. +Somit passiert $\cos( N~\cos^{-1}(w))$ im Intervall $[-1, 1]$ $N$ Nullstellen, wie dargestellt in Abbildung \ref{ellfilter:fig:arccos2}. \begin{figure} \centering \input{papers/ellfilter/tikz/arccos2.tikz.tex} \caption{ $z_1=N \cos^{-1}(w)$-Ebene der Tschebyscheff-Funktion. - Die eingefärbten Pfade sind Verläufe von $w~\forall~[-\infty, \infty]$ für verschiedene Ordnungen $N$. + Die eingefärbten Pfade sind Verläufe von $w~\forall~[-\infty, \infty]$ für $N = 4$. Je grösser die Ordnung $N$ gewählt wird, desto mehr Nullstellen werden passiert. } \label{ellfilter:fig:arccos2} \end{figure} -Somit passert $\cos( N~\cos^{-1}(w))$ im Intervall $[-1, 1]$ $N$ Nullstellen. Durch die spezielle Anordnung der Nullstellen hat die Funktion Equirippel-Verhalten und ist dennoch ein Polynom, was sich perfekt für linear Filter eignet. -- cgit v1.2.1 From 2cf30b784f1cf73cd4ae8c9924435f236f351470 Mon Sep 17 00:00:00 2001 From: tim30b Date: Wed, 10 Aug 2022 21:51:06 +0200 Subject: =?UTF-8?q?Korrekturen=20von=20M=C3=BCller=20umgesetzt?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- buch/papers/kreismembran/teil0.tex | 16 +++++++++------- buch/papers/kreismembran/teil4.tex | 29 ++++++++++++++--------------- 2 files changed, 23 insertions(+), 22 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil0.tex b/buch/papers/kreismembran/teil0.tex index 6f55358..a0a4152 100644 --- a/buch/papers/kreismembran/teil0.tex +++ b/buch/papers/kreismembran/teil0.tex @@ -5,18 +5,18 @@ % \section{Einleitung\label{kreismembran:section:teil0}} \rhead{Membran} -Eine Membran oder selten ein Schwingblatt ist laut Duden \cite{kreismembran:Duden:Membran} ein ``dünnes Blättchen aus Metall, Papier o. Ä., das durch seine Schwingungsfähigkeit geeignet ist, Schallwellen zu übertragen ...''. +Eine Membran oder selten ein Schwingblatt ist laut Duden \cite{kreismembran:Duden:Membran} ein ``dünnes Blättchen aus Metall, Papier o. Ä., das durch seine Schwingungsfähigkeit geeignet ist, Schallwellen zu übertragen \dots''. Ein dünnes Blättchen aus Metall zeig jedoch nicht die selben dynamischen Eigenschaften wie ein gespanntes Stück Papier. Beschreibt man das dynamische Verhalten, muss zwischen einer dünnen Platte und einer Membrane unterschieden werden \cite{kreismembran:membrane_vs_thin_plate}. Eine dünne Platte zum Beispiel aus Metall, wirkt selbst entgegen ihrer Deformation sobald sie gekrümmt wird. Eine Membran auf der anderen Seite besteht aus einem Material, welches sich ohne Kraftaufwand verbiegen lässt wie zum Beispiel Papier. -Bevor Papier als schwingende Membran betrachtet werden kann, wird jedoch noch eine Spannung $ T $ benötigt welche das Material daran hindert aus der Ruhelage gebracht zu werden. +Bevor Papier als schwingende Membran betrachtet werden kann, wird jedoch noch eine Spannung $ T $ benötigt welche das Material daran hindert, aus der Ruhelage gebracht zu werden. Ein geläufiges Beispiel einer Kreismembran ist eine runde Trommel. Sie besteht herkömmlicherweise aus einem Leder (Fell), welches auf einen offenen Zylinder (Zargen) aufgespannt wird. Das Leder alleine erzeugt nach einem Aufschlag keine hörbaren Schwingungen. Sobald das Fell jedoch über den Zargen gespannt wird, kann das Fell auf verschiedensten Weisen weiter schwingen, was für den Klang der Trommel verantwortlich ist. -Wie genau diese Schwingungen untersucht werden können wird in der folgenden Arbeit diskutiert. +Wie genau diese Schwingungen untersucht werden können, wird in der folgenden Arbeit diskutiert. \subsection{Annahmen} \label{kreimembran:annahmen} @@ -48,9 +48,10 @@ Die Verbindung zwischen Membran und Saite ist intuitiv ersichtlich, stellt man s \end{center} \end{figure} -Abbildung \ref{kreismembran:im:Saite} ist ein infinitesimales Stück einer Saite mit Länge $ dx $ skizziert. -Wie für die Membran ist die Annahme iii) gültig, keine Bewegung in die Richtung $ \hat{x} $. -Um dies zu erfüllen muss der Punkt $ P_1 $ gleich stark in Richtung $ -\hat{x} $ gezogen werden wie der Punkt $ P_2 $ in Richtung $ \hat{x} $ gezogen wird. Ist $ T_1 $ die Kraft welche mit Winkel $ \alpha $ auf Punkt $ P_1 $ wirkt sowie $ T_2 $ und $ \beta$ das analoge für Punkt $ P_2 $ ist, so können die Kräfte +In Abbildung \ref{kreismembran:im:Saite} ist ein infinitesimales Stück einer Saite mit Länge $ dx $ skizziert. +Wie für die Membran ist die Annahme iii) gültig, keine Bewegung entlang der $ x $-Achse. +Um dies zu erfüllen, muss der Punkt $ P_1 $ gleich stark entgegen der $ x $-Achse gezogen werden wie der Punkt $ P_2 $ in Richtung der $ x $-Achse gezogen wird. +Ist $ T_1 $ die Kraft, welche mit Winkel $ \alpha $ auf Punkt $ P_1 $ wirkt sowie $ T_2 $ und $ \beta$ das analoge für Punkt $ P_2 $ ist, so können die Kräfte \begin{equation}\label{kreismembran:eq:no_translation} T_1 \cos \alpha = T_2 \cos \beta = T \end{equation} @@ -81,7 +82,8 @@ Durch die Division mit $ dx $ entsteht \begin{equation*} \frac{1}{dx} \left[\frac{\partial u}{\partial x} \bigg|_{x_0 + dx} - \frac{\partial u}{\partial x} \bigg|_{x_0}\right] = \frac{\rho}{T}\frac{\partial^2 u}{\partial t^2}. \end{equation*} -Auf der linken Seite der Gleichung wird die Differenz der Steigungen durch die Intervalllänge geteilt, in anderen Worten die zweite Ableitung von $ u(x,t) $ nach $ x $ berechnet. +Auf der linken Seite der Gleichung wird die Differenz der Steigungen durch die Intervalllänge geteilt. +Wenn $ dx $ als unendlich kleines Stück betrachtet wird, ergibt sich als Grenzwert die zweite Ableitung von $ u(x,t) $ nach $ x $. Der Term $ \frac{\rho}{T} $ wird durch $ c^2 $ ersetzt, da der Bruch für eine gegebene Membran eine positive Konstante sein muss. Somit resultiert die in der Literatur gebräuchliche Form \begin{equation} diff --git a/buch/papers/kreismembran/teil4.tex b/buch/papers/kreismembran/teil4.tex index 74bb87d..95cb516 100644 --- a/buch/papers/kreismembran/teil4.tex +++ b/buch/papers/kreismembran/teil4.tex @@ -67,37 +67,37 @@ Die Filtermaske kann dann auf jedes Element einzeln angewendet werden mit einer \subsection{Simulation: Kreisförmige Membran} Als Beispiel soll nun eine schwingende kreisförmige Membran simuliert werden. -\paragraph{Initialisierung} -Die Anzahl der simulierten Elementen soll $ m \times n $ was dementsprechend die Dimensionen von $ U $ und $ V $ vorgibt. +\subsubsection{Initialisierung} +Die Anzahl der simulierten Elemente soll $ m \times n $ sein, was die Dimensionen von $ U $ und $ V $ vorgibt. Als Anfangsbedingung wird eine Membran gewählt, welche bei $ t=0 $ mit einer Gauss-Kurve ausgelenkt wird. Die Membran soll sich zu Beginn nicht bewegen, also wird $ V[0] $ mit Nullen initialisiert. Die Auslenkung kann kompakt erreicht werden, wenn $ U[0] $ als Null-Matrix mit einer $ 1 $ in der Mitte initialisiert wird. Diese Matrix wird anschliessend mit einer Filtermaske in Form einer Gauss-Glocke gefaltet. -Die Faltung mit einer Gauss-Glocke ist in Programmen wie Matlab eine Standartfunktion, da dies einm Tiefpassfilter in der Bildverarbeitung entspricht. +Die Faltung mit einer Gauss-Glocke ist in Programmen wie Matlab eine Standartfunktion, da dies einem Tiefpassfilter in der Bildverarbeitung entspricht. -\paragraph{Rand} +\subsubsection{Rand} Bislang ist die definierte Matrix rechteckig. Um eine kreisförmige Membran zu simulieren muss der Rand angepasst werden. Da in den meisten Programme keine Möglichkeit besteht, mit runden Matrizen zu rechnen, wird der Rand in der Berechnung des Folgezustandes implementiert. Der Rand bedeutet, das Membran-Elemente auf dem Rand sich nicht Bewegen können. -Die Position sowie die Geschwindigkeit aller Elemente welche nicht auf der definierten Membran sind müssen zu beliebiger Zeit $0$ entsprechen. +Die Position sowie die Geschwindigkeit aller Elemente, welche nicht auf der definierten Membran sind, müssen zu beliebiger Zeit $0$ sein. Hierzu wird eine Maske $M$ erstellt. Diese Maske besteht aus einer binären Matrix von identischer Dimension wie $ U $ und $ V $. Ist in der Matrix $M$ eine $1$ abgebildet so ist an jener stelle ein Element der Membran, ist es eine $0$ so befindet sich dieses Element auf dem Rand oder ausserhalb der Membran. In dieser Anwendung ist $M$ eine Matrix mit einem Kreis voller $1$ umgeben von $0$ bis an den Rand der Matrix. -Die Maske wird angewendet indem das Resultat des nächsten Zustandes noch mit der Maske elementweise multipliziert wird. +Die Maske wird angewendet, indem das Resultat des nächsten Zustandes noch mit der Maske elementweise multipliziert wird. Der Folgezustand kann also mit den Gleichungen \begin{align} \label{kreismembran:eq:folge_U} - U[w+1] &= (U[w] + dt \cdot V[w])*M\\ + U[w+1] &= (U[w] + dt \cdot V[w])\odot M\\ \label{kreismembran:eq:folge_V} - V[w+1] &= (V[w] + dt \cdot \Delta_h u \cdot c^2)*M + V[w+1] &= (V[w] + dt \cdot \Delta_h u \cdot c^2)\odot M \end{align} berechnet werden. -\paragraph{Simulation} +\subsubsection{Simulation} Mit den gegebenen Gleichungen \ref{kreismembran:eq:folge_U} und \ref{kreismembran:eq:folge_V} das Verhalten der Membran mit einem Loop über das zu untersuchende Zeitintervall berechnet werden. In der Abbildung \ref{kreismembran:im:simres_rund} sind Simulationsresultate zu sehen. -Die Erste Figur zeigt die Ausgangslage gefolgt von den Auslenkungen nach jeweils $ 50 $ weiteren Iterationsschritten. +Die erste Figur zeigt die Ausgangslage gefolgt von den Auslenkungen nach jeweils $ 50 $ weiteren Iterationsschritten. Es ist zu erkennen, wie sich die Störung vom Zentrum an den Rand ausbreitet. Erreicht die Störung den Rand wird sie reflektiert und nähert sich dem Zentrum. \begin{figure} @@ -123,13 +123,13 @@ Wenn anschliessend nur das Verhalten im Zentrum, bei der Störung beobachtet wir Dies aber nur bis die Störung am Rand reflektiert wird und wieder das innere zu beobachtende Zentrum beeinflusst. Soll erst gar keine Reflexion entstehen, muss ein Absorber modelliert werden welcher die Störung möglichst ohne Reflexion aufnimmt. -\paragraph{Absorber} +\subsubsection{Absorber} Sehr knapp formuliert entstehen Reflexionen, wenn eine Welle von einem Material in ein anderes Material mit unterschiedlichen Eigenschaften eindringen möchte. Je unterschiedlicher und abrupter der Übergang zwischen den Materialien umso ausgeprägter die Reflexion. In diesem Fall sind die Eigenschaften vorgegeben. Im Zentrum soll sich die Membran verhalten, wie von der DGL vorgegeben, am Rand jedoch muss sich jedes Membran-Element in der Ausgangslage befinden. Der Spielraum welcher dem Absorber übrig bleibt ist die Art der Überganges. -Bei der endlichen kreisförmigen Membran hat die Maske $M$ ein binärer Übergang von Membran zu Rand bezweckt. +Bei der endlichen kreisförmigen Membran hat die Maske $M$ einen binären Übergang von Membran zu Rand bezweckt. Anstelle dieses abrupten Wechsels wird nun eine Maske definiert, welche graduell von Membran $1$ zu Rand-Element $0$ wechselt. Die Elemente werden auf Basis ihres Abstand $r$ zum Zentrum definiert. Der Abstand entspricht @@ -156,11 +156,10 @@ In der Abbildung \ref{kreismembran:im:masks} ist der Unterschied der beiden Mask \label{kreismembran:im:masks} \end{center} \end{figure} -\paragraph{Simulation} +\subsubsection{Simulation} Bis auf die Absorber-Maske kann nun identisch zur endlichen Membran simuliert werden. Auch hier wurde eine Gauss-Glocke als Anfangsbedingung gewählt. Die Simulationsresultate von Abbildung \ref{kreismembran:im:simres_unendlich} - \begin{figure} \begin{center} @@ -183,7 +182,7 @@ Dieses Verhalten spricht für den Absorber-Ansatz, es soll jedoch erwähnt sein, Die DGL \ref{kreismembran:Ausgang_DGL} welche simuliert wird geht jedoch von der Annahme \ref{kreimembran:annahmen} iv) aus, dass die Membran keine Art von Dämpfung erfährt. \section{Schlusswort} -Auch wenn ein Physikalisches Verhalten bereits durch Annahmen und Annäherungen deutlich vereinfacht wird, bestehen auch dann noch eine Vielzahl von Lösungsansätzen. +Auch wenn ein physikalisches Verhalten bereits durch Annahmen und Annäherungen deutlich vereinfacht wird, bestehen auch dann noch eine Vielzahl von Lösungsansätzen. Lösungen einer unendlich grosse Membran scheinen fern der Realität zu sein, doch dies darf es im Sinne der Mathematik. Und wer weis, für eine Ameise auf einem Trampolin ist eine unendliche Membran vielleicht eine ganz gute Annäherung. -- cgit v1.2.1 From 16f447cb8a9df0d271f29b1aecb24532948bea8c Mon Sep 17 00:00:00 2001 From: Nicolas Tobler Date: Wed, 10 Aug 2022 23:52:40 +0200 Subject: working on elliptic rational functions --- buch/papers/ellfilter/elliptic.tex | 76 ++++++++++++++++++++------------------ 1 file changed, 41 insertions(+), 35 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/ellfilter/elliptic.tex b/buch/papers/ellfilter/elliptic.tex index 8c60e46..793fd6c 100644 --- a/buch/papers/ellfilter/elliptic.tex +++ b/buch/papers/ellfilter/elliptic.tex @@ -1,15 +1,15 @@ \section{Elliptische rationale Funktionen} -Kommen wir nun zum eigentlichen Teil dieses Papers, den elliptischen rationalen Funktionen +Kommen wir nun zum eigentlichen Teil dieses Papers, den elliptischen rationalen Funktionen \ref{ellfilter:bib:orfanidis} \begin{align} - R_N(\xi, w) &= \cd \left(N~f_1(\xi)~\cd^{-1}(w, 1/\xi), f_2(\xi)\right) \\ + R_N(\xi, w) &= \cd \left(N~f_1(\xi)~\cd^{-1}(w, 1/\xi), f_2(\xi)\right) \label{ellfilter:eq:elliptic}\\ &= \cd \left(N~\frac{K_1}{K}~\cd^{-1}(w, k), k_1)\right) , \quad k= 1/\xi, k_1 = 1/f(\xi) \\ &= \cd \left(N~K_1~z , k_1 \right), \quad w= \cd(z K, k) \end{align} Beim Betrachten dieser Definition, fällt die Ähnlichkeit zur trigonometrische Darstellung der Tschebyschef-Polynome \eqref{ellfilter:eq:chebychef_polynomials} auf. Anstelle vom Kosinus kommt hier die $\cd$-Funktion zum Einsatz. Die Ordnungszahl $N$ kommt auch als Faktor for. -Zusätzlich werden noch zwei verschiedene elliptische Module $k$ und $k_1$ gebraucht. +Zusätzlich werden noch zwei verschiedene elliptische Moduli $k$ und $k_1$ gebraucht. Bei $k = k_1 = 0$ wird der $\cd$ zum Kosinus und wir erhalten in diesem Spezialfall die Tschebyschef-Polynome. Durch das Konzept vom fundamentalen Rechteck, siehe Abbildung \ref{buch:elliptisch:fig:ellall} können für alle inversen Jacobi elliptischen Funktionen die Positionen der Null- und Polstellen anhand eines Diagramms ermittelt werden. @@ -24,21 +24,25 @@ Die $\cd^{-1}(w, k)$-Funktion ist um $K$ verschoben zur $\sn^{-1}(w, k)$-Funktio \label{ellfilter:fig:cd} \end{figure} Auffallend an der $w = \cd(z, k)$-Funktion ist, dass sich $w$ auf der reellen Achse wie der Kosinus immer zwischen $-1$ und $1$ bewegt, während bei $\mathrm{Im(z) = K^\prime}$ die Werte zwischen $\pm 1/k$ und $\pm \infty$ verlaufen. -Die Funktion hat also Equirippel-Verhalten um $w=0$ und um $w=\pm \infty$. %TODO Check -Falls es möglich ist diese Werte abzufahren im Stil der Tschebyscheff-Polynome, kann ein Filter gebaut werden, dass Equirippel-Verhalten im Durchlass- und Sperrbereich aufweist. - -Analog zu Abbildung \ref{ellfilter:fig:arccos2} können wir auch bei den elliptisch rationalen Funktionen die komplexe $z$-Ebene betrachten, wie ersichtlich in Abbildung \ref{ellfilter:fig:cd2}, um die besser zu verstehen. +Die Idee des elliptischen Filter ist es, diese zwei Equirippel-Zonen abzufahren, wie ersichtlich in Abbildung \ref{ellfilter:fig:cd2}, welche Analog zu Abbildung \ref{ellfilter:fig:arccos2} gesehen werden kann. \begin{figure} \centering \input{papers/ellfilter/tikz/cd2.tikz.tex} \caption{ $z_1$-Ebene der elliptischen rationalen Funktionen. - Je grösser die Ordnung $N$ gewählt wird, desto mehr Nullstellen passiert. + Je grösser die Ordnung $N$ gewählt wird, desto mehr Nullstellen werden passiert. } \label{ellfilter:fig:cd2} \end{figure} -% Da die $\cd^{-1}$-Funktion - +Das elliptische Filter hat im Gegensatz zum Tschebyscheff-Filter drei Zonen. +Im Durchlassbereich werden wie beim Tschebyscheff-Filter die Nullstellen durchlaufen. +Statt dass $z_1$ für alle $w>1$ in die imaginäre Richtung geht, bewegen wir uns im Sperrbereich wieder in reeller Richtung, wo Pole durchlaufen werden. +Aus dieser Sicht kann der Sperrbereich vom Tschebyscheff-Filter als unendlich langer Übergangsbereich angesehen werden. +% Falls es möglich ist diese Werte abzufahren im Stil der Tschebyscheff-Polynome, kann ein Filter gebaut werden, dass Equirippel-Verhalten im Durchlass- und Sperrbereich aufweist. +Da sich die Funktion im Übergangsbereich nur zur nächsten Reihe bewegt ist der Übergangsbereich monoton steigend. +Theoretisch könnte eine gleiches Durchlass- und Sperrbereichverhalten erreicht werden, wenn die Funktion auf eine andere Reihe ansteigen würde. +Dies würde jedoch zu Oszillationen zwischen $1$ und $1/k$ im Übergangsbereich führen. +Abbildung \ref{ellfilter:fig:elliptic_freq} zeigt eine elliptisch rationale Funktion und die Frequenzantwort des daraus resultierenden Filters. \begin{figure} \centering \input{papers/ellfilter/python/elliptic.pgf} @@ -48,43 +52,45 @@ Analog zu Abbildung \ref{ellfilter:fig:arccos2} können wir auch bei den ellipti \subsection{Gradgleichung} -Der $\cd^{-1}$ Term muss so verzogen werden, dass die umgebene $\cd$-Funktion die Nullstellen und Pole trifft. -Dies trifft ein wenn die Gradengleichung erfüllt ist. - -\begin{equation} - N \frac{K^\prime}{K} = \frac{K^\prime_1}{K_1} -\end{equation} - - -Leider ist das lösen dieser Gleichung nicht trivial. -Die Rechnung wird in \ref{ellfilter:bib:orfanidis} im Detail angeschaut. - -$K$ und $K^\prime$ sind voneinender abhängig. - -Das Problem lässt sich grafisch darstellen. - +Damit die Pol- und Nullstellen genau in dieser Konstellation durchfahren werden, müssen die elliptischen Moduli des inneren und äusseren $\cd$ aufeinander abgestimmt werden. +In der reellen Richtung müssen sich die Periodizitäten $K$ und $K_1$ um den Faktor $N$ unterscheiden, während die imagiäre Periodizitäten $K^\prime$ und $K^\prime_1$ gleich bleiben müssen. +Zur Erinnerung, $K$ und $K^\prime$ sind durch elliptische Integrale definiert und vom Modul $k$ abhängig wie ersichtlich in Abbildung \ref{ellfilter:fig:kprime}. \begin{figure} \centering \input{papers/ellfilter/python/k.pgf} \caption{Die Periodizitäten in realer und imaginärer Richtung in Abhängigkeit vom elliptischen Modul $k$.} + \label{ellfilter:fig:kprime} \end{figure} - -%TODO combine figures? -\begin{figure} - \centering - \input{papers/ellfilter/tikz/elliptic_transform1.tikz} - \caption{Die Gradgleichung als geometrisches Problem.} -\end{figure} +$K$ und $K^\prime$ sind durch die Ortskurve $K + jK^\prime$ aneinander Gebunden und benötigen den Zusatzfaktor $K_1/K$ in \eqref{ellfilter:eq:elliptic}, um die genanten Forderungen einzuhalten. +Abbildung \ref{ellfilter:fig:degree_eq} zeigt das Problem geometrisch auf, wobei zwei Punkte auf der Ortskurve gesucht sind. \begin{figure} \centering \input{papers/ellfilter/tikz/elliptic_transform2.tikz} - \caption{Die Gradgleichung als geometrisches Problem.} + \caption{Die Gradgleichung als geometrisches Problem ($N=3$).} + \label{ellfilter:fig:degree_eq} \end{figure} +Algebraisch kann so die Gradgleichung +\begin{equation} + N \frac{K^\prime}{K} = \frac{K^\prime_1}{K_1} +\end{equation} +aufgestellt werden, dessen Lösung ist gegeben durch +\begin{equation} %TODO check +k_1 = k^N \prod_{i=1}^L \sn^4 \Bigg( \frac{2i - 1}{N} K, k \Bigg), +\quad \text{wobei} \quad +N = 2L+r. +\end{equation} +Die Herleitung ist sehr umfassend und wird in \ref{ellfilter:bib:orfanidis} im Detail angeschaut. + +% \begin{figure} +% \centering +% \input{papers/ellfilter/tikz/elliptic_transform1.tikz} +% \caption{Die Gradgleichung als geometrisches Problem.} +% \end{figure} -\subsection{Polynome?} +\subsection{Darstellung als rationale Funktion} Bei den Tschebyscheff-Polynomen haben wir gesehen, dass die Trigonometrische Formel zu einfachen Polynomen umgewandelt werden kann. -Im gegensatz zum $\cos^{-1}$ hat der $\cd^{-1}$ nicht nur Nullstellen sondern auch Pole. +Im Gegensatz zum $\cos^{-1}$ hat der $\cd^{-1}$ nicht nur Nullstellen sondern auch Pole. Somit entstehen bei den elliptischen rationalen Funktionen, wie es der name auch deutet, rationale Funktionen, also ein Bruch von zwei Polynomen. Da Transformationen einer rationalen Funktionen mit Grundrechenarten, wie es in \eqref{ellfilter:eq:h_omega} der Fall ist, immer noch rationale Funktionen ergeben, stellt dies kein Problem für die Implementierung dar. -- cgit v1.2.1 From 330038bafaaf6ed6462a3efdcf9869b6ecf645ce Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Thu, 11 Aug 2022 14:34:39 +0200 Subject: Added mu calculation to both fourier examples. --- buch/papers/sturmliouville/standalone.tex | 31 ------- .../sturmliouville/waermeleitung_beispiel.tex | 99 ++++++++++++++++++++-- 2 files changed, 94 insertions(+), 36 deletions(-) delete mode 100644 buch/papers/sturmliouville/standalone.tex (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/standalone.tex b/buch/papers/sturmliouville/standalone.tex deleted file mode 100644 index cd0e8dc..0000000 --- a/buch/papers/sturmliouville/standalone.tex +++ /dev/null @@ -1,31 +0,0 @@ -\documentclass{book} - -\def\IncludeBookCover{0} -\input{common/packages.tex} - -% additional packages used by the individual papers, add a line for -% each paper -\input{papers/common/addpackages.tex} - -% workaround for biblatex bug -\makeatletter -\def\blx@maxline{77} -\makeatother -\addbibresource{chapters/references.bib} - -% Bibresources for each article -\input{papers/common/addbibresources.tex} - -% make sure the last index starts on an odd page -\AtEndDocument{\clearpage\ifodd\value{page}\else\null\clearpage\fi} -\makeindex - -%\pgfplotsset{compat=1.12} -\setlength{\headheight}{15pt} % fix headheight warning -\DeclareGraphicsRule{*}{mps}{*}{} - -\begin{document} - \input{common/macros.tex} - \def\chapterauthor#1{{\large #1}\bigskip\bigskip} - \input{papers/sturmliouville/main.tex} -\end{document} diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 92ecc49..89d158c 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -216,7 +216,7 @@ und durch umformen somit \mu A\sin(\alpha x) + \mu B\cos(\beta x). \] -Durch Koeffizientenvergleich von +Mittels Koeffizientenvergleich von \[ \begin{aligned} -\alpha^{2}A\sin(\alpha x) @@ -238,16 +238,105 @@ Dazu werden nochmals die Randbedingungen Zu bemerken ist, dass die Randbedingungen nur Anforderungen in $x$ stellen und somit direkt für $X(x)$ übernomen werden können. -Daraus ergibt sich für einen Stab mit Enden auf konstanter Temperatur +Da die Koeffizienten $A$ und $B$, sowie die Parameter $\alpha$ uns $\beta$ im +allgemeninen ungleich $0$ sind, müssen die Randbedingungen durch die +trigonometrischen Funktionen erfüllt werden. +Es werden nun die Randbedingungen +\eqref{eq:slp-example-fourier-boundary-condition-ends-constant} für einen Stab +mit Enden auf konstanter Temperatur in die Gleichung +\eqref{eq:slp-example-fourier-separated-x} eingesetzt. +Betrachten wir zunächst die Bedingung für $x = 0$. +Dies fürht zu +\[ + X(0) + = + A \sin(0 \alpha) + B \cos(0 \beta) + = + 0. +\] +Da $\cos(0) \neq 0$ ist, muss in diesem Fall $B = 0$ gelten. +Für den ersten Summanden ist wegen $\sin(0) = 0$ die Randbedingung erfüllt. + +Wird nun die zweite Randbedingung für $x = l$ mit $B = 0$ eingesetzt, ergibt +sich +\[ + X(l) + = + A \sin(\alpha l) + 0 \cos(\beta l) + = + A \sin(\alpha l) + = 0. +\] + +$\alpha$ muss also so gewählt werden, dass $\sin(\alpha l) = 0$ gilt. +Es gilt nun nach $\alpha$ aufzulösen: +\[ +\begin{aligned} + \sin(\alpha l) &= 0 \\ + \alpha l &= n \pi \qquad n \in \mathbb{N} \\ + \alpha &= \frac{n \pi}{l} \qquad n \in \mathbb{N} +\end{aligned} +\] + +Es folgt nun wegen $\mu = -\alpha^{2}$, dass +\begin{equation} + \mu_1 = -\alpha^{2} = -\frac{n^{2}\pi^{2}}{l^{2}} +\end{equation} +sein muss. +Ausserdem ist zu bemerken, dass dies auch gleich $-\beta^{2}$ ist. +Da aber $B = 0$ gilt und der Summand mit $\beta$ verschwindet, ist dies keine +Verletzung der Randbedingungen. + +Durch alanoges Vorgehen kann nun auch das Problem mit isolierten Enden gelöst +werden. +Setzen wir nun die Randbedingungen +\eqref{eq:slp-example-fourier-boundary-condition-ends-isolated} in $X^{\prime}$ +ein, beginnend für $x = 0$. Es ergibt sich +\[ + X^{\prime}(0) + = + \alpha A \cos(\alpha 0) - \beta B \sin(\beta 0) + = 0. +\] +In diesem Fall muss $A = 0$ gelten. +Zusammen mit der Bedignung für $x = l$ +folgt nun +\[ + X^{\prime}(l) + = + \alpha A \cos(\alpha l) - \beta B \sin(\beta l) + = + -\beta B \sin(\beta l) + = 0. +\] + +Wiedrum muss über die $\sin$-Funktion sicher gestellt werden, dass der Ausdruck +den Randbedingungen entspricht. Es folgt nun +\[ +\begin{aligned} + \sin(\beta l) &= 0 \\ + \beta l &= n \pi \qquad n \in \mathbb{N} \\ + \beta &= \frac{n \pi}{l} \qquad n \in \mathbb{N} +\end{aligned} +\] +und somit +\[ + \mu_2 = -\alpha^{2} = -\frac{n^{2}\pi^{2}}{l^{2}}. +\] + +Es ergibt sich also sowohl für einen Stab mit Enden auf konstanter Temperatur +wie auch mit isolierten Enden \begin{equation} \label{eq:slp-example-fourier-mu-solution} \mu = - -\frac{n^{2}\pi^{2}}{l^{2}} + -\frac{n^{2}\pi^{2}}{l^{2}}. \end{equation} -Betrachten wir nun die zweite Gleichung der Separation + + +Betrachten wir zuletzt die zweite Gleichung der Separation \eqref{eq:slp-example-fourier-separated-t}. Diese Lösen wir über das charakteristische Polynom \[ @@ -263,7 +352,7 @@ Lösung e^{-\kappa \mu t} \] führt. -Und mit dem Resultat \eqref{eq:slp-example-fourier-mu-solution} die Lösung +Und mit dem Resultat \eqref{eq:slp-example-fourier-mu-solution} \[ T(t) = -- cgit v1.2.1 From cc1f753efdfe46d546b1769e2f61d9765380373d Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Thu, 11 Aug 2022 14:47:19 +0200 Subject: Corrected some grammar. --- buch/papers/sturmliouville/waermeleitung_beispiel.tex | 14 ++++++++------ 1 file changed, 8 insertions(+), 6 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 89d158c..1b267cb 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -270,7 +270,7 @@ sich \] $\alpha$ muss also so gewählt werden, dass $\sin(\alpha l) = 0$ gilt. -Es gilt nun nach $\alpha$ aufzulösen: +Es bleibt noch nach $\alpha$ aufzulösen: \[ \begin{aligned} \sin(\alpha l) &= 0 \\ @@ -296,7 +296,7 @@ ein, beginnend für $x = 0$. Es ergibt sich \[ X^{\prime}(0) = - \alpha A \cos(\alpha 0) - \beta B \sin(\beta 0) + \alpha A \cos(0 \alpha) - \beta B \sin(0 \beta) = 0. \] In diesem Fall muss $A = 0$ gelten. @@ -305,14 +305,15 @@ folgt nun \[ X^{\prime}(l) = - \alpha A \cos(\alpha l) - \beta B \sin(\beta l) + 0 \alpha \cos(\alpha l) - \beta B \sin(\beta l) = -\beta B \sin(\beta l) = 0. \] -Wiedrum muss über die $\sin$-Funktion sicher gestellt werden, dass der Ausdruck -den Randbedingungen entspricht. Es folgt nun +Wiedrum muss über die $ \sin $-Funktion sicher gestellt werden, dass der +Ausdruck den Randbedingungen entspricht. +Es folgt nun \[ \begin{aligned} \sin(\beta l) &= 0 \\ @@ -322,7 +323,7 @@ den Randbedingungen entspricht. Es folgt nun \] und somit \[ - \mu_2 = -\alpha^{2} = -\frac{n^{2}\pi^{2}}{l^{2}}. + \mu_2 = -\beta^{2} = -\frac{n^{2}\pi^{2}}{l^{2}}. \] Es ergibt sich also sowohl für einen Stab mit Enden auf konstanter Temperatur @@ -334,6 +335,7 @@ wie auch mit isolierten Enden -\frac{n^{2}\pi^{2}}{l^{2}}. \end{equation} +%%%% Koeffizienten a_n und b_n mittels skalarprodukt. %%%%%%%%%%%%%%%%%%%%%%%%%% Betrachten wir zuletzt die zweite Gleichung der Separation -- cgit v1.2.1 From d8b0e6f27ac13c684bf829f4f73c11f4408945a5 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Thu, 11 Aug 2022 15:29:32 +0200 Subject: Added Tschebyscheff file. --- buch/papers/sturmliouville/beispiele.tex | 5 ++++- buch/papers/sturmliouville/tschebyscheff_beispiel.tex | 7 +++++++ buch/papers/sturmliouville/waermeleitung_beispiel.tex | 15 +++++++++++++++ 3 files changed, 26 insertions(+), 1 deletion(-) create mode 100644 buch/papers/sturmliouville/tschebyscheff_beispiel.tex (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/beispiele.tex b/buch/papers/sturmliouville/beispiele.tex index b23593e..94082cf 100644 --- a/buch/papers/sturmliouville/beispiele.tex +++ b/buch/papers/sturmliouville/beispiele.tex @@ -8,4 +8,7 @@ \rhead{Beispiele} % Fourier: Erik work -\input{papers/sturmliouville/waermeleitung_beispiel.tex} \ No newline at end of file +\input{papers/sturmliouville/waermeleitung_beispiel.tex} + +% Tschebyscheff +\input{papers/sturmliouville/tschebyscheff_beispiel.tex} \ No newline at end of file diff --git a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex new file mode 100644 index 0000000..54f13d4 --- /dev/null +++ b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex @@ -0,0 +1,7 @@ +% +% tschebyscheff_beispiel.tex +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% + +\subsection{Tschebyscheff} \ No newline at end of file diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 1b267cb..14fca40 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -337,6 +337,21 @@ wie auch mit isolierten Enden %%%% Koeffizienten a_n und b_n mittels skalarprodukt. %%%%%%%%%%%%%%%%%%%%%%%%%% +Bisher wurde über die Koeffizienten $A$ und $B$ noch nicht viel ausgesagt. +Zunächst ist wegen vorhergehender Rechnung ersichtlich, dass es sich bei +$A$ und $B$ nicht um einzelne Koeffizienten handelt. +Stattdessen können die Koeffizienten für jedes $n \in \mathbb{N}$ +unterschiedlich sein. +Schreiben wir also die Lösung $X(x)$ um zu +\[ + X(x) + = + a_n\sin\left(\frac{n\pi}{l}x\right) + + + b_n\cos\left(\frac{n\pi}{l}x\right) +\] +was für jedes $n$ wiederum eine Linearkombination aus orthogonalen Funktionen +ist. Betrachten wir zuletzt die zweite Gleichung der Separation \eqref{eq:slp-example-fourier-separated-t}. -- cgit v1.2.1 From e27b521c00cdde53f0cbc0f0051881b5242adadc Mon Sep 17 00:00:00 2001 From: haddoucher Date: Thu, 11 Aug 2022 18:47:14 +0200 Subject: Beispiel & einleitung beispiel angefangen und einleitung korrigiert --- buch/papers/sturmliouville/einleitung.tex | 6 ++- .../sturmliouville/tschebyscheff_beispiel.tex | 51 +++++++++++++++++++++- 2 files changed, 54 insertions(+), 3 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/einleitung.tex b/buch/papers/sturmliouville/einleitung.tex index 44c3192..78c1800 100644 --- a/buch/papers/sturmliouville/einleitung.tex +++ b/buch/papers/sturmliouville/einleitung.tex @@ -10,7 +10,7 @@ Gemeinsam haben sie in der mathematischen Physik die Sturm-Liouville-Theorie ent Normalerweise betrachtet man für das Strum-Liouville-Problem eine gewöhnliche Differentialgleichung 2. Ordnung, und wenn es sich um eine partielle Differentialgleichung handelt, kann man sie mit Hilfe einiger Methoden in mehrere gewöhnliche Differentialgleichungen umwandeln, wie z. B. den Separationsansatz, die partielle Differentialgleichung mit mehreren Variablen. \begin{definition} - \index{Sturm-Liouville-Gleichung} + \index{Sturm-Liouville-Gleichung}% Angenommen man hat die lineare homogene Differentialgleichung \begin{equation} \frac{d^2y}{dx^2} + a(x)\frac{dy}{dx} + b(x)y = 0 @@ -20,7 +20,7 @@ und schreibt die Gleichung um in: \label{eq:sturm-liouville-equation} \frac{d}{dx}\lbrack p(x) \frac{dy}{dx} \rbrack + \lbrack q(x) + \lambda w(x) \rbrack y = 0 \end{equation} -, diese Gleichung wird dann Sturm-liouville-Gleichung bezeichnet. +, diese Gleichung wird dann Sturm-Liouville-Gleichung bezeichnet. \end{definition} Alle homogene 2.Ordnung lineare gewöhnliche Differentialgleichungen können in die Form der Gleichung \ref{eq:sturm-liouville-equation} umgeformt werden. @@ -71,6 +71,7 @@ Die Funktionen für das reguläre und das singuläre Sturm-Liouville-Problem sin \subsection{Das reguläre Sturm-Liouville-Problem\label{sub:reguläre_sturm_liouville_problem}} Damit es sich um ein reguläres Sturm-Liouville-Problem handelt, müssen einige Bedingungen beachtet werden. \begin{definition} + \label{def:reguläres_sturm-liouville-problem} \index{regläres Sturm-Liouville-Problem} Die Bedingungen für ein reguläres Sturm-Liouville-Problem sind: \begin{itemize} @@ -91,6 +92,7 @@ Bei einem regulären Sturm-Liouville-Problem geht es darum, ohne genaue Kenntnis \subsection{Das singuläre Sturm-Liouville-Problem\label{sub:singuläre_sturm_liouville_problem}} Von einem singulären Sturm-Liouville-Problem spricht man, wenn die Bedingungen des regulärem Problem nicht erfüllt sind. \begin{definition} + \label{def:singulär_sturm-liouville-problem} \index{singuläres Sturm-Liouville-Problem} Es handelt sich um ein singuläres Sturm-Liouville-Problem, wenn: \begin{itemize} diff --git a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex index 54f13d4..391841a 100644 --- a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex +++ b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex @@ -4,4 +4,53 @@ % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\subsection{Tschebyscheff} \ No newline at end of file +\subsection{Tschebyscheff-Polynome\label{sub:tschebyscheff-polynome}} +Im Kapitel \ref{sub:beispiele_sturm_liouville_problem} sind die Koeffizientenfunktionen die man braucht schon aufgeliste, und zwar mit +\begin{align*} + w(x) &= \frac{1}{\sqrt{1-x^2}} \\ + p(x) &= \sqrt{1-x^2} \\ + q(x) &= 0 +\end{align*}. +Da die Sturm-Liouville-Gleichung +\begin{equation} + \label{eq:sturm-liouville-equation} + \frac{d}{dx}\lbrack \sqrt{1-x^2} \frac{dy}{dx} \rbrack + \lbrack 0 + \lambda \frac{1}{\sqrt{1-x^2}} \rbrack y = 0 +\end{equation} +nun mit den Koeffizientenfunktionen aufgestellt werden kann, bleibt die Frage, ob es sich um ein reguläres oder singuläres Sturm-Liouville-Problem handelt. +Für das reguläre Problem laut der Definition \ref{def:reguläres_sturm-liouville-problem} muss die funktion $p(x) = \sqrt{1-x^2}$, $p'(x) = -2x$, $q(x) = 0$ und $w(x) = \frac{1}{\sqrt{1-x^2}}$ stetig und reell sein - und sie sind es auch. +Auf dem Intervall $(-1,1)$ sind die Tschebyscheff-Polynome erster Art mit Hilfe von Hyperbelfunktionen +\begin{equation} + T_n(x) = \cos n (\arccos x) +\end{equation}. +Für $x>1$ und $x<-1$ sehen die Polynome wie folgt aus: +\begin{equation} + T_n(x) = \left\{\begin{array}{ll} \cosh (n \arccos x), & x > 1\\ + (-1)^n \cosh (n \arccos (-x)), & x<-1 \end{array}\right. +\end{equation}, +jedoch ist die Orthogonalität nur auf dem Intervall $\[ -1, 1\]$ sichergestellt. +Die nächste Bedingung beinhaltet, dass die Funktion $p(x)^-1$ und $w(x)>0$ sein müssen. +Die Funktion +\begin{equation*} + p(x)^-1 = \frac{1}{\sqrt{1-x^2}} +\end{equation*} +ist die gleiche wie $w(x)$. + +Für die Verifizierung der Randbedingungen benötigt man erneut $p(x)$. +Da sich die Polynome nur auf dem Intervall $\[ -1,1 \]$ orthogonal verhalten, sind $a = -1$ und $b = 1$ gesetzt. +Beim einsetzen in die Randbedingung \ref{eq:randbedingungen}, erhält man +\begin{equation} +\begin{aligned} + k_a y(-1) + h_a y'(-1) &= h_a +\end{aligned} +\end{equation} + + + + + + + + + + + -- cgit v1.2.1 From c2d2d48156ab7cfb0d69541e58f54c3a55b2daf9 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Thu, 11 Aug 2022 19:23:32 +0200 Subject: Added start to coefficient calculation. --- .../sturmliouville/waermeleitung_beispiel.tex | 75 ++++++++++++++++++++-- 1 file changed, 71 insertions(+), 4 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 14fca40..58569e9 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -346,12 +346,79 @@ Schreiben wir also die Lösung $X(x)$ um zu \[ X(x) = - a_n\sin\left(\frac{n\pi}{l}x\right) + a_0 + - b_n\cos\left(\frac{n\pi}{l}x\right) + \sum_{n = 1}^{\infty} a_n\sin\left(\frac{n\pi}{l}x\right) + + + \sum_{n = 1}^{\infty} b_n\cos\left(\frac{n\pi}{l}x\right). +\] + +Um eine eindeutige Lösung für $ X(x) $ zu erhalten werden noch weitere +Bedingungen benötigt. +Diese sind die Startbedingungen oder $u(0, x) = X(x)$ für $t = 0$. +Es gilt also nun die Gleichung +\begin{equation} + \label{eq:slp-example-fourier-initial-conditions} + u(0, x) + = + a_0 + + + \sum_{n = 1}^{\infty} a_n\sin\left(\frac{n\pi}{l}x\right) + + + \sum_{n = 1}^{\infty} b_n\cos\left(\frac{n\pi}{l}x\right) +\end{equation} +nach allen $a_n$ und $b_n$ aufzulösen. +Da aber $a_n$ und $b_n$ jeweils als Faktor zu einer trigonometrischen Funktion +gehört, von der wir wissen, dass sie orthogonal zu allen anderen +trigonometrischen Funktionen der Lösung ist, kann direkt das Skalarprodukt +verwendet werden um die Koeffizienten $a_n$ und $b_n$ zu bestimmen. +Es wird also die Tatsache ausgenutzt, dass die Gleichheit in +\eqref{eq:slp-example-fourier-initial-conditions} nach Anwendung des +Skalarproduktes immernoch gelten muss und dass das Skalaprodukt mit einer +Basisfunktion sämtliche Summanden auf der rechten Seite auslöscht. + +Zur Berechnung von $a_m$ mit $ m \in \mathbb{N} $ wird beidseitig das +Skalarprodukt mit der Basisfunktion $ sin\left(\frac{m \pi}{l}x\right)$ +gebildet: +\[ + \langle u(0, x), sin\left(\frac{m \pi}{l}x\right) \rangle + = + \langle a_0 + + + \sum_{n = 1}^{\infty} a_n\sin\left(\frac{n\pi}{l}x\right) + + + \sum_{n = 1}^{\infty} b_n\cos\left(\frac{n\pi}{l}x\right), + sin\left(\frac{m \pi}{l}x\right)\rangle +\] + +Bevor diese Form in die Integralform umgeschrieben werden kann, muss überlegt +sein, welche Integralgrenzen zu verwenden sind. +In diesem Fall haben die $ \sin $ und $ \cos $ Terme beispielsweise keine ganze +Periode im Intervall $ x \in [0, l] $ für ungerade $ n $ und $ m $. +Um die + +\[ +\begin{aligned} + \int_{-l}^{l}\hat{u}(0, x)sin\left(\frac{m \pi}{l}x\right)dx + =& + \int_{-l}^{l} \left[a_0 + + + \sum_{n = 1}^{\infty} a_n\sin\left(\frac{n\pi}{l}x\right) + + + \sum_{n = 1}^{\infty} b_n\cos\left(\frac{n\pi}{l}x\right)\right] + sin\left(\frac{m \pi}{l}x\right) dx + \\ + =& + a_0 \int_{-l}^{l}sin\left(\frac{m \pi}{l}x\right) dx + + + \sum_{n = 1}^{\infty}\left[a_n\int_{-l}^{l}\sin\left(\frac{n\pi}{l}x\right) + sin\left(\frac{m \pi}{l}x\right)dx\right] + \\ + &+ + \sum_{n = 1}^{\infty}\left[b_n\int_{-l}^{l} \cos\left(\frac{n\pi}{l}x\right) + sin\left(\frac{m \pi}{l}x\right)dx\right] +\end{aligned} \] -was für jedes $n$ wiederum eine Linearkombination aus orthogonalen Funktionen -ist. Betrachten wir zuletzt die zweite Gleichung der Separation \eqref{eq:slp-example-fourier-separated-t}. -- cgit v1.2.1 From fc17a8247db60871ce49b23f1bbbb9b5523d8473 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Thu, 11 Aug 2022 20:05:16 +0200 Subject: Corrected Coefficient names. --- .../sturmliouville/waermeleitung_beispiel.tex | 104 ++++++++++----------- 1 file changed, 52 insertions(+), 52 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 58569e9..fb5f331 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -181,7 +181,7 @@ Die Lösungen für $X(x)$ sind also von der Form \[ X(x) = - A \sin \left( \alpha x\right) + B \cos \left( \beta x\right). + A \cos \left( \alpha x\right) + B \sin \left( \beta x\right). \] Dieser Ansatz wird nun solange differenziert, bis alle in Gleichung @@ -191,41 +191,41 @@ Man erhält also \[ X^{\prime}(x) = - \alpha A \cos \left( \alpha x \right) - - \beta B \sin \left( \beta x \right) + - \alpha A \sin \left( \alpha x \right) + + \beta B \cos \left( \beta x \right) \] und \[ X^{\prime \prime}(x) = - -\alpha^{2} A \sin \left( \alpha x \right) - - \beta^{2} B \cos \left( \beta x \right). + -\alpha^{2} A \cos \left( \alpha x \right) - + \beta^{2} B \sin \left( \beta x \right). \] Eingesetzt in Gleichung \eqref{eq:slp-example-fourier-separated-x} ergibt dies \[ - -\alpha^{2}A\sin(\alpha x) - \beta^{2}B\cos(\beta x) - - \mu\left(A\sin(\alpha x) + B\cos(\beta x)\right) + -\alpha^{2}A\cos(\alpha x) - \beta^{2}B\sin(\beta x) - + \mu\left(A\cos(\alpha x) + B\sin(\beta x)\right) = 0 \] und durch umformen somit \[ - -\alpha^{2}A\sin(\alpha x) - \beta^{2}B\cos(\beta x) + -\alpha^{2}A\cos(\alpha x) - \beta^{2}B\sin(\beta x) = - \mu A\sin(\alpha x) + \mu B\cos(\beta x). + \mu A\cos(\alpha x) + \mu B\sin(\beta x). \] Mittels Koeffizientenvergleich von \[ \begin{aligned} - -\alpha^{2}A\sin(\alpha x) + -\alpha^{2}A\cos(\alpha x) &= - \mu A\sin(\alpha x) + \mu A\cos(\alpha x) \\ - -\beta^{2}B\cos(\beta x) + -\beta^{2}B\sin(\beta x) &= - \mu B\cos(\beta x) + \mu B\sin(\beta x) \end{aligned} \] ist schnell ersichtlich, dass $ \mu = -\alpha^{2} = -\beta^{2} $ gelten muss für @@ -251,41 +251,41 @@ Dies fürht zu \[ X(0) = - A \sin(0 \alpha) + B \cos(0 \beta) + A \cos(0 \alpha) + B \sin(0 \beta) = 0. \] -Da $\cos(0) \neq 0$ ist, muss in diesem Fall $B = 0$ gelten. -Für den ersten Summanden ist wegen $\sin(0) = 0$ die Randbedingung erfüllt. +Da $\cos(0) \neq 0$ ist, muss in diesem Fall $A = 0$ gelten. +Für den zweiten Summanden ist wegen $\sin(0) = 0$ die Randbedingung erfüllt. -Wird nun die zweite Randbedingung für $x = l$ mit $B = 0$ eingesetzt, ergibt +Wird nun die zweite Randbedingung für $x = l$ mit $A = 0$ eingesetzt, ergibt sich \[ X(l) = - A \sin(\alpha l) + 0 \cos(\beta l) + 0 \cos(\alpha l) + B \sin(\beta l) = - A \sin(\alpha l) + B \sin(\beta l) = 0. \] -$\alpha$ muss also so gewählt werden, dass $\sin(\alpha l) = 0$ gilt. -Es bleibt noch nach $\alpha$ aufzulösen: +$\beta$ muss also so gewählt werden, dass $\sin(\beta l) = 0$ gilt. +Es bleibt noch nach $\beta$ aufzulösen: \[ \begin{aligned} - \sin(\alpha l) &= 0 \\ - \alpha l &= n \pi \qquad n \in \mathbb{N} \\ - \alpha &= \frac{n \pi}{l} \qquad n \in \mathbb{N} + \sin(\beta l) &= 0 \\ + \beta l &= n \pi \qquad n \in \mathbb{N} \\ + \beta &= \frac{n \pi}{l} \qquad n \in \mathbb{N} \end{aligned} \] -Es folgt nun wegen $\mu = -\alpha^{2}$, dass +Es folgt nun wegen $\mu = -\beta^{2}$, dass \begin{equation} - \mu_1 = -\alpha^{2} = -\frac{n^{2}\pi^{2}}{l^{2}} + \mu_1 = -\beta^{2} = -\frac{n^{2}\pi^{2}}{l^{2}} \end{equation} sein muss. -Ausserdem ist zu bemerken, dass dies auch gleich $-\beta^{2}$ ist. -Da aber $B = 0$ gilt und der Summand mit $\beta$ verschwindet, ist dies keine +Ausserdem ist zu bemerken, dass dies auch gleich $-\alpha^{2}$ ist. +Da aber $A = 0$ gilt und der Summand mit $\alpha$ verschwindet, ist dies keine Verletzung der Randbedingungen. Durch alanoges Vorgehen kann nun auch das Problem mit isolierten Enden gelöst @@ -296,18 +296,18 @@ ein, beginnend für $x = 0$. Es ergibt sich \[ X^{\prime}(0) = - \alpha A \cos(0 \alpha) - \beta B \sin(0 \beta) + -\alpha A \sin(0 \alpha) + \beta B \cos(0 \beta) = 0. \] -In diesem Fall muss $A = 0$ gelten. +In diesem Fall muss $B = 0$ gelten. Zusammen mit der Bedignung für $x = l$ folgt nun \[ X^{\prime}(l) = - 0 \alpha \cos(\alpha l) - \beta B \sin(\beta l) + - \alpha A \sin(\alpha l) + 0 \beta \cos(\beta l) = - -\beta B \sin(\beta l) + - \alpha A \sin(\alpha l) = 0. \] @@ -316,14 +316,14 @@ Ausdruck den Randbedingungen entspricht. Es folgt nun \[ \begin{aligned} - \sin(\beta l) &= 0 \\ - \beta l &= n \pi \qquad n \in \mathbb{N} \\ - \beta &= \frac{n \pi}{l} \qquad n \in \mathbb{N} + \sin(\alpha l) &= 0 \\ + \alpha l &= n \pi \qquad n \in \mathbb{N} \\ + \alpha &= \frac{n \pi}{l} \qquad n \in \mathbb{N} \end{aligned} \] und somit \[ - \mu_2 = -\beta^{2} = -\frac{n^{2}\pi^{2}}{l^{2}}. + \mu_2 = -\alpha^{2} = -\frac{n^{2}\pi^{2}}{l^{2}}. \] Es ergibt sich also sowohl für einen Stab mit Enden auf konstanter Temperatur @@ -348,9 +348,9 @@ Schreiben wir also die Lösung $X(x)$ um zu = a_0 + - \sum_{n = 1}^{\infty} a_n\sin\left(\frac{n\pi}{l}x\right) + \sum_{n = 1}^{\infty} a_n\cos\left(\frac{n\pi}{l}x\right) + - \sum_{n = 1}^{\infty} b_n\cos\left(\frac{n\pi}{l}x\right). + \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right). \] Um eine eindeutige Lösung für $ X(x) $ zu erhalten werden noch weitere @@ -363,9 +363,9 @@ Es gilt also nun die Gleichung = a_0 + - \sum_{n = 1}^{\infty} a_n\sin\left(\frac{n\pi}{l}x\right) + \sum_{n = 1}^{\infty} a_n\cos\left(\frac{n\pi}{l}x\right) + - \sum_{n = 1}^{\infty} b_n\cos\left(\frac{n\pi}{l}x\right) + \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right) \end{equation} nach allen $a_n$ und $b_n$ aufzulösen. Da aber $a_n$ und $b_n$ jeweils als Faktor zu einer trigonometrischen Funktion @@ -378,17 +378,17 @@ Skalarproduktes immernoch gelten muss und dass das Skalaprodukt mit einer Basisfunktion sämtliche Summanden auf der rechten Seite auslöscht. Zur Berechnung von $a_m$ mit $ m \in \mathbb{N} $ wird beidseitig das -Skalarprodukt mit der Basisfunktion $ sin\left(\frac{m \pi}{l}x\right)$ +Skalarprodukt mit der Basisfunktion $ \cos\left(\frac{m \pi}{l}x\right)$ gebildet: \[ - \langle u(0, x), sin\left(\frac{m \pi}{l}x\right) \rangle + \langle u(0, x), \cos\left(\frac{m \pi}{l}x\right) \rangle = \langle a_0 + \sum_{n = 1}^{\infty} a_n\sin\left(\frac{n\pi}{l}x\right) + \sum_{n = 1}^{\infty} b_n\cos\left(\frac{n\pi}{l}x\right), - sin\left(\frac{m \pi}{l}x\right)\rangle + \cos\left(\frac{m \pi}{l}x\right)\rangle \] Bevor diese Form in die Integralform umgeschrieben werden kann, muss überlegt @@ -399,24 +399,24 @@ Um die \[ \begin{aligned} - \int_{-l}^{l}\hat{u}(0, x)sin\left(\frac{m \pi}{l}x\right)dx + \int_{-l}^{l}\hat{u}(0, x)\cos\left(\frac{m \pi}{l}x\right)dx =& \int_{-l}^{l} \left[a_0 + - \sum_{n = 1}^{\infty} a_n\sin\left(\frac{n\pi}{l}x\right) + \sum_{n = 1}^{\infty} a_n\cos\left(\frac{n\pi}{l}x\right) + - \sum_{n = 1}^{\infty} b_n\cos\left(\frac{n\pi}{l}x\right)\right] - sin\left(\frac{m \pi}{l}x\right) dx + \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right)\right] + \cos\left(\frac{m \pi}{l}x\right) dx \\ =& - a_0 \int_{-l}^{l}sin\left(\frac{m \pi}{l}x\right) dx + a_0 \int_{-l}^{l}cos\left(\frac{m \pi}{l}x\right) dx + - \sum_{n = 1}^{\infty}\left[a_n\int_{-l}^{l}\sin\left(\frac{n\pi}{l}x\right) - sin\left(\frac{m \pi}{l}x\right)dx\right] + \sum_{n = 1}^{\infty}\left[a_n\int_{-l}^{l}\cos\left(\frac{n\pi}{l}x\right) + \cos\left(\frac{m \pi}{l}x\right)dx\right] \\ &+ - \sum_{n = 1}^{\infty}\left[b_n\int_{-l}^{l} \cos\left(\frac{n\pi}{l}x\right) - sin\left(\frac{m \pi}{l}x\right)dx\right] + \sum_{n = 1}^{\infty}\left[b_n\int_{-l}^{l} \sin\left(\frac{n\pi}{l}x\right) + \cos\left(\frac{m \pi}{l}x\right)dx\right] \end{aligned} \] -- cgit v1.2.1 From 37861bde4183d5134147df65dc06236d6878b36b Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Thu, 11 Aug 2022 21:19:44 +0200 Subject: Added periodically continued function u-hat. --- .../sturmliouville/waermeleitung_beispiel.tex | 23 +++++++++++++++++----- 1 file changed, 18 insertions(+), 5 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index fb5f331..fa96eff 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -385,9 +385,9 @@ gebildet: = \langle a_0 + - \sum_{n = 1}^{\infty} a_n\sin\left(\frac{n\pi}{l}x\right) + \sum_{n = 1}^{\infty} a_n\cos\left(\frac{n\pi}{l}x\right) + - \sum_{n = 1}^{\infty} b_n\cos\left(\frac{n\pi}{l}x\right), + \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right), \cos\left(\frac{m \pi}{l}x\right)\rangle \] @@ -395,8 +395,21 @@ Bevor diese Form in die Integralform umgeschrieben werden kann, muss überlegt sein, welche Integralgrenzen zu verwenden sind. In diesem Fall haben die $ \sin $ und $ \cos $ Terme beispielsweise keine ganze Periode im Intervall $ x \in [0, l] $ für ungerade $ n $ und $ m $. -Um die - +Um die skalarprodukte aber korrekt zu berechnen, muss über die ganze Periode +integriert werden. +Dazu werden die Integralgrenzen $-l$ und $l$ verwendet und es wird ausserdem +eine neue Funktion +\[ + \hat{u}(0, x) + = + \begin{cases} + u(0, x + l) & -l \leq x < 0 + \\ + u(0, x) & 0 \leq x \leq l + \end{cases} +\] +angenomen, welche $u(0, x)$ auf dem Intervall $[-l, l]$ periodisch fortsetzt. +Es kann nun das Skalarodukt geschrieben werden als \[ \begin{aligned} \int_{-l}^{l}\hat{u}(0, x)\cos\left(\frac{m \pi}{l}x\right)dx @@ -416,7 +429,7 @@ Um die \\ &+ \sum_{n = 1}^{\infty}\left[b_n\int_{-l}^{l} \sin\left(\frac{n\pi}{l}x\right) - \cos\left(\frac{m \pi}{l}x\right)dx\right] + \cos\left(\frac{m \pi}{l}x\right)dx\right]. \end{aligned} \] -- cgit v1.2.1 From 6887191ba574292b6a9009867c0e16e66831ca17 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Thu, 11 Aug 2022 22:01:25 +0200 Subject: Added titles to specific solutions. --- buch/papers/sturmliouville/waermeleitung_beispiel.tex | 13 ++++++------- 1 file changed, 6 insertions(+), 7 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index fa96eff..1bfdaef 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -409,7 +409,7 @@ eine neue Funktion \end{cases} \] angenomen, welche $u(0, x)$ auf dem Intervall $[-l, l]$ periodisch fortsetzt. -Es kann nun das Skalarodukt geschrieben werden als +Das Skalarodukt kann nun geschrieben werden als \[ \begin{aligned} \int_{-l}^{l}\hat{u}(0, x)\cos\left(\frac{m \pi}{l}x\right)dx @@ -428,7 +428,7 @@ Es kann nun das Skalarodukt geschrieben werden als \cos\left(\frac{m \pi}{l}x\right)dx\right] \\ &+ - \sum_{n = 1}^{\infty}\left[b_n\int_{-l}^{l} \sin\left(\frac{n\pi}{l}x\right) + \sum_{n = 1}^{\infty}\left[b_n\int_{-l}^{l}\sin\left(\frac{n\pi}{l}x\right) \cos\left(\frac{m \pi}{l}x\right)dx\right]. \end{aligned} \] @@ -457,22 +457,21 @@ Und mit dem Resultat \eqref{eq:slp-example-fourier-mu-solution} \] ergibt. -% TODO: Rechenweg -TODO: Rechenweg... Enden auf konstanter Temperatur: +\subsubsection{Lösung für einen Stab mit Enden auf konstanter Temperatur} \[ \begin{aligned} u(t,x) &= - \sum_{n=1}^{\infty}a_{n}e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t} + \sum_{n=1}^{\infty}b_{n}e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t} \sin\left(\frac{n\pi}{l}x\right) \\ - a_{n} + b_{n} &= \frac{2}{l}\int_{0}^{l}u(0,x)sin\left(\frac{n\pi}{l}x\right) dx \end{aligned} \] -TODO: Rechenweg... Enden isoliert: +\subsubsection{Lösung für einen Stab mit isolierten Enden} \[ \begin{aligned} u(t,x) -- cgit v1.2.1 From 964db187eaf5512601a04c6326094d6a1975d941 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Thu, 11 Aug 2022 22:11:59 +0200 Subject: Rewrote everything in passive form. --- .../sturmliouville/waermeleitung_beispiel.tex | 25 ++++++++++++---------- 1 file changed, 14 insertions(+), 11 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 1bfdaef..868f241 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -6,8 +6,8 @@ \subsection{Wärmeleitung in einem Homogenen Stab} -In diesem Abschnitt betrachten wir das Problem der Wärmeleitung in einem -homogenen Stab und wie das Sturm-Liouville-Problem bei der Beschreibung dieses +In diesem Abschnitt wird das Problem der Wärmeleitung in einem homogenen Stab +betrachtet und wie das Sturm-Liouville-Problem bei der Beschreibung dieses physikalischen Phänomenes auftritt. Zunächst wird ein eindimensionaler homogener Stab der Länge $l$ und @@ -141,8 +141,9 @@ erfüllt sein und es muss ausserdem \end{equation} gelten. -Um zu verifizieren, ob die Randbedingungen erfüllt sind, benötigen wir zunächst -$p(x)$. +Um zu verifizieren, ob die Randbedingungen erfüllt sind, wird zunächst +$p(x)$ +benötigt. Dazu wird die Gleichung \eqref{eq:slp-example-fourier-separated-x} mit der Sturm-Liouville-Form \eqref{eq:sturm-liouville-equation} verglichen, was zu $p(x) = 1$ führt. @@ -169,7 +170,7 @@ Analog dazu kann gezeit werden, dass die Randbedingungen für einen Stab mit isolierten Enden ebenfalls die Sturm-Liouville-Randbedingungen erfüllen und somit auch zu orthogonalen Lösungen führen. -Widmen wir uns zunächst der ersten Gleichung. +Als erstes wird auf die erste erste Gleichung eingegangen. Aufgrund der Struktur der Gleichung \[ X^{\prime \prime}(x) - \mu X(x) @@ -290,7 +291,7 @@ Verletzung der Randbedingungen. Durch alanoges Vorgehen kann nun auch das Problem mit isolierten Enden gelöst werden. -Setzen wir nun die Randbedingungen +Setzt man nun die Randbedingungen \eqref{eq:slp-example-fourier-boundary-condition-ends-isolated} in $X^{\prime}$ ein, beginnend für $x = 0$. Es ergibt sich \[ @@ -342,7 +343,7 @@ Zunächst ist wegen vorhergehender Rechnung ersichtlich, dass es sich bei $A$ und $B$ nicht um einzelne Koeffizienten handelt. Stattdessen können die Koeffizienten für jedes $n \in \mathbb{N}$ unterschiedlich sein. -Schreiben wir also die Lösung $X(x)$ um zu +Die Lösung $X(x)$ wird nun umgeschrieben zu \[ X(x) = @@ -433,14 +434,16 @@ Das Skalarodukt kann nun geschrieben werden als \end{aligned} \] -Betrachten wir zuletzt die zweite Gleichung der Separation -\eqref{eq:slp-example-fourier-separated-t}. -Diese Lösen wir über das charakteristische Polynom +Zuletzt wird die zweite Gleichung der Separation +\eqref{eq:slp-example-fourier-separated-t} betrachtet. +Diese wird über das charakteristische Polynom \[ \lambda - \kappa \mu = - 0. + 0 \] +gelöst. + Es ist direkt ersichtlich, dass $\lambda = \kappa \mu$ gelten muss, was zur Lösung \[ -- cgit v1.2.1 From ff04ad95214c0ecdf8343fa8cd0aaa74dda45715 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Fri, 12 Aug 2022 14:22:03 +0200 Subject: Corrected error with continuation of u hat. --- .../sturmliouville/waermeleitung_beispiel.tex | 52 +++++++++++++++++----- 1 file changed, 40 insertions(+), 12 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 868f241..cfa7386 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -381,7 +381,8 @@ Basisfunktion sämtliche Summanden auf der rechten Seite auslöscht. Zur Berechnung von $a_m$ mit $ m \in \mathbb{N} $ wird beidseitig das Skalarprodukt mit der Basisfunktion $ \cos\left(\frac{m \pi}{l}x\right)$ gebildet: -\[ +\begin{equation} + \label{eq:slp-dot-product-cosine} \langle u(0, x), \cos\left(\frac{m \pi}{l}x\right) \rangle = \langle a_0 @@ -390,30 +391,56 @@ gebildet: + \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right), \cos\left(\frac{m \pi}{l}x\right)\rangle -\] +\end{equation} Bevor diese Form in die Integralform umgeschrieben werden kann, muss überlegt sein, welche Integralgrenzen zu verwenden sind. In diesem Fall haben die $ \sin $ und $ \cos $ Terme beispielsweise keine ganze Periode im Intervall $ x \in [0, l] $ für ungerade $ n $ und $ m $. -Um die skalarprodukte aber korrekt zu berechnen, muss über die ganze Periode -integriert werden. -Dazu werden die Integralgrenzen $-l$ und $l$ verwendet und es wird ausserdem -eine neue Funktion +Um die Skalarprodukte aber korrekt zu berechnen, muss über ein ganzzahliges +Vielfaches der Periode der triginimetrischen Funktionen integriert werden. +Dazu werden die Integralgrenzen $-l$ und $l$ verwendet und es werden ausserdem +neue Funktionen $ \hat{u}_c(0, x) $ für die Berechnung mit Cosinus und +$ \hat{u}_s(0, x) $ für die Berechnung mit Sinus angenomen, welche $ u(0, t) $ +gerade, respektive ungerade auf $[-l, l]$ fortsetzen: \[ - \hat{u}(0, x) - = +\begin{aligned} + \hat{u}_c(0, x) + &= \begin{cases} - u(0, x + l) & -l \leq x < 0 + u(0, -x) & -l \leq x < 0 \\ u(0, x) & 0 \leq x \leq l \end{cases} + \\ + \hat{u}_s(0, x) + &= + \begin{cases} + -u(0, -x) & -l \leq x < 0 + \\ + u(0, x) & 0 \leq x \leq l + \end{cases}. +\end{aligned} \] -angenomen, welche $u(0, x)$ auf dem Intervall $[-l, l]$ periodisch fortsetzt. -Das Skalarodukt kann nun geschrieben werden als + +Die Konsequenz davon ist, dass nun das Resultat der Integrale um den Faktor zwei +skalliert wurde, also gilt nun +\[ +\begin{aligned} + \int_{-l}^{l}\hat{u}_c(0, x)\cos\left(\frac{m \pi}{l}x\right)dx + &= + 2\int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx + \\ + \int_{-l}^{l}\hat{u}_s(0, x)\sin\left(\frac{m \pi}{l}x\right)dx + &= + 2\int_{0}^{l}u(0, x)\sin\left(\frac{m \pi}{l}x\right)dx. +\end{aligned} +\] + +Zunächst wird nun das Skalaprodukt \eqref{eq:slp-dot-product-cosine} berechnet: \[ \begin{aligned} - \int_{-l}^{l}\hat{u}(0, x)\cos\left(\frac{m \pi}{l}x\right)dx + \int_{-l}^{l}\hat{u}_c(0, x)\cos\left(\frac{m \pi}{l}x\right)dx =& \int_{-l}^{l} \left[a_0 + @@ -422,6 +449,7 @@ Das Skalarodukt kann nun geschrieben werden als \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right)\right] \cos\left(\frac{m \pi}{l}x\right) dx \\ + 2\int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx =& a_0 \int_{-l}^{l}cos\left(\frac{m \pi}{l}x\right) dx + -- cgit v1.2.1 From a961142ba09e0e9a962aaba4d90e1613e0ff97b0 Mon Sep 17 00:00:00 2001 From: Fabian <@> Date: Fri, 12 Aug 2022 15:06:08 +0200 Subject: 1. Ueberarbeitung --- buch/papers/0f1/teil1.tex | 22 +++++++++++----------- buch/papers/0f1/teil2.tex | 21 ++++++++++----------- buch/papers/0f1/teil3.tex | 18 +++++++++--------- 3 files changed, 30 insertions(+), 31 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/0f1/teil1.tex b/buch/papers/0f1/teil1.tex index 2ca9647..f697f45 100644 --- a/buch/papers/0f1/teil1.tex +++ b/buch/papers/0f1/teil1.tex @@ -6,12 +6,12 @@ \section{Mathematischer Hintergrund \label{0f1:section:mathHintergrund}} \rhead{Mathematischer Hintergrund} -Basierend auf den Herleitungen des vorhergehenden Kapitels \ref{buch:rekursion:section:hypergeometrische-funktion}, werden im nachfolgenden Abschnitt nochmals die Resultate +Basierend auf den Herleitungen des vorhergehenden Abschnittes \ref{buch:rekursion:section:hypergeometrische-funktion}, werden im nachfolgenden Abschnitt nochmals die Resultate beschrieben. \subsection{Hypergeometrische Funktion \label{0f1:subsection:hypergeometrisch}} -Als Grundlage der umgesetzten Algorithmen dient die Hypergeometrische Funktion $\mathstrut_0F_1$. Diese ist eine Unterfunktion der allgemein definierten Funktion $\mathstrut_pF_q$. +Als Grundlage der umgesetzten Algorithmen dient die hypergeometrische Funktion $\mathstrut_0F_1$. Diese ist eine Anwendung der allgemein definierten Funktion $\mathstrut_pF_q$. \begin{definition} \label{0f1:math:qFp:def} @@ -42,7 +42,7 @@ Angewendet auf die Funktion $\mathstrut_pF_q$ ergibt sich für $\mathstrut_0F_1$ \mathstrut_0F_1 \biggl( \begin{matrix} - \\ + \\- b_1 \end{matrix} ; @@ -60,22 +60,22 @@ Angewendet auf die Funktion $\mathstrut_pF_q$ ergibt sich für $\mathstrut_0F_1$ \subsection{Airy Funktion \label{0f1:subsection:airy}} -Die Airy-Funktion $Ai(x)$ und die verwandte Funktion $Bi(x)$ werden als Airy-Funktion bezeichnet. Sie werden zur Lösung verschiedener physikalischer Probleme benutzt, wie zum Beispiel zur Lösung der Schrödinger-Gleichung. \cite{0f1:wiki-airyFunktion} +Die Funktion Ai(x) und die verwandte Funktion Bi(x) werden als Airy-Funktion bezeichnet. Sie werden zur Lösung verschiedener physikalischer Probleme benutzt, wie zum Beispiel zur Lösung der Schrödinger-Gleichung \cite{0f1:wiki-airyFunktion}. \begin{definition} \label{0f1:airy:differentialgleichung:def} Die Differentialgleichung $y'' - xy = 0$ - heisst die {\em Airy-Differentialgleichung}. \cite{0f1:wiki-airyFunktion} + heisst die {\em Airy-Differentialgleichung}. \end{definition} -Die Airy Funktion lässt sich auf verschiedene Arten darstellen. \cite{0f1:wiki-airyFunktion} -Als hypergeometrische Funktion berechnet, ergibt sich wie in Kapitel \ref{buch:differentialgleichungen:section:hypergeometrisch} hergeleitet, folgende Lösungen der Airy-Differentialgleichung zu den Anfangsbedingungen $A(0)=1$ und $A'(0)=0$, sowie $B(0)=0$ und $B'(0)=0$. +Die Airy Funktion lässt sich auf verschiedene Arten darstellen. +Als hypergeometrische Funktion berechnet, ergibt sich wie in Abschnitt \ref{buch:differentialgleichungen:section:hypergeometrisch} hergeleitet, folgende Lösungen der Airy-Differentialgleichung zu den Anfangsbedingungen $Ai(0)=1$ und $Ai'(0)=0$, sowie $Bi(0)=0$ und $Bi'(0)=0$. \begin{align} \label{0f1:airy:hypergeometrisch:eq} Ai(x) -= +=& \sum_{k=0}^\infty \frac{1}{(\frac23)_k} \frac{1}{k!}\biggl(\frac{x^3}{9}\biggr)^k = @@ -84,7 +84,7 @@ Ai(x) \biggr). \\ Bi(x) -= +=& \sum_{k=0}^\infty \frac{1}{(\frac43)_k} \frac{1}{k!}\biggl(\frac{x^3}{9}\biggr)^k = @@ -95,7 +95,7 @@ x\cdot\mathstrut_0F_1\biggl( \qedhere \end{align} -In diesem speziellem Fall wird die Airy Funktion $Ai(x)$ \eqref{0f1:airy:hypergeometrisch:eq} -benutzt, um die Stabilität der Algorithmen zu $\mathstrut_0F_1$ zu überprüfen. +Um die Stabilität der Algorithmen zu $\mathstrut_0F_1$ zu überprüfen, wird in diesem speziellem Fall die Airy Funktion $Ai(x)$ \eqref{0f1:airy:hypergeometrisch:eq} +benutzt. diff --git a/buch/papers/0f1/teil2.tex b/buch/papers/0f1/teil2.tex index 9269961..15a1c44 100644 --- a/buch/papers/0f1/teil2.tex +++ b/buch/papers/0f1/teil2.tex @@ -6,12 +6,12 @@ \section{Umsetzung \label{0f1:section:teil2}} \rhead{Umsetzung} -Zur Umsetzung wurden drei verschiedene Ansätze gewählt.\cite{0f1:code} Dabei wurde der Schwerpunkt auf die Funktionalität und eine gute Lesbarkeit des Codes gelegt. -Die Unterprogramme wurde jeweils, wie die GNU Scientific Library, in C geschrieben. Die Zwischenresultate wurden vom Hauptprogramm in einem CSV-File gespeichert. Anschliessen wurde mit der Matplot-Libray in Python die Resultate geplottet. +Zur Umsetzung wurden drei verschiedene Ansätze gewählt \cite{0f1:code}. Dabei wurde der Schwerpunkt auf die Funktionalität und eine gute Lesbarkeit des Codes gelegt. +Die Unterprogramme wurde jeweils, wie die GNU Scientific Library, in C geschrieben. Die Zwischenresultate wurden vom Hauptprogramm in einem CSV-File gespeichert. Anschliessen wurde mit der Matplot-Library in Python die Resultate geplottet. \subsection{Potenzreihe \label{0f1:subsection:potenzreihe}} -Die naheliegendste Lösung ist die Programmierung der Potenzreihe. Allerdings ist ein Problem dieser Umsetzung \ref{0f1:listing:potenzreihe}, dass die Fakultät im Nenner schnell grosse Werte annimmt und so der Bruch gegen Null strebt. Spätesten ab $k=167$ stösst diese Umsetzung \eqref{0f1:umsetzung:0f1:eq} an ihre Grenzen, da die Fakultät von $168$ eine Bereichsüberschreitung des \textit{double} Bereiches darstellt. \cite{0f1:double} +Die naheliegendste Lösung ist die Programmierung der Potenzreihe. Allerdings ist ein Problem dieser Umsetzung \ref{0f1:listing:potenzreihe}, dass die Fakultät im Nenner schnell grosse Werte annimmt und so der Bruch gegen Null strebt. Spätesten ab $k=167$ stösst diese Umsetzung \eqref{0f1:umsetzung:0f1:eq} an ihre Grenzen, da die Fakultät von $168$ eine Bereichsüberschreitung des \textit{double} Bereiches darstellt \cite{0f1:double}. \begin{align} \label{0f1:umsetzung:0f1:eq} @@ -34,23 +34,22 @@ Ein endlicher Kettenbruch ist ein Bruch der Form \begin{equation*} a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cdots}}} \end{equation*} -in welchem $a_0, a_1,\dots,a_n$ und $b_1,b_2,\dots,b_n$ ganze Zahlen darstellen. +in welchem $a_0, a_1,\dots,a_n$ und $b_1,b_2,\dots,b_n$ ganze Zahlen sind. Die Kurzschreibweise für einen allgemeinen Kettenbruch ist \begin{equation*} a_0 + \frac{a_1|}{|b_1} + \frac{a_2|}{|b_2} + \frac{a_3|}{|b_3} + \cdots \end{equation*} -und ist somit verknüpfbar mit der Potenzreihe. -\cite{0f1:wiki-kettenbruch} -Angewendet auf die Funktion $\mathstrut_0F_1$ bedeutet dies\cite{0f1:wiki-fraction}: +\cite{0f1:wiki-kettenbruch}. +Angewendet auf die Funktion $\mathstrut_0F_1$ bedeutet dies \cite{0f1:wiki-fraction}: \begin{equation*} \mathstrut_0F_1(;c;z) = 1 + \frac{z}{c\cdot1!} + \frac{z^2}{c(c+1)\cdot2!} + \frac{z^3}{c(c+1)(c+2)\cdot3!} + \cdots \end{equation*} -Nach allen Umformungen ergibt sich folgender, irregulärer Kettenbruch \eqref{0f1:math:kettenbruch:0f1:eq} +Umgeformt ergibt sich folgender Kettenbruch \begin{equation} \label{0f1:math:kettenbruch:0f1:eq} \mathstrut_0F_1(;c;z) = 1 + \cfrac{\cfrac{z}{c}}{1+\cfrac{-\cfrac{z}{2(c+1)}}{1+\cfrac{z}{2(c+1)}+\cfrac{-\cfrac{z}{3(c+2)}}{1+\cfrac{z}{5(c+4)} + \cdots}}}, \end{equation} -der als Code \ref{0f1:listing:kettenbruchIterativ} umgesetzt wurde. +der als Code (siehe: Listing \ref{0f1:listing:kettenbruchIterativ}) umgesetzt wurde. \cite{0f1:wolfram-0f1} \lstinputlisting[style=C,float,caption={Iterativ umgesetzter Kettenbruch.},label={0f1:listing:kettenbruchIterativ}, firstline=8]{papers/0f1/listings/kettenbruchIterativ.c} @@ -138,7 +137,7 @@ Nach vollständiger Induktion ergibt sich für den Schritt $k$, die Matrix \end{equation} Und Schlussendlich kann der Näherungsbruch \[ -\frac{Ak}{Bk} +\frac{A_k}{B_k} \] berechnet werden. @@ -166,7 +165,7 @@ B_{k+1} &= B_{k-1} \cdot b_k + B_k \cdot a_k Näherungsbruch: \qquad$\displaystyle\frac{A_k}{B_k}$ \end{itemize} -Ein grosser Vorteil dieser Umsetzung \ref{0f1:listing:kettenbruchRekursion} ist, dass im Vergleich zum Code \ref{0f1:listing:kettenbruchIterativ} eine Division gespart werden kann und somit weniger Rundungsfehler entstehen können. +Ein grosser Vorteil dieser Umsetzung als Rekursionsformel ist \ref{0f1:listing:kettenbruchRekursion}, dass im Vergleich zum Code \ref{0f1:listing:kettenbruchIterativ} eine Division gespart werden kann und somit weniger Rundungsfehler entstehen können. %Code \lstinputlisting[style=C,float,caption={Rekursionsformel für Kettenbruch.},label={0f1:listing:kettenbruchRekursion}, firstline=8]{papers/0f1/listings/kettenbruchRekursion.c} \ No newline at end of file diff --git a/buch/papers/0f1/teil3.tex b/buch/papers/0f1/teil3.tex index 2855e26..72b1b21 100644 --- a/buch/papers/0f1/teil3.tex +++ b/buch/papers/0f1/teil3.tex @@ -6,19 +6,19 @@ \section{Auswertung \label{0f1:section:teil3}} \rhead{Resultate} -Im Verlauf des Seminares hat sich gezeigt, +Im Verlauf dieser Arbeit hat sich gezeigt, das ein einfacher mathematischer Algorithmus zu implementieren gar nicht so einfach ist. So haben alle drei umgesetzten Ansätze Probleme mit grossen negativen $z$ in der Funktion $\mathstrut_0F_1(;c;z)$. -Ebenso kann festgestellt werden,dass je grösser der Wert $z$ in $\mathstrut_0F_1(;c;z)$ wird, desto mehr weichen die berechneten Resultate von den Erwarteten ab. \cite{0f1:wolfram-0f1} +Ebenso kann festgestellt werden, dass je grösser der Wert $z$ in $\mathstrut_0F_1(;c;z)$ wird, desto mehr weichen die berechneten Resultate von den Erwarteten ab \cite{0f1:wolfram-0f1}. \subsection{Konvergenz \label{0f1:subsection:konvergenz}} Es zeigt sich in Abbildung \ref{0f1:ausblick:plot:airy:konvergenz}, dass schon nach drei Iterationen ($k = 3$) die Funktionen schon genaue Resultate im Bereich von -2 bis 2 liefert. Ebenso kann festgestellt werden, dass der Kettenbruch schneller konvergiert und im positiven Bereich sogar mit der Referenzfunktion $Ai(x)$ übereinstimmt. Da die Rekursionsformel \ref{0f1:listing:kettenbruchRekursion} eine Abwandlung des Kettenbruches ist, verhalten sich die Funktionen in diesem Fall gleich. -Erst wenn mehrere Durchläufe gemacht werden, um die Genauigkeit zu verbessern, ist der Kettenbruch den anderen zwei Algorithmen, bezüglich Konvergenz überlegen. -Interessant ist auch, dass die Rekursionsformel nahezu gleich schnell wie die Potenzreihe konvergiert, aber sich danach einschwingt. Dieses Verhalten ist auch bei grösseren $z$ zu beobachten, allerdings ist dann die Differenz zwischen dem ersten lokalen Minimum von k bis zum Abbruch kleiner. -\ref{0f1:ausblick:plot:konvergenz:positiv} -Dieses Phänomen ist auf die Lösung der Rekursionsformel zurück zu führen.\ref{0f1:math:loesung:eq} Da im Gegensatz die ganz kleinen Werte nicht zu einer Konvergenz wie beim Kettenbruch führen, sondern sich noch eine Zeit lang durch die Multiplikation aufschwingen. +Erst wenn mehrerer Iterationen gemacht werden, um die Genauigkeit zu verbessern, ist der Kettenbruch den anderen zwei Algorithmen, bezüglich Konvergenz überlegen. +Interessant ist auch, dass die Rekursionsformel nahezu gleich schnell wie die Potenzreihe konvergiert, aber sich danach einschwingt \ref{0f1:ausblick:plot:konvergenz:positiv}. Dieses Verhalten ist auch bei grösseren $z$ zu beobachten, allerdings ist dann die Differenz zwischen dem ersten lokalen Minimum von k bis zum Abbruch kleiner +\ref{0f1:ausblick:plot:konvergenz:positiv}. +Dieses Phänomen ist auf die Lösung der Rekursionsformel zurück zu führen\eqref{0f1:math:loesung:eq}. Da im Gegensatz die ganz kleinen Werte nicht zu einer Konvergenz wie beim Kettenbruch führen, sondern sich noch eine Zeit lang durch die Multiplikation aufschwingen. Ist $z$ negativ wie im Abbild \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu einer Gegenseitigen Kompensation von negativen und positiven Termen so bricht die Rekursionsformel hier zusammen mit der Potenzreihe ab. Die ansteigende Differenz mit anschliessender, ist aufgrund der sich alternierenden Termen mit wechselnden Vorzeichens zu erklären. @@ -27,10 +27,10 @@ Die ansteigende Differenz mit anschliessender, ist aufgrund der sich alternieren \label{0f1:subsection:Stabilitaet}} Verändert sich der Wert von z in $\mathstrut_0F_1(;c;z)$ gegen grössere positive Werte, wie zum Beispiel $c = 800$ liefert die Kettenbruch-Funktion \ref{0f1:listing:kettenbruchIterativ} \verb+inf+ zurück. Dies könnte durch ein Abbruchkriterien abgefangen werden. Allerdings würde das, bei grossen Werten zulasten der Genauigkeit gehen. Trotzdem könnte, je nach Anwendung, auf ein paar Nachkommastellen verzichtet werden. -Wohingegen die Potenzreihe \ref{0f1:listing:potenzreihe} das Problem hat, dass je mehr Terme berechnet werden, desto schneller wächst die Fakultät und irgendwann gibt es eine Bereichsüberschreitung von \verb+double+. Schlussendlich gibt das Unterprogramm das Resultat \verb+-nan(ind)+ zurück. -Die Rekursionformel \ref{0f1:listing:kettenbruchRekursion} liefert für sehr grosse positive Werte die genausten Ergebnisse, verglichen mit der GNU Scientific Library. Wie schon vermutet ist die Rekursionsformel, im positivem Bereich, der stabilste Algorithmus. Um die Stabilität zu gewährleisten, muss wie in \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, die Iterationstiefe $k$ genug gross gewählt werden. +Wohingegen die Potenzreihe \eqref{0f1:listing:potenzreihe} das Problem hat, dass je mehr Terme berechnet werden, desto schneller wächst die Fakultät und irgendwann gibt es eine Bereichsüberschreitung von \verb+double+. Schlussendlich gibt das Unterprogramm das Resultat \verb+-nan(ind)+ zurück. +Die Rekursionformel \eqref{0f1:listing:kettenbruchRekursion} liefert für sehr grosse positive Werte die genausten Ergebnisse, verglichen mit der GNU Scientific Library. Wie schon vermutet ist die Rekursionsformel, im positivem Bereich, der stabilste Algorithmus. Um die Stabilität zu gewährleisten, muss wie in Abbild \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, die Iterationstiefe $k$ genug gross gewählt werden. -Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Grund dafür ist die Fakultät im Nenner, was zum Phänomen der Auslöschung führt.\cite{0f1:SeminarNumerik} Schön zu beobachten ist dies in der Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} mit der Airy-Funktion als Test. So sind sowohl der Kettenbruch, als auch die Rekursionsformel bis ungefähr $\frac{-15^3}{9}$ stabil. Dies macht auch Sinn, da beide auf der gleichen mathematischen Grundlage basieren. Danach verhält sich allerdings die Instabilität unterschiedlich. Das unterschiedliche Verhalten kann damit erklärt werden, dass beim Kettenbruch jeweils eine zusätzliche Division stattfindet. Diese Unterschiede sind auch in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} festzustellen. +Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Grund dafür ist die Potenz von z, was zum Phänomen der Auslöschung führt \cite{0f1:SeminarNumerik}. Schön zu beobachten ist dies in der Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} mit der Airy-Funktion als Test. So sind sowohl der Kettenbruch, als auch die Rekursionsformel bis ungefähr $\frac{-15^3}{9}$ stabil. Dies macht auch Sinn, da beide auf der gleichen mathematischen Grundlage basieren. Danach verhält sich allerdings die Instabilität unterschiedlich. Das unterschiedliche Verhalten kann damit erklärt werden, dass beim Kettenbruch jeweils eine zusätzliche Division stattfindet. Diese Unterschiede sind auch in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} festzustellen. -- cgit v1.2.1 From d9c6ead18aae68a14ce72b893d9c671156a1d6b3 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Fri, 12 Aug 2022 18:03:55 +0200 Subject: Full calculation for a_m explained. --- .../sturmliouville/waermeleitung_beispiel.tex | 58 ++++++++++++++++++++++ 1 file changed, 58 insertions(+) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index cfa7386..5c246f2 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -462,6 +462,64 @@ Zunächst wird nun das Skalaprodukt \eqref{eq:slp-dot-product-cosine} berechnet: \end{aligned} \] +Betrachtet man nun die Summanden auf der rechten Seite stellt man fest, dass +nahezu alle Terme verschinden, denn +\[ + \int_{-l}^{l}cos\left(\frac{m \pi}{l}x\right) dx + = + 0 +\] +da hier über ein ganzzahliges Vielfaches der Periode integriert wird, +\[ + \int_{-l}^{l}\cos\left(\frac{n\pi}{l}x\right) + \cos\left(\frac{m \pi}{l}x\right)dx + = + 0 +\] +für $m\neq n$, da Cosinus-Funktionen mit verschiedenen Kreisfrequenzen +orthogonal zueinander stehen und +\[ + \int_{-l}^{l}\sin\left(\frac{n\pi}{l}x\right) + \cos\left(\frac{m \pi}{l}x\right)dx + = + 0 +\] +da Sinus- und Cosinus-Funktionen ebenfalls orthogonal zueinander sin. + +Es bleibt also lediglich der Summand für $a_m$ stehen, was die Gleichung zu +\[ + 2\int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx + = + a_m\int_{-l}^{l}\cos^2\left(\frac{m\pi}{l}x\right)dx +\] +vereinfacht. Im nächsten Schritt wird nun das Integral auf der rechten Seite +berechnet und dann nach $a_m$ aufgelöst. Am einnfachsten geht dies, wenn zuerst +mit $u = \frac{m \pi}{l}x$ substituiert wird: +\[ + \begin{aligned} + 2\int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx + &= + a_m\frac{l}{m\pi}\int_{-m\pi}^{m\pi}\cos^2\left(u\right)du + \\ + &= + a_m\frac{l}{m\pi}\left[\frac{u}{2} + + \frac{\sin\left(2u\right)}{4}\right]_{u=-m\pi}^{m\pi} + \\ + &= + a_m\frac{l}{m\pi}\left(\frac{m\pi}{2} + + \underbrace{\frac{\sin\left(2m\pi\right)}{4}}_{\displaystyle = 0} - + \frac{-m\pi}{2} - + \underbrace{\frac{\sin\left(-2m\pi\right)}{4}}_{\displaystyle = 0}\right) + \\ + &= + a_m l + \\ + a_m + &= + \frac{2}{l} \int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx + \end{aligned} +\] + Zuletzt wird die zweite Gleichung der Separation \eqref{eq:slp-example-fourier-separated-t} betrachtet. Diese wird über das charakteristische Polynom -- cgit v1.2.1 From b1f2ce6c7f7b277558e7fd18cedae9a0a06aefde Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Sat, 13 Aug 2022 12:33:04 +0200 Subject: Finished first draft of fourier example. --- .../sturmliouville/waermeleitung_beispiel.tex | 74 +++++++++++++++++++++- 1 file changed, 73 insertions(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 5c246f2..5bd5ce2 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -170,6 +170,7 @@ Analog dazu kann gezeit werden, dass die Randbedingungen für einen Stab mit isolierten Enden ebenfalls die Sturm-Liouville-Randbedingungen erfüllen und somit auch zu orthogonalen Lösungen führen. +\subsubsection{Lösund der Differentialgleichung in x} Als erstes wird auf die erste erste Gleichung eingegangen. Aufgrund der Struktur der Gleichung \[ @@ -463,7 +464,7 @@ Zunächst wird nun das Skalaprodukt \eqref{eq:slp-dot-product-cosine} berechnet: \] Betrachtet man nun die Summanden auf der rechten Seite stellt man fest, dass -nahezu alle Terme verschinden, denn +nahezu alle Terme verschwinden, denn \[ \int_{-l}^{l}cos\left(\frac{m \pi}{l}x\right) dx = @@ -520,6 +521,74 @@ mit $u = \frac{m \pi}{l}x$ substituiert wird: \end{aligned} \] +Analog dazu kann durch das Bilden des Skalarproduktes mit +$ \sin\left(\frac{m \pi}{l}x\right) $ gezeigt werden, dass +\[ + b_m + = + \frac{2}{l} \int_{0}^{l}u(0, x)\sin\left(\frac{m \pi}{l}x\right)dx +\] +gilt. + +Etwas anders ist es allerdings bei $a_0$. +Wie der Name bereits suggeriert, handelt es sich hierbei um den Koeffizienten +zur Basisfunktion $ \cos\left(\frac{0 \pi}{l}x\right) $ beziehungsweise der +konstanten Funktion $1$. +Um einen Ausdruck für $ a_0 $ zu erhalten, wird wiederum auf beiden Seiten +der Gleichung \eqref{eq:slp-example-fourier-initial-conditions} das +Skalarprodukt mit der konstanten Basisfunktion $ 1 $ gebildet: +\[ +\begin{aligned} + \int_{-l}^{l}\hat{u}_c(0, x)dx + &= + \int_{-l}^{l} a_0 + + + \sum_{n = 1}^{\infty} a_n\cos\left(\frac{n\pi}{l}x\right) + + + \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right)dx + \\ + 2\int_{0}^{l}u(0, x)dx + &= + a_0 \int_{-l}^{l}dx + + + \sum_{n = 1}^{\infty}\left[a_n\int_{-l}^{l}\cos\left(\frac{n\pi}{l}x\right) + dx\right] + + \sum_{n = 1}^{\infty}\left[b_n\int_{-l}^{l}\sin\left(\frac{n\pi}{l}x\right) + dx\right]. +\end{aligned} +\] + +Hier fallen nun alle Terme, die $\sin$ oder $\cos$ beinhalten weg, da jeweils +über ein Vielfaches der Periode integriert wird. +Es bleibt also noch +\[ + 2\int_{0}^{l}u(0, x)dx + = + a_0 \int_{-l}^{l}dx +\] +, was sich wie folgt nach $a_0$ auflösen lässt: +\[ +\begin{aligned} + 2\int_{0}^{l}u(0, x)dx + &= + a_0 \int_{-l}^{l}dx + \\ + &= + a_0 \left[x\right]_{x=-l}^{l} + \\ + &= + a_0(l - (-l)) + \\ + &= + a_0 \cdot 2l + \\ + a_0 + &= + \frac{1}{l} \int_{0}^{l}u(0, x)dx +\end{aligned} +\] + +\subsubsection{Lösund der Differentialgleichung in t} Zuletzt wird die zweite Gleichung der Separation \eqref{eq:slp-example-fourier-separated-t} betrachtet. Diese wird über das charakteristische Polynom @@ -546,6 +615,9 @@ Und mit dem Resultat \eqref{eq:slp-example-fourier-mu-solution} \] ergibt. +Dieses Resultat kann nun mit allen vorhergehenden Resultaten zudammengesetzt +werden um die vollständige Lösung für das Stab-Problem zu erhalten. + \subsubsection{Lösung für einen Stab mit Enden auf konstanter Temperatur} \[ \begin{aligned} -- cgit v1.2.1 From 3a530cc844c8213dade9fcf70d3ea7715f5c2a1b Mon Sep 17 00:00:00 2001 From: Fabian <@> Date: Sat, 13 Aug 2022 15:21:13 +0200 Subject: 2. Ueberarbeitung, Referenzen --- buch/papers/0f1/references.bib | 7 +------ buch/papers/0f1/teil0.tex | 6 +++--- buch/papers/0f1/teil1.tex | 17 +++++++++-------- buch/papers/0f1/teil2.tex | 21 ++++++++++----------- buch/papers/0f1/teil3.tex | 27 +++++++++++++-------------- 5 files changed, 36 insertions(+), 42 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/0f1/references.bib b/buch/papers/0f1/references.bib index ca1b558..f9a358b 100644 --- a/buch/papers/0f1/references.bib +++ b/buch/papers/0f1/references.bib @@ -69,12 +69,7 @@ @book{0f1:SeminarNumerik, title = {Mathematisches Seminar Numerik}, - author = {Andreas Müller, Benjamin Bouhafs-Keller, Daniel Bucher, Manuel Cattaneo -Patrick Elsener, Reto Fritsche, Niccolò Galliani, Tobias Grab -Thomas Kistler, Fabio Marti, Joël Rechsteiner, Cédric Renda -Michael Schmid, Mike Schmid, Michael Schneeberger -Martin Stypinski, Manuel Tischhauser, Nicolas Tobler -Raphael Unterer, Severin Weiss}, + author = {Andreas Müller et al.}, publisher = {Andreas Müller}, year = {2022}, } diff --git a/buch/papers/0f1/teil0.tex b/buch/papers/0f1/teil0.tex index adccac7..9aca368 100644 --- a/buch/papers/0f1/teil0.tex +++ b/buch/papers/0f1/teil0.tex @@ -5,11 +5,11 @@ % \section{Ausgangslage\label{0f1:section:ausgangslage}} \rhead{Ausgangslage} -Die Hypergeometrische Funktion $\mathstrut_0F_1$ wird in vielen Funktionen als Basisfunktion benutzt, +Die hypergeometrische Funktion $\mathstrut_0F_1$ wird in vielen Funktionen als Basisfunktion benutzt, zum Beispiel um die Airy Funktion zu berechnen. In der GNU Scientific Library \cite{0f1:library-gsl} ist die Funktion $\mathstrut_0F_1$ vorhanden. -Allerdings wirft die Funktion, bei negativen Übergabenwerten wie zum Beispiel \verb+gsl_sf_hyperg_0F1(1, -1)+, eine Exception. +Allerdings wirft die Funktion bei negativen Übergabenwerten wie zum Beispiel \verb+gsl_sf_hyperg_0F1(1, -1)+ eine Exception. Bei genauerer Untersuchung hat sich gezeigt, dass die Funktion je nach Betriebssystem funktioniert oder eben nicht. So kann die Funktion unter Windows fehlerfrei aufgerufen werden, beim Mac OS und Linux sind negative Übergabeparameter im Moment nicht möglich. -Ziel dieser Arbeit war es zu evaluieren, ob es mit einfachen mathematischen Operationen möglich ist, die Hypergeometrische Funktion $\mathstrut_0F_1$ zu implementieren. +Ziel dieser Arbeit war es zu evaluieren, ob es mit einfachen mathematischen Operationen möglich ist, die hypergeometrische Funktion $\mathstrut_0F_1$ zu implementieren. diff --git a/buch/papers/0f1/teil1.tex b/buch/papers/0f1/teil1.tex index f697f45..50198fc 100644 --- a/buch/papers/0f1/teil1.tex +++ b/buch/papers/0f1/teil1.tex @@ -6,12 +6,12 @@ \section{Mathematischer Hintergrund \label{0f1:section:mathHintergrund}} \rhead{Mathematischer Hintergrund} -Basierend auf den Herleitungen des vorhergehenden Abschnittes \ref{buch:rekursion:section:hypergeometrische-funktion}, werden im nachfolgenden Abschnitt nochmals die Resultate +Basierend auf den Herleitungen des Abschnittes \ref{buch:rekursion:section:hypergeometrische-funktion}, werden im nachfolgenden Abschnitt nochmals die Resultate beschrieben. \subsection{Hypergeometrische Funktion \label{0f1:subsection:hypergeometrisch}} -Als Grundlage der umgesetzten Algorithmen dient die hypergeometrische Funktion $\mathstrut_0F_1$. Diese ist eine Anwendung der allgemein definierten Funktion $\mathstrut_pF_q$. +Als Grundlage der umgesetzten Algorithmen dient die hypergeometrische Funktion $\mathstrut_0F_1$. Diese ist ein Speziallfall der allgemein definierten Funktion $\mathstrut_pF_q$. \begin{definition} \label{0f1:math:qFp:def} @@ -42,7 +42,8 @@ Angewendet auf die Funktion $\mathstrut_pF_q$ ergibt sich für $\mathstrut_0F_1$ \mathstrut_0F_1 \biggl( \begin{matrix} - \\- + \text{---} + \\\ b_1 \end{matrix} ; @@ -60,7 +61,7 @@ Angewendet auf die Funktion $\mathstrut_pF_q$ ergibt sich für $\mathstrut_0F_1$ \subsection{Airy Funktion \label{0f1:subsection:airy}} -Die Funktion Ai(x) und die verwandte Funktion Bi(x) werden als Airy-Funktion bezeichnet. Sie werden zur Lösung verschiedener physikalischer Probleme benutzt, wie zum Beispiel zur Lösung der Schrödinger-Gleichung \cite{0f1:wiki-airyFunktion}. +Die Funktion $\operatorname{Ai}(x)$ und die verwandte Funktion $\operatorname{Bi}(x)$ werden als Airy-Funktion bezeichnet. Sie werden zur Lösung verschiedener physikalischer Probleme benutzt, wie zum Beispiel zur Lösung der Schrödinger-Gleichung \cite{0f1:wiki-airyFunktion}. \begin{definition} \label{0f1:airy:differentialgleichung:def} @@ -70,11 +71,11 @@ Die Funktion Ai(x) und die verwandte Funktion Bi(x) werden als Airy-Funktion bez \end{definition} Die Airy Funktion lässt sich auf verschiedene Arten darstellen. -Als hypergeometrische Funktion berechnet, ergibt sich wie in Abschnitt \ref{buch:differentialgleichungen:section:hypergeometrisch} hergeleitet, folgende Lösungen der Airy-Differentialgleichung zu den Anfangsbedingungen $Ai(0)=1$ und $Ai'(0)=0$, sowie $Bi(0)=0$ und $Bi'(0)=0$. +Als hypergeometrische Funktion berechnet, ergibt sich wie in Abschnitt \ref{buch:differentialgleichungen:section:hypergeometrisch} hergeleitet, folgende Lösungen der Airy-Differentialgleichung zu den Anfangsbedingungen $\operatorname{Ai}(0)=1$ und $\operatorname{Ai}'(0)=0$, sowie $\operatorname{Bi}(0)=0$ und $\operatorname{Bi}'(0)=1$. \begin{align} \label{0f1:airy:hypergeometrisch:eq} -Ai(x) +\operatorname{Ai}(x) =& \sum_{k=0}^\infty \frac{1}{(\frac23)_k} \frac{1}{k!}\biggl(\frac{x^3}{9}\biggr)^k @@ -83,7 +84,7 @@ Ai(x) \begin{matrix}\text{---}\\\frac23\end{matrix};\frac{x^3}{9} \biggr). \\ -Bi(x) +\operatorname{Bi}(x) =& \sum_{k=0}^\infty \frac{1}{(\frac43)_k} \frac{1}{k!}\biggl(\frac{x^3}{9}\biggr)^k @@ -95,7 +96,7 @@ x\cdot\mathstrut_0F_1\biggl( \qedhere \end{align} -Um die Stabilität der Algorithmen zu $\mathstrut_0F_1$ zu überprüfen, wird in diesem speziellem Fall die Airy Funktion $Ai(x)$ \eqref{0f1:airy:hypergeometrisch:eq} +Um die Stabilität der Algorithmen zu $\mathstrut_0F_1$ zu überprüfen, wird in diesem speziellem Fall die Airy Funktion $\operatorname{Ai}(x)$ \eqref{0f1:airy:hypergeometrisch:eq} benutzt. diff --git a/buch/papers/0f1/teil2.tex b/buch/papers/0f1/teil2.tex index 15a1c44..587f63b 100644 --- a/buch/papers/0f1/teil2.tex +++ b/buch/papers/0f1/teil2.tex @@ -6,12 +6,12 @@ \section{Umsetzung \label{0f1:section:teil2}} \rhead{Umsetzung} -Zur Umsetzung wurden drei verschiedene Ansätze gewählt \cite{0f1:code}. Dabei wurde der Schwerpunkt auf die Funktionalität und eine gute Lesbarkeit des Codes gelegt. +Zur Umsetzung wurden drei verschiedene Ansätze gewählt, die in vollständiger Form auf Github \cite{0f1:code} zu finden sind. Dabei wurde der Schwerpunkt auf die Funktionalität und eine gute Lesbarkeit des Codes gelegt. Die Unterprogramme wurde jeweils, wie die GNU Scientific Library, in C geschrieben. Die Zwischenresultate wurden vom Hauptprogramm in einem CSV-File gespeichert. Anschliessen wurde mit der Matplot-Library in Python die Resultate geplottet. \subsection{Potenzreihe \label{0f1:subsection:potenzreihe}} -Die naheliegendste Lösung ist die Programmierung der Potenzreihe. Allerdings ist ein Problem dieser Umsetzung \ref{0f1:listing:potenzreihe}, dass die Fakultät im Nenner schnell grosse Werte annimmt und so der Bruch gegen Null strebt. Spätesten ab $k=167$ stösst diese Umsetzung \eqref{0f1:umsetzung:0f1:eq} an ihre Grenzen, da die Fakultät von $168$ eine Bereichsüberschreitung des \textit{double} Bereiches darstellt \cite{0f1:double}. +Die naheliegendste Lösung ist die Programmierung der Potenzreihe \eqref{0f1:umsetzung:0f1:eq}. Allerdings ist ein Problem dieser Umsetzung (Listing \ref{0f1:listing:potenzreihe}), dass die Fakultät im Nenner schnell grosse Werte annimmt. Dies führt zu einer Bereichsüberschreitung des \verb+double+ Bereiches \cite{0f1:double}. Spätesten ab $k=167$ tritt dieser Falle ein. \begin{align} \label{0f1:umsetzung:0f1:eq} @@ -30,7 +30,7 @@ Die naheliegendste Lösung ist die Programmierung der Potenzreihe. Allerdings is \subsection{Kettenbruch \label{0f1:subsection:kettenbruch}} -Ein endlicher Kettenbruch ist ein Bruch der Form +Ein endlicher Kettenbruch \cite{0f1:wiki-kettenbruch} ist ein Bruch der Form \begin{equation*} a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cdots}}} \end{equation*} @@ -39,24 +39,23 @@ Die Kurzschreibweise für einen allgemeinen Kettenbruch ist \begin{equation*} a_0 + \frac{a_1|}{|b_1} + \frac{a_2|}{|b_2} + \frac{a_3|}{|b_3} + \cdots \end{equation*} -\cite{0f1:wiki-kettenbruch}. Angewendet auf die Funktion $\mathstrut_0F_1$ bedeutet dies \cite{0f1:wiki-fraction}: \begin{equation*} \mathstrut_0F_1(;c;z) = 1 + \frac{z}{c\cdot1!} + \frac{z^2}{c(c+1)\cdot2!} + \frac{z^3}{c(c+1)(c+2)\cdot3!} + \cdots \end{equation*} -Umgeformt ergibt sich folgender Kettenbruch +Umgeformt ergibt sich folgender Kettenbruch \cite{0f1:wolfram-0f1} \begin{equation} \label{0f1:math:kettenbruch:0f1:eq} \mathstrut_0F_1(;c;z) = 1 + \cfrac{\cfrac{z}{c}}{1+\cfrac{-\cfrac{z}{2(c+1)}}{1+\cfrac{z}{2(c+1)}+\cfrac{-\cfrac{z}{3(c+2)}}{1+\cfrac{z}{5(c+4)} + \cdots}}}, \end{equation} -der als Code (siehe: Listing \ref{0f1:listing:kettenbruchIterativ}) umgesetzt wurde. -\cite{0f1:wolfram-0f1} +der als Code (Listing \ref{0f1:listing:kettenbruchIterativ}) umgesetzt wurde. + \lstinputlisting[style=C,float,caption={Iterativ umgesetzter Kettenbruch.},label={0f1:listing:kettenbruchIterativ}, firstline=8]{papers/0f1/listings/kettenbruchIterativ.c} \subsection{Rekursionsformel \label{0f1:subsection:rekursionsformel}} -Wesentlich stabiler zur Berechnung eines Kettenbruches ist die Rekursionsformel. Nachfolgend wird die verkürzte Herleitung vom Kettenbruch zur Rekursionsformel aufgezeigt. Eine vollständige Schritt für Schritt Herleitung ist im Seminarbuch Numerik, im Kapitel Kettenbrüche zu finden. \cite{0f1:kettenbrueche} +Wesentlich stabiler zur Berechnung eines Kettenbruches ist die Rekursionsformel. Nachfolgend wird die verkürzte Herleitung vom Kettenbruch zur Rekursionsformel aufgezeigt. Eine vollständige Schritt für Schritt Herleitung ist im Seminarbuch Numerik, im Kapitel Kettenbrüche \cite{0f1:kettenbrueche} zu finden. \subsubsection{Herleitung} Ein Näherungsbruch in der Form @@ -135,7 +134,7 @@ Nach vollständiger Induktion ergibt sich für den Schritt $k$, die Matrix a_k \end{pmatrix}. \end{equation} -Und Schlussendlich kann der Näherungsbruch +Und schlussendlich kann der Näherungsbruch \[ \frac{A_k}{B_k} \] @@ -143,7 +142,7 @@ berechnet werden. \subsubsection{Lösung} -Die Berechnung von $A_k, B_k$ \eqref{0f1:math:matrix:ende:eq} kann man auch ohne die Matrizenschreibweise aufschreiben: \cite{0f1:wiki-fraction} +Die Berechnung von $A_k, B_k$ \eqref{0f1:math:matrix:ende:eq} kann man auch ohne die Matrizenschreibweise \cite{0f1:kettenbrueche} aufschreiben: \begin{itemize} \item Startbedingungen: \begin{align*} @@ -165,7 +164,7 @@ B_{k+1} &= B_{k-1} \cdot b_k + B_k \cdot a_k Näherungsbruch: \qquad$\displaystyle\frac{A_k}{B_k}$ \end{itemize} -Ein grosser Vorteil dieser Umsetzung als Rekursionsformel ist \ref{0f1:listing:kettenbruchRekursion}, dass im Vergleich zum Code \ref{0f1:listing:kettenbruchIterativ} eine Division gespart werden kann und somit weniger Rundungsfehler entstehen können. +Ein grosser Vorteil dieser Umsetzung als Rekursionsformel \eqref{0f1:listing:kettenbruchRekursion} ist, dass im Vergleich zum Code (Listing \ref{0f1:listing:kettenbruchIterativ}) eine Division gespart werden kann und somit weniger Rundungsfehler entstehen können. %Code \lstinputlisting[style=C,float,caption={Rekursionsformel für Kettenbruch.},label={0f1:listing:kettenbruchRekursion}, firstline=8]{papers/0f1/listings/kettenbruchRekursion.c} \ No newline at end of file diff --git a/buch/papers/0f1/teil3.tex b/buch/papers/0f1/teil3.tex index 72b1b21..00d4182 100644 --- a/buch/papers/0f1/teil3.tex +++ b/buch/papers/0f1/teil3.tex @@ -9,56 +9,55 @@ Im Verlauf dieser Arbeit hat sich gezeigt, das ein einfacher mathematischer Algorithmus zu implementieren gar nicht so einfach ist. So haben alle drei umgesetzten Ansätze Probleme mit grossen negativen $z$ in der Funktion $\mathstrut_0F_1(;c;z)$. -Ebenso kann festgestellt werden, dass je grösser der Wert $z$ in $\mathstrut_0F_1(;c;z)$ wird, desto mehr weichen die berechneten Resultate von den Erwarteten ab \cite{0f1:wolfram-0f1}. +Ebenso kann festgestellt werden, dass je grösser der Wert $z$ in $\mathstrut_0F_1(;c;z)$ wird, desto mehr weichen die berechneten Resultate von den Erwarteten \cite{0f1:wolfram-0f1} ab. \subsection{Konvergenz \label{0f1:subsection:konvergenz}} -Es zeigt sich in Abbildung \ref{0f1:ausblick:plot:airy:konvergenz}, dass schon nach drei Iterationen ($k = 3$) die Funktionen schon genaue Resultate im Bereich von -2 bis 2 liefert. Ebenso kann festgestellt werden, dass der Kettenbruch schneller konvergiert und im positiven Bereich sogar mit der Referenzfunktion $Ai(x)$ übereinstimmt. Da die Rekursionsformel \ref{0f1:listing:kettenbruchRekursion} eine Abwandlung des Kettenbruches ist, verhalten sich die Funktionen in diesem Fall gleich. +Es zeigt sich in Abbildung \ref{0f1:ausblick:plot:airy:konvergenz}, dass schon nach drei Iterationen ($k = 3$) die Funktionen schon genaue Resultate im Bereich von $-2$ bis $2$ liefert. Ebenso kann festgestellt werden, dass der Kettenbruch schneller konvergiert und im positiven Bereich sogar mit der Referenzfunktion $\operatorname{Ai}(x)$ übereinstimmt. Da die Rekursionsformel eine Abwandlung des Kettenbruches ist, verhalten sich die Funktionen in diesem Fall gleich. Erst wenn mehrerer Iterationen gemacht werden, um die Genauigkeit zu verbessern, ist der Kettenbruch den anderen zwei Algorithmen, bezüglich Konvergenz überlegen. -Interessant ist auch, dass die Rekursionsformel nahezu gleich schnell wie die Potenzreihe konvergiert, aber sich danach einschwingt \ref{0f1:ausblick:plot:konvergenz:positiv}. Dieses Verhalten ist auch bei grösseren $z$ zu beobachten, allerdings ist dann die Differenz zwischen dem ersten lokalen Minimum von k bis zum Abbruch kleiner -\ref{0f1:ausblick:plot:konvergenz:positiv}. -Dieses Phänomen ist auf die Lösung der Rekursionsformel zurück zu führen\eqref{0f1:math:loesung:eq}. Da im Gegensatz die ganz kleinen Werte nicht zu einer Konvergenz wie beim Kettenbruch führen, sondern sich noch eine Zeit lang durch die Multiplikation aufschwingen. +Interessant ist auch, dass die Rekursionsformel nahezu gleich schnell wie die Potenzreihe konvergiert, aber sich danach, wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} zu beobachten ist, einschwingt. Dieses Verhalten ist auch bei grösseren $z$ zu beobachten, allerdings ist dann die Differenz zwischen dem ersten lokalen Minimum von k bis zum Abbruch kleiner. +Dieses Phänomen ist auf die Lösung der Rekursionsformel \eqref{0f1:math:loesung:eq} zurück zu führen. Da im Gegensatz die ganz kleinen Werte nicht zu einer Konvergenz wie beim Kettenbruch führen, sondern sich noch eine Zeit lang durch die Multiplikation aufschwingen. -Ist $z$ negativ wie im Abbild \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu einer Gegenseitigen Kompensation von negativen und positiven Termen so bricht die Rekursionsformel hier zusammen mit der Potenzreihe ab. +Ist $z$ negativ wie im Abbildung \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu einer Gegenseitigen Kompensation von negativen und positiven Termen so bricht die Rekursionsformel hier zusammen mit der Potenzreihe ab. Die ansteigende Differenz mit anschliessender, ist aufgrund der sich alternierenden Termen mit wechselnden Vorzeichens zu erklären. \subsection{Stabilität \label{0f1:subsection:Stabilitaet}} -Verändert sich der Wert von z in $\mathstrut_0F_1(;c;z)$ gegen grössere positive Werte, wie zum Beispiel $c = 800$ liefert die Kettenbruch-Funktion \ref{0f1:listing:kettenbruchIterativ} \verb+inf+ zurück. Dies könnte durch ein Abbruchkriterien abgefangen werden. Allerdings würde das, bei grossen Werten zulasten der Genauigkeit gehen. Trotzdem könnte, je nach Anwendung, auf ein paar Nachkommastellen verzichtet werden. +Verändert sich der Wert von z in $\mathstrut_0F_1(;c;z)$ gegen grössere positive Werte, wie zum Beispiel $c = 800$ liefert die Kettenbruch-Funktion (Listing \ref{0f1:listing:kettenbruchIterativ}) \verb+inf+ zurück. Dies könnte durch ein Abbruchkriterien abgefangen werden. Allerdings würde das, bei grossen Werten zulasten der Genauigkeit gehen. Trotzdem könnte, je nach Anwendung, auf ein paar Nachkommastellen verzichtet werden. -Wohingegen die Potenzreihe \eqref{0f1:listing:potenzreihe} das Problem hat, dass je mehr Terme berechnet werden, desto schneller wächst die Fakultät und irgendwann gibt es eine Bereichsüberschreitung von \verb+double+. Schlussendlich gibt das Unterprogramm das Resultat \verb+-nan(ind)+ zurück. -Die Rekursionformel \eqref{0f1:listing:kettenbruchRekursion} liefert für sehr grosse positive Werte die genausten Ergebnisse, verglichen mit der GNU Scientific Library. Wie schon vermutet ist die Rekursionsformel, im positivem Bereich, der stabilste Algorithmus. Um die Stabilität zu gewährleisten, muss wie in Abbild \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, die Iterationstiefe $k$ genug gross gewählt werden. +Wohingegen die Potenzreihe (Listing \ref{0f1:listing:potenzreihe}) das Problem hat, dass je mehr Terme berechnet werden, desto schneller wächst die Fakultät und irgendwann gibt es eine Bereichsüberschreitung von \verb+double+. Schlussendlich gibt das Unterprogramm das Resultat \verb+-nan(ind)+ zurück. +Die Rekursionformel \eqref{0f1:listing:kettenbruchRekursion} liefert für sehr grosse positive Werte die genausten Ergebnisse, verglichen mit der GNU Scientific Library. Wie schon vermutet ist die Rekursionsformel, im positivem Bereich, der stabilste Algorithmus. Um die Stabilität zu gewährleisten, muss wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, die Iterationstiefe $k$ genug gross gewählt werden. -Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Grund dafür ist die Potenz von z, was zum Phänomen der Auslöschung führt \cite{0f1:SeminarNumerik}. Schön zu beobachten ist dies in der Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} mit der Airy-Funktion als Test. So sind sowohl der Kettenbruch, als auch die Rekursionsformel bis ungefähr $\frac{-15^3}{9}$ stabil. Dies macht auch Sinn, da beide auf der gleichen mathematischen Grundlage basieren. Danach verhält sich allerdings die Instabilität unterschiedlich. Das unterschiedliche Verhalten kann damit erklärt werden, dass beim Kettenbruch jeweils eine zusätzliche Division stattfindet. Diese Unterschiede sind auch in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} festzustellen. +Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Grund dafür ist die Potenz von z, was zum Phänomen der Auslöschung \cite{0f1:SeminarNumerik} führt. Schön zu beobachten ist dies in der Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} mit der Airy-Funktion als Test. So sind sowohl der Kettenbruch, als auch die Rekursionsformel bis ungefähr $\frac{-15^3}{9}$ stabil. Dies macht auch Sinn, da beide auf der gleichen mathematischen Grundlage basieren. Danach verhält sich allerdings die Instabilität unterschiedlich. Das unterschiedliche Verhalten kann damit erklärt werden, dass beim Kettenbruch jeweils eine zusätzliche Division stattfindet. Diese Unterschiede sind auch in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} festzustellen. \begin{figure} \centering \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzAiry.pdf} - \caption{Konvergenz nach drei Iterationen, dargestellt anhand der Airy Funktion zu den Anfangsbedingungen $Ai(0)=1$ und $Ai'(0)=0$. + \caption{Konvergenz nach drei Iterationen, dargestellt anhand der Airy Funktion zu den Anfangsbedingungen $\operatorname{Ai}(0)=1$ und $\operatorname{Ai}'(0)=0$. \label{0f1:ausblick:plot:airy:konvergenz}} \end{figure} \begin{figure} \centering \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzPositiv.pdf} - \caption{Konvergenz: Logarithmisch dargestellte Differenz vom erwarteten Endresultat. + \caption{Konvergenz mit positivem z; Logarithmisch dargestellte Differenz vom erwarteten Endresultat. \label{0f1:ausblick:plot:konvergenz:positiv}} \end{figure} \begin{figure} \centering \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzNegativ.pdf} - \caption{Konvergenz: Logarithmisch dargestellte Differenz vom erwarteten Endresultat. + \caption{Konvergenz mit negativem z; Logarithmisch dargestellte Differenz vom erwarteten Endresultat. \label{0f1:ausblick:plot:konvergenz:negativ}} \end{figure} \begin{figure} \centering \includegraphics[width=1\textwidth]{papers/0f1/images/stabilitaet.pdf} - \caption{Stabilität der 3 Algorithmen verglichen mit der Referenz Funktion $Ai(x)$. + \caption{Stabilität der 3 Algorithmen verglichen mit der Referenz Funktion $\operatorname{Ai}(x)$. \label{0f1:ausblick:plot:airy:stabilitaet}} \end{figure} -- cgit v1.2.1 From 0a59be0b3c470a0f7d71ba2e39fb6ec323d89f84 Mon Sep 17 00:00:00 2001 From: "samuel.niederer" Date: Sat, 13 Aug 2022 18:48:50 +0200 Subject: add content --- buch/papers/kra/anwendung.tex | 45 +++++++++-------------------------- buch/papers/kra/einleitung.tex | 16 ++++++------- buch/papers/kra/loesung.tex | 53 ++++++++++++++++++++++++++++++++++++------ buch/papers/kra/references.bib | 15 ++++++++++++ 4 files changed, 80 insertions(+), 49 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kra/anwendung.tex b/buch/papers/kra/anwendung.tex index 4d4d351..0deaf3c 100644 --- a/buch/papers/kra/anwendung.tex +++ b/buch/papers/kra/anwendung.tex @@ -1,45 +1,40 @@ -\section{Anwendungen \label{kra:section:anwendung}} -\rhead{Anwendungen} +\section{Anwendung \label{kra:section:anwendung}} +\rhead{Anwendung} \newcommand{\dt}[0]{\frac{d}{dt}} Die Matrix-Riccati Differentialgleichung findet unter anderem Anwendung in der Regelungstechnik beim RQ- und RQG-Regler oder aber auch beim Kalmanfilter. -Im folgenden Abschnitt möchten wir uns an einem Beispiel anschauen wie wir mit Hilfe der Matrix-Riccati Differentialgleichung (\ref{kra:matrixriccati}) ein Feder-Masse-System untersuchen können. +Im folgenden Abschnitt möchten wir uns an einem Beispiel anschauen wie wir mit Hilfe der Matrix-Riccati Differentialgleichung (\ref{kra:equation:matrixriccati}) ein Feder-Masse-System untersuchen können \cite{kra:riccati}. \subsection{Feder-Masse-System} -Die Einfachste Form eines Feder-Masse-Systems ist dargestellt in Abbildung \ref{kra:fig:simple_mass_spring}. -Es besteht aus einer Masse $m$ welche reibungsfrei gelagert ist und einer Feder mit der Federkonstante $k$. +Die einfachste Form eines Feder-Masse-Systems ist dargestellt in Abbildung \ref{kra:fig:simple_mass_spring}. +Es besteht aus einer reibungsfrei gelagerten Masse $m$ ,welche an eine Feder mit der Federkonstante $k$ gekoppelt ist. Die im System wirkenden Kräfte teilen sich auf in die auf dem hookeschen Gesetz basierenden Rückstellkraft $F_R = k \Delta_x$ und der auf dem Aktionsprinzip basierenden Kraft $F_a = am = \ddot{x} m$. Das Kräftegleichgewicht fordert $F_R = F_a$ woraus folgt, dass \begin{equation*} k \Delta_x = \ddot{x} m \Leftrightarrow \ddot{x} = \frac{k \Delta_x}{m} \end{equation*} -Die funktion die diese Differentialgleichung löst ist die harmonische Schwingung +Die Funktion die diese Differentialgleichung löst, ist die harmonische Schwingung \begin{equation} x(t) = A \cos(\omega_0 t + \Phi), \quad \omega_0 = \sqrt{\frac{k}{m}} \end{equation} - - \begin{figure} \input{papers/kra/images/simple_mass_spring.tex} \caption{Einfaches Feder-Masse-System.} \label{kra:fig:simple_mass_spring} \end{figure} - \begin{figure} \input{papers/kra/images/multi_mass_spring.tex} \caption{Feder-Masse-System mit zwei Massen und drei Federn.} \label{kra:fig:multi_mass_spring} \end{figure} - \subsection{Hamilton-Funktion} Die Bewegung der Masse $m$ kann mit Hilfe der hamiltonschen Mechanik im Phasenraum untersucht werden. -Die hamiltonschen Gleichungen verwenden dafür die veralgemeinerten Ortskoordinaten +Die hamiltonschen Gleichungen verwenden dafür die verallgemeinerten Ortskoordinaten $q = (q_{1}, q_{2}, ..., q_{n})$ und die verallgemeinerten Impulskoordinaten $p = (p_{1}, p_{2}, ..., p_{n})$, wobei der Impuls definiert ist als $p_k = m_k \cdot v_k$. Liegen keine zeitabhängigen Zwangsbedingungen vor, so entspricht die Hamitlon-Funktion der Gesamtenergie des Systems \cite{kra:hamilton}. Im Falle des einfachen Feder-Masse-Systems, Abbildung \ref{kra:fig:simple_mass_spring}, setzt sich die Hamilton-Funktion aus kinetischer und potentieller Energie zusammen. - \begin{equation} \label{kra:harmonischer_oszillator} \begin{split} @@ -47,7 +42,6 @@ Im Falle des einfachen Feder-Masse-Systems, Abbildung \ref{kra:fig:simple_mass_s &= \underbrace{\frac{p^2}{2m}}_{E_{kin}} + \underbrace{\frac{k q^2}{2}}_{E_{pot}} \end{split} \end{equation} - Die Hamiltonschen Bewegungsgleichungen liefern \cite{kra:kanonischegleichungen} \begin{equation} \label{kra:hamilton:bewegungsgleichung} @@ -55,17 +49,13 @@ Die Hamiltonschen Bewegungsgleichungen liefern \cite{kra:kanonischegleichungen} \qquad \dot{p_{k}} = -\frac{\partial \mathcal{H}}{\partial q_k} \end{equation} - daraus folgt - \[ \dot{q} = \frac{p}{m} \qquad \dot{p} = -kq \] - in Matrixschreibweise erhalten wir also - \[ \begin{pmatrix} \dot{q} \\ @@ -81,11 +71,9 @@ in Matrixschreibweise erhalten wir also p \end{pmatrix} \] - Für das erweiterte Federmassesystem, Abbildung \ref{kra:fig:multi_mass_spring}, können wir analog vorgehen. Die kinetische Energie setzt sich nun aus den kinetischen Energien der einzelnen Massen $m_1$ und $m_2$ zusammen. Die Potentielle Energie erhalten wir aus der Summe der kinetischen Energien der einzelnen Federn mit den Federkonstanten $k_1$, $k_c$ und $k_2$. - \begin{align*} \begin{split} T &= T_1 + T_2 \\ @@ -97,16 +85,13 @@ Die Potentielle Energie erhalten wir aus der Summe der kinetischen Energien der &= \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2} \end{split} \end{align*} - Die Hamilton-Funktion ist also - \begin{align*} \begin{split} \mathcal{H} &= T + V \\ &= \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \frac{k_1 q_1^2}{2} + \frac{k_c (q_2 - q_1)^2}{2} + \frac{k_2 q_2^2}{2} \end{split} \end{align*} - Die Bewegungsgleichungen \ref{kra:hamilton:bewegungsgleichung} liefern \begin{align*} \frac{\partial \mathcal{H}}{\partial p_k} & = \dot{q_k} @@ -127,9 +112,7 @@ Die Bewegungsgleichungen \ref{kra:hamilton:bewegungsgleichung} liefern \end{alignedat} \right. \end{align*} - In Matrixschreibweise erhalten wir - \begin{equation} \label{kra:hamilton:multispringmass} \begin{pmatrix} @@ -171,7 +154,7 @@ In Matrixschreibweise erhalten wir \end{equation} \subsection{Phasenraum} -Der Phasenraum erlaubt die eindeutige Beschreibung aller möglichen Bewegungszustände eines mechanischen System durch einen Punkt. +Der Phasenraum erlaubt die eindeutige Beschreibung aller möglichen Bewegungszustände eines mechanischen Systems durch einen Punkt. Die Phasenraumdarstellung eignet sich somit sehr gut für die systematische Untersuchung der Feder-Masse-Systeme. \subsubsection{Harmonischer Oszillator} @@ -181,7 +164,6 @@ Die Hamiltonfunktion des harmonischen Oszillators \ref{kra:harmonischer_oszillat \end{equation*} die Phasenraumtrajektorien bilden also Ellipsen mit Zentrum $q=0, p=0$ und Halbachsen $A$ und $m \omega A$. Abbildung \ref{kra:fig:phasenraum} zeigt Phasenraumtrajektorien mit den Energien $E_{x \in \{A, B, C, D\}}$ und verschiedenen Werten von $\omega$. - \begin{figure} \input{papers/kra/images/phase_space.tex} \caption{Phasenraumdarstellung des einfachen Feder-Masse-Systems.} @@ -191,7 +173,6 @@ Abbildung \ref{kra:fig:phasenraum} zeigt Phasenraumtrajektorien mit den Energien \subsubsection{Erweitertes Feder-Masse-System} Wir intressieren uns nun dafür wie der Phasenwinkel $U = PQ^{-1}$ von der Zeit abhängt, wir suchen also die Grösse $\Theta = \dt U$. - Ersetzten wir in der Gleichung \ref{kra:hamilton:multispringmass} die Matrix $G$ mit $\tilde{G}$ so erhalten wir \begin{equation} \dt @@ -211,9 +192,7 @@ Ersetzten wir in der Gleichung \ref{kra:hamilton:multispringmass} die Matrix $G$ P \end{pmatrix} \end{equation} - Mit einsetzten folgt - \begin{align*} \dot{Q} = AQ + BP \\ \dot{P} = CQ + DP @@ -227,9 +206,7 @@ Mit einsetzten folgt &= C + DU - UA - UBU \end{split} \end{equation} +was uns auf die Matrix-Riccati Gleichung \ref{kra:equation:matrixriccati} führt. -was uns auf die Matrix-Riccati Gleichung \ref{kra:matrixriccati} führt. - - -\subsection{Fazit} -% @TODO +% @TODO Einfluss auf anfangsbedingungen, plots? +% @TODO Fazit ? diff --git a/buch/papers/kra/einleitung.tex b/buch/papers/kra/einleitung.tex index 1a347a8..cde2e66 100644 --- a/buch/papers/kra/einleitung.tex +++ b/buch/papers/kra/einleitung.tex @@ -1,14 +1,14 @@ \section{Einleitung} \label{kra:section:einleitung} \rhead{Einleitung} -Die riccatische Differentialgleichung ist eine nichtlineare gewöhnliche Differentialgleichunge erster Ordnung der form +Die riccatische Differentialgleichung ist eine nicht lineare gewöhnliche Differentialgleichung erster Ordnung der Form \begin{equation} - \label{kra:riccati} - y'(x) = f(x)y^2(x) + g(x)y(x) + h(x) + \label{kra:equation:riccati} + y' = f(x)y + g(x)y^2 + h(x) \end{equation} -Sie ist bennant nach dem italienischen Grafen Jacopo Francesco Riccati (1676–1754) der sich mit der Klassifizierung von Differentialgleichungen befasste und Methoden zur Verringerung der Ordnung von Gleichungen entwickelte. -Als Riccati Gleichung werden auch Matrixgleichugen der Form +Sie ist benannt nach dem italienischen Grafen Jacopo Francesco Riccati (1676–1754) der sich mit der Klassifizierung von Differentialgleichungen befasste. +Als Riccati Gleichung werden auch Matrixgleichungen der Form \begin{equation} - \label{kra:matrixriccati} - \dot{U}(t) = DU(t) - UA(t) - U(t)BU(t) % +Q ? + \label{kra:equation:matrixriccati} + \dot{X}(t) = C + DX(t) - X(t)A -X(t)BX(t) \end{equation} -bezeichnet, welche aufgrund ihres quadratischen Terms eine gewisse ähnlichkeit aufweisen. \ No newline at end of file +bezeichnet, welche aufgrund ihres quadratischen Terms eine gewisse Ähnlichkeit aufweisen \cite{kra:ethz} \cite{kra:riccati}. diff --git a/buch/papers/kra/loesung.tex b/buch/papers/kra/loesung.tex index ece0f15..4e0da1c 100644 --- a/buch/papers/kra/loesung.tex +++ b/buch/papers/kra/loesung.tex @@ -1,11 +1,53 @@ \section{Lösungsmethoden} \label{kra:section:loesung} \rhead{Lösungsmethoden} -% @TODO Lösung normal riccati -Lösung der Riccatischen Differentialgleichung \ref{kra:riccati}. +\subsection{Riccatische Differentialgleichung} \label{kra:loesung:riccati} +Eine allgemeine analytische Lösung der Riccati Differentialgleichung ist nicht möglich. +Es gibt aber Spezialfälle, in denen sich die Gleichung vereinfachen lässt und so eine analytische Lösung gefunden werden kann. +Diese wollen wir im folgenden Abschnitt genauer anschauen. +\subsubsection{Fall 1: Konstante Koeffizienten} +Sind die Koeffizienten $f(x), g(x), h(x)$ Konstanten, so lässt sich die DGL separieren und reduziert sich auf die Lösung des Integrals \ref{kra:equation:case1_int}. +\begin{equation} + y' = fy^2 + gy + h +\end{equation} +\begin{equation} + \frac{dy}{dx} = fy^2 + gy + h +\end{equation} +\begin{equation} \label{kra:equation:case1_int} + \int \frac{dy}{fy^2 + gy + h} = \int dx +\end{equation} + +\subsubsection{Fall 2: Bekannte spezielle Lösung} +Kennt man eine spezielle Lösung $y_p$ so kann die riccatische DGL mit Hilfe einer Substitution auf eine lineare Gleichung reduziert werden. +Wir wählen als Substitution +\begin{equation} \label{kra:equation:substitution} + z = \frac{1}{y - y_p} +\end{equation} +durch Umstellen von \ref{kra:equation:substitution} folgt +\begin{equation} + y = y_p + \frac{1}{z^2} \label{kra:equation:backsubstitution} +\end{equation} +\begin{equation} + y' = y_p' - \frac{1}{z^2}z' +\end{equation} +mit Einsetzten in die DGL \ref{kra:equation:riccati} folgt +\begin{equation} + y_p' - \frac{1}{z^2}z' = f(x)(y_p + \frac{1}{z}) + g(x)(y_p + \frac{1}{z})^2 + h(x) +\end{equation} +\begin{equation} + -z^{2}y_p' + z' = -z^2\underbrace{(y_{p}f(x) + g(x)y_p^2 + h(x))}_{y_p'} - z(f(x) + 2y_{p}g(x)) - g(x) +\end{equation} +was uns direkt auf eine lineare Differentialgleichung 1.Ordnung führt. +\begin{equation} + z' = -z(f(x) + 2y_{p}g(x)) - g(x) +\end{equation} +Diese kann nun mit den Methoden zur Lösung von linearen Differentialgleichungen 1.Ordnung gelöst werden. +Durch die Rücksubstitution \ref{kra:equation:backsubstitution} erhält man dann die Lösung von \ref{kra:equation:riccati}. + +\subsection{Matrix-Riccati Differentialgleichung} \label{kra:loesung:riccati} % Lösung matrix riccati -Die Lösung der Matrix-Riccati Gleichung \ref{kra:matrixriccati} erhalten wir nach \cite{kra:kalmanisae} folgendermassen +Die Lösung der Matrix-Riccati Gleichung \ref{kra:equation:matrixriccati} erhalten wir nach \cite{kra:kalmanisae} folgendermassen \begin{equation} \label{kra:matrixriccati-solution} \begin{pmatrix} @@ -28,7 +70,6 @@ Die Lösung der Matrix-Riccati Gleichung \ref{kra:matrixriccati} erhalten wir na U_0(t) \end{pmatrix} \end{equation} - \begin{equation} U(t) = \begin{pmatrix} @@ -39,9 +80,7 @@ Die Lösung der Matrix-Riccati Gleichung \ref{kra:matrixriccati} erhalten wir na \end{pmatrix} ^{-1} \end{equation} - -wobei $\Phi(t, t_0)$ die sogennante Zustandsübergangsmatrix ist. - +wobei $\Phi(t, t_0)$ die sogenannte Zustandsübergangsmatrix ist. \begin{equation} \Phi(t_0, t) = e^{H(t - t_0)} \end{equation} diff --git a/buch/papers/kra/references.bib b/buch/papers/kra/references.bib index 7f972ec..a9a8ede 100644 --- a/buch/papers/kra/references.bib +++ b/buch/papers/kra/references.bib @@ -4,6 +4,19 @@ % (c) 2020 Autor, Hochschule Rapperswil % +@misc{kra:riccati, +title = {Riccatische Differentialgleichung}, +url = {https://de.wikipedia.org/wiki/Riccatische_Differentialgleichung}, +date = {2022-05-26} +} + +@misc{kra:ethz, +author = {Ch. Roduner}, +title = {Die-Riccati-Gleichung}, +url = {https://www.imrtweb.ethz.ch/users/geering/Riccati.pdf}, +date = {2022-05-26} +} + @online{kra:hamilton, title = {Hamilton-Funktion}, url = {https://de.wikipedia.org/wiki/Hamilton-Funktion}, @@ -28,3 +41,5 @@ url = {https://pagespro.isae-supaero.fr/IMG/pdf/introKalman_e_151211.pdf}, date = {2022-05-26} } + + -- cgit v1.2.1 From ac66147d7ac9b65ead1946ea4e72d681fc4abcf4 Mon Sep 17 00:00:00 2001 From: "samuel.niederer" Date: Sat, 13 Aug 2022 18:55:22 +0200 Subject: remove dev file --- buch/papers/kra/test.tex | 12 ------------ 1 file changed, 12 deletions(-) delete mode 100644 buch/papers/kra/test.tex (limited to 'buch/papers') diff --git a/buch/papers/kra/test.tex b/buch/papers/kra/test.tex deleted file mode 100644 index ebe0aa0..0000000 --- a/buch/papers/kra/test.tex +++ /dev/null @@ -1,12 +0,0 @@ -\begin{figure} - \input{papers/kra/images/phase_space.tex} - % \begin{minipage}{.45\textwidth} - % \input{papers/kra/images/phase_space_small_omega.tex} - % \end{minipage} - % \begin{minipage}{.45\textwidth} - % \input{papers/kra/images/phase_space_large_omega.tex} - % \end{minipage} - % \begin{minipage}[.5\textwidth] - % \input{papers/kra/images/phase_space_large_omega.tex} - % \end{minipage} -\end{figure} \ No newline at end of file -- cgit v1.2.1 From efa82f7edc7345c29c2d44674d8c8d8ad8741548 Mon Sep 17 00:00:00 2001 From: Nicolas Tobler Date: Sat, 13 Aug 2022 19:32:21 +0200 Subject: corrections --- buch/papers/ellfilter/einleitung.tex | 35 +++++++++----- buch/papers/ellfilter/elliptic.tex | 22 +++++---- buch/papers/ellfilter/jacobi.tex | 73 +++++++++++++++++------------ buch/papers/ellfilter/tikz/arccos.tikz.tex | 9 +++- buch/papers/ellfilter/tikz/arccos2.tikz.tex | 19 +++++++- buch/papers/ellfilter/tikz/cd.tikz.tex | 4 +- buch/papers/ellfilter/tikz/cd2.tikz.tex | 15 ++++++ buch/papers/ellfilter/tikz/filter.tikz.tex | 26 ++++++---- buch/papers/ellfilter/tikz/sn.tikz.tex | 4 +- buch/papers/ellfilter/tschebyscheff.tex | 25 +++++----- 10 files changed, 153 insertions(+), 79 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/ellfilter/einleitung.tex b/buch/papers/ellfilter/einleitung.tex index 5bc2ead..cf57698 100644 --- a/buch/papers/ellfilter/einleitung.tex +++ b/buch/papers/ellfilter/einleitung.tex @@ -1,37 +1,48 @@ \section{Einleitung} -Filter sind womöglich eines der wichtigsten Element in der Signalverarbeitung und finden Anwendungen in der digitalen und analogen Elektrotechnik. +Filter sind womöglich eines der wichtigsten Elementen in der Signalverarbeitung und finden Anwendungen in der digitalen und analogen Elektrotechnik. Besonders hilfreich ist die Untergruppe der linearen Filter. Elektronische Schaltungen mit linearen Bauelementen wie Kondensatoren, Spulen und Widerständen führen immer zu linearen zeitinvarianten Systemen (LTI-System von englich \textit{time-invariant system}). Durch die Linearität werden beim das Filtern keine neuen Frequenzanteile erzeugt, was es erlaubt, einen Frequenzanteil eines Signals verzerrungsfrei herauszufiltern. %TODO review sentence Diese Eigenschaft macht es Sinnvoll, lineare Filter im Frequenzbereich zu beschreiben. -Die Übertragungsfunktion eines linearen Filters im Frequenzbereich $H(\Omega)$ ist dabei immer eine rationale Funktion, also eine Division von zwei Polynomen. -Dabei ist $\Omega = 2 \pi f$ die analoge Frequenzeinheit. +Die Übertragungsfunktion eines linearen Filters im Frequenzbereich $H(\Omega)$ ist dabei immer eine rationale Funktion, also ein Quotient von zwei Polynomen. +Dabei ist $\Omega = 2 \pi f$ die Frequenzeinheit. Die Polynome haben dabei immer reelle oder komplex-konjugierte Nullstellen. -Ein breit angewendeter Filtertyp ist das Tiefpassfilter, welches beabsichtigt alle Frequenzen eines Signals über der Grenzfrequenz $\Omega_p$ auszulöschen. +Ein breit angewendeter Filtertyp ist das Tiefpassfilter, welches beabsichtigt alle Frequenzen eines Signals oberhalb der Grenzfrequenz $\Omega_p$ auszulöschen. Der Rest soll dabei unverändert passieren. -Ein solches Filter hat idealerweise eine Frequenzantwort +Aus dem Tiefpassifilter können dann durch Transformationen auch Hochpassfilter, Bandpassfilter und Bandsperren realisiert werden. +Ein solches Filter hat idealerweise die Frequenzantwort \begin{equation} \label{ellfilter:eq:h_omega} H(\Omega) = \begin{cases} 1 & \Omega < \Omega_p \\ 0 & \Omega < \Omega_p - \end{cases}. + \end{cases}, \end{equation} +wie dargestellt in Abbildung \ref{ellfilter:fig:lp} +\begin{figure} + \centering + \input{papers/ellfilter/tikz/filter.tikz.tex} + \caption{Frequenzantwort eines Tiefpassfilters.} + \label{ellfilter:fig:lp} +\end{figure} Leider ist eine solche Funktion nicht als rationale Funktion darstellbar. Aus diesem Grund sind realisierbare Approximationen gesucht. -Jede Approximation wird einen kontinuierlichen übergang zwischen Durchlassbereich und Sperrbereich aufweisen. +Jede Approximation wird einen kontinuierlichen Übergang zwischen Durchlassbereich und Sperrbereich aufweisen. Oft wird dabei der Faktor $1/\sqrt{2}$ als Schwelle zwischen den beiden Bereichen gewählt. Somit lassen sich lineare Tiefpassfilter mit folgender Funktion zusammenfassen: \begin{equation} \label{ellfilter:eq:h_omega} | H(\Omega)|^2 = \frac{1}{1 + \varepsilon_p^2 F_N^2(w)}, \quad w=\frac{\Omega}{\Omega_p}, \end{equation} -%TODO figure? wobei $F_N(w)$ eine rationale Funktion ist, $|F_N(w)| \leq 1 ~\forall~ |w| \leq 1$ erfüllt und für $|w| \geq 1$ möglichst schnell divergiert. Des weiteren müssen alle Nullstellen und Pole von $F_N$ auf der linken Halbebene liegen, damit das Filter implementierbar und stabil ist. -$N \in \mathbb{N} $ gibt dabei die Ordnung des Filters vor, also die maximale Anzahl Pole oder Nullstellen, die zur Komplexitätsmilderung klein gehalten werden soll. -Eine einfache Funktion für $F_N$ ist das Polynom $w^N$. +$w$ ist die normalisierte Frequenz, die es erlaubt ein Filter unabhängig von der Grenzfrequenz zu beschrieben. +Bei $w=1$ hat das Filter eine Dämpfung von $1/(1+\varepsilon^2)$. +$N \in \mathbb{N} $ gibt die Ordnung des Filters vor, also die maximale Anzahl Pole oder Nullstellen. +Je hoher $N$ gewählt wird, desto steiler ist der Übergang in denn Sperrbereich. +Grössere $N$ sind erfordern jedoch aufwendigere Implementierungen und haben mehr Phasenverschiebung. +Eine einfache Funktion, die für $F_N$ eingesetzt werden kann, ist das Polynom $w^N$. Tatsächlich erhalten wir damit das Butterworth Filter, wie in Abbildung \ref{ellfilter:fig:butterworth} ersichtlich. \begin{figure} \centering @@ -46,10 +57,10 @@ Eine Reihe von rationalen Funktionen können für $F_N$ eingesetzt werden, um Ti w^N & \text{Butterworth} \\ T_N(w) & \text{Tschebyscheff, Typ 1} \\ [k_1 T_N (k^{-1} w^{-1})]^{-1} & \text{Tschebyscheff, Typ 2} \\ - R_N(w, \xi) & \text{Elliptisch (Cauer)} \\ + R_N(w, \xi) & \text{Elliptisch} \\ \end{cases} \end{align} -Mit der Ausnahme vom Butterworth filter sind alle Filter nach speziellen Funktionen benannt. +Mit der Ausnahme vom Butterworth-Filter sind alle Filter nach speziellen Funktionen benannt. Alle diese Filter sind optimal hinsichtlich einer Eigenschaft. Das Butterworth-Filter, zum Beispiel, ist maximal flach im Durchlassbereich. Das Tschebyscheff-1 Filter ist maximal steil für eine definierte Welligkeit im Durchlassbereich, währendem es im Sperrbereich monoton abfallend ist. diff --git a/buch/papers/ellfilter/elliptic.tex b/buch/papers/ellfilter/elliptic.tex index 793fd6c..89a2d7a 100644 --- a/buch/papers/ellfilter/elliptic.tex +++ b/buch/papers/ellfilter/elliptic.tex @@ -3,10 +3,10 @@ Kommen wir nun zum eigentlichen Teil dieses Papers, den elliptischen rationalen Funktionen \ref{ellfilter:bib:orfanidis} \begin{align} R_N(\xi, w) &= \cd \left(N~f_1(\xi)~\cd^{-1}(w, 1/\xi), f_2(\xi)\right) \label{ellfilter:eq:elliptic}\\ - &= \cd \left(N~\frac{K_1}{K}~\cd^{-1}(w, k), k_1)\right) , \quad k= 1/\xi, k_1 = 1/f(\xi) \\ + &= \cd \left(N~\frac{K_1}{K}~\cd^{-1}(w, k), k_1\right) , \quad k= 1/\xi, k_1 = 1/f(\xi) \\ &= \cd \left(N~K_1~z , k_1 \right), \quad w= \cd(z K, k) \end{align} -Beim Betrachten dieser Definition, fällt die Ähnlichkeit zur trigonometrische Darstellung der Tschebyschef-Polynome \eqref{ellfilter:eq:chebychef_polynomials} auf. +Beim Betrachten dieser Definition, fällt die Ähnlichkeit zur trigonometrische Darstellung der Tsche\-byschef-Polynome \eqref{ellfilter:eq:chebychef_polynomials} auf. Anstelle vom Kosinus kommt hier die $\cd$-Funktion zum Einsatz. Die Ordnungszahl $N$ kommt auch als Faktor for. Zusätzlich werden noch zwei verschiedene elliptische Moduli $k$ und $k_1$ gebraucht. @@ -29,8 +29,9 @@ Die Idee des elliptischen Filter ist es, diese zwei Equirippel-Zonen abzufahren, \centering \input{papers/ellfilter/tikz/cd2.tikz.tex} \caption{ - $z_1$-Ebene der elliptischen rationalen Funktionen. + $z_1=N\frac{K_1}{K}\cd^{-1}(w, k)$-Ebene der elliptischen rationalen Funktionen. Je grösser die Ordnung $N$ gewählt wird, desto mehr Nullstellen werden passiert. + Als Vereinfachung ist die Funktion nur für $w>0$ dargestellt. } \label{ellfilter:fig:cd2} \end{figure} @@ -39,7 +40,7 @@ Im Durchlassbereich werden wie beim Tschebyscheff-Filter die Nullstellen durchla Statt dass $z_1$ für alle $w>1$ in die imaginäre Richtung geht, bewegen wir uns im Sperrbereich wieder in reeller Richtung, wo Pole durchlaufen werden. Aus dieser Sicht kann der Sperrbereich vom Tschebyscheff-Filter als unendlich langer Übergangsbereich angesehen werden. % Falls es möglich ist diese Werte abzufahren im Stil der Tschebyscheff-Polynome, kann ein Filter gebaut werden, dass Equirippel-Verhalten im Durchlass- und Sperrbereich aufweist. -Da sich die Funktion im Übergangsbereich nur zur nächsten Reihe bewegt ist der Übergangsbereich monoton steigend. +Da sich die Funktion im Übergangsbereich nur zur nächsten Reihe bewegt, ist der Übergangsbereich monoton steigend. Theoretisch könnte eine gleiches Durchlass- und Sperrbereichverhalten erreicht werden, wenn die Funktion auf eine andere Reihe ansteigen würde. Dies würde jedoch zu Oszillationen zwischen $1$ und $1/k$ im Übergangsbereich führen. Abbildung \ref{ellfilter:fig:elliptic_freq} zeigt eine elliptisch rationale Funktion und die Frequenzantwort des daraus resultierenden Filters. @@ -61,8 +62,8 @@ Zur Erinnerung, $K$ und $K^\prime$ sind durch elliptische Integrale definiert un \caption{Die Periodizitäten in realer und imaginärer Richtung in Abhängigkeit vom elliptischen Modul $k$.} \label{ellfilter:fig:kprime} \end{figure} -$K$ und $K^\prime$ sind durch die Ortskurve $K + jK^\prime$ aneinander Gebunden und benötigen den Zusatzfaktor $K_1/K$ in \eqref{ellfilter:eq:elliptic}, um die genanten Forderungen einzuhalten. -Abbildung \ref{ellfilter:fig:degree_eq} zeigt das Problem geometrisch auf, wobei zwei Punkte auf der Ortskurve gesucht sind. +$K$ und $K^\prime$ sind durch die Ortskurve $K + jK^\prime$ aneinander gebunden und benötigen den Zusatzfaktor $K_1/K$ in \eqref{ellfilter:eq:elliptic}, um die genanten Forderungen einzuhalten. +Abbildung \ref{ellfilter:fig:degree_eq} zeigt das Problem geometrisch auf, wobei zwei Punkte $K+jK^\prime$ und $K_1+jK_1^\prime$ auf der Ortskurve gesucht sind. \begin{figure} \centering \input{papers/ellfilter/tikz/elliptic_transform2.tikz} @@ -87,10 +88,13 @@ Die Herleitung ist sehr umfassend und wird in \ref{ellfilter:bib:orfanidis} im D % \caption{Die Gradgleichung als geometrisches Problem.} % \end{figure} -\subsection{Darstellung als rationale Funktion} +\subsection{Schlussfolgerung} +Die elliptischen Filter können als direkte Erweiterung der Tschebyscheff-Filter verstanden werden. Bei den Tschebyscheff-Polynomen haben wir gesehen, dass die Trigonometrische Formel zu einfachen Polynomen umgewandelt werden kann. -Im Gegensatz zum $\cos^{-1}$ hat der $\cd^{-1}$ nicht nur Nullstellen sondern auch Pole. +Im elliptischen Fall entstehen so rationale Funktionen mit Nullstellen und auch Pole. Somit entstehen bei den elliptischen rationalen Funktionen, wie es der name auch deutet, rationale Funktionen, also ein Bruch von zwei Polynomen. -Da Transformationen einer rationalen Funktionen mit Grundrechenarten, wie es in \eqref{ellfilter:eq:h_omega} der Fall ist, immer noch rationale Funktionen ergeben, stellt dies kein Problem für die Implementierung dar. +% Da Transformationen einer rationalen Funktionen mit Grundrechenarten, wie es in \eqref{ellfilter:eq:h_omega} der Fall ist, immer noch rationale Funktionen ergeben, stellt dies kein Problem für die Implementierung dar. + + diff --git a/buch/papers/ellfilter/jacobi.tex b/buch/papers/ellfilter/jacobi.tex index 3940171..fae6b31 100644 --- a/buch/papers/ellfilter/jacobi.tex +++ b/buch/papers/ellfilter/jacobi.tex @@ -13,7 +13,7 @@ Zum Beispiel gibt es analog zum Sinus den elliptischen $\sn(z, k)$. Im Gegensatz zum den trigonometrischen Funktionen haben die elliptischen Funktionen zwei parameter. Den \textit{elliptische Modul} $k$, der die Exzentrizität der Ellipse parametrisiert und das Winkelargument $z$. Im Kreis ist der Radius für alle Winkel konstant, bei Ellipsen ändert sich das. -Dies hat zur Folge, dass bei einer Ellipse die Kreisbodenstrecke nicht linear zum Winkel verläuft. +Dies hat zur Folge, dass bei einer Ellipse die Kreisbogenlänge nicht linear zum Winkel verläuft. Darum kann hier nicht der gewohnte Winkel verwendet werden. Das Winkelargument $z$ kann durch das elliptische Integral erster Art \begin{equation} @@ -95,37 +95,40 @@ Mithilfe von $F^{-1}$ kann zum Beispiel $sn^{-1}$ mit dem Elliptischen integral = \sn(z, k) = - w + w. \end{equation} -\begin{equation} %TODO remove unnecessary equations - \phi - = - F^{-1}(z, k) - = - \sin^{-1} \big( \sn (z, k ) \big) - = - \sin^{-1} ( w ) -\end{equation} +% \begin{equation} %TODO remove unnecessary equations +% \phi +% = +% F^{-1}(z, k) +% = +% \sin^{-1} \big( \sn (z, k ) \big) +% = +% \sin^{-1} ( w ) +% \end{equation} -\begin{equation} - F(\phi, k) - = - z - = - F( \sin^{-1} \big( \sn (z, k ) \big) , k) - = - F( \sin^{-1} ( w ), k) -\end{equation} +% \begin{equation} +% F(\phi, k) +% = +% z +% = +% F( \sin^{-1} \big( \sn (z, k ) \big) , k) +% = +% F( \sin^{-1} ( w ), k) +% \end{equation} -\begin{equation} - \sn^{-1}(w, k) - = - F(\phi, k), - \quad - \phi = \sin^{-1}(w) -\end{equation} +% \begin{equation} +% \sn^{-1}(w, k) +% = +% F(\phi, k), +% \quad +% \phi = \sin^{-1}(w) +% \end{equation} +Beim Tschebyscheff-Filter konnten wir mit Betrachten des Arcuscosinus die Funktionalität erklären. +Für das Elliptische Filter machen wir die gleiche Betrachtung mit der $\sn^{-1}$-Funktion. +Der $\sn^{-1}$ ist durch das elliptische Integral \begin{align} \sn^{-1}(w, k) & = @@ -150,12 +153,22 @@ Mithilfe von $F^{-1}$ kann zum Beispiel $sn^{-1}$ mit dem Elliptischen integral } } \end{align} - +beschrieben. +Dazu betrachten wir wieder den Integranden +\begin{equation} + \frac{ + 1 + }{ + \sqrt{ + (1-t^2)(1-k^2 t^2) + } + }. +\end{equation} Beim $\cos^{-1}(x)$ haben wir gesehen, dass die analytische Fortsetzung bei $x < -1$ und $x > 1$ rechtwinklig in die Komplexen zahlen wandert. -Wenn man das gleiche mit $\sn^{-1}(w, k)$ macht, erkennt man zwei interessante Stellen. +Wenn man das Gleiche mit $\sn^{-1}(w, k)$ macht, erkennt man zwei interessante Stellen. Die erste ist die gleiche wie beim $\cos^{-1}(x)$ nämlich bei $t = \pm 1$. Der erste Term unter der Wurzel wird dann negativ, während der zweite noch positiv ist, da $k \leq 1$. -Ab diesem Punkt verläuft knickt die Funktion in die imaginäre Richtung ab. +Ab diesem Punkt knickt die Funktion in die imaginäre Richtung ab. Bei $t = 1/k$ ist auch der zweite Term negativ und die Funktion verläuft in die negative reelle Richtung. Abbildung \label{ellfilter:fig:sn} zeigt den Verlauf der Funktion in der komplexen Ebene. \begin{figure} diff --git a/buch/papers/ellfilter/tikz/arccos.tikz.tex b/buch/papers/ellfilter/tikz/arccos.tikz.tex index a139fc4..b11c25d 100644 --- a/buch/papers/ellfilter/tikz/arccos.tikz.tex +++ b/buch/papers/ellfilter/tikz/arccos.tikz.tex @@ -52,9 +52,14 @@ \end{scope} \node[zero] at (4,2) (n) {}; - \node[anchor=west] at (n.east) {Zero}; + \node[anchor=west] at (n.east) {Nullstelle}; - \begin{scope}[yshift=-3cm] + \begin{scope}[yshift=-3.25cm] + + \draw[->, thick](0,0) -- node[anchor=center, fill=white]{$z = \cos^{-1}(w)$} (0,1); + + \end{scope} + \begin{scope}[yshift=-4cm] \draw[gray, ->] (-5,0) -- (5,0) node[anchor=west]{$w$}; diff --git a/buch/papers/ellfilter/tikz/arccos2.tikz.tex b/buch/papers/ellfilter/tikz/arccos2.tikz.tex index c3f11bb..2cec75f 100644 --- a/buch/papers/ellfilter/tikz/arccos2.tikz.tex +++ b/buch/papers/ellfilter/tikz/arccos2.tikz.tex @@ -54,6 +54,23 @@ \end{scope} \node[zero] at (6.5,2) (n) {}; - \node[anchor=west] at (n.east) {Zero}; + \node[anchor=west] at (n.east) {Nullstelle}; + + \begin{scope}[xshift=2.75cm, yshift=-2cm] + + \draw[gray, ->] (-5,0) -- (5,0) node[anchor=west]{$w$}; + + \draw[ultra thick, ->, blue] (-4, 0) -- (-2, 0); + \draw[ultra thick, ->, cyan] (-2, 0) -- (0, 0); + \draw[ultra thick, ->, darkgreen] (0, 0) -- (2, 0); + \draw[ultra thick, ->, orange] (2, 0) -- (4, 0); + + \node[anchor=south] at (-4,0) {$-\infty$}; + \node[anchor=south] at (-2,0) {$-1$}; + \node[anchor=south] at (0,0) {$0$}; + \node[anchor=south] at (2,0) {$1$}; + \node[anchor=south] at (4,0) {$\infty$}; + + \end{scope} \end{tikzpicture} \ No newline at end of file diff --git a/buch/papers/ellfilter/tikz/cd.tikz.tex b/buch/papers/ellfilter/tikz/cd.tikz.tex index cc5852c..0cf2417 100644 --- a/buch/papers/ellfilter/tikz/cd.tikz.tex +++ b/buch/papers/ellfilter/tikz/cd.tikz.tex @@ -67,9 +67,9 @@ \end{scope} \node[zero] at (4,3) (n) {}; - \node[anchor=west] at (n.east) {Zero}; + \node[anchor=west] at (n.east) {Nullstelle}; \node[pole, below=0.25cm of n] (n) {}; - \node[anchor=west] at (n.east) {Pole}; + \node[anchor=west] at (n.east) {Polstelle}; \begin{scope}[yshift=-4cm, xscale=0.75] diff --git a/buch/papers/ellfilter/tikz/cd2.tikz.tex b/buch/papers/ellfilter/tikz/cd2.tikz.tex index bba5789..d4187c4 100644 --- a/buch/papers/ellfilter/tikz/cd2.tikz.tex +++ b/buch/papers/ellfilter/tikz/cd2.tikz.tex @@ -76,4 +76,19 @@ \end{scope} + \begin{scope}[xshift=1cm , yshift=-3cm, xscale=0.75] + + \draw[gray, ->] (-1,0) -- (6,0) node[anchor=west]{$w$}; + + \draw[ultra thick, ->, darkgreen] (0, 0) -- (2, 0); + \draw[ultra thick, ->, orange] (2, 0) -- (3, 0); + \draw[ultra thick, ->, red] (3, 0) -- (5, 0); + + \node[anchor=south] at (0,0) {$0$}; + \node[anchor=south] at (2,0) {$1$}; + \node[anchor=south] at (3,0) {$1/k$}; + \node[anchor=south] at (5,0) {$\infty$}; + + \end{scope} + \end{tikzpicture} \ No newline at end of file diff --git a/buch/papers/ellfilter/tikz/filter.tikz.tex b/buch/papers/ellfilter/tikz/filter.tikz.tex index 05b59b9..769602a 100644 --- a/buch/papers/ellfilter/tikz/filter.tikz.tex +++ b/buch/papers/ellfilter/tikz/filter.tikz.tex @@ -4,22 +4,28 @@ \tikzset{pole/.style={cross out, draw=black, minimum size=(0.15cm-\pgflinewidth), inner sep=0pt, outer sep=0pt}} - \begin{scope}[xscale=2, yscale=2] + \begin{scope}[xscale=3, yscale=2.5] - \fill[ gray!20] (0,0) rectangle (1,0.707); + \fill[darkgreen!15] (0,0) rectangle (1,1); + \node[darkgreen] at (0.5,0.5) {Durchlassbereich}; + \fill[orange!15] (1,0) rectangle (2.5,1); + \node[orange] at (1.75,0.5) {Sperrbereich}; - \draw[gray, ->] (0,-0.25) -- (0,1.25) node[anchor=south]{$|H(\Omega)|$}; - \draw[gray, ->] (-0.25,0) -- (3,0) node[anchor=west]{$\Omega$}; + \draw[gray, ->] (0,0) -- (0,1.25) node[anchor=south]{$|H(\Omega)|$}; + \draw[gray, ->] (0,0) -- (2.75,0) node[anchor=west]{$\Omega$}; - \draw[fill = gray!20] (0,0.707) node[left] {$\sqrt{\frac{1}{1+\varepsilon^2}}$} -| (1,0) node[below] {$\Omega_p$}; + \draw[dashed] (0,0.707) node[left] {$\sqrt{\frac{1}{1+\varepsilon^2}}$} -| (1,0) node[below] {$\Omega_p$}; + \draw[dashed] (0,0.707) node[left] {$\sqrt{\frac{1}{1+\varepsilon^2}}$} -| (1,0) node[below] {$\Omega_p$}; - \draw[fill = gray!20] (0,0.707) node[left] {$\sqrt{\frac{1}{1+\varepsilon^2}}$} -| (1,0) node[below] {$\Omega_p$}; + \node[left] at(0,1) {$1$}; - \begin{scope}[] - \draw[thick, domain=0:2.5, variable=\x, smooth, samples=200] plot - ({\x}, {sqrt(abs(1/ (1 + \x^10)))}); + \draw[red, thick] (0,1) -- (1,1) -- (1,0) -- (2.5,0); - \end{scope} + \node[anchor=north, red] at (0.5,1) {Ideal}; + + \draw[thick, domain=0:2.5, variable=\x, smooth, samples=200] plot + ({\x}, {sqrt(abs(1/ (1 + \x^10)))}); + \node[anchor=south] at (0.5,1) {Butterworth ($N=5$)}; \end{scope} diff --git a/buch/papers/ellfilter/tikz/sn.tikz.tex b/buch/papers/ellfilter/tikz/sn.tikz.tex index c3df8d1..0546fda 100644 --- a/buch/papers/ellfilter/tikz/sn.tikz.tex +++ b/buch/papers/ellfilter/tikz/sn.tikz.tex @@ -70,9 +70,9 @@ \end{scope} \node[zero] at (4,3) (n) {}; - \node[anchor=west] at (n.east) {Zero}; + \node[anchor=west] at (n.east) {Nullstelle}; \node[pole, below=0.25cm of n] (n) {}; - \node[anchor=west] at (n.east) {Pole}; + \node[anchor=west] at (n.east) {Polstelle}; \begin{scope}[yshift=-4cm, xscale=0.75] diff --git a/buch/papers/ellfilter/tschebyscheff.tex b/buch/papers/ellfilter/tschebyscheff.tex index 8a82c5f..639c87c 100644 --- a/buch/papers/ellfilter/tschebyscheff.tex +++ b/buch/papers/ellfilter/tschebyscheff.tex @@ -1,8 +1,8 @@ \section{Tschebyscheff-Filter} -Als Einstieg betrachten wir das Tschebyscheff-Filter, welches sehr verwand ist mit dem elliptischen Filter. -Genauer ausgedrückt sind die Tschebyscheff-1 und -2 Filter Spezialfälle davon. -Der Name des Filters deutet schon an, dass die Tschebyscheff-Polynome $T_N$ für das Filter relevant sind: +Als Einstieg betrachten wir das Tschebyscheff-Filter, welches sehr verwandt ist mit dem elliptischen Filter. +Genauer ausgedrückt erhält man die Tschebyscheff-1 und -2 Filter bei Grenzwerten von Parametern beim elliptischen Filter. +Der Name des Filters deutet schon an, dass die Tschebyscheff-Polynome $T_N$ (siehe auch Kapitel \label{buch:polynome:section:tschebyscheff}) für das Filter relevant sind: \begin{align} T_{0}(x)&=1\\ T_{1}(x)&=x\\ @@ -27,7 +27,7 @@ Abbildung \ref{ellfilter:fig:chebychef_polynomials} zeigt einige Tschebyscheff-P Da der Kosinus begrenzt zwischen $-1$ und $1$ ist, sind auch die Tschebyscheff-Polynome begrenzt. Geht man aber über das Intervall $[-1, 1]$ hinaus, divergieren die Funktionen mit zunehmender Ordnung immer steiler gegen $\pm \infty$. Diese Eigenschaft ist sehr nützlich für ein Filter. -Wenn wir die Tschebyscheff-Polynome quadrieren, passen sie perfekt in die Voraussetzungen für Filterfunktionen, wie es Abbildung \ref{ellfiter:fig:chebychef} demonstriert. +Wenn wir die Tschebyscheff-Polynome quadrieren, passen sie perfekt in die Forderungen für Filterfunktionen, wie es Abbildung \ref{ellfiter:fig:chebychef} demonstriert. \begin{figure} \centering \input{papers/ellfilter/python/F_N_chebychev.pgf} @@ -61,9 +61,9 @@ Die invertierte Funktion des Kosinus kann als bestimmtes Integral dargestellt we } } ~dz - + \frac{\pi}{2} + + \frac{\pi}{2}. \end{align} -Der Integrand oder auch die Ableitung +Der Integrand oder auch die Ableitung von $\cos^{-1}(x)$ \begin{equation} \frac{ -1 @@ -73,13 +73,13 @@ Der Integrand oder auch die Ableitung } } \end{equation} -bestimmt dabei die Richtung, in der die Funktion verläuft. +bestimmt dabei die Richtung, in welche die Funktion verläuft. Der reelle Arcuscosinus is bekanntlich nur für $|z| \leq 1$ definiert. Hier bleibt der Wert unter der Wurzel positiv und das Integral liefert reelle Werte. Doch wenn $|z|$ über 1 hinausgeht, wird der Term unter der Wurzel negativ. Durch die Quadratwurzel entstehen für den Integranden zwei rein komplexe Lösungen. Der Wert des Arcuscosinus verlässt also bei $z= \pm 1$ den reellen Zahlenstrahl und knickt in die komplexe Ebene ab. -Abbildung \ref{ellfilter:fig:arccos} zeigt den $\arccos$ in der komplexen Ebene. +Abbildung \ref{ellfilter:fig:arccos} zeigt den Arcuscosinus in der komplexen Ebene. \begin{figure} \centering \input{papers/ellfilter/tikz/arccos.tikz.tex} @@ -98,9 +98,12 @@ Somit passiert $\cos( N~\cos^{-1}(w))$ im Intervall $[-1, 1]$ $N$ Nullstellen, w \input{papers/ellfilter/tikz/arccos2.tikz.tex} \caption{ $z_1=N \cos^{-1}(w)$-Ebene der Tschebyscheff-Funktion. - Die eingefärbten Pfade sind Verläufe von $w~\forall~[-\infty, \infty]$ für $N = 4$. - Je grösser die Ordnung $N$ gewählt wird, desto mehr Nullstellen werden passiert. + Die eingefärbten Pfade sind Verläufe von $w\in(-\infty, \infty)$ für $N = 4$. + Je grösser die Ordnung $N$ gewählt wird, desto mehr Nullstellen werden passiert die zu Equirippel-Verhalten führen. + Die vertikalen Segmente der Funktion sorgen für das Ansteigen der Funktion gegen $\infty$ nach der Grenzfrequenz. + Die eingezeichneten Nullstellen sind vom zurücktransformierenden Kosinus. } \label{ellfilter:fig:arccos2} \end{figure} -Durch die spezielle Anordnung der Nullstellen hat die Funktion Equirippel-Verhalten und ist dennoch ein Polynom, was sich perfekt für linear Filter eignet. +Durch die spezielle Anordnung der Nullstellen hat die Funktion auf der reellen Achse Equirippel-Verhalten und ist dennoch ein Polynom, was sich perfekt für linear Filter eignet. +Equirippel bedeutet, dass alle lokalen Maxima der Betragsfunktion gleich gross sind. -- cgit v1.2.1 From 1a65f1e2cc20e1dfe5d0d88cf42ee7355c20b1ff Mon Sep 17 00:00:00 2001 From: Fabian <@> Date: Sat, 13 Aug 2022 22:27:32 +0200 Subject: 2. Ueberarbeitung --- buch/papers/0f1/listings/kettenbruchIterativ.c | 60 +++++++------------------ buch/papers/0f1/listings/kettenbruchRekursion.c | 60 ++++++++++++++++++------- buch/papers/0f1/teil1.tex | 2 +- buch/papers/0f1/teil2.tex | 12 +++-- buch/papers/0f1/teil3.tex | 20 ++++----- 5 files changed, 79 insertions(+), 75 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/0f1/listings/kettenbruchIterativ.c b/buch/papers/0f1/listings/kettenbruchIterativ.c index d897b8f..3caaf43 100644 --- a/buch/papers/0f1/listings/kettenbruchIterativ.c +++ b/buch/papers/0f1/listings/kettenbruchIterativ.c @@ -1,53 +1,27 @@ /** - * @brief Calculates the Hypergeometric Function 0F1(;b;z) - * @param b0 in 0F1(;b0;z) - * @param z in 0F1(;b0;z) - * @param n number of itertions (precision) + * @brief Calculates the Hypergeometric Function 0F1(;c;z) + * @param c in 0F1(;c;z) + * @param z in 0F1(;c;z) + * @param k number of itertions (precision) * @return Result */ -static double fractionRekursion0f1(const double c, const double z, unsigned int n) +static double fractionIter0f1(const double c, const double z, unsigned int k) { //declaration double a = 0.0; double b = 0.0; - double Ak = 0.0; - double Bk = 0.0; - double Ak_1 = 0.0; - double Bk_1 = 0.0; - double Ak_2 = 0.0; - double Bk_2 = 0.0; + double abk = 0.0; + double temp = 0.0; - for (unsigned int k = 0; k <= n; ++k) + for (; k > 0; --k) { - if (k == 0) - { - a = 1.0; //a0 - //recursion fomula for A0, B0 - Ak = a; - Bk = 1.0; - } - else if (k == 1) - { - a = 1.0; //a1 - b = z/c; //b1 - //recursion fomula for A1, B1 - Ak = a * Ak_1 + b * 1.0; - Bk = a * Bk_1; - } - else - { - a = 1 + (z / (k * ((k - 1) + c)));//ak - b = -(z / (k * ((k - 1) + c))); //bk - //recursion fomula for Ak, Bk - Ak = a * Ak_1 + b * Ak_2; - Bk = a * Bk_1 + b * Bk_2; - } - //save old values - Ak_2 = Ak_1; - Bk_2 = Bk_1; - Ak_1 = Ak; - Bk_1 = Bk; + abk = z / (k * ((k - 1) + c)); //abk = ak, bk + + a = k > 1 ? (1 + abk) : 1; //a0, a1 + b = k > 1 ? -abk : abk; //b1 + + temp = b / (a + temp); //bk / (ak + last result) } - //approximation fraction - return Ak/Bk; -} + + return a + temp; //a0 + temp +} \ No newline at end of file diff --git a/buch/papers/0f1/listings/kettenbruchRekursion.c b/buch/papers/0f1/listings/kettenbruchRekursion.c index 3caaf43..d897b8f 100644 --- a/buch/papers/0f1/listings/kettenbruchRekursion.c +++ b/buch/papers/0f1/listings/kettenbruchRekursion.c @@ -1,27 +1,53 @@ /** - * @brief Calculates the Hypergeometric Function 0F1(;c;z) - * @param c in 0F1(;c;z) - * @param z in 0F1(;c;z) - * @param k number of itertions (precision) + * @brief Calculates the Hypergeometric Function 0F1(;b;z) + * @param b0 in 0F1(;b0;z) + * @param z in 0F1(;b0;z) + * @param n number of itertions (precision) * @return Result */ -static double fractionIter0f1(const double c, const double z, unsigned int k) +static double fractionRekursion0f1(const double c, const double z, unsigned int n) { //declaration double a = 0.0; double b = 0.0; - double abk = 0.0; - double temp = 0.0; + double Ak = 0.0; + double Bk = 0.0; + double Ak_1 = 0.0; + double Bk_1 = 0.0; + double Ak_2 = 0.0; + double Bk_2 = 0.0; - for (; k > 0; --k) + for (unsigned int k = 0; k <= n; ++k) { - abk = z / (k * ((k - 1) + c)); //abk = ak, bk - - a = k > 1 ? (1 + abk) : 1; //a0, a1 - b = k > 1 ? -abk : abk; //b1 - - temp = b / (a + temp); //bk / (ak + last result) + if (k == 0) + { + a = 1.0; //a0 + //recursion fomula for A0, B0 + Ak = a; + Bk = 1.0; + } + else if (k == 1) + { + a = 1.0; //a1 + b = z/c; //b1 + //recursion fomula for A1, B1 + Ak = a * Ak_1 + b * 1.0; + Bk = a * Bk_1; + } + else + { + a = 1 + (z / (k * ((k - 1) + c)));//ak + b = -(z / (k * ((k - 1) + c))); //bk + //recursion fomula for Ak, Bk + Ak = a * Ak_1 + b * Ak_2; + Bk = a * Bk_1 + b * Bk_2; + } + //save old values + Ak_2 = Ak_1; + Bk_2 = Bk_1; + Ak_1 = Ak; + Bk_1 = Bk; } - - return a + temp; //a0 + temp -} \ No newline at end of file + //approximation fraction + return Ak/Bk; +} diff --git a/buch/papers/0f1/teil1.tex b/buch/papers/0f1/teil1.tex index 50198fc..c0f857d 100644 --- a/buch/papers/0f1/teil1.tex +++ b/buch/papers/0f1/teil1.tex @@ -96,7 +96,7 @@ x\cdot\mathstrut_0F_1\biggl( \qedhere \end{align} -Um die Stabilität der Algorithmen zu $\mathstrut_0F_1$ zu überprüfen, wird in diesem speziellem Fall die Airy Funktion $\operatorname{Ai}(x)$ \eqref{0f1:airy:hypergeometrisch:eq} +Um die Stabilität der Algorithmen zu $\mathstrut_0F_1$ zu überprüfen, wird in dieser Arbeit die Airy Funktion $\operatorname{Ai}(x)$ \eqref{0f1:airy:hypergeometrisch:eq} benutzt. diff --git a/buch/papers/0f1/teil2.tex b/buch/papers/0f1/teil2.tex index 587f63b..06ac53e 100644 --- a/buch/papers/0f1/teil2.tex +++ b/buch/papers/0f1/teil2.tex @@ -11,7 +11,7 @@ Die Unterprogramme wurde jeweils, wie die GNU Scientific Library, in C geschrieb \subsection{Potenzreihe \label{0f1:subsection:potenzreihe}} -Die naheliegendste Lösung ist die Programmierung der Potenzreihe \eqref{0f1:umsetzung:0f1:eq}. Allerdings ist ein Problem dieser Umsetzung (Listing \ref{0f1:listing:potenzreihe}), dass die Fakultät im Nenner schnell grosse Werte annimmt. Dies führt zu einer Bereichsüberschreitung des \verb+double+ Bereiches \cite{0f1:double}. Spätesten ab $k=167$ tritt dieser Falle ein. +Die naheliegendste Lösung ist die Programmierung der Potenzreihe \eqref{0f1:umsetzung:0f1:eq}. \begin{align} \label{0f1:umsetzung:0f1:eq} @@ -30,6 +30,9 @@ Die naheliegendste Lösung ist die Programmierung der Potenzreihe \eqref{0f1:ums \subsection{Kettenbruch \label{0f1:subsection:kettenbruch}} +Eine weitere Variante zur Berechnung von $\mathstrut_0F_1(;c;z)$ ist die Umsetzung als Kettenbruch. +Der Vorteil einer Umsetzung als Kettenbruch gegenüber der Potenzreihe, ist die schnellere Konvergenz. + Ein endlicher Kettenbruch \cite{0f1:wiki-kettenbruch} ist ein Bruch der Form \begin{equation*} a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cdots}}} @@ -44,6 +47,7 @@ Angewendet auf die Funktion $\mathstrut_0F_1$ bedeutet dies \cite{0f1:wiki-fract \mathstrut_0F_1(;c;z) = 1 + \frac{z}{c\cdot1!} + \frac{z^2}{c(c+1)\cdot2!} + \frac{z^3}{c(c+1)(c+2)\cdot3!} + \cdots \end{equation*} Umgeformt ergibt sich folgender Kettenbruch \cite{0f1:wolfram-0f1} +{\color{red}TODO Herleitung} \begin{equation} \label{0f1:math:kettenbruch:0f1:eq} \mathstrut_0F_1(;c;z) = 1 + \cfrac{\cfrac{z}{c}}{1+\cfrac{-\cfrac{z}{2(c+1)}}{1+\cfrac{z}{2(c+1)}+\cfrac{-\cfrac{z}{3(c+2)}}{1+\cfrac{z}{5(c+4)} + \cdots}}}, @@ -115,7 +119,7 @@ an, ergibt sich folgende Matrixdarstellungen: \begin{pmatrix} b_k\\ a_k - \end{pmatrix} + \end{pmatrix}. \end{align*} Nach vollständiger Induktion ergibt sich für den Schritt $k$, die Matrix \begin{equation} @@ -142,7 +146,7 @@ berechnet werden. \subsubsection{Lösung} -Die Berechnung von $A_k, B_k$ \eqref{0f1:math:matrix:ende:eq} kann man auch ohne die Matrizenschreibweise \cite{0f1:kettenbrueche} aufschreiben: +Die Berechnung von $A_k, B_k$ gemäss \eqref{0f1:math:matrix:ende:eq} kann man auch ohne die Matrizenschreibweise \cite{0f1:kettenbrueche} aufschreiben: \begin{itemize} \item Startbedingungen: \begin{align*} @@ -161,7 +165,7 @@ B_{k+1} &= B_{k-1} \cdot b_k + B_k \cdot a_k \end{aligned} \] \item -Näherungsbruch: \qquad$\displaystyle\frac{A_k}{B_k}$ +Näherungsbruch: \qquad$\displaystyle\frac{A_k}{B_k}$. \end{itemize} Ein grosser Vorteil dieser Umsetzung als Rekursionsformel \eqref{0f1:listing:kettenbruchRekursion} ist, dass im Vergleich zum Code (Listing \ref{0f1:listing:kettenbruchIterativ}) eine Division gespart werden kann und somit weniger Rundungsfehler entstehen können. diff --git a/buch/papers/0f1/teil3.tex b/buch/papers/0f1/teil3.tex index 00d4182..2942a0b 100644 --- a/buch/papers/0f1/teil3.tex +++ b/buch/papers/0f1/teil3.tex @@ -13,25 +13,25 @@ Ebenso kann festgestellt werden, dass je grösser der Wert $z$ in $\mathstrut_0F \subsection{Konvergenz \label{0f1:subsection:konvergenz}} -Es zeigt sich in Abbildung \ref{0f1:ausblick:plot:airy:konvergenz}, dass schon nach drei Iterationen ($k = 3$) die Funktionen schon genaue Resultate im Bereich von $-2$ bis $2$ liefert. Ebenso kann festgestellt werden, dass der Kettenbruch schneller konvergiert und im positiven Bereich sogar mit der Referenzfunktion $\operatorname{Ai}(x)$ übereinstimmt. Da die Rekursionsformel eine Abwandlung des Kettenbruches ist, verhalten sich die Funktionen in diesem Fall gleich. +Es zeigt sich in Abbildung \ref{0f1:ausblick:plot:airy:konvergenz}, dass nach drei Iterationen ($k = 3$) die Funktionen genaue Resultate im Bereich von $-2$ bis $2$ liefert. Ebenso kann festgestellt werden, dass der Kettenbruch schneller konvergiert und im positiven Bereich sogar mit der Referenzfunktion $\operatorname{Ai}(x)$ übereinstimmt. Da die Rekursionsformel eine Abwandlung des Kettenbruches ist, verhalten sich die Funktionen in diesem Fall gleich. -Erst wenn mehrerer Iterationen gemacht werden, um die Genauigkeit zu verbessern, ist der Kettenbruch den anderen zwei Algorithmen, bezüglich Konvergenz überlegen. -Interessant ist auch, dass die Rekursionsformel nahezu gleich schnell wie die Potenzreihe konvergiert, aber sich danach, wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} zu beobachten ist, einschwingt. Dieses Verhalten ist auch bei grösseren $z$ zu beobachten, allerdings ist dann die Differenz zwischen dem ersten lokalen Minimum von k bis zum Abbruch kleiner. -Dieses Phänomen ist auf die Lösung der Rekursionsformel \eqref{0f1:math:loesung:eq} zurück zu führen. Da im Gegensatz die ganz kleinen Werte nicht zu einer Konvergenz wie beim Kettenbruch führen, sondern sich noch eine Zeit lang durch die Multiplikation aufschwingen. +Erst wenn mehrerer Iterationen gemacht werden, um die Genauigkeit zu verbessern, ist der Kettenbruch den anderen zwei Algorithmen bezüglich Konvergenz überlegen. +Interessant ist auch, dass die Rekursionsformel nahezu gleich schnell wie die Potenzreihe konvergiert, aber sich danach, wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} zu beobachten ist, einschwingt. Dieses Verhalten ist auch bei grösseren $z$ zu beobachten, allerdings ist dann die Differenz zwischen dem ersten lokalen Minimum von $k$ bis zum Abbruch kleiner. +Dieses Phänomen ist auf die Lösung der Rekursionsformel \eqref{0f1:math:matrix:ende:eq} zurück zu führen. Da im Gegensatz die ganz kleinen Werte nicht zu einer Konvergenz wie beim Kettenbruch führen, sondern sich noch eine Zeit lang durch die Multiplikation aufschwingen. -Ist $z$ negativ wie im Abbildung \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu einer Gegenseitigen Kompensation von negativen und positiven Termen so bricht die Rekursionsformel hier zusammen mit der Potenzreihe ab. -Die ansteigende Differenz mit anschliessender, ist aufgrund der sich alternierenden Termen mit wechselnden Vorzeichens zu erklären. +Ist $z$ negativ wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu einer gegenseitigen Kompensation von negativen und positiven Termen. Dies führt dazu, dass die Rekursionsformel zusammen mit der Potenzreihe abbricht. +Die ansteigende Differenz mit anschliessendendem Einschwingen, ist aufgrund der sich alternierenden Termen mit wechselnden Vorzeichens zu erklären. \subsection{Stabilität \label{0f1:subsection:Stabilitaet}} -Verändert sich der Wert von z in $\mathstrut_0F_1(;c;z)$ gegen grössere positive Werte, wie zum Beispiel $c = 800$ liefert die Kettenbruch-Funktion (Listing \ref{0f1:listing:kettenbruchIterativ}) \verb+inf+ zurück. Dies könnte durch ein Abbruchkriterien abgefangen werden. Allerdings würde das, bei grossen Werten zulasten der Genauigkeit gehen. Trotzdem könnte, je nach Anwendung, auf ein paar Nachkommastellen verzichtet werden. +Verändert sich der Wert von $z$ in $\mathstrut_0F_1(;c;z)$ gegen grössere positive Werte, wie zum Beispiel $c = 800$ liefert die Kettenbruch-Funktion (Listing \ref{0f1:listing:kettenbruchIterativ}) \verb+inf+ zurück. Dies könnte durch ein Abbruchkriterien abgefangen werden. Allerdings würde das, bei grossen Werten zulasten der Genauigkeit gehen. Trotzdem könnte, je nach Anwendung, auf ein paar Nachkommastellen verzichtet werden. -Wohingegen die Potenzreihe (Listing \ref{0f1:listing:potenzreihe}) das Problem hat, dass je mehr Terme berechnet werden, desto schneller wächst die Fakultät und irgendwann gibt es eine Bereichsüberschreitung von \verb+double+. Schlussendlich gibt das Unterprogramm das Resultat \verb+-nan(ind)+ zurück. +Wohingegen die Potenzreihe (Listing \ref{0f1:listing:potenzreihe}) das Problem hat, dass je mehr Terme berechnet werden, desto schneller wächst die Fakultät im Nenner. Dies führt zu einer Bereichsüberschreitung des \verb+double+ Bereiches \cite{0f1:double}, der spätesten ab $k=167$ eintritt. Schlussendlich gibt das Unterprogramm das Resultat \verb+-nan(ind)+ zurück. Die Rekursionformel \eqref{0f1:listing:kettenbruchRekursion} liefert für sehr grosse positive Werte die genausten Ergebnisse, verglichen mit der GNU Scientific Library. Wie schon vermutet ist die Rekursionsformel, im positivem Bereich, der stabilste Algorithmus. Um die Stabilität zu gewährleisten, muss wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, die Iterationstiefe $k$ genug gross gewählt werden. -Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Grund dafür ist die Potenz von z, was zum Phänomen der Auslöschung \cite{0f1:SeminarNumerik} führt. Schön zu beobachten ist dies in der Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} mit der Airy-Funktion als Test. So sind sowohl der Kettenbruch, als auch die Rekursionsformel bis ungefähr $\frac{-15^3}{9}$ stabil. Dies macht auch Sinn, da beide auf der gleichen mathematischen Grundlage basieren. Danach verhält sich allerdings die Instabilität unterschiedlich. Das unterschiedliche Verhalten kann damit erklärt werden, dass beim Kettenbruch jeweils eine zusätzliche Division stattfindet. Diese Unterschiede sind auch in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} festzustellen. - +{\color{red}TODO Abb. 20.3} +Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Grund dafür ist die Potenz von $z$, was zum Phänomen der Auslöschung \cite{0f1:SeminarNumerik} führt. Schön zu beobachten ist dies in der Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} mit der Airy-Funktion als Test. So sind sowohl der Kettenbruch, als auch die Rekursionsformel bis ungefähr $\frac{-15^3}{9}$ stabil. Dies macht auch Sinn, da beide auf der gleichen mathematischen Grundlage basieren. Danach verhält sich allerdings die Instabilität unterschiedlich. Das unterschiedliche Verhalten kann damit erklärt werden, dass beim Kettenbruch jeweils eine zusätzliche Division stattfindet. Diese Unterschiede sind auch in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} festzustellen. \begin{figure} \centering -- cgit v1.2.1 From bc0c70fdd1bd92d48fc38b17877d6d8515253225 Mon Sep 17 00:00:00 2001 From: Nicolas Tobler Date: Sun, 14 Aug 2022 15:42:31 +0200 Subject: corrections --- buch/papers/ellfilter/einleitung.tex | 4 ++-- buch/papers/ellfilter/elliptic.tex | 4 ++-- buch/papers/ellfilter/jacobi.tex | 15 ++------------- 3 files changed, 6 insertions(+), 17 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/ellfilter/einleitung.tex b/buch/papers/ellfilter/einleitung.tex index cf57698..ae7127f 100644 --- a/buch/papers/ellfilter/einleitung.tex +++ b/buch/papers/ellfilter/einleitung.tex @@ -13,7 +13,7 @@ Ein breit angewendeter Filtertyp ist das Tiefpassfilter, welches beabsichtigt al Der Rest soll dabei unverändert passieren. Aus dem Tiefpassifilter können dann durch Transformationen auch Hochpassfilter, Bandpassfilter und Bandsperren realisiert werden. Ein solches Filter hat idealerweise die Frequenzantwort -\begin{equation} \label{ellfilter:eq:h_omega} +\begin{equation} H(\Omega) = \begin{cases} 1 & \Omega < \Omega_p \\ @@ -32,7 +32,7 @@ Aus diesem Grund sind realisierbare Approximationen gesucht. Jede Approximation wird einen kontinuierlichen Übergang zwischen Durchlassbereich und Sperrbereich aufweisen. Oft wird dabei der Faktor $1/\sqrt{2}$ als Schwelle zwischen den beiden Bereichen gewählt. Somit lassen sich lineare Tiefpassfilter mit folgender Funktion zusammenfassen: -\begin{equation} \label{ellfilter:eq:h_omega} +\begin{equation} | H(\Omega)|^2 = \frac{1}{1 + \varepsilon_p^2 F_N^2(w)}, \quad w=\frac{\Omega}{\Omega_p}, \end{equation} wobei $F_N(w)$ eine rationale Funktion ist, $|F_N(w)| \leq 1 ~\forall~ |w| \leq 1$ erfüllt und für $|w| \geq 1$ möglichst schnell divergiert. diff --git a/buch/papers/ellfilter/elliptic.tex b/buch/papers/ellfilter/elliptic.tex index 89a2d7a..67bcca0 100644 --- a/buch/papers/ellfilter/elliptic.tex +++ b/buch/papers/ellfilter/elliptic.tex @@ -1,6 +1,6 @@ \section{Elliptische rationale Funktionen} -Kommen wir nun zum eigentlichen Teil dieses Papers, den elliptischen rationalen Funktionen \ref{ellfilter:bib:orfanidis} +Kommen wir nun zum eigentlichen Teil dieses Papers, den elliptischen rationalen Funktionen \cite{ellfilter:bib:orfanidis} \begin{align} R_N(\xi, w) &= \cd \left(N~f_1(\xi)~\cd^{-1}(w, 1/\xi), f_2(\xi)\right) \label{ellfilter:eq:elliptic}\\ &= \cd \left(N~\frac{K_1}{K}~\cd^{-1}(w, k), k_1\right) , \quad k= 1/\xi, k_1 = 1/f(\xi) \\ @@ -80,7 +80,7 @@ k_1 = k^N \prod_{i=1}^L \sn^4 \Bigg( \frac{2i - 1}{N} K, k \Bigg), \quad \text{wobei} \quad N = 2L+r. \end{equation} -Die Herleitung ist sehr umfassend und wird in \ref{ellfilter:bib:orfanidis} im Detail angeschaut. +Die Herleitung ist sehr umfassend und wird in \cite{ellfilter:bib:orfanidis} im Detail angeschaut. % \begin{figure} % \centering diff --git a/buch/papers/ellfilter/jacobi.tex b/buch/papers/ellfilter/jacobi.tex index fae6b31..567bbcc 100644 --- a/buch/papers/ellfilter/jacobi.tex +++ b/buch/papers/ellfilter/jacobi.tex @@ -1,7 +1,5 @@ \section{Jacobische elliptische Funktionen} -%TODO $z$ or $u$ for parameter? - Für das elliptische Filter werden, wie es der Name bereits deutet, elliptische Funktionen gebraucht. Wie die trigonometrischen Funktionen Zusammenhänge eines Kreises darlegen, beschreiben die elliptischen Funktionen Ellipsen. Es ist daher naheliegend, dass Kosinus des Tschebyscheff-Filters mit einem elliptischen Pendant ausgetauscht werden könnte. @@ -29,15 +27,6 @@ Das Winkelargument $z$ kann durch das elliptische Integral erster Art 1-k^2 \sin^2 \theta } } - % = - % \int_{0}^{\phi} - % \frac{ - % dt - % }{ - % \sqrt{ - % (1-t^2)(1-k^2 t^2) - % } - % } %TODO which is right? are both functions from phi? \end{equation} mit dem Winkel $\phi$ in Verbindung gebracht werden. @@ -170,7 +159,7 @@ Die erste ist die gleiche wie beim $\cos^{-1}(x)$ nämlich bei $t = \pm 1$. Der erste Term unter der Wurzel wird dann negativ, während der zweite noch positiv ist, da $k \leq 1$. Ab diesem Punkt knickt die Funktion in die imaginäre Richtung ab. Bei $t = 1/k$ ist auch der zweite Term negativ und die Funktion verläuft in die negative reelle Richtung. -Abbildung \label{ellfilter:fig:sn} zeigt den Verlauf der Funktion in der komplexen Ebene. +Abbildung \ref{ellfilter:fig:sn} zeigt den Verlauf der Funktion in der komplexen Ebene. \begin{figure} \centering \input{papers/ellfilter/tikz/sn.tikz.tex} @@ -180,7 +169,7 @@ Abbildung \label{ellfilter:fig:sn} zeigt den Verlauf der Funktion in der komplex } \label{ellfilter:fig:sn} \end{figure} -In der reellen Richtung ist sie $4K(k)$-periodisch und in der imaginären Richtung $4K^\prime(k)$-periodisch, wobei $K^\prime$ das komplemenäre vollständige Elliptische Integral ist: +In der reellen Richtung ist sie $4K(k)$-periodisch und in der imaginären Richtung $4K^\prime(k)$-periodisch, wobei $K^\prime$ das komplementäre vollständige Elliptische Integral ist: \begin{equation} K^\prime(k) = -- cgit v1.2.1 From a1a811ef08f16f61382f4f7eecc45fd71bd1e1d6 Mon Sep 17 00:00:00 2001 From: tim30b Date: Mon, 15 Aug 2022 00:50:56 +0200 Subject: gegengelesene Fehler angepasst --- buch/papers/kreismembran/teil0.tex | 10 +++++----- buch/papers/kreismembran/teil1.tex | 2 +- buch/papers/kreismembran/teil2.tex | 8 ++++---- buch/papers/kreismembran/teil3.tex | 6 +++--- buch/papers/kreismembran/teil4.tex | 16 ++++++++-------- 5 files changed, 21 insertions(+), 21 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil0.tex b/buch/papers/kreismembran/teil0.tex index a0a4152..c6dac06 100644 --- a/buch/papers/kreismembran/teil0.tex +++ b/buch/papers/kreismembran/teil0.tex @@ -10,7 +10,7 @@ Ein dünnes Blättchen aus Metall zeig jedoch nicht die selben dynamischen Eigen Beschreibt man das dynamische Verhalten, muss zwischen einer dünnen Platte und einer Membrane unterschieden werden \cite{kreismembran:membrane_vs_thin_plate}. Eine dünne Platte zum Beispiel aus Metall, wirkt selbst entgegen ihrer Deformation sobald sie gekrümmt wird. Eine Membran auf der anderen Seite besteht aus einem Material, welches sich ohne Kraftaufwand verbiegen lässt wie zum Beispiel Papier. -Bevor Papier als schwingende Membran betrachtet werden kann, wird jedoch noch eine Spannung $ T $ benötigt welche das Material daran hindert, aus der Ruhelage gebracht zu werden. +Bevor Papier als schwingende Membran betrachtet werden kann, wird jedoch noch eine Spannung $ T $ benötigt, welche das Material daran hindert, aus der Ruhelage gebracht zu werden. Ein geläufiges Beispiel einer Kreismembran ist eine runde Trommel. Sie besteht herkömmlicherweise aus einem Leder (Fell), welches auf einen offenen Zylinder (Zargen) aufgespannt wird. @@ -36,8 +36,8 @@ Das untersuchte Modell erfüllt folgende Eigenschaften: \end{enumerate} -\subsection{Wellengleichung} Um die Wellengleichung einer Membran herzuleiten wird vorerst eine schwingende Saite betrachtet. -Es lohnt sich das Verhalten einer Saite zu beschreiben, da eine Saite das selbe Verhalten wie eine Membran aufweist mit dem Unterschied einer fehlenden Dimension. +\subsection{Wellengleichung} Um die Wellengleichung einer Membran herzuleiten, wird vorerst eine schwingende Saite betrachtet. +Es lohnt sich, das Verhalten einer Saite zu beschreiben, da eine Saite dasselbe Verhalten wie eine Membran aufweist, mit dem Unterschied einer fehlenden Dimension. Die Verbindung zwischen Membran und Saite ist intuitiv ersichtlich, stellt man sich einen Querschnitt einer Trommel vor. \begin{figure} @@ -49,7 +49,7 @@ Die Verbindung zwischen Membran und Saite ist intuitiv ersichtlich, stellt man s \end{figure} In Abbildung \ref{kreismembran:im:Saite} ist ein infinitesimales Stück einer Saite mit Länge $ dx $ skizziert. -Wie für die Membran ist die Annahme iii) gültig, keine Bewegung entlang der $ x $-Achse. +Wie für die Membran ist die Annahme iii) gültig, es entsteht keine Bewegung entlang der $ x $-Achse. Um dies zu erfüllen, muss der Punkt $ P_1 $ gleich stark entgegen der $ x $-Achse gezogen werden wie der Punkt $ P_2 $ in Richtung der $ x $-Achse gezogen wird. Ist $ T_1 $ die Kraft, welche mit Winkel $ \alpha $ auf Punkt $ P_1 $ wirkt sowie $ T_2 $ und $ \beta$ das analoge für Punkt $ P_2 $ ist, so können die Kräfte \begin{equation}\label{kreismembran:eq:no_translation} @@ -85,7 +85,7 @@ Durch die Division mit $ dx $ entsteht Auf der linken Seite der Gleichung wird die Differenz der Steigungen durch die Intervalllänge geteilt. Wenn $ dx $ als unendlich kleines Stück betrachtet wird, ergibt sich als Grenzwert die zweite Ableitung von $ u(x,t) $ nach $ x $. Der Term $ \frac{\rho}{T} $ wird durch $ c^2 $ ersetzt, da der Bruch für eine gegebene Membran eine positive Konstante sein muss. -Somit resultiert die in der Literatur gebräuchliche Form +Damit resultiert die in der Literatur gebräuchliche Form \begin{equation} \label{kreismembran:Ausgang_DGL} \frac{1}{c^2}\frac{\partial^2u}{\partial t^2} = \Delta u. diff --git a/buch/papers/kreismembran/teil1.tex b/buch/papers/kreismembran/teil1.tex index a872ed1..f6ba7d1 100644 --- a/buch/papers/kreismembran/teil1.tex +++ b/buch/papers/kreismembran/teil1.tex @@ -83,7 +83,7 @@ Die Gleichung für $F$ hat die Gestalt (Verweis auf \label{buch:differentialglei r^2F''(r) + rF'(r) + (\kappa^2 r^2 - n^2)F(r) = 0 \label{eq:2nd_degree_PDE} \end{align} -Wir bereits in Kapitel \ref{buch:differntialgleichungen:section:bessel} gezeigt, sind die Bessel-Funktionen +Wie bereits in Kapitel \ref{buch:differntialgleichungen:section:bessel} gezeigt, sind die Bessel-Funktionen \begin{equation*} J_{\nu}(x) = r^\nu \displaystyle\sum_{m=0}^{\infty} \frac{(-1)^m x^{2m}}{2^{2m+\nu}m! \Gamma (\nu + m+1)} \end{equation*} diff --git a/buch/papers/kreismembran/teil2.tex b/buch/papers/kreismembran/teil2.tex index 133ee31..ec27bd3 100644 --- a/buch/papers/kreismembran/teil2.tex +++ b/buch/papers/kreismembran/teil2.tex @@ -7,7 +7,7 @@ Hermann Hankel (1839--1873) war ein deutscher Mathematiker, der für seinen Beitrag zur mathematischen Analysis und insbesondere für die nach ihm benannte Transformation bekannt ist. Diese Transformation tritt bei der Untersuchung von Funktionen auf, die nur von der Entfernung des Ursprungs abhängen. -Er studierte auch Funktionen, jetzt Hankel- oder Bessel- Funktionen genannt, der dritten Art. +Er untersuchte auch Funktionen, jetzt Hankel- oder Bessel- Funktionen genannt, der dritten Art. Die Hankel-Transformation, die die Bessel-Funktion enthält, taucht natürlich bei achsensymmetrischen Problemen auf, die in zylindrischen Polarkoordinaten formuliert sind. In diesem Abschnitt werden die Theorie der Transformation und einige Eigenschaften der Grundoperationen erläutert. @@ -17,12 +17,12 @@ Wir führen die Definition der Hankel-Transformation \cite{lokenath_debnath_inte \mathscr{F}\{f(x,y)\} & = F(k,l)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i( \bm{\kappa}\cdot \mathbf{r})}f(x,y) \; dx \; dy,\label{equation:fourier_transform}\\ \mathscr{F}^{-1}\{F(x,y)\} & = f(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{i(\bm{\kappa}\cdot \mathbf{r})}F(k,l) \; dx \; dy \label{equation:inv_fourier_transform} \end{align} -wo $\mathbf{r}=(x,y)$ und $\bm{\kappa}=(k,l)$. Polarkoordinaten sind für diese Art von Problem am besten geeignet, mit $(x,y)=r(\cos\theta,\sin\theta)$ und $(k,l)=\kappa(\cos\phi,\sin\phi)$ findet man $\bm{\kappa}\cdot\mathbf{r}=\kappa r(\cos(\theta-\phi))$ und danach: +definiert ist, wobei $\mathbf{r}=(x,y)$ und $\bm{\kappa}=(k,l)$. Polarkoordinaten sind für diese Art von Problem am besten geeignet. Mit $(x,y)=r(\cos\theta,\sin\theta)$ und $(k,l)=\kappa(\cos\phi,\sin\phi)$ findet man $\bm{\kappa}\cdot\mathbf{r}=\kappa r(\cos(\theta-\phi))$ und danach: \begin{align} F(k,\phi)=\frac{1}{2\pi}\int_{0}^{\infty}r \; dr \int_{0}^{2\pi}e^{-ikr\cos(\theta-\phi)}f(r,\theta) \; d\phi. \label{equation:F_ohne_variable_wechsel} \end{align} -Dann wird angenommen dass, $f(r,\theta)=e^{in\theta}f(r)$, was keine strenge Einschränkung ist, weil die \textit{Fourier-Theorie} besagt, dass sich jede Funktion durch Überlagerung solcher Terme darstellen lässt. Es wird auch eine Änderung der Variabeln vorgenommen $\theta-\phi=\alpha-\frac{\pi}{2}$, um \eqref{equation:F_ohne_variable_wechsel} zu reduzieren: +Dann wird angenommen, dass $f(r,\theta)=e^{in\theta}f(r)$, was keine strenge Einschränkung ist, weil die \textit{Fourier-Theorie} besagt, dass sich jede Funktion durch Überlagerung solcher Terme darstellen lässt. Es wird auch eine Änderung der Variabeln vorgenommen $\theta-\phi=\alpha-\frac{\pi}{2}$, um \eqref{equation:F_ohne_variable_wechsel} zu reduzieren: \begin{align} F(k,\phi)=\frac{1}{2\pi}\int_{0}^{\infty}rf(r) \; dr \int_{\phi_{0}}^{2\pi+\phi_{0}}e^{in(\phi-\frac{\pi}{2})+i(n\alpha-kr\sin\alpha)} \; d\alpha, \label{equation:F_ohne_bessel} @@ -69,7 +69,7 @@ verwendet werden, um die Hankel-Transformation \eqref{equation:hankel} und ihre Insbesondere die Hankel-Transformation der nullten Ordnung ($n=0$) und der ersten Ordnung ($n=1$) sind häufig nützlich, um Lösungen für Probleme mit der Laplace Gleichung in einer achsensymmetrischen zylindrischen Geometrie zu finden. \subsection{Operatoreigenschaften der Hankel-Transformation \label{sub:op_properties_hankel}} -In diesem Kapitel werden die operativen Eigenschaften der Hankel-Transformation aufgeführt. Die Beweise für ihre Gültigkeit werden jedoch nicht analysiert, dies ist in Buch \textit{Integral Tansforms and Their Applications} \cite{lokenath_debnath_integral_2015} zu finden. +In diesem Kapitel werden die operativen Eigenschaften der Hankel-Transformation aufgeführt. Die Beweise für ihre Gültigkeit werden jedoch nicht analysiert, diese sind im Buch \textit{Integral Tansforms and Their Applications} \cite{lokenath_debnath_integral_2015} zu finden. \begin{satz}{Skalierung:} Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann gilt: diff --git a/buch/papers/kreismembran/teil3.tex b/buch/papers/kreismembran/teil3.tex index 468ee24..a9dcd95 100644 --- a/buch/papers/kreismembran/teil3.tex +++ b/buch/papers/kreismembran/teil3.tex @@ -17,7 +17,7 @@ Führt man also das Konzept einer unendlichen und achsensymmetrischen Membran ei + \frac{1}{r} \frac{\partial u}{\partial r} \right), \quad 00 \label{eq:PDE_inf_membane} \\ - u(r,0)=f(r), \quad u_t(r,0) = g(r), \quad \text{für} \quad 0 Date: Mon, 15 Aug 2022 06:54:23 +0200 Subject: fix physics dependency --- buch/papers/kra/anwendung.tex | 5 ++++- buch/papers/kra/images/Makefile | 9 +++++++++ buch/papers/kra/images/simple.pdf | Bin 0 -> 23130 bytes buch/papers/kra/images/simple.tex | 24 ++++++++++++++++++++++++ buch/papers/kra/images/simple_mass_spring.tex | 12 ++++++------ buch/papers/kra/packages.tex | 6 +----- 6 files changed, 44 insertions(+), 12 deletions(-) create mode 100644 buch/papers/kra/images/Makefile create mode 100644 buch/papers/kra/images/simple.pdf create mode 100644 buch/papers/kra/images/simple.tex (limited to 'buch/papers') diff --git a/buch/papers/kra/anwendung.tex b/buch/papers/kra/anwendung.tex index 0deaf3c..6383984 100644 --- a/buch/papers/kra/anwendung.tex +++ b/buch/papers/kra/anwendung.tex @@ -19,7 +19,10 @@ Die Funktion die diese Differentialgleichung löst, ist die harmonische Schwingu x(t) = A \cos(\omega_0 t + \Phi), \quad \omega_0 = \sqrt{\frac{k}{m}} \end{equation} \begin{figure} - \input{papers/kra/images/simple_mass_spring.tex} + % move image to standalone because the physics package is + % incompatible with underbrace + \includegraphics{papers/kra/images/simple.pdf} + %\input{papers/kra/images/simple_mass_spring.tex} \caption{Einfaches Feder-Masse-System.} \label{kra:fig:simple_mass_spring} \end{figure} diff --git a/buch/papers/kra/images/Makefile b/buch/papers/kra/images/Makefile new file mode 100644 index 0000000..ef226a9 --- /dev/null +++ b/buch/papers/kra/images/Makefile @@ -0,0 +1,9 @@ +# +# Makefile -- build standalone images +# +# (c) 2022 Prof Dr Andreas Müller +# +all: simple.pdf + +simple.pdf: simple.tex simple_mass_spring.tex + pdflatex simple.tex diff --git a/buch/papers/kra/images/simple.pdf b/buch/papers/kra/images/simple.pdf new file mode 100644 index 0000000..4351518 Binary files /dev/null and b/buch/papers/kra/images/simple.pdf differ diff --git a/buch/papers/kra/images/simple.tex b/buch/papers/kra/images/simple.tex new file mode 100644 index 0000000..3bdde27 --- /dev/null +++ b/buch/papers/kra/images/simple.tex @@ -0,0 +1,24 @@ +% +% tikztemplate.tex -- template for standalon tikz images +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{times} +\usepackage{txfonts} +\usepackage{pgfplots} +\pgfplotsset{compat=1.16} +\usepackage[outline]{contour} +\usepackage{csvsimple} +\usepackage{physics} +\usetikzlibrary{arrows,intersections,math} +\usetikzlibrary{patterns} +\usetikzlibrary{snakes} +\usetikzlibrary{arrows.meta} +\usetikzlibrary{decorations} +\usetikzlibrary{decorations.markings} +\begin{document} +\input{simple_mass_spring.tex} +\end{document} + diff --git a/buch/papers/kra/images/simple_mass_spring.tex b/buch/papers/kra/images/simple_mass_spring.tex index e0e869a..868362d 100644 --- a/buch/papers/kra/images/simple_mass_spring.tex +++ b/buch/papers/kra/images/simple_mass_spring.tex @@ -6,7 +6,7 @@ \tikzstyle{mass}=[line width=0.6,red!30!black,fill=red!40!black!10,rounded corners=1,top color=red!40!black!20,bottom color=red!40!black!10,shading angle=20] \tikzstyle{spring}=[line width=0.8,blue!7!black!80,snake=coil,segment amplitude=5,line cap=round] -\begin{tikzpicture}[scale=2] +\begin{tikzpicture}[scale=2,>=latex] \newcommand{\ticks}[2] { % arguments: x, y coordinates @@ -47,9 +47,9 @@ % create springs \draw[spring, segment length=(\xMass1 - \wWall) * \springscale] (\originX1, \originY1) ++ - (\wWall, \wWall + \hMass / 2) --++ (\xMass1 - \wWall, 0) node[midway,above=0.2] {$k$}; + (\wWall, \wWall + \hMass / 2) --++ (\xMass1 - \wWall, 0) node[midway,above=3.5] {$k$}; \draw[spring, segment length=(\xMass2 - \wWall) * \springscale] (\originX2, \originY2) ++ - (\wWall, \wWall + \hMass / 2) --++ (\xMass2 - \wWall, 0) node[midway,above=0.2] {$k$}; + (\wWall, \wWall + \hMass / 2) --++ (\xMass2 - \wWall, 0) node[midway,above=3.5] {$k$}; % create vertical measurement line \draw[vmline] (\xMass1, \xAxisYpos) --+(0, \originY1 + \wWall); @@ -57,10 +57,10 @@ \draw[vmline] (\wWall, \originY1+\wWall) --(\wWall, \originY2 + \hWall); % create horizontal measurement line - \draw[hmline] (\wWall, \xAxisYpos + 0.2) -- (\xMass1, \xAxisYpos + 0.2) node[midway,fill=white,inner sep=0] {$\ell_0$}; + \draw[hmline] (\wWall, \xAxisYpos + 0.2) -- (\xMass1, \xAxisYpos + 0.2) node[midway,fill=white,inner sep=0] {$l_0$}; \draw[hmline] (\xMass1, \xAxisYpos + 0.2) -- (\xMass2, \xAxisYpos + 0.2) node[midway,fill=white,inner sep=0] {$\Delta_{x}$}; - \draw[hmline] (\wWall, \xAxisYpos - 0.3) -- (\xMass2, \xAxisYpos - 0.3) node[midway,fill=white,inner sep=0] {$\ell_{1}$}; + \draw[hmline] (\wWall, \xAxisYpos - 0.3) -- (\xMass2, \xAxisYpos - 0.3) node[midway,fill=white,inner sep=0] {$l_{1}$}; % create force arrow \draw[->,blue, very thick,line cap=round] (\xMass2 + \wMass / 2, \originY2 + \wWall + \hMass + 0.15) node[above] {$\vb{F_{R}}$} --+ (-0.5, 0); -\end{tikzpicture} \ No newline at end of file +\end{tikzpicture} diff --git a/buch/papers/kra/packages.tex b/buch/papers/kra/packages.tex index b16f074..56c48d9 100644 --- a/buch/papers/kra/packages.tex +++ b/buch/papers/kra/packages.tex @@ -8,15 +8,11 @@ % following example %\usepackage{packagename} -\usepackage{physics} -\usepackage{pgfplots} -\usepackage{tikz} +%\usepackage{physics} \usepackage[outline]{contour} \pgfplotsset{compat=1.16} \usetikzlibrary{patterns} \usetikzlibrary{snakes} -\usetikzlibrary{math} \usetikzlibrary{arrows.meta} \usetikzlibrary{decorations} \usetikzlibrary{decorations.markings} -\usetikzlibrary{calc} \ No newline at end of file -- cgit v1.2.1 From 1b634d9be2a8536dbc55b3ac3b60efda6a5a16c8 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 15 Aug 2022 09:46:33 +0200 Subject: Corrected some errors. --- .../sturmliouville/waermeleitung_beispiel.tex | 45 +++++++++++----------- 1 file changed, 23 insertions(+), 22 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 5bd5ce2..5d178c2 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -11,8 +11,8 @@ betrachtet und wie das Sturm-Liouville-Problem bei der Beschreibung dieses physikalischen Phänomenes auftritt. Zunächst wird ein eindimensionaler homogener Stab der Länge $l$ und -Wärmeleitkoeffizient $\kappa$. -Somit ergibt sich für das Wärmeleitungsproblem +Wärmeleitkoeffizient $\kappa$ betrachtet. +Es ergibt sich für das Wärmeleitungsproblem die partielle Differentialgleichung \begin{equation} \label{eq:slp-example-fourier-heat-equation} @@ -26,13 +26,14 @@ Stab beschreibt, werden zusätzliche Bedingungen benötigt, um beispielsweise die Lösung für einen Stab zu finden, bei dem die Enden auf konstanter Tempreatur gehalten werden. -%%%%%%%%%%%%% Randbedingungen für Stab mit konstanten Endtemperaturen %%%%%%%%% - -\subsubsection{Stab mit Enden auf konstanter Temperatur} +% +% Randbedingungen für Stab mit konstanten Endtemperaturen +% +\subsubsection{Randbedingungen für Stab mit Enden auf konstanter Temperatur} Die Enden des Stabes auf konstanter Temperatur zu halten bedeutet, dass die Lösungsfunktion $u(t,x)$ bei $x = 0$ und $x = l$ nur die vorgegebene -Temperatur zurückgeben darf. +Temperatur zurückgeben darf. Diese wird einfachheitshalber als $0$ angenomen. Es folgen nun \begin{equation} \label{eq:slp-example-fourier-boundary-condition-ends-constant} @@ -44,12 +45,14 @@ Es folgen nun \end{equation} als Randbedingungen. -%%%%%%%%%%%%% Randbedingungen für Stab mit isolierten Enden %%%%%%%%%%%%%%%%%%% +% +% Randbedingungen für Stab mit isolierten Enden +% -\subsubsection{Stab mit isolierten Enden} +\subsubsection{Randbedingungen für Stab mit isolierten Enden} Bei isolierten Enden des Stabes können belibige Temperaturen für $x = 0$ und -$x = l$ auftreten. In diesem Fall nicht erlaubt ist es, dass Wärme vom Stab +$x = l$ auftreten. In diesem Fall ist es nicht erlaubt, dass Wärme vom Stab an die Umgebung oder von der Umgebung an den Stab abgegeben wird. Aus der Physik ist bekannt, dass Wärme immer von der höheren zur tieferen @@ -72,9 +75,6 @@ als Randbedingungen. \subsubsection{Lösung der Differenzialgleichung} -% TODO: Referenz Separationsmethode -% TODO: Formeln sauber in Text einbinden. - Da die Lösungsfunktion von zwei Variablen abhängig ist, wird als Lösungsansatz die Separationsmethode verwendet. Dazu wird @@ -83,7 +83,8 @@ Dazu wird = T(t)X(x) \] -in die ursprüngliche Differenzialgleichung eingesetzt. +in die partielle Differenzialgleichung +\eqref{eq:slp-example-fourier-heat-equation} eingesetzt. Daraus ergibt sich \[ T^{\prime}(t)X(x) @@ -95,13 +96,13 @@ als neue Form. Nun können alle von $t$ abhängigen Ausdrücke auf die linke Seite, sowie alle von $x$ abhängigen Ausdrücke auf die rechte Seite gebracht werden und mittels der neuen Variablen $\mu$ gekoppelt werden: -\begin{equation} +\[ \frac{T^{\prime}(t)}{\kappa T(t)} = \frac{X^{\prime \prime}(x)}{X(x)} = \mu -\end{equation} +\] Durch die Einführung von $\mu$ kann das Problem nun in zwei separate Differenzialgleichungen aufgeteilt werden: \begin{equation} @@ -123,12 +124,14 @@ Erfüllen die Randbedingungen des Stab-Problems auch die Randbedingungen des Sturm-Liouville-Problems, kann bereits die Aussage getroffen werden, dass alle Lösungen für die Gleichung in $x$ orthogonal sein werden. +Da die Bedingungen des Stab-Problem nur Anforderungen an $x$ stellen, können +diese direkt für $X(x)$ übernomen werden. Es gilt also $X(0) = X(l) = 0$. Damit die Lösungen von $X$ orthogonal sind, müssen also die Gleichungen \begin{equation} \begin{aligned} \label{eq:slp-example-fourier-randbedingungen} - k_a y(a) + h_a p(a) y'(a) &= 0 \\ - k_b y(b) + h_b p(b) y'(b) &= 0 + k_a X(a) + h_a p(a) X'(a) &= 0 \\ + k_b X(b) + h_b p(b) X'(b) &= 0 \end{aligned} \end{equation} erfüllt sein und es muss ausserdem @@ -237,8 +240,6 @@ bestimmen. Dazu werden nochmals die Randbedingungen \eqref{eq:slp-example-fourier-boundary-condition-ends-constant} und \eqref{eq:slp-example-fourier-boundary-condition-ends-isolated} benötigt. -Zu bemerken ist, dass die Randbedingungen nur Anforderungen in $x$ stellen und -somit direkt für $X(x)$ übernomen werden können. Da die Koeffizienten $A$ und $B$, sowie die Parameter $\alpha$ uns $\beta$ im allgemeninen ungleich $0$ sind, müssen die Randbedingungen durch die @@ -282,9 +283,9 @@ Es bleibt noch nach $\beta$ aufzulösen: \] Es folgt nun wegen $\mu = -\beta^{2}$, dass -\begin{equation} +\[ \mu_1 = -\beta^{2} = -\frac{n^{2}\pi^{2}}{l^{2}} -\end{equation} +\] sein muss. Ausserdem ist zu bemerken, dass dies auch gleich $-\alpha^{2}$ ist. Da aber $A = 0$ gilt und der Summand mit $\alpha$ verschwindet, ist dies keine @@ -485,7 +486,7 @@ orthogonal zueinander stehen und = 0 \] -da Sinus- und Cosinus-Funktionen ebenfalls orthogonal zueinander sin. +da Sinus- und Cosinus-Funktionen ebenfalls orthogonal zueinander sind. Es bleibt also lediglich der Summand für $a_m$ stehen, was die Gleichung zu \[ -- cgit v1.2.1 From d80f928a8c5248d4fb92d04ed81cdaeec61bc10a Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 15 Aug 2022 09:51:21 +0200 Subject: Added comments to source. --- .../papers/sturmliouville/waermeleitung_beispiel.tex | 20 ++++++++++++++++++-- 1 file changed, 18 insertions(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 5d178c2..14c0d9a 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -71,7 +71,9 @@ Somit folgen \end{equation} als Randbedingungen. -%%%%%%%%%%% Lösung der Differenzialgleichung %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +% +% Lösung der Differenzialgleichung mittels Separation +% \subsubsection{Lösung der Differenzialgleichung} @@ -118,6 +120,10 @@ Differenzialgleichungen aufgeteilt werden: 0 \end{equation} +% +% Überprüfung Orthogonalität der Lösungen +% + Es ist an dieser Stelle zu bemerken, dass die Gleichung in $x$ in Sturm-Liouville-Form ist. Erfüllen die Randbedingungen des Stab-Problems auch die Randbedingungen des @@ -173,6 +179,10 @@ Analog dazu kann gezeit werden, dass die Randbedingungen für einen Stab mit isolierten Enden ebenfalls die Sturm-Liouville-Randbedingungen erfüllen und somit auch zu orthogonalen Lösungen führen. +% +% Lösung von X(x), Teil mu +% + \subsubsection{Lösund der Differentialgleichung in x} Als erstes wird auf die erste erste Gleichung eingegangen. Aufgrund der Struktur der Gleichung @@ -338,7 +348,9 @@ wie auch mit isolierten Enden -\frac{n^{2}\pi^{2}}{l^{2}}. \end{equation} -%%%% Koeffizienten a_n und b_n mittels skalarprodukt. %%%%%%%%%%%%%%%%%%%%%%%%%% +% +% Lösung von X(x), Teil: Koeffizienten a_n und b_n mittels skalarprodukt. +% Bisher wurde über die Koeffizienten $A$ und $B$ noch nicht viel ausgesagt. Zunächst ist wegen vorhergehender Rechnung ersichtlich, dass es sich bei @@ -589,6 +601,10 @@ Es bleibt also noch \end{aligned} \] +% +% Lösung von T(t) +% + \subsubsection{Lösund der Differentialgleichung in t} Zuletzt wird die zweite Gleichung der Separation \eqref{eq:slp-example-fourier-separated-t} betrachtet. -- cgit v1.2.1 From b06a9e5b30562e550540ea4b13b4e449970e9b2d Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 15 Aug 2022 10:03:42 +0200 Subject: Updated upstrem. --- buch/papers/sturmliouville/waermeleitung_beispiel.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 14c0d9a..b466d15 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -602,7 +602,7 @@ Es bleibt also noch \] % -% Lösung von T(t) +% Lösung von T(t) % \subsubsection{Lösund der Differentialgleichung in t} -- cgit v1.2.1 From 97e85459986381371236d1b9529d67064ac226c8 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 15 Aug 2022 11:58:27 +0200 Subject: Started properties of solutions. --- buch/papers/sturmliouville/eigenschaften.tex | 16 ++++++++++++++-- 1 file changed, 14 insertions(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex index 9f20070..6e6a26f 100644 --- a/buch/papers/sturmliouville/eigenschaften.tex +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -1,9 +1,21 @@ % -% teil1.tex -- Beispiel-File für das Paper +% eigenschaften.tex -- Eigenschaften der Lösungen % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % \section{Eigenschaften von Lösungen \label{sturmliouville:section:solution-properties}} \rhead{Eigenschaften von Lösungen} -% Erik work + +Im weiteren werden nun die Eigenschaften der Lösungen eines +Sturm-Liouville-Problems diskutiert und aufgezeigt, wie diese Eigenschaften +zustande kommen. + +Dazu wird der Operator $L_0$ welcher bereits in Kapitel +\ref{buch:integrale:subsection:sturm-liouville-problem} betrachtet wurde, +noch etwas genauer angeschaut. Es wird also +\[ + L_0 + = + -\frac{d}{dx}p(x) +\] \ No newline at end of file -- cgit v1.2.1 From 23d12ac04f38d75c3a904fd99cf6586efc7ea267 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 15 Aug 2022 13:13:59 +0200 Subject: Finished first version of solution properties. --- buch/papers/sturmliouville/eigenschaften.tex | 60 ++++++++++++++++++++++++++-- 1 file changed, 57 insertions(+), 3 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex index 6e6a26f..1552f7f 100644 --- a/buch/papers/sturmliouville/eigenschaften.tex +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -13,9 +13,63 @@ zustande kommen. Dazu wird der Operator $L_0$ welcher bereits in Kapitel \ref{buch:integrale:subsection:sturm-liouville-problem} betrachtet wurde, -noch etwas genauer angeschaut. Es wird also +noch etwas genauer angeschaut. Es wird also im Folgenden \[ L_0 = - -\frac{d}{dx}p(x) -\] \ No newline at end of file + -\frac{d}{dx}p(x)\frac{d}{dx} +\] +zusammen mit den Randbedingungen +\[ + \begin{aligned} + k_a y(a) + h_a p(a) y'(a) &= 0 \\ + k_b y(b) + h_b p(b) y'(b) &= 0 + \end{aligned} +\] +verwendet. Wie im Kapitel +\ref{buch:integrale:subsection:sturm-liouville-problem} bereits gezeigt, +resultieren die Randbedingungen aus der Anforderung den Operator $L_0$ +selbsadjungiert zu machen. +Es wurde allerdings noch nicht darauf eingegangen, welche Eigenschaften dies +für die Lösungen des Sturm-Liouville-Problems zur Folge hat. + +\subsubsection{Exkurs zum Spektralsatz} + +Um zu verstehen was für Eigenschaften der selbstadjungierte Operator $L_0$ in +den Lösungen hervorbringt, wird der Spektralsatz benötigt. + +Dieser wird in der linearen Algebra oft verwendet um zu zeigen, dass eine Matrix +diagonalisierbar ist, beziehungsweise dass eine Orthonormalbasis existiert. +Dazu wird zunächst gezeigt, dass eine gegebene $n\times n$-Matrix $A$ aus einem +endlichdimensionalem $\mathbb{K}$-Vektorraum selbstadungiert ist, also dass +\[ + \langle Av, w \rangle + = + \langle v, Aw \rangle +\] +für $ v, w \in \mathbb{K}^n$ gilt. +Ist dies der Fall, folgt direkt, dass $A$ auch normal ist. +Dann wird die Aussage des Spektralsatzes verwended, welche besagt, dass für +Endomorphismen genau dann eine Orthonormalbasis aus Eigenvektoren existiert, +wenn sie normal sind und nur Eigenwerte aus $\mathbb{K}$ besitzten. + +Dies ist allerdings nicht die Einzige Version des Spektralsatzes. +Unter anderen gibt es den Spektralsatz für kompakte Operatoren. +Dieser besagt, dass wenn ein linearer kompakter Operator in +$\mathbb{R}$ selbstadjungiert ist, ein (eventuell endliches) +Orthonormalsystem existiert. + +\subsubsection{Anwendung des Spektralsatzes auf $L_0$} + +Der Spektralsatz besagt also, dass, weil $ L_0 $ selbstadjungiert ist, eine +Orthonormalbasis aus Eigenvektoren existiert. +Genauer bedeutet dies, dass alle Eigenvektoren beziehungsweise alle Lösungen +des Sturm-Liouville-Problems orthogonal zueinander sind bezüglich dem +Skalarprodukt, in dem $ L_0 $ selbstadjungiert ist. + +Erfüllt also eine Differenzialgleichung die in Abschnitt +\ref{sturmliouville:section:teil0} präsentierten Eigenschaften und erfüllen +die Randbedingungen der Differentialgleichung die Randbedingungen +des Sturm-Liouville-Problems, kann bereits geschlossen werden, dass die +Lösungsfunktion des Problems eine Linearkombination aus orthogonalen +Basisfunktionen ist. \ No newline at end of file -- cgit v1.2.1 From c253055febe85abf5379e416f9731a1115a817b1 Mon Sep 17 00:00:00 2001 From: daHugen Date: Mon, 15 Aug 2022 13:21:08 +0200 Subject: Changed font color for some words in subsection \ref{lambertw:subsection:LoesAnalys} and defined a new color named applegreen. --- buch/papers/lambertw/teil4.tex | 6 ++++-- 1 file changed, 4 insertions(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/lambertw/teil4.tex b/buch/papers/lambertw/teil4.tex index ba32696..5a7c5ca 100644 --- a/buch/papers/lambertw/teil4.tex +++ b/buch/papers/lambertw/teil4.tex @@ -212,10 +212,12 @@ Nun haben wir eine Lösung, aber wie es immer mit Lösungen ist, stellt sich die \subsection{Lösung analysieren \label{lambertw:subsection:LoesAnalys}} +\definecolor{applegreen}{rgb}{0.55, 0.71, 0.0} + \begin{figure} \centering \includegraphics{papers/lambertw/Bilder/VerfolgungskurveBsp.png} - \caption[Graph der Verfolgungskurve]{Graph der Verfolgungskurve wobei, ({\color{red}rot}) die Funktion \ensuremath{y(x)} ist, ({\color{darkgreen}grün}) der quadratische Teil und ({\color{blue}blau}) dem \ensuremath{\operatorname{ln}(x)}-Teil entspricht. + \caption[Graph der Verfolgungskurve]{Graph der Verfolgungskurve wobei, ({\color{red}rot}) die Funktion \ensuremath{y(x)} ist, ({\color{applegreen}grün}) der quadratische Teil und ({\color{blue}blau}) dem \ensuremath{\operatorname{ln}(x)}-Teil entspricht. \label{lambertw:BildFunkLoes} } \end{figure} @@ -224,7 +226,7 @@ Das Resultat, wie ersichtlich, ist die Funktion \begin{equation} {\color{red}{y(x)}} = - C_1 + C_2 {\color{darkgreen}{x^2}} {\color{blue}{-}} \frac{\color{blue}{\operatorname{ln}(x)}}{8 \cdot C_2}, + C_1 + C_2 {\color{applegreen}{x^2}} {\color{blue}{-}} \frac{\color{blue}{\operatorname{ln}(x)}}{8 \cdot C_2}, \label{lambertw:funkLoes} \end{equation} für welche die Koeffizienten \(C_1\) und \(C_2\) aus den Anfangsbedingungen bestimmt werden können. Zuerst soll aber eine qualitative Intuition oder Idee für das Aussehen der Funktion \(y(x)\) geschaffen werden: -- cgit v1.2.1 From c0bbcf891e2e02a760eb640b735b2da80d2dc286 Mon Sep 17 00:00:00 2001 From: Andrea Mozzini Vellen Date: Mon, 15 Aug 2022 13:41:03 +0200 Subject: korrektur 15.08 --- buch/papers/kreismembran/references.bib | 6 ++++++ buch/papers/kreismembran/teil1.tex | 27 ++++++++++++++++----------- buch/papers/kreismembran/teil2.tex | 8 ++++---- buch/papers/kreismembran/teil3.tex | 22 +++++++++++++--------- 4 files changed, 39 insertions(+), 24 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/references.bib b/buch/papers/kreismembran/references.bib index 3d9d0c1..65173f8 100644 --- a/buch/papers/kreismembran/references.bib +++ b/buch/papers/kreismembran/references.bib @@ -89,4 +89,10 @@ type = {Dissertation}, author = {{Eric John Ruggiero Doctor of Philosophy In Mechanical Engineering}}, date = {2005}, +} + +@online{noauthor_laplace_nodate, + title = {Laplace Transform of Bessel Function of the First Kind of Order Zero - {ProofWiki}}, + url = {https://proofwiki.org/wiki/Laplace_Transform_of_Bessel_Function_of_the_First_Kind_of_Order_Zero}, + urldate = {2022-08-15}, } \ No newline at end of file diff --git a/buch/papers/kreismembran/teil1.tex b/buch/papers/kreismembran/teil1.tex index f6ba7d1..a9db48f 100644 --- a/buch/papers/kreismembran/teil1.tex +++ b/buch/papers/kreismembran/teil1.tex @@ -7,7 +7,7 @@ \section{Lösungsmethode 1: Separationsmethode  \label{kreismembran:section:teil1}} \rhead{Lösungsmethode 1: Separationsmethode} -An diesem Punkt bleibt also nur noch die Lösung der partiellen Differentialgleichung. In diesem Abschnitt wird sie mit Hilfe der Separationsmethode gelöst. +An diesem Punkt bleibt also "nur" noch die Lösung der partiellen Differentialgleichung. In diesem Abschnitt wird sie mit Hilfe der Separationsmethode gelöst. \subsection{Aufgabestellung\label{sub:aufgabestellung}} Wie im vorherigen Abschnitt gezeigt, lautet die partielle Differentialgleichung, die die Schwingungen einer Membran beschreibt: @@ -30,7 +30,7 @@ Da es sich um eine Kreisscheibe handelt, werden Polarkoordinaten verwendet, so d ergibt. Es wird eine runde elastische Membran berücksichtigt, die das Gebiet $\Omega$ abdeckt und am Rand $\Gamma$ befestigt ist. -Es wirken keine äusseren Kräfte. Es handelt sich somit von einer kreisförmligen eingespannten homogenen schwingenden Membran nach den Annahmen von \ref{kreimembran:annahmen}. +Es wirken keine äusseren Kräfte. Es handelt sich somit von einer kreisförmligen eingespannten homogenen schwingenden Membran nach den Annahmen von Abschnitt \ref{kreimembran:annahmen}. Daher ist die Membranabweichung im Punkt $(r,\varphi)$ $\in$ $\overline{\rm \Omega}$ zum Zeitpunkt $t$: \begin{align*} @@ -50,9 +50,9 @@ Nun wird das in Abschnitt \ref{sub:aufgabestellung} vorgestellte Problem mit Hil \subsubsection{Ansatz der Separation der Variablen\label{subsub:ansatz_separation}} Hierfür wird folgenden Ansatz gemacht: \begin{equation*} - u(r,\varphi, t) = F(r)G(\varphi)T(t) + u(r,\varphi, t) = F(r)G(\varphi)T(t). \end{equation*} -Dank der Randbedingungen kann gefordert werden, dass $F(R)=0$ ist, und natürlich, dass $G(\varphi)$ $2\pi$ periodisch ist. Eingesetzt in der Differenzialgleichung ergibt sich: +Dank der Randbedingungen kann gefordert werden, dass $F(R)=0$ ist, und natürlich, dass $G(\varphi)$ $2\pi$-periodisch ist. Eingesetzt in der Differenzialgleichung ergibt sich nach Division durch $u$: \begin{equation*} \frac{1}{c^2}\frac{T''(t)}{T(t)}=-\kappa^2=\frac{F''(r)}{F(r)}+\frac{1}{r}\frac{F'(r)}{F(r)}+\frac{1}{r^2}\frac{G''(\varphi)}{G(\varphi)}. \end{equation*} @@ -71,9 +71,9 @@ In der zweiten Gleichung hängt die linke Seite nur von $r$ ab, während die rec \end{align*} \subsubsection{Lösung für $G(\varphi)$\label{subsub:lösung_G}} -Da für die zweite Gleichung Lösungen von Schwingungen erwartet werden, für die $G''(\varphi)=-\omega^2 G(\varphi)$ gilt, schreibt man die gemeinsame Konstante als $\nu=-\omega^2$, was die Formeln später vereinfacht. Also: +Da für die zweite Gleichung Lösungen von Schwingungen erwartet werden, für die $G''(\varphi)=-n^2 G(\varphi)$ gilt, schreibt man die gemeinsame Konstante als $\nu=-n^2$, was die Formeln später vereinfacht. $n$ muss auch eine ganze Zahl sein, weil $G(\varphi)$ sonst nicht $2\pi$-periodisch ist. Also: \begin{equation*} - G(\varphi) = C_n \cos(\nu\varphi) + D_n \sin(\nu\varphi) + G(\varphi) = C_n \cos(n\varphi) + D_n \sin(n\varphi) \label{eq:cos_sin_überlagerung} \end{equation*} @@ -85,17 +85,20 @@ Die Gleichung für $F$ hat die Gestalt (Verweis auf \label{buch:differentialglei \end{align} Wie bereits in Kapitel \ref{buch:differntialgleichungen:section:bessel} gezeigt, sind die Bessel-Funktionen \begin{equation*} - J_{\nu}(x) = r^\nu \displaystyle\sum_{m=0}^{\infty} \frac{(-1)^m x^{2m}}{2^{2m+\nu}m! \Gamma (\nu + m+1)} + J_{n}(x) = r^n \displaystyle\sum_{m=0}^{\infty} \frac{(-1)^m x^{2m}}{2^{2m+n}m! \Gamma (n + m+1)} \end{equation*} Lösungen der Besselschen Differenzialgleichung \begin{equation*} - x^2 y'' + xy' + (\kappa^2 - \nu^2)y = 0 + x^2 y'' + xy' + (\kappa^2 - n^2)y = 0 \end{equation*} Die Funktionen $F(r) = J_n(\kappa r)$ lösen die Differentialgleichung \eqref{eq:2nd_degree_PDE}. \subsubsection{Lösung für $T(t)$\label{subsub:lösung_T}} -Die Differenzialgleichung $T''(t) + c^2\kappa^2T(t) = 0$, wird auf ähnliche Weise gelöst wie $G(\varphi)$. - +Die Differenzialgleichung $T''(t) + c^2\kappa^2T(t) = 0$, wird auf ähnliche Weise gelöst wie $G(\varphi)$. Um eine Einschränkung der möglichen Frequenzen zu erhalten und die Lösung als Reihe schreiben zu können, muss die folgende homogene Randbedingung definiert werden: +\begin{equation*} + u\big|_{\Gamma} = 0 \quad \text{für} \quad 0 \leq \varphi \leq 2\pi,\quad t \geq 0, +\end{equation*} +welche die $\kappa$ auf mögliche werte $\kappa_{mn}$ einschränkt. \subsubsection{Zusammenfassung der Lösungen\label{subsub:zusammenfassung_lösungen}} Durch Überlagerung aller Ergebnisse erhält man die Lösung \begin{align} @@ -120,5 +123,7 @@ für die Anzahl der Knotenlinien steht. Es gibt bestimmte Bereiche auf der Membr \label{buch:pde:kreis:fig:pauke}} \end{figure} - +\begin{center} + * \quad *\quad * +\end{center} An diesem Punkt stellte sich die Frage, ob es möglich wäre, die partielle Differentialgleichung mit einer anderen Methode als der der Trennung der Variablen zu lösen. Nach einer kurzen Recherche wurde festgestellt, dass eine weitere Methode die Transformationsmethode ist, genauer gesagt die Anwendung der Hankel-Transformation. Im nächsten Kapitel wird daher diese Integraltransformation vorgestellt und entwickelt, und es wird erläutert, warum sie für diese Art von Problem geeignet ist. diff --git a/buch/papers/kreismembran/teil2.tex b/buch/papers/kreismembran/teil2.tex index ec27bd3..4ceeb84 100644 --- a/buch/papers/kreismembran/teil2.tex +++ b/buch/papers/kreismembran/teil2.tex @@ -7,12 +7,12 @@ Hermann Hankel (1839--1873) war ein deutscher Mathematiker, der für seinen Beitrag zur mathematischen Analysis und insbesondere für die nach ihm benannte Transformation bekannt ist. Diese Transformation tritt bei der Untersuchung von Funktionen auf, die nur von der Entfernung des Ursprungs abhängen. -Er untersuchte auch Funktionen, jetzt Hankel- oder Bessel- Funktionen genannt, der dritten Art. +Er untersuchte auch Funktionen, jetzt Hankel- oder Bessel-Funktionen genannt, der dritten Art. Die Hankel-Transformation, die die Bessel-Funktion enthält, taucht natürlich bei achsensymmetrischen Problemen auf, die in zylindrischen Polarkoordinaten formuliert sind. In diesem Abschnitt werden die Theorie der Transformation und einige Eigenschaften der Grundoperationen erläutert. \subsubsection{Definition der Hankel-Transformation \label{subsub:hankel_tansformation}} -Wir führen die Definition der Hankel-Transformation \cite{lokenath_debnath_integral_2015} aus der zweidimensionalen Fourier-Transformation und ihrer Umkehrung ein, die durch: +Wir führen die Definition der Hankel-Transformation \cite{lokenath_debnath_integral_2015} aus der zweidimensionalen Fourier-Trans\-formation und ihrer Umkehrung ein, die durch: \begin{align} \mathscr{F}\{f(x,y)\} & = F(k,l)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i( \bm{\kappa}\cdot \mathbf{r})}f(x,y) \; dx \; dy,\label{equation:fourier_transform}\\ \mathscr{F}^{-1}\{F(x,y)\} & = f(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{i(\bm{\kappa}\cdot \mathbf{r})}F(k,l) \; dx \; dy \label{equation:inv_fourier_transform} @@ -49,13 +49,13 @@ wo $\tilde{f}_n(\kappa)$ ist die \textit{Hankel-Transformation} von $f(r)$ und i \subsubsection{Inverse Hankel-Transformation \label{subsub:inverse_hankel_tansformation}} Wie bei der Entwicklung der Hankel-Transformation können auch für die Umkehrformel Analogien zur Fourier-Transformation hergestellt werden. Vergleicht man die beiden Transformationen, so stellt man fest, dass sie sehr ähnlich sind, wenn man den Term $J_n(\kappa r)$ der Hankel-Transformation durch $e^{-i( \bm{\kappa}\cdot \mathbf{r})}$ der Fourier-Transformation ersetzt. Diese beide Funktionen sind orthogonal, und bei orthogonalen Matrizen genügt bekanntlich die Transponierung, um sie zu invertieren. Da das Skalarprodukt der Bessel-Funktionen jedoch nicht dasselbe ist wie das der Exponentialfunktionen, muss man durch $\kappa\; d\kappa$ statt nur durch $d\kappa$ integrieren, um die Umkehrfunktion zu erhalten. -Von \eqref{equation:hankel} also ist, die inverse \textit{Hankel-Transformation} so definiert: +Die inverse \textit{Hankel-Transformation} ist also als \begin{align} \mathscr{H}^{-1}_n\{\tilde{f}_n(\kappa)\}=f(r)=\int_{0}^{\infty}\kappa J_n(\kappa r) \tilde{f}_n(\kappa) \; d\kappa. \label{equation:inv_hankel} \end{align} +definiert. -Anstelle von $\tilde{f}_n(\kappa)$, wird häufig einfach $\tilde{f}(\kappa)$ für die Hankel-Transformation verwendet, indem die Ordnung angegeben wird. Die Integrale \eqref{equation:hankel} und \eqref{equation:inv_hankel} existieren für bestimmte grosse Klassen von Funktionen, die normalerweise in physikalischen Anwendungen vorkommen. Alternativ dazu kann die berühmte Hankel-Integralformel diff --git a/buch/papers/kreismembran/teil3.tex b/buch/papers/kreismembran/teil3.tex index a9dcd95..d143ec7 100644 --- a/buch/papers/kreismembran/teil3.tex +++ b/buch/papers/kreismembran/teil3.tex @@ -60,19 +60,23 @@ so dass $\tilde{g}(\kappa)\equiv 0$ und \tilde{f}(\kappa)=Aa\int_{0}^{\infty}r(a^2 + r^2)^{-\frac{1}{2}} J_0 (\kappa r) \; dr=\frac{Aa}{\kappa}e^{-a\kappa}. \end{equation*} -Aus der Laplace-Transformation und unter Verwendung der Skalierungseigenschaft ergibt sich, dass +\noindent Die formale Lösung \eqref{eq:formale_lösung} lautet also +\begin{align} + u(r,t)=Aa\int_{0}^{\infty}e^{-a\kappa} J_0(\kappa r)\cos(c\kappa t) \; dk=AaRe\int_{0}^{\infty}e^{-\kappa(a+ict)} J_0(\kappa r) \; dk. + \label{form_lösung2_step1} +\end{align} +\noindent Aus der Laplace-Transformation und unter Verwendung der Skalierungseigenschaft \cite{noauthor_laplace_nodate} ergibt sich, dass \begin{align*} - \int_{0}^{\infty}e^{-px} J_0(\kappa x) \; dx = \frac{1}{\sqrt{\kappa^2 + p^2}}. + \int_{0}^{\infty}e^{-px} J_0(\kappa x) \; dx = \frac{1}{\sqrt{\kappa^2 + p^2}}, \end{align*} -Die formale Lösung \eqref{eq:formale_lösung} lautet also -\begin{align*} - u(r,t)&=Aa\int_{0}^{\infty}e^{-a\kappa} J_0(\kappa r)\cos(c\kappa t) \; dk=AaRe\int_{0}^{\infty}e^{-\kappa(a+ict)} J_0(\kappa r) \; dk\\ - &=AaRe\left\{r^2+\left(a+ict\right)^2\right\}^{-\frac{1}{2}}. -\end{align*} +\noindent \eqref{form_lösung2_step1} kann somit vereinfacht werden in: +\begin{equation*} + u(r,t)=AaRe\left\{r^2+\left(a+ict\right)^2\right\}^{-\frac{1}{2}}. +\end{equation*} -Nimmt man jedoch die allgemeine Lösung durch Überlagerung, +\noindent Nimmt man jedoch die allgemeine Lösung durch Überlagerung, \begin{align} u(r, t) = \displaystyle\sum_{m=1}^{\infty} J_0 (k_{m}r)[a_{m}\cos(c \kappa_{m} t)+b_{m}\sin(c \kappa_{m} t)] @@ -84,6 +88,6 @@ kann man die Lösungsmethoden 1 und 2 vergleichen. \label{kreismembran:vergleich}} Bei der Analyse der Gleichungen \eqref{eq:lösung_endliche_generelle} und \eqref{eq:lösung_unendliche_generelle} fällt sofort auf, dass die Gleichung \eqref{eq:lösung_unendliche_generelle} nicht mehr von $m$ und $n$ abhängt, sondern nur noch von $n$ \cite{nishanth_p_vibrations_2018}. Das macht Sinn, denn $n$ beschreibt die Anzahl der Knotenlinien, welche unter der Annahme einer rotationssymmetrischen Lösung nicht vorhanden sein können. Tatsächlich werden $a_{m0}$, $b_{m0}$ und $\kappa_{m0}$ in $a_m$, $b_m$ bzw. $\kappa_m$ umbenannt. Die beiden Termen $\cos(n\varphi)$ und $\sin(n\varphi)$ verschwinden ebenfalls, da für $n=0$ der $\cos(n\varphi)$ gleich 1 und der $\sin(n \varphi)$ gleich 0 ist. -Die Funktion hängt also nicht mehr von der Besselfunktionen $n$-ter Ordnung ab, sondern nur von der nullter Ordnung. +Die Funktion hängt also nicht mehr von der Bessel-Funktionen $n$-ter Ordnung ab, sondern nur von der nullter Ordnung. -- cgit v1.2.1 From 8c6898303fc394c4f132664ef0b15fe484e9c5d9 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 15 Aug 2022 13:42:16 +0200 Subject: Added reference for "Spektralsatz". --- buch/papers/sturmliouville/eigenschaften.tex | 6 ++++-- buch/papers/sturmliouville/references.bib | 13 +++++++++++++ 2 files changed, 17 insertions(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex index 1552f7f..f972cd5 100644 --- a/buch/papers/sturmliouville/eigenschaften.tex +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -49,12 +49,14 @@ endlichdimensionalem $\mathbb{K}$-Vektorraum selbstadungiert ist, also dass \] für $ v, w \in \mathbb{K}^n$ gilt. Ist dies der Fall, folgt direkt, dass $A$ auch normal ist. -Dann wird die Aussage des Spektralsatzes verwended, welche besagt, dass für +Dann wird die Aussage des Spektralsatzes +\cite{sturmliouville:spektralsatz-wiki} verwended, welche besagt, dass für Endomorphismen genau dann eine Orthonormalbasis aus Eigenvektoren existiert, wenn sie normal sind und nur Eigenwerte aus $\mathbb{K}$ besitzten. Dies ist allerdings nicht die Einzige Version des Spektralsatzes. -Unter anderen gibt es den Spektralsatz für kompakte Operatoren. +Unter anderen gibt es den Spektralsatz für kompakte Operatoren +\cite{sturmliouville:spektralsatz-wiki}. Dieser besagt, dass wenn ein linearer kompakter Operator in $\mathbb{R}$ selbstadjungiert ist, ein (eventuell endliches) Orthonormalsystem existiert. diff --git a/buch/papers/sturmliouville/references.bib b/buch/papers/sturmliouville/references.bib index f66a74d..0c4724b 100644 --- a/buch/papers/sturmliouville/references.bib +++ b/buch/papers/sturmliouville/references.bib @@ -4,6 +4,19 @@ % (c) 2020 Autor, Hochschule Rapperswil % +@online{sturmliouville:spektralsatz-wiki, + title = {Spektralsatz}, + url = {https://de.wikipedia.org/wiki/Spektralsatz}, + date = {2020-08-15}, + year = {2020}, + month = {8}, + day = {15} +} + +% +% examples (not referenced in book) +% + @online{sturmliouville:bibtex, title = {BibTeX}, url = {https://de.wikipedia.org/wiki/BibTeX}, -- cgit v1.2.1 From 8b01c81f362cf20246e3d8319edfda15c18ff83f Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 15 Aug 2022 13:45:34 +0200 Subject: Improved code formatting. --- buch/papers/sturmliouville/eigenschaften.tex | 6 ++++-- 1 file changed, 4 insertions(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex index f972cd5..4c14630 100644 --- a/buch/papers/sturmliouville/eigenschaften.tex +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -13,7 +13,8 @@ zustande kommen. Dazu wird der Operator $L_0$ welcher bereits in Kapitel \ref{buch:integrale:subsection:sturm-liouville-problem} betrachtet wurde, -noch etwas genauer angeschaut. Es wird also im Folgenden +noch etwas genauer angeschaut. +Es wird also im Folgenden \[ L_0 = @@ -26,7 +27,8 @@ zusammen mit den Randbedingungen k_b y(b) + h_b p(b) y'(b) &= 0 \end{aligned} \] -verwendet. Wie im Kapitel +verwendet. +Wie im Kapitel \ref{buch:integrale:subsection:sturm-liouville-problem} bereits gezeigt, resultieren die Randbedingungen aus der Anforderung den Operator $L_0$ selbsadjungiert zu machen. -- cgit v1.2.1 From 7e26e1b395d7c793962caad2b78ffc6c6d588463 Mon Sep 17 00:00:00 2001 From: daHugen Date: Mon, 15 Aug 2022 13:47:06 +0200 Subject: Changed font in figure \ref{lambertw:BildFunkLoes} to match the text and the other figures --- .../papers/lambertw/Bilder/VerfolgungskurveBsp.png | Bin 356399 -> 318960 bytes 1 file changed, 0 insertions(+), 0 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png b/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png index e6e7c1e..dc4720a 100644 Binary files a/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png and b/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png differ -- cgit v1.2.1 From 1ac37227a82c02817b25d85638fbc1768fd38753 Mon Sep 17 00:00:00 2001 From: Kuster Yanik Date: Mon, 15 Aug 2022 14:53:23 +0200 Subject: polishing --- buch/papers/lambertw/teil1.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/lambertw/teil1.tex b/buch/papers/lambertw/teil1.tex index 8c30375..8025830 100644 --- a/buch/papers/lambertw/teil1.tex +++ b/buch/papers/lambertw/teil1.tex @@ -54,7 +54,7 @@ Wenn der Verfolger im ersten Quadranten startet, dann kann $v(t)$ mit den Gleich \text{,} \end{align} % -die Verfolgungskurve beschrieben werden. +beschrieben werden. Der Verfolger ist durch \begin{equation} v(t) -- cgit v1.2.1 From 987a5b51eaf65c4074c50ba12a3b21c2d2957260 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 15 Aug 2022 15:06:11 +0200 Subject: Corrected small mistake in psolution roperties. --- buch/papers/sturmliouville/eigenschaften.tex | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex index 4c14630..8553238 100644 --- a/buch/papers/sturmliouville/eigenschaften.tex +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -37,7 +37,7 @@ für die Lösungen des Sturm-Liouville-Problems zur Folge hat. \subsubsection{Exkurs zum Spektralsatz} -Um zu verstehen was für Eigenschaften der selbstadjungierte Operator $L_0$ in +Um zu verstehen was für Eigenschaften der selbstadjungierte Operator $ L_0 $ in den Lösungen hervorbringt, wird der Spektralsatz benötigt. Dieser wird in der linearen Algebra oft verwendet um zu zeigen, dass eine Matrix @@ -67,7 +67,7 @@ Orthonormalsystem existiert. Der Spektralsatz besagt also, dass, weil $ L_0 $ selbstadjungiert ist, eine Orthonormalbasis aus Eigenvektoren existiert. -Genauer bedeutet dies, dass alle Eigenvektoren beziehungsweise alle Lösungen +Genauer bedeutet dies, dass alle Eigenvektoren, beziehungsweise alle Lösungen des Sturm-Liouville-Problems orthogonal zueinander sind bezüglich dem Skalarprodukt, in dem $ L_0 $ selbstadjungiert ist. -- cgit v1.2.1 From 53cc7f1baf28448cb6196ba6ddf305e1b1403e7d Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 15 Aug 2022 15:45:11 +0200 Subject: Changed reference to conform with convetion. --- buch/papers/sturmliouville/eigenschaften.tex | 23 +++-- .../sturmliouville/waermeleitung_beispiel.tex | 107 +++++++++++---------- 2 files changed, 67 insertions(+), 63 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex index 8553238..fda8be6 100644 --- a/buch/papers/sturmliouville/eigenschaften.tex +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -11,9 +11,9 @@ Im weiteren werden nun die Eigenschaften der Lösungen eines Sturm-Liouville-Problems diskutiert und aufgezeigt, wie diese Eigenschaften zustande kommen. -Dazu wird der Operator $L_0$ welcher bereits in Kapitel -\ref{buch:integrale:subsection:sturm-liouville-problem} betrachtet wurde, -noch etwas genauer angeschaut. +Dazu wird der Operator $L_0$ welcher bereits in +Kapitel~\ref{buch:integrale:subsection:sturm-liouville-problem} betrachtet +wurde, noch etwas genauer angeschaut. Es wird also im Folgenden \[ L_0 @@ -28,16 +28,15 @@ zusammen mit den Randbedingungen \end{aligned} \] verwendet. -Wie im Kapitel -\ref{buch:integrale:subsection:sturm-liouville-problem} bereits gezeigt, -resultieren die Randbedingungen aus der Anforderung den Operator $L_0$ +Wie im Kapitel~\ref{buch:integrale:subsection:sturm-liouville-problem} bereits +gezeigt, resultieren die Randbedingungen aus der Anforderung den Operator $L_0$ selbsadjungiert zu machen. Es wurde allerdings noch nicht darauf eingegangen, welche Eigenschaften dies für die Lösungen des Sturm-Liouville-Problems zur Folge hat. \subsubsection{Exkurs zum Spektralsatz} -Um zu verstehen was für Eigenschaften der selbstadjungierte Operator $ L_0 $ in +Um zu verstehen was für Eigenschaften der selbstadjungierte Operator $L_0$ in den Lösungen hervorbringt, wird der Spektralsatz benötigt. Dieser wird in der linearen Algebra oft verwendet um zu zeigen, dass eine Matrix @@ -65,15 +64,15 @@ Orthonormalsystem existiert. \subsubsection{Anwendung des Spektralsatzes auf $L_0$} -Der Spektralsatz besagt also, dass, weil $ L_0 $ selbstadjungiert ist, eine +Der Spektralsatz besagt also, dass, weil $L_0$ selbstadjungiert ist, eine Orthonormalbasis aus Eigenvektoren existiert. Genauer bedeutet dies, dass alle Eigenvektoren, beziehungsweise alle Lösungen des Sturm-Liouville-Problems orthogonal zueinander sind bezüglich dem -Skalarprodukt, in dem $ L_0 $ selbstadjungiert ist. +Skalarprodukt, in dem $L_0$ selbstadjungiert ist. -Erfüllt also eine Differenzialgleichung die in Abschnitt -\ref{sturmliouville:section:teil0} präsentierten Eigenschaften und erfüllen -die Randbedingungen der Differentialgleichung die Randbedingungen +Erfüllt also eine Differenzialgleichung die in +Abschnitt~\ref{sturmliouville:section:teil0} präsentierten Eigenschaften und +erfüllen die Randbedingungen der Differentialgleichung die Randbedingungen des Sturm-Liouville-Problems, kann bereits geschlossen werden, dass die Lösungsfunktion des Problems eine Linearkombination aus orthogonalen Basisfunktionen ist. \ No newline at end of file diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index b466d15..b22d5f5 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -15,7 +15,7 @@ Wärmeleitkoeffizient $\kappa$ betrachtet. Es ergibt sich für das Wärmeleitungsproblem die partielle Differentialgleichung \begin{equation} - \label{eq:slp-example-fourier-heat-equation} + \label{sturmliouville:eq:example-fourier-heat-equation} \frac{\partial u}{\partial t} = \kappa \frac{\partial^{2}u}{{\partial x}^{2}} \end{equation} @@ -36,7 +36,7 @@ Lösungsfunktion $u(t,x)$ bei $x = 0$ und $x = l$ nur die vorgegebene Temperatur zurückgeben darf. Diese wird einfachheitshalber als $0$ angenomen. Es folgen nun \begin{equation} - \label{eq:slp-example-fourier-boundary-condition-ends-constant} + \label{sturmliouville:eq:example-fourier-boundary-condition-ends-constant} u(t,0) = u(t,l) @@ -62,7 +62,7 @@ dass die partiellen Ableitungen von $u(t,x)$ nach $x$ bei $x = 0$ und $x = l$ verschwinden. Somit folgen \begin{equation} - \label{eq:slp-example-fourier-boundary-condition-ends-isolated} + \label{sturmliouville:eq:example-fourier-boundary-condition-ends-isolated} \frac{\partial}{\partial x} u(t, 0) = \frac{\partial}{\partial x} u(t, l) @@ -85,8 +85,9 @@ Dazu wird = T(t)X(x) \] -in die partielle Differenzialgleichung -\eqref{eq:slp-example-fourier-heat-equation} eingesetzt. +in die partielle +Differenzialgleichung~\eqref{sturmliouville:eq:example-fourier-heat-equation} +eingesetzt. Daraus ergibt sich \[ T^{\prime}(t)X(x) @@ -108,13 +109,13 @@ der neuen Variablen $\mu$ gekoppelt werden: Durch die Einführung von $\mu$ kann das Problem nun in zwei separate Differenzialgleichungen aufgeteilt werden: \begin{equation} - \label{eq:slp-example-fourier-separated-x} + \label{sturmliouville:eq:example-fourier-separated-x} X^{\prime \prime}(x) - \mu X(x) = 0 \end{equation} \begin{equation} - \label{eq:slp-example-fourier-separated-t} + \label{sturmliouville:eq:example-fourier-separated-t} T^{\prime}(t) - \kappa \mu T(t) = 0 @@ -135,7 +136,7 @@ diese direkt für $X(x)$ übernomen werden. Es gilt also $X(0) = X(l) = 0$. Damit die Lösungen von $X$ orthogonal sind, müssen also die Gleichungen \begin{equation} \begin{aligned} - \label{eq:slp-example-fourier-randbedingungen} + \label{sturmliouville:eq:example-fourier-randbedingungen} k_a X(a) + h_a p(a) X'(a) &= 0 \\ k_b X(b) + h_b p(b) X'(b) &= 0 \end{aligned} @@ -143,7 +144,7 @@ Damit die Lösungen von $X$ orthogonal sind, müssen also die Gleichungen erfüllt sein und es muss ausserdem \begin{equation} \begin{aligned} - \label{eq:slp-example-fourier-coefficient-constraints} + \label{sturmliouville:eq:example-fourier-coefficient-constraints} |k_a|^2 + |h_a|^2 &\neq 0\\ |k_b|^2 + |h_b|^2 &\neq 0\\ \end{aligned} @@ -153,13 +154,15 @@ gelten. Um zu verifizieren, ob die Randbedingungen erfüllt sind, wird zunächst $p(x)$ benötigt. -Dazu wird die Gleichung \eqref{eq:slp-example-fourier-separated-x} mit der -Sturm-Liouville-Form \eqref{eq:sturm-liouville-equation} verglichen, was zu +Dazu wird die Gleichung~\eqref{sturmliouville:eq:example-fourier-separated-x} +mit der +Sturm-Liouville-Form~\eqref{eq:sturm-liouville-equation} verglichen, was zu $p(x) = 1$ führt. -Werden nun $p(x)$ und die Randbedingungen -\eqref{eq:slp-example-fourier-boundary-condition-ends-constant} in -\eqref{eq:slp-example-fourier-randbedingungen} eigesetzt, erhält man +Werden nun $p(x)$ und die +Randbedingungen~\eqref{sturmliouville:eq:example-fourier-boundary-condition-ends-constant} +in \eqref{sturmliouville:eq:example-fourier-randbedingungen} eigesetzt, erhält +man \[ \begin{aligned} k_a y(0) + h_a y'(0) &= h_a y'(0) = 0 \\ @@ -167,10 +170,10 @@ Werden nun $p(x)$ und die Randbedingungen \end{aligned} \] Damit die Gleichungen erfüllt sind, müssen $h_a = 0$ und $h_b = 0$ sein. -Zusätzlich müssen aber die Bedingungen -\eqref{eq:slp-example-fourier-coefficient-constraints} erfüllt sein und -da $y(0) = 0$ und $y(l) = 0$ sind, können belibige $k_a \neq 0$ und $k_b \neq 0$ -gewählt werden. +Zusätzlich müssen aber die +Bedingungen~\eqref{sturmliouville:eq:example-fourier-coefficient-constraints} +erfüllt sein und da $y(0) = 0$ und $y(l) = 0$ sind, können belibige $k_a \neq 0$ +und $k_b \neq 0$ gewählt werden. Somit ist gezeigt, dass die Randbedingungen des Stab-Problems für Enden auf konstanter Temperatur auch die Sturm-Liouville-Randbedingungen erfüllen und @@ -199,9 +202,9 @@ Die Lösungen für $X(x)$ sind also von der Form A \cos \left( \alpha x\right) + B \sin \left( \beta x\right). \] -Dieser Ansatz wird nun solange differenziert, bis alle in Gleichung -\eqref{eq:slp-example-fourier-separated-x} enthaltenen Ableitungen vorhanden -sind. +Dieser Ansatz wird nun solange differenziert, bis alle in +Gleichung~\eqref{sturmliouville:eq:example-fourier-separated-x} enthaltenen +Ableitungen vorhanden sind. Man erhält also \[ X^{\prime}(x) @@ -217,7 +220,8 @@ und \beta^{2} B \sin \left( \beta x \right). \] -Eingesetzt in Gleichung \eqref{eq:slp-example-fourier-separated-x} ergibt dies +Eingesetzt in Gleichung~\eqref{sturmliouville:eq:example-fourier-separated-x} +ergibt dies \[ -\alpha^{2}A\cos(\alpha x) - \beta^{2}B\sin(\beta x) - \mu\left(A\cos(\alpha x) + B\sin(\beta x)\right) @@ -247,18 +251,19 @@ ist schnell ersichtlich, dass $ \mu = -\alpha^{2} = -\beta^{2} $ gelten muss fü $ A \neq 0 $ oder $ B \neq 0 $. Zur Berechnung von $ \mu $ bleiben also noch $ \alpha $ und $ \beta $ zu bestimmen. -Dazu werden nochmals die Randbedingungen -\eqref{eq:slp-example-fourier-boundary-condition-ends-constant} und -\eqref{eq:slp-example-fourier-boundary-condition-ends-isolated} benötigt. +Dazu werden nochmals die +Randbedingungen~\eqref{sturmliouville:eq:example-fourier-boundary-condition-ends-constant} +und \eqref{sturmliouville:eq:example-fourier-boundary-condition-ends-isolated} +benötigt. Da die Koeffizienten $A$ und $B$, sowie die Parameter $\alpha$ uns $\beta$ im allgemeninen ungleich $0$ sind, müssen die Randbedingungen durch die trigonometrischen Funktionen erfüllt werden. -Es werden nun die Randbedingungen -\eqref{eq:slp-example-fourier-boundary-condition-ends-constant} für einen Stab -mit Enden auf konstanter Temperatur in die Gleichung -\eqref{eq:slp-example-fourier-separated-x} eingesetzt. +Es werden nun die +Randbedingungen~\eqref{sturmliouville:eq:example-fourier-boundary-condition-ends-constant} +für einen Stab mit Enden auf konstanter Temperatur in die +Gleichung~\eqref{sturmliouville:eq:example-fourier-separated-x} eingesetzt. Betrachten wir zunächst die Bedingung für $x = 0$. Dies fürht zu \[ @@ -303,9 +308,9 @@ Verletzung der Randbedingungen. Durch alanoges Vorgehen kann nun auch das Problem mit isolierten Enden gelöst werden. -Setzt man nun die Randbedingungen -\eqref{eq:slp-example-fourier-boundary-condition-ends-isolated} in $X^{\prime}$ -ein, beginnend für $x = 0$. Es ergibt sich +Setzt man nun die +Randbedingungen~\eqref{sturmliouville:eq:example-fourier-boundary-condition-ends-isolated} +in $X^{\prime}$ ein, beginnend für $x = 0$. Es ergibt sich \[ X^{\prime}(0) = @@ -324,7 +329,7 @@ folgt nun = 0. \] -Wiedrum muss über die $ \sin $-Funktion sicher gestellt werden, dass der +Wiedrum muss über die $\sin$-Funktion sicher gestellt werden, dass der Ausdruck den Randbedingungen entspricht. Es folgt nun \[ @@ -342,7 +347,7 @@ und somit Es ergibt sich also sowohl für einen Stab mit Enden auf konstanter Temperatur wie auch mit isolierten Enden \begin{equation} - \label{eq:slp-example-fourier-mu-solution} + \label{sturmliouville:eq:example-fourier-mu-solution} \mu = -\frac{n^{2}\pi^{2}}{l^{2}}. @@ -368,12 +373,12 @@ Die Lösung $X(x)$ wird nun umgeschrieben zu \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right). \] -Um eine eindeutige Lösung für $ X(x) $ zu erhalten werden noch weitere +Um eine eindeutige Lösung für $X(x)$ zu erhalten werden noch weitere Bedingungen benötigt. Diese sind die Startbedingungen oder $u(0, x) = X(x)$ für $t = 0$. Es gilt also nun die Gleichung \begin{equation} - \label{eq:slp-example-fourier-initial-conditions} + \label{sturmliouville:eq:example-fourier-initial-conditions} u(0, x) = a_0 @@ -388,7 +393,7 @@ gehört, von der wir wissen, dass sie orthogonal zu allen anderen trigonometrischen Funktionen der Lösung ist, kann direkt das Skalarprodukt verwendet werden um die Koeffizienten $a_n$ und $b_n$ zu bestimmen. Es wird also die Tatsache ausgenutzt, dass die Gleichheit in -\eqref{eq:slp-example-fourier-initial-conditions} nach Anwendung des +\eqref{sturmliouville:eq:example-fourier-initial-conditions} nach Anwendung des Skalarproduktes immernoch gelten muss und dass das Skalaprodukt mit einer Basisfunktion sämtliche Summanden auf der rechten Seite auslöscht. @@ -396,7 +401,7 @@ Zur Berechnung von $a_m$ mit $ m \in \mathbb{N} $ wird beidseitig das Skalarprodukt mit der Basisfunktion $ \cos\left(\frac{m \pi}{l}x\right)$ gebildet: \begin{equation} - \label{eq:slp-dot-product-cosine} + \label{sturmliouville:eq:dot-product-cosine} \langle u(0, x), \cos\left(\frac{m \pi}{l}x\right) \rangle = \langle a_0 @@ -409,13 +414,13 @@ gebildet: Bevor diese Form in die Integralform umgeschrieben werden kann, muss überlegt sein, welche Integralgrenzen zu verwenden sind. -In diesem Fall haben die $ \sin $ und $ \cos $ Terme beispielsweise keine ganze -Periode im Intervall $ x \in [0, l] $ für ungerade $ n $ und $ m $. +In diesem Fall haben die $\sin$ und $\cos$ Terme beispielsweise keine ganze +Periode im Intervall $x \in [0, l]$ für ungerade $n$ und $m$. Um die Skalarprodukte aber korrekt zu berechnen, muss über ein ganzzahliges Vielfaches der Periode der triginimetrischen Funktionen integriert werden. Dazu werden die Integralgrenzen $-l$ und $l$ verwendet und es werden ausserdem -neue Funktionen $ \hat{u}_c(0, x) $ für die Berechnung mit Cosinus und -$ \hat{u}_s(0, x) $ für die Berechnung mit Sinus angenomen, welche $ u(0, t) $ +neue Funktionen $\hat{u}_c(0, x)$ für die Berechnung mit Cosinus und +$\hat{u}_s(0, x)$ für die Berechnung mit Sinus angenomen, welche $u(0, t)$ gerade, respektive ungerade auf $[-l, l]$ fortsetzen: \[ \begin{aligned} @@ -451,7 +456,8 @@ skalliert wurde, also gilt nun \end{aligned} \] -Zunächst wird nun das Skalaprodukt \eqref{eq:slp-dot-product-cosine} berechnet: +Zunächst wird nun das Skalaprodukt~\eqref{sturmliouville:eq:dot-product-cosine} +berechnet: \[ \begin{aligned} \int_{-l}^{l}\hat{u}_c(0, x)\cos\left(\frac{m \pi}{l}x\right)dx @@ -545,11 +551,11 @@ gilt. Etwas anders ist es allerdings bei $a_0$. Wie der Name bereits suggeriert, handelt es sich hierbei um den Koeffizienten -zur Basisfunktion $ \cos\left(\frac{0 \pi}{l}x\right) $ beziehungsweise der +zur Basisfunktion $\cos\left(\frac{0 \pi}{l}x\right)$ beziehungsweise der konstanten Funktion $1$. -Um einen Ausdruck für $ a_0 $ zu erhalten, wird wiederum auf beiden Seiten -der Gleichung \eqref{eq:slp-example-fourier-initial-conditions} das -Skalarprodukt mit der konstanten Basisfunktion $ 1 $ gebildet: +Um einen Ausdruck für $a_0$ zu erhalten, wird wiederum auf beiden Seiten +der Gleichung~\eqref{sturmliouville:eq:example-fourier-initial-conditions} das +Skalarprodukt mit der konstanten Basisfunktion $1$ gebildet: \[ \begin{aligned} \int_{-l}^{l}\hat{u}_c(0, x)dx @@ -606,8 +612,8 @@ Es bleibt also noch % \subsubsection{Lösund der Differentialgleichung in t} -Zuletzt wird die zweite Gleichung der Separation -\eqref{eq:slp-example-fourier-separated-t} betrachtet. +Zuletzt wird die zweite Gleichung der +Separation~\eqref{sturmliouville:eq:example-fourier-separated-t} betrachtet. Diese wird über das charakteristische Polynom \[ \lambda - \kappa \mu @@ -623,8 +629,7 @@ Lösung = e^{-\kappa \mu t} \] -führt. -Und mit dem Resultat \eqref{eq:slp-example-fourier-mu-solution} +führt und mit dem Resultat~\eqref{sturmliouville:eq:example-fourier-mu-solution} \[ T(t) = -- cgit v1.2.1 From 2f1c9ad7d59e33f2c1e95a321947f608b5b06587 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 15 Aug 2022 15:58:10 +0200 Subject: Added renamed files to Makefile.inc and removed old ones. --- buch/papers/sturmliouville/Makefile.inc | 14 +++++++------- 1 file changed, 7 insertions(+), 7 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/Makefile.inc b/buch/papers/sturmliouville/Makefile.inc index e2039ce..7ffdad2 100644 --- a/buch/papers/sturmliouville/Makefile.inc +++ b/buch/papers/sturmliouville/Makefile.inc @@ -3,12 +3,12 @@ # # (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule # -dependencies-sturmliouville = \ +dependencies-sturmliouville = \ papers/sturmliouville/packages.tex \ - papers/sturmliouville/main.tex \ + papers/sturmliouville/main.tex \ papers/sturmliouville/references.bib \ - papers/sturmliouville/teil0.tex \ - papers/sturmliouville/teil1.tex \ - papers/sturmliouville/teil2.tex \ - papers/sturmliouville/teil3.tex - + papers/sturmliouville/einleitung.tex \ + papers/sturmliouville/eigenschaften.tex \ + papers/sturmliouville/beispiele.tex \ + papers/sturmliouville/waermeleitung_beispiel.tex \ + papers/sturmliouville/tschebyscheff_beispiel.tex -- cgit v1.2.1 From 4975028b529788a885c44a62808dc938e2b9d50d Mon Sep 17 00:00:00 2001 From: daHugen Date: Mon, 15 Aug 2022 16:05:00 +0200 Subject: Last update with syntax corrections --- buch/papers/lambertw/teil4.tex | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/lambertw/teil4.tex b/buch/papers/lambertw/teil4.tex index 5a7c5ca..36fb7e6 100644 --- a/buch/papers/lambertw/teil4.tex +++ b/buch/papers/lambertw/teil4.tex @@ -147,7 +147,7 @@ Wir wissen, dass sich der Verfolger mit Geschwindigkeit 1 bewegt, also legt er i \end{equation} verbunden werden. -Nicht gerade auffällig ist die Richtung, in welche hier integriert wird. Wenn der Verfolger sich wie vorgesehen am Anfang im ersten Quadranten befindet, dann muss sich dieser nach links bewegen, was nicht der üblichen Integrationsrichtung entspricht. Um eine Integration wie üblich von links nach rechts ausführen zu können, müssen die Integrationsgenerzen vertauscht werden, was in einem Vorzeichenwechsel resultiert. +Nicht gerade auffällig ist die Richtung, in welche hier integriert wird. Wenn der Verfolger sich wie vorgesehen am Anfang im ersten Quadranten befindet, dann muss sich dieser nach links bewegen, was nicht der üblichen Integrationsrichtung entspricht. Um eine Integration wie üblich von links nach rechts ausführen zu können, müssen die Integrationsgrenzen vertauscht werden, was in einem Vorzeichenwechsel resultiert. Wenn man nun \eqref{lambertw:eqZuBogenlaenge} in die DGL \eqref{lambertw:DGLmitT} einfügt, dann ergibt sich der neue Ausdruck \begin{equation} @@ -324,7 +324,7 @@ Nun sind wir soweit, dass wir eine \(y(x)\)-Beziehung für beliebige Anfangswert \subsection{Funktion nach der Zeit \label{lambertw:subsection:FunkNachT}} -In diesem Abschnitt werden algebraischen Umformungen ein wenig detaillierter als zuvor beschrieben. Dies hat auch einen bestimmten Grund: Den Einsatz einer speziellen Funktion aufzeigen, sowie auch wann und wieso diese vorkommt. Welche spezielle Funktion? Fragt man sich wahrscheinlich in diesem Moment. Nun, um diese Frage kurz zu beantworten, es ist ``YouTube's favorite special function'' laut dem Mathematiker Michael Penn, die Lambert-\(W\)-Funktion \(W(x)\) welche im Kapitel \ref{buch:section:lambertw} bereits beschrieben wurde. +In diesem Abschnitt werden algebraische Umformungen ein wenig detaillierter als zuvor beschrieben. Dies hat auch einen bestimmten Grund: Den Einsatz einer speziellen Funktion aufzeigen, sowie auch wann und wieso diese vorkommt. Welche spezielle Funktion? Fragt man sich wahrscheinlich in diesem Moment. Nun, um diese Frage kurz zu beantworten, es ist ``YouTube's favorite special function'' laut dem Mathematiker Michael Penn, die Lambert-\(W\)-Funktion \(W(x)\) welche im Kapitel \ref{buch:section:lambertw} bereits beschrieben wurde. \subsubsection{Zeitabhängigkeit wiederherstellen \label{lambertw:subsubsection:ZeitabhWiederherst}} -- cgit v1.2.1 From 7ab3ba297ddca9b1c920a4161fda2548211b4ac1 Mon Sep 17 00:00:00 2001 From: tim30b Date: Mon, 15 Aug 2022 16:08:35 +0200 Subject: korrektur 15.8 Tim --- buch/papers/kreismembran/teil0.tex | 14 +++++++------- buch/papers/kreismembran/teil1.tex | 2 +- buch/papers/kreismembran/teil4.tex | 30 +++++++++++++++--------------- 3 files changed, 23 insertions(+), 23 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil0.tex b/buch/papers/kreismembran/teil0.tex index c6dac06..27c6f0f 100644 --- a/buch/papers/kreismembran/teil0.tex +++ b/buch/papers/kreismembran/teil0.tex @@ -7,9 +7,9 @@ \rhead{Membran} Eine Membran oder selten ein Schwingblatt ist laut Duden \cite{kreismembran:Duden:Membran} ein ``dünnes Blättchen aus Metall, Papier o. Ä., das durch seine Schwingungsfähigkeit geeignet ist, Schallwellen zu übertragen \dots''. Ein dünnes Blättchen aus Metall zeig jedoch nicht die selben dynamischen Eigenschaften wie ein gespanntes Stück Papier. -Beschreibt man das dynamische Verhalten, muss zwischen einer dünnen Platte und einer Membrane unterschieden werden \cite{kreismembran:membrane_vs_thin_plate}. -Eine dünne Platte zum Beispiel aus Metall, wirkt selbst entgegen ihrer Deformation sobald sie gekrümmt wird. -Eine Membran auf der anderen Seite besteht aus einem Material, welches sich ohne Kraftaufwand verbiegen lässt wie zum Beispiel Papier. +Beschreibt man das dynamische Verhalten, muss zwischen einer dünnen Platte und einer Membran unterschieden werden \cite{kreismembran:membrane_vs_thin_plate}. +Eine dünne Platte zum Beispiel aus Metall, wirkt selbst entgegen ihrer Deformation, sobald sie gekrümmt wird. +Eine Membran auf der anderen Seite besteht aus einem Material, welches sich ohne Kraftaufwand verbiegen lässt, wie zum Beispiel Papier. Bevor Papier als schwingende Membran betrachtet werden kann, wird jedoch noch eine Spannung $ T $ benötigt, welche das Material daran hindert, aus der Ruhelage gebracht zu werden. Ein geläufiges Beispiel einer Kreismembran ist eine runde Trommel. @@ -28,11 +28,11 @@ Das untersuchte Modell erfüllt folgende Eigenschaften: Durch die konstante Elastizität ist die ganze Membran unter gleichmässiger Spannung $ T $. \item Die Membran ist perfekt flexibel. Damit ist gemeint, dass die Membran ohne Kraftaufwand verbogen werden kann. - Die Membran ist dadurch nicht allein stehend schwingfähig, hierzu muss sie gespannt werden mit einer Kraft $ T $. + Die Membran ist dadurch nicht allein stehend schwingfähig, hierzu muss sie mit einer Kraft $ T $ gespannt werden. \item Die Membran kann sich nur in Richtung ihrer Normalen in kleinem Ausmass auslenken. Auslenkungen in der Ebene der Membran sind nicht möglich. \item Die Membran erfährt keine Art von Dämpfung. - Die Membran wird also nicht durch ihr umliegendes Medium abgebremst noch erfährt sie Wärmeverluste durch Deformation. + Die Membran wird also nicht durch ihr umliegendes Medium abgebremst noch erfährt sie Reibungsverluste durch Deformation. \end{enumerate} @@ -64,7 +64,7 @@ befolgen. Die senkrecht wirkenden Kräfte werden mit $ T_1 $ und $ T_2 $ ausgedr \begin{equation*} T_2 \sin \beta - T_1 \sin \alpha = \rho dx \frac{\partial^2 u}{\partial t^2} . \end{equation*} -Die Gleichung wird durch $ T $ dividiert, wobei $ T $ nach \ref{kreismembran:eq:no_translation} geschickt gewählt wird. Somit kann +Die Gleichung wird durch $ T $ dividiert, wobei $ T $ nach \eqref{kreismembran:eq:no_translation} geschickt gewählt wird. Somit kann \begin{equation*} \frac{T_2 \sin \beta}{T_2 \cos \beta} - \frac{T_1 \sin \alpha}{T_1 \cos \alpha} = \frac{\rho dx}{T} \frac{\partial^2 u}{\partial t^2} \end{equation*} @@ -91,4 +91,4 @@ Damit resultiert die in der Literatur gebräuchliche Form \frac{1}{c^2}\frac{\partial^2u}{\partial t^2} = \Delta u. \end{equation} In dieser Form ist die Gleichung auch gültig für eine Membran. -Für den Fall einer Membran muss lediglich der Laplace-Operator $\Delta$ in zwei Dimensionen gerechnet werden. \ No newline at end of file +Für den Fall einer Membran muss lediglich der Laplace-Operator $\Delta$ in zwei Dimensionen verwendet werden. \ No newline at end of file diff --git a/buch/papers/kreismembran/teil1.tex b/buch/papers/kreismembran/teil1.tex index a9db48f..a9b2fad 100644 --- a/buch/papers/kreismembran/teil1.tex +++ b/buch/papers/kreismembran/teil1.tex @@ -7,7 +7,7 @@ \section{Lösungsmethode 1: Separationsmethode  \label{kreismembran:section:teil1}} \rhead{Lösungsmethode 1: Separationsmethode} -An diesem Punkt bleibt also "nur" noch die Lösung der partiellen Differentialgleichung. In diesem Abschnitt wird sie mit Hilfe der Separationsmethode gelöst. +An diesem Punkt bleibt also ``nur'' noch die Lösung der partiellen Differentialgleichung. In diesem Abschnitt wird sie mit Hilfe der Separationsmethode gelöst. \subsection{Aufgabestellung\label{sub:aufgabestellung}} Wie im vorherigen Abschnitt gezeigt, lautet die partielle Differentialgleichung, die die Schwingungen einer Membran beschreibt: diff --git a/buch/papers/kreismembran/teil4.tex b/buch/papers/kreismembran/teil4.tex index 01a6029..3b174e0 100644 --- a/buch/papers/kreismembran/teil4.tex +++ b/buch/papers/kreismembran/teil4.tex @@ -8,13 +8,13 @@ Um numerisch das Verhalten einer Membran zu ermitteln, muss eine numerische Darstellung definiert werden. Die Membran wird hier in Form der Matrix $ U $ digitalisiert. -Jedes Element $ U_{ij} $ steh für die Auslenkung der Membran $ u(x,y,t) $ an der Stelle $ \{x,y\}=\{i,j\} $. -Zwischen benachbarten Elementen in der Matrix $ U $ liegt immer der Abstand $ dh $, eine Inkrementierung von $ i $ oder $ j $ entspricht somit einem Schritt in Richtung $ x $ oder $ y $ von Länge $ dh $ auf der Membran. -Die zeitliche Dimension wird in Form des Array $ U[] $ aus $ z \times U $ Matrizen dargestellt, wobei $ z $ der Anzahl Zeitschritten entspricht. -Das Element auf Zeile $ i $, Spalte $ j $ der $ w $-ten Matrix von $ U[] $ also $ U[w]_{ij} $ entspricht somit der Auslenkung $ u(i,j,w) $. +Jedes Element $ U_{ij} $ steht für die Auslenkung der Membran $ u(x,y,t) $ an der Stelle $ \{x,y\}=\{i,j\} $. +Zwischen benachbarten Elementen in der Matrix $ U $ liegt immer der Abstand $ dh $, eine Inkrementierung von $ i $ oder $ j $ ist somit einem Schritt in Richtung $ x $ oder $ y $ von Länge $ dh $ auf der Membran. +Die zeitliche Dimension wird in Form des Array $ U[] $ aus $ z \times U $ Matrizen dargestellt, wobei $ z $ die Anzahl von Zeitschritten ist. +Das Element auf Zeile $ i $, Spalte $ j $ der $ w $-ten Matrix von $ U[] $ also $ U[w]_{ij} $ ist somit die Auslenkung $ u(i,j,w) $. Da die DGL von zweiter Ordnung ist, reicht eine Zustandsvariabel pro Membran-Element nicht aus. Es wird neben der Auslenkung auch die Geschwindigkeit jedes Membran-Elementes benötigt um den Zustand eindeutig zu beschreiben. -Dazu existiert neben $ U[] $ ein analoger Array $ V[] $ welcher die Geschwindigkeiten aller Membran-Elementen repräsentiert. +Dazu existiert neben $ U[] $ ein analoger Array $ V[] $ welcher die Geschwindigkeiten aller Membran-Elemente repräsentiert. $ V[w]_{ij} $ entspricht also $ \dot{u}(i,j,w) $. Der Zustand einer Membran zum Zeitpunkt $ w $ wird mit $ X[w] $ beschrieben, was $ U[w] $ und $ V[w] $ beinhaltet. @@ -25,7 +25,7 @@ Die Folgeposition $ U[w+1] $ ergibt sich als \begin{equation} U[w+1] = U[w] + dt \cdot V[w], \end{equation} -also die Ausgangslage $ + $ die Strecke welche während des Zeitintervall mit der Geschwindigkeit des Elementes zurückgelegt wurde. +also die Ausgangslage plus die Strecke welche während des Zeitintervall mit der Geschwindigkeit des Elementes zurückgelegt wurde. Neben der Position muss auch die Geschwindigkeit aktualisiert werden. Analog zur Folgeposition wird \begin{equation*} @@ -40,7 +40,7 @@ Die Geschwindigkeit des Folgezustandes kann somit mit V[w+1] = V[w] + dt \cdot \Delta_h U \cdot c^2 \end{equation} berechnet werden. -Während $ c^2 $ lediglich eine Material spezifische Konstante ist, muss noch erläutert werden, wie der diskrete Laplace-Operator für $ \Delta_h u $ definiert ist. +Während $ c^2 $ lediglich eine Material spezifische Konstante ist, muss noch erläutert werden, wie der diskrete Laplace-Operator für $ \Delta_h u $ definiert ist. Dieses Verfahren wird Euler-Methode genannt. \subsection{Diskreter Laplace-Operator $\Delta_h$} Die diskrete Ableitung zweiter Ordnung kann mit Hilfe der Taylor-Reihen-Entwicklung als @@ -93,9 +93,9 @@ Der Folgezustand kann also mit den Gleichungen \label{kreismembran:eq:folge_V} V[w+1] &= (V[w] + dt \cdot \Delta_h u \cdot c^2)\odot M \end{align} -berechnet werden. +berechnet werden. Das Symbol \cdot steht hier für eine elementweise Matrixmultiplikation (Hadamard-Produkt) \subsubsection{Simulation} -Mit den gegebenen Gleichungen \ref{kreismembran:eq:folge_U} und \ref{kreismembran:eq:folge_V} das Verhalten der Membran mit einem Loop über das zu untersuchende Zeitintervall berechnet werden. +Mit den gegebenen Gleichungen \eqref{kreismembran:eq:folge_U} und \eqref{kreismembran:eq:folge_V} das Verhalten der Membran mit einem Loop über das zu untersuchende Zeitintervall berechnet werden. In der Abbildung \ref{kreismembran:im:simres_rund} sind Simulationsresultate zu sehen. Die erste Figur zeigt die Ausgangslage gefolgt von den Auslenkungen nach jeweils $ 50 $ weiteren Iterationsschritten. Es ist zu erkennen, wie sich die Störung vom Zentrum an den Rand ausbreitet. @@ -120,7 +120,7 @@ Erreicht die Störung den Rand, wird sie reflektiert und nähert sich dem Zentru Um eine unendlich grosse Membran zu simulieren, könnte der unpraktische Weg gewählt werden, die Matrix unendlich gross zu definieren, dies wird jedoch spätestens bei der numerischen Berechnung seine Probleme mit sich bringen. Etwas geeigneter ist es, die Matrix so gross wie möglich zu definieren, wie es die Kapazitäten erlauben. Wenn anschliessend nur das Verhalten im Zentrum, bei der Störung beobachtet wird, verhaltet sich die Membran wie eine unendliche. -Dies aber nur bis die Störung am Rand reflektiert wird und wieder das innere zu beobachtende Zentrum beeinflusst. +Dies aber nur bis die Störung am Rand reflektiert wird und wieder das Zentrum beeinflusst. Soll erst gar keine Reflexion entstehen, muss ein Absorber modelliert werden welcher die Störung möglichst ohne Reflexion aufnimmt. \subsubsection{Absorber} @@ -132,15 +132,15 @@ Der Spielraum welcher dem Absorber übrig bleibt ist die Art der Überganges. Bei der endlichen kreisförmigen Membran hat die Maske $M$ einen binären Übergang von Membran zu Rand bezweckt. Anstelle dieses abrupten Wechsels wird nun eine Maske definiert, welche graduell von Membran $1$ zu Rand-Element $0$ wechselt. Die Elemente werden auf Basis ihres Abstand $r$ zum Zentrum definiert. -Der Abstand entspricht +Der Abstand ist \begin{equation*} r(i,j) = \sqrt{|i-\frac{m}{2}|^2+|j-\frac{n}{2}|^2}, \end{equation*} -wobei $ m $ und $n$ den Dimensionen der Matrix entsprechen. -Für einen Stufenlosen Übergang werden die Elemente der Maske auf +wobei $ m $ und $n$ die Dimensionen der Matrix sind. +Für einen stufenlosen Übergang werden die Elemente der Maske auf \begin{align} - M_{ij} = \begin{cases} 1-e^{(r(i,j)-b)a} & \text{wenn $x > b$} \\ + M_{ij} = \begin{cases} 1-e^{(r(i,j)-b)a} & \text{$x > b$} \\ 0 & \text{sonst} \end{cases} \end{align} gesetzt. @@ -184,7 +184,7 @@ Die DGL \ref{kreismembran:Ausgang_DGL} welche simuliert wird geht jedoch von der \section{Schlusswort} Auch wenn ein physikalisches Verhalten bereits durch Annahmen und Annäherungen deutlich vereinfacht wird, bestehen auch dann noch eine Vielzahl von Lösungsansätzen. Lösungen einer unendlich grosse Membran scheinen fern der Realität zu sein, doch dies darf es im Sinne der Mathematik. -Und wer weis, für eine Ameise auf einem Trampolin ist eine unendliche Membran vielleicht eine ganz gute Annäherung. +Und wer weiss, für eine Ameise auf einem Trampolin ist eine unendliche Membran vielleicht eine ganz gute Annäherung. -- cgit v1.2.1 From 2f2665e1ee770f7813fe5406ba27a480cff2f541 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Mon, 15 Aug 2022 16:21:39 +0200 Subject: Corrected some error to make tschebyscheff example compile. --- buch/papers/sturmliouville/tschebyscheff_beispiel.tex | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex index 391841a..8561479 100644 --- a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex +++ b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex @@ -13,7 +13,7 @@ Im Kapitel \ref{sub:beispiele_sturm_liouville_problem} sind die Koeffizientenfun \end{align*}. Da die Sturm-Liouville-Gleichung \begin{equation} - \label{eq:sturm-liouville-equation} + \label{eq:sturm-liouville-equation-tscheby} \frac{d}{dx}\lbrack \sqrt{1-x^2} \frac{dy}{dx} \rbrack + \lbrack 0 + \lambda \frac{1}{\sqrt{1-x^2}} \rbrack y = 0 \end{equation} nun mit den Koeffizientenfunktionen aufgestellt werden kann, bleibt die Frage, ob es sich um ein reguläres oder singuläres Sturm-Liouville-Problem handelt. @@ -27,7 +27,7 @@ Für $x>1$ und $x<-1$ sehen die Polynome wie folgt aus: T_n(x) = \left\{\begin{array}{ll} \cosh (n \arccos x), & x > 1\\ (-1)^n \cosh (n \arccos (-x)), & x<-1 \end{array}\right. \end{equation}, -jedoch ist die Orthogonalität nur auf dem Intervall $\[ -1, 1\]$ sichergestellt. +jedoch ist die Orthogonalität nur auf dem Intervall $[ -1, 1]$ sichergestellt. Die nächste Bedingung beinhaltet, dass die Funktion $p(x)^-1$ und $w(x)>0$ sein müssen. Die Funktion \begin{equation*} @@ -36,7 +36,7 @@ Die Funktion ist die gleiche wie $w(x)$. Für die Verifizierung der Randbedingungen benötigt man erneut $p(x)$. -Da sich die Polynome nur auf dem Intervall $\[ -1,1 \]$ orthogonal verhalten, sind $a = -1$ und $b = 1$ gesetzt. +Da sich die Polynome nur auf dem Intervall $[ -1,1 ]$ orthogonal verhalten, sind $a = -1$ und $b = 1$ gesetzt. Beim einsetzen in die Randbedingung \ref{eq:randbedingungen}, erhält man \begin{equation} \begin{aligned} -- cgit v1.2.1 From 74d360e00d3c2dc97d257b0b3dfa8d4c4c5d9417 Mon Sep 17 00:00:00 2001 From: Kuster Yanik Date: Mon, 15 Aug 2022 16:59:11 +0200 Subject: last corrections --- buch/papers/lambertw/teil1.tex | 20 ++++++++++---------- 1 file changed, 10 insertions(+), 10 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/lambertw/teil1.tex b/buch/papers/lambertw/teil1.tex index 8025830..c4b2d05 100644 --- a/buch/papers/lambertw/teil1.tex +++ b/buch/papers/lambertw/teil1.tex @@ -34,33 +34,32 @@ Die Parametrisierung von $v(t)$ ist von den Startbedingungen abhängig. Deshalb % \subsection{Anfangsbedingung im ersten Quadranten} % -Wenn der Verfolger im ersten Quadranten startet, dann kann $v(t)$ mit den Gleichungen aus \eqref{lambertw:eqFunkXNachT}, welche sind +Wenn der Verfolger im ersten Quadranten startet, dann kann $v(t)$ mit den Gleichungen aus \eqref{lambertw:eqFunkXNachT}, welche \begin{align} x\left(t\right) &= - x_0\cdot\sqrt{\frac{1}{\chi}W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right) \right)} \\ + x_0\cdot\sqrt{\frac{1}{\chi}W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right) \right)} \text{,}\\ y(t) &= - \frac{1}{4}\biggl(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\biggl(\left(\frac{x(t)}{x_0}\right)^2\biggr)-r_0+3y_0\biggr)\\ + \frac{1}{4}\biggl(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\biggl(\left(\frac{x(t)}{x_0}\right)^2\biggr)-r_0+3y_0\biggr) \text{,}\\ \chi &= - \frac{r_0+y_0}{r_0-y_0}, \quad + \frac{r_0+y_0}{r_0-y_0}\text{,} \quad \eta = - \left(\frac{x}{x_0}\right)^2,\quad + \left(\frac{x}{x_0}\right)^2 \quad\text{und}\quad r_0 = \sqrt{x_0^2+y_0^2} - \text{,} \end{align} % +sind, beschrieben werden. Der Verfolger ist durch \begin{equation} v(t) = \left( \begin{array}{c} x(t) \\ y(t) \end{array} \right) - \text{.} \end{equation} % parametrisiert, wobei $y(t)$ viel komplexer ist als $x(t)$. @@ -238,14 +237,15 @@ Die Ortsvektoren der Punkte können wiederum mit \begin{align} v &= - t\cdot\left(\begin{array}{c} \cos (\alpha) \\ \sin (\alpha) \end{array}\right) +\left(\begin{array}{c} x_0 \\ y_0 \end{array}\right) + t\cdot\left(\begin{array}{c} \cos (\alpha) \\ \sin (\alpha) \end{array}\right) +\left(\begin{array}{c} x_0 \\ 0 \end{array}\right) \\ z &= \left(\begin{array}{c} 0 \\ t \end{array}\right) \end{align} beschrieben werden. -Da der Abstand +$x_0$ ist der Abstand bei $t=0$, damit alle möglichen Fälle untersucht werden können. +Da der Abstand allgemein \begin{equation} a = @@ -266,7 +266,7 @@ Der Abstand im Quadrat abgeleitet nach der Zeit ist \begin{equation} \frac{d a^2}{d t} = - 2(t\cdot\cos (\alpha)+x_0)\cdot\cos(\alpha)(\alpha)+2t(\sin(\alpha)-1)^2 + 2(t\cdot\cos (\alpha)+x_0)\cdot\cos(\alpha)+2t(\sin(\alpha)-1)^2 \text{.} \end{equation} Da nur die unmittelbar benachbarten Punkten von Interesse sind, wird die Ableitung für $t=0$ untersucht. Dabei kann die Ableitung in -- cgit v1.2.1 From 9830947147dbe750081f4d8801c4172424283001 Mon Sep 17 00:00:00 2001 From: Fabian <@> Date: Mon, 15 Aug 2022 17:25:40 +0200 Subject: 2. Ueberarbeitung, Inhalt --- buch/papers/0f1/teil3.tex | 16 +++++++--------- 1 file changed, 7 insertions(+), 9 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/0f1/teil3.tex b/buch/papers/0f1/teil3.tex index 2942a0b..b283b07 100644 --- a/buch/papers/0f1/teil3.tex +++ b/buch/papers/0f1/teil3.tex @@ -7,7 +7,7 @@ \label{0f1:section:teil3}} \rhead{Resultate} Im Verlauf dieser Arbeit hat sich gezeigt, -das ein einfacher mathematischer Algorithmus zu implementieren gar nicht so einfach ist. +das einen einfachen mathematischen Algorithmus zu implementieren gar nicht so einfach ist. So haben alle drei umgesetzten Ansätze Probleme mit grossen negativen $z$ in der Funktion $\mathstrut_0F_1(;c;z)$. Ebenso kann festgestellt werden, dass je grösser der Wert $z$ in $\mathstrut_0F_1(;c;z)$ wird, desto mehr weichen die berechneten Resultate von den Erwarteten \cite{0f1:wolfram-0f1} ab. @@ -19,19 +19,17 @@ Erst wenn mehrerer Iterationen gemacht werden, um die Genauigkeit zu verbessern, Interessant ist auch, dass die Rekursionsformel nahezu gleich schnell wie die Potenzreihe konvergiert, aber sich danach, wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} zu beobachten ist, einschwingt. Dieses Verhalten ist auch bei grösseren $z$ zu beobachten, allerdings ist dann die Differenz zwischen dem ersten lokalen Minimum von $k$ bis zum Abbruch kleiner. Dieses Phänomen ist auf die Lösung der Rekursionsformel \eqref{0f1:math:matrix:ende:eq} zurück zu führen. Da im Gegensatz die ganz kleinen Werte nicht zu einer Konvergenz wie beim Kettenbruch führen, sondern sich noch eine Zeit lang durch die Multiplikation aufschwingen. -Ist $z$ negativ wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu einer gegenseitigen Kompensation von negativen und positiven Termen. Dies führt dazu, dass die Rekursionsformel zusammen mit der Potenzreihe abbricht. -Die ansteigende Differenz mit anschliessendendem Einschwingen, ist aufgrund der sich alternierenden Termen mit wechselnden Vorzeichens zu erklären. +Ist $z$ negativ wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu aufgrund des Vorzeichens zu alternierenden Termen. So steigt bei allen Algorithmen zuerst die Differenz zum erwarteten Endwert. Erst nach genügend Iterationen sind die Terme genügend klein, so dass sie das Endresultat nicht mehr signifikant beeinflussen. +Auch hier konvergiert der Kettenbruch am schnellsten von allen Algorithmen. Ebenso bricht die Rekursionsformel nahezu gleichzeitig mit der Potenzreihe ab. \subsection{Stabilität \label{0f1:subsection:Stabilitaet}} Verändert sich der Wert von $z$ in $\mathstrut_0F_1(;c;z)$ gegen grössere positive Werte, wie zum Beispiel $c = 800$ liefert die Kettenbruch-Funktion (Listing \ref{0f1:listing:kettenbruchIterativ}) \verb+inf+ zurück. Dies könnte durch ein Abbruchkriterien abgefangen werden. Allerdings würde das, bei grossen Werten zulasten der Genauigkeit gehen. Trotzdem könnte, je nach Anwendung, auf ein paar Nachkommastellen verzichtet werden. Wohingegen die Potenzreihe (Listing \ref{0f1:listing:potenzreihe}) das Problem hat, dass je mehr Terme berechnet werden, desto schneller wächst die Fakultät im Nenner. Dies führt zu einer Bereichsüberschreitung des \verb+double+ Bereiches \cite{0f1:double}, der spätesten ab $k=167$ eintritt. Schlussendlich gibt das Unterprogramm das Resultat \verb+-nan(ind)+ zurück. -Die Rekursionformel \eqref{0f1:listing:kettenbruchRekursion} liefert für sehr grosse positive Werte die genausten Ergebnisse, verglichen mit der GNU Scientific Library. Wie schon vermutet ist die Rekursionsformel, im positivem Bereich, der stabilste Algorithmus. Um die Stabilität zu gewährleisten, muss wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, die Iterationstiefe $k$ genug gross gewählt werden. +Die Rekursionformel \eqref{0f1:listing:kettenbruchRekursion} liefert für sehr grosse positive Werte die genausten Ergebnisse, verglichen mit der GNU Scientific Library. Wie schon vermutet ist die Rekursionsformel, im positivem Bereich, der stabilste Algorithmus. Um die Konvergenz zu gewährleisten, muss wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, die Iterationstiefe $k$ genug gross gewählt werden. -{\color{red}TODO Abb. 20.3} - -Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Grund dafür ist die Potenz von $z$, was zum Phänomen der Auslöschung \cite{0f1:SeminarNumerik} führt. Schön zu beobachten ist dies in der Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} mit der Airy-Funktion als Test. So sind sowohl der Kettenbruch, als auch die Rekursionsformel bis ungefähr $\frac{-15^3}{9}$ stabil. Dies macht auch Sinn, da beide auf der gleichen mathematischen Grundlage basieren. Danach verhält sich allerdings die Instabilität unterschiedlich. Das unterschiedliche Verhalten kann damit erklärt werden, dass beim Kettenbruch jeweils eine zusätzliche Division stattfindet. Diese Unterschiede sind auch in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} festzustellen. +Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Grund dafür ist die Potenz von $z$, was zum Phänomen der Auslöschung \cite{0f1:SeminarNumerik} führt. Schön zu beobachten ist dies in der Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} mit der Airy-Funktion als Test. So sind sowohl die Potenzreihe, der Kettenbruch, als auch die Rekursionsformel bis ungefähr $\frac{-15^3}{9}$ stabil. Dies macht auch Sinn, da alle Algorithmen auf der gleichen mathematischen Grundlage basieren. Danach verhält sich allerdings die Instabilität unterschiedlich. Diese programmiertechnischen Unterschiede sind auch in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} und \ref{0f1:ausblick:plot:konvergenz:negativ} festzustellen. \begin{figure} \centering @@ -43,14 +41,14 @@ Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Gru \begin{figure} \centering \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzPositiv.pdf} - \caption{Konvergenz mit positivem z; Logarithmisch dargestellte Differenz vom erwarteten Endresultat. + \caption{Konvergenz mit positivem z; Logarithmisch, vorzeichenlose dargestellte Differenz vom erwarteten Endresultat. \label{0f1:ausblick:plot:konvergenz:positiv}} \end{figure} \begin{figure} \centering \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzNegativ.pdf} - \caption{Konvergenz mit negativem z; Logarithmisch dargestellte Differenz vom erwarteten Endresultat. + \caption{Konvergenz mit negativem z; Logarithmisch, vorzeichenlose dargestellte Differenz vom erwarteten Endresultat. \label{0f1:ausblick:plot:konvergenz:negativ}} \end{figure} -- cgit v1.2.1 From 059dd7a0ec72d91ed7879201c10e0abfb8cea3ef Mon Sep 17 00:00:00 2001 From: Fabian <@> Date: Mon, 15 Aug 2022 20:10:10 +0200 Subject: 2. Ueberarbeitung, done --- buch/papers/0f1/teil2.tex | 31 ++++++++++++++++++++++++++----- 1 file changed, 26 insertions(+), 5 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/0f1/teil2.tex b/buch/papers/0f1/teil2.tex index 06ac53e..0c2f1e6 100644 --- a/buch/papers/0f1/teil2.tex +++ b/buch/papers/0f1/teil2.tex @@ -38,16 +38,37 @@ Ein endlicher Kettenbruch \cite{0f1:wiki-kettenbruch} ist ein Bruch der Form a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cdots}}} \end{equation*} in welchem $a_0, a_1,\dots,a_n$ und $b_1,b_2,\dots,b_n$ ganze Zahlen sind. -Die Kurzschreibweise für einen allgemeinen Kettenbruch ist + +Nimmt man nun folgenden Gleichung \cite{0f1:wiki-fraction}: \begin{equation*} - a_0 + \frac{a_1|}{|b_1} + \frac{a_2|}{|b_2} + \frac{a_3|}{|b_3} + \cdots + f_{i-1} - f_i = k_i z f_{i+1}, \end{equation*} -Angewendet auf die Funktion $\mathstrut_0F_1$ bedeutet dies \cite{0f1:wiki-fraction}: +wo $f_i$ analytische Funktionen sind und $i > 0$ ist, sowie $k_i$ konstant. +Ergibt sich folgender Zusammenhang: \begin{equation*} + \cfrac{f_i}{f_{i-1}} = \cfrac{1}{1+k_iz\cfrac{f_{i+1}}{f_i}} +\end{equation*} + +Angewendet auf die Funktion $\mathstrut_0F_1$ bedeutet dies: +\begin{equation} + \label{0f1:math:potenzreihe:0f1:eq} \mathstrut_0F_1(;c;z) = 1 + \frac{z}{c\cdot1!} + \frac{z^2}{c(c+1)\cdot2!} + \frac{z^3}{c(c+1)(c+2)\cdot3!} + \cdots +\end{equation} +Durch Substitution kann bewiesen werden, dass die nachfolgende Formel eine Relation zur obigen Potenzreihe \eqref{0f1:math:potenzreihe:0f1:eq} ist: +\begin{equation*} + \mathstrut_0F_1(;c-1;z) - \mathstrut_0F_1(;c;z) = \frac{z}{c(c-1)} \cdot \mathstrut_0F_1(;c+1;z). \end{equation*} -Umgeformt ergibt sich folgender Kettenbruch \cite{0f1:wolfram-0f1} -{\color{red}TODO Herleitung} +Wenn man für $f_i$ und $k_i$ folgende Annahme trifft: +\begin{align*} + f_i =& \mathstrut_0F_1(;c+1;z)\\ + k_i =& \frac{1}{(c+1)(c+i-1)} +\end{align*} +erhält man: +\begin{equation*} + \cfrac{\mathstrut_0F_1(;c+1;z)}{\mathstrut_0F_1(;c;z)} = \cfrac{1}{1+\cfrac{\cfrac{z}{c(c+1)}}{1+\cfrac{\cfrac{z}{(c+1)(c+2)}}{1+\cfrac{z}{(c+2)(c+3)} + \cdots}}}. +\end{equation*} + +Mit weiteren Relationen ergibt sich nach Wolfram Alpha \cite{0f1:wolfram-0f1} folgender Kettenbruch \begin{equation} \label{0f1:math:kettenbruch:0f1:eq} \mathstrut_0F_1(;c;z) = 1 + \cfrac{\cfrac{z}{c}}{1+\cfrac{-\cfrac{z}{2(c+1)}}{1+\cfrac{z}{2(c+1)}+\cfrac{-\cfrac{z}{3(c+2)}}{1+\cfrac{z}{5(c+4)} + \cdots}}}, -- cgit v1.2.1 From 2e1c6aecc9e99334b84a10e0da9597e03f2de3c4 Mon Sep 17 00:00:00 2001 From: Fabian <@> Date: Mon, 15 Aug 2022 20:14:36 +0200 Subject: 2.Uerbarbeitung, bruch --- buch/papers/0f1/teil2.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/0f1/teil2.tex b/buch/papers/0f1/teil2.tex index 0c2f1e6..ef9f55e 100644 --- a/buch/papers/0f1/teil2.tex +++ b/buch/papers/0f1/teil2.tex @@ -65,7 +65,7 @@ Wenn man für $f_i$ und $k_i$ folgende Annahme trifft: \end{align*} erhält man: \begin{equation*} - \cfrac{\mathstrut_0F_1(;c+1;z)}{\mathstrut_0F_1(;c;z)} = \cfrac{1}{1+\cfrac{\cfrac{z}{c(c+1)}}{1+\cfrac{\cfrac{z}{(c+1)(c+2)}}{1+\cfrac{z}{(c+2)(c+3)} + \cdots}}}. + \cfrac{\mathstrut_0F_1(;c+1;z)}{\mathstrut_0F_1(;c;z)} = \cfrac{1}{1+\cfrac{\cfrac{z}{c(c+1)}}{1+\cfrac{\cfrac{z}{(c+1)(c+2)}}{1+\cfrac{\cfrac{z}{(c+2)(c+3)}}{\cdots}}}}. \end{equation*} Mit weiteren Relationen ergibt sich nach Wolfram Alpha \cite{0f1:wolfram-0f1} folgender Kettenbruch -- cgit v1.2.1 From e9f63ddb1d4de82392ca66eb162ecdc3474e5190 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Andreas=20M=C3=BCller?= Date: Tue, 16 Aug 2022 06:46:12 +0200 Subject: fix missing $$ in kreismembran/teil4.tex --- buch/papers/kreismembran/teil4.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil4.tex b/buch/papers/kreismembran/teil4.tex index 3b174e0..0b6299e 100644 --- a/buch/papers/kreismembran/teil4.tex +++ b/buch/papers/kreismembran/teil4.tex @@ -93,7 +93,7 @@ Der Folgezustand kann also mit den Gleichungen \label{kreismembran:eq:folge_V} V[w+1] &= (V[w] + dt \cdot \Delta_h u \cdot c^2)\odot M \end{align} -berechnet werden. Das Symbol \cdot steht hier für eine elementweise Matrixmultiplikation (Hadamard-Produkt) +berechnet werden. Das Symbol $\cdot$ steht hier für eine elementweise Matrixmultiplikation (Hadamard-Produkt) \subsubsection{Simulation} Mit den gegebenen Gleichungen \eqref{kreismembran:eq:folge_U} und \eqref{kreismembran:eq:folge_V} das Verhalten der Membran mit einem Loop über das zu untersuchende Zeitintervall berechnet werden. In der Abbildung \ref{kreismembran:im:simres_rund} sind Simulationsresultate zu sehen. -- cgit v1.2.1 From e3b55b863915e45287bebe8bff6027c468359fe6 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Andreas=20M=C3=BCller?= Date: Tue, 16 Aug 2022 10:25:25 +0200 Subject: fix a typo --- buch/papers/kreismembran/teil4.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/teil4.tex b/buch/papers/kreismembran/teil4.tex index 0b6299e..d6aa54f 100644 --- a/buch/papers/kreismembran/teil4.tex +++ b/buch/papers/kreismembran/teil4.tex @@ -93,7 +93,7 @@ Der Folgezustand kann also mit den Gleichungen \label{kreismembran:eq:folge_V} V[w+1] &= (V[w] + dt \cdot \Delta_h u \cdot c^2)\odot M \end{align} -berechnet werden. Das Symbol $\cdot$ steht hier für eine elementweise Matrixmultiplikation (Hadamard-Produkt) +berechnet werden. Das Symbol $\odot$ steht hier für eine elementweise Matrixmultiplikation (Hadamard-Produkt) \subsubsection{Simulation} Mit den gegebenen Gleichungen \eqref{kreismembran:eq:folge_U} und \eqref{kreismembran:eq:folge_V} das Verhalten der Membran mit einem Loop über das zu untersuchende Zeitintervall berechnet werden. In der Abbildung \ref{kreismembran:im:simres_rund} sind Simulationsresultate zu sehen. -- cgit v1.2.1 From e63dbdeedff258f834311a449d935e7945d6c9db Mon Sep 17 00:00:00 2001 From: Joshua Baer Date: Tue, 16 Aug 2022 10:28:32 +0200 Subject: signal m draw --- buch/papers/fm/03_bessel.tex | 8 +-- buch/papers/fm/Python animation/Bessel-FM.ipynb | 73 +++++++++++++++++++++++-- 2 files changed, 71 insertions(+), 10 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/fm/03_bessel.tex b/buch/papers/fm/03_bessel.tex index 5f85dc6..45f2dfd 100644 --- a/buch/papers/fm/03_bessel.tex +++ b/buch/papers/fm/03_bessel.tex @@ -157,14 +157,14 @@ jedoch so \(-1 \cdot J_{-n}(\beta) = J_n(\beta)\) und daraus wird dann: \sum_{n=- \infty}^{-1} J_{n}(\beta) \cos((\omega_c + n \omega_m) t) \,+\, \sum_{n=1}^\infty J_{n}(\beta) \cos((\omega_c + n\omega_m) t) \end{align*} -Da \(n\) immer ungerade ist und \(0\) nicht zu den ungeraden zahlen zählt, kann man dies so vereinfacht +Da \(n\) immer ungerade ist und \(0\) nicht zu den ungeraden Zahlen zählt, kann man dies so vereinfacht \[ s(t) = - \sum_{n\, \text{ungerade}} -1 \cdot J_{n}(\beta) \cos((\omega_c + n\omega_m) t). + \sum_{n\, \text{ungerade}} J_{n}(\beta) \cos((\omega_c + n\omega_m) t). \label{fm:eq:ungerade} \] -schreiben. +, mit allen positiven und negativen Ganzzahlen schreiben. %------------------------------------------------------------------------------------------ \subsubsection{Summe Zusammenführen} Beide Teile \eqref{fm:eq:gerade} Gerade @@ -179,7 +179,7 @@ ergeben zusammen \[ \cos(\omega_ct+\beta\sin(\omega_mt)) = - \sum_{k= -\infty}^\infty J_{k}(\beta) \cos((\omega_c+k\omega_m)t). + \sum_{k= -\infty}^\infty J_{n}(\beta) \cos((\omega_c+ n\omega_m)t). \] Somit ist \eqref{fm:eq:proof} bewiesen. \newpage diff --git a/buch/papers/fm/Python animation/Bessel-FM.ipynb b/buch/papers/fm/Python animation/Bessel-FM.ipynb index 74f1011..ef6300b 100644 --- a/buch/papers/fm/Python animation/Bessel-FM.ipynb +++ b/buch/papers/fm/Python animation/Bessel-FM.ipynb @@ -2,7 +2,7 @@ "cells": [ { "cell_type": "code", - "execution_count": 4, + "execution_count": 1, "metadata": {}, "outputs": [], "source": [ @@ -11,10 +11,11 @@ "from scipy.fft import fft, ifft, fftfreq\n", "import scipy.special as sc\n", "import scipy.fftpack\n", + "import matplotlib.pyplot as plt\n", "import matplotlib as mpl\n", "# Use the pgf backend (must be set before pyplot imported)\n", - "mpl.use('pgf')\n", - "import matplotlib.pyplot as plt\n", + "# mpl.use('pgf')\n", + "\n", "from matplotlib.widgets import Slider\n", "def fm(beta):\n", " # Number of samplepoints\n", @@ -42,7 +43,7 @@ }, { "cell_type": "code", - "execution_count": 114, + "execution_count": 9, "metadata": {}, "outputs": [ { @@ -56,6 +57,18 @@ "needs_background": "light" }, "output_type": "display_data" + }, + { + "data": { + "image/png": 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", + "text/plain": [ + "
" + ] + }, + "metadata": { + "needs_background": "light" + }, + "output_type": "display_data" + } + ], + "source": [ + "ratio = 2\n", + "first = 1\n", + "(length,) = x.shape\n", + "slop = int(length/6)\n", + "second = ratio-first\n", + "odd = ratio % 2\n", + "\n", + "first = int(first * length/ratio) + odd\n", + "second = int( second * length/ratio)\n", + "slop = np.array(np.append(np.zeros(first-slop) , (np.arange(slop))/slop))\n", + "#steep = np.ones(int(first * length/ratio)+ odd) - np.exp(-np.arange(int(first * length/ratio) + odd)/200)\n", + "steep = (np.ones(first) + slop)*0.5\n", + "\n", + "step = np.append(steep, np.ones(second))\n", + "m = np.sin(30 * 2.0 * np.pi * x) * step \n", + "plt.plot(x, step, '-')\n", + "plt.plot(x, m, '-')" + ] } ], "metadata": { -- cgit v1.2.1 From f7c0dfbd20c97ae0e617aec796f2adc6d81369dc Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Tue, 16 Aug 2022 14:20:30 +0200 Subject: Merged tschebyscheff section. --- .../sturmliouville/tschebyscheff_beispiel.tex | 23 ++++++++++++++++++---- 1 file changed, 19 insertions(+), 4 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex index 8561479..a18684f 100644 --- a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex +++ b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex @@ -28,10 +28,10 @@ Für $x>1$ und $x<-1$ sehen die Polynome wie folgt aus: (-1)^n \cosh (n \arccos (-x)), & x<-1 \end{array}\right. \end{equation}, jedoch ist die Orthogonalität nur auf dem Intervall $[ -1, 1]$ sichergestellt. -Die nächste Bedingung beinhaltet, dass die Funktion $p(x)^-1$ und $w(x)>0$ sein müssen. +Die nächste Bedingung beinhaltet, dass die Funktion $p(x)^{-1}$ und $w(x)>0$ sein müssen. Die Funktion \begin{equation*} - p(x)^-1 = \frac{1}{\sqrt{1-x^2}} + p(x)^{-1} = \frac{1}{\sqrt{1-x^2}} \end{equation*} ist die gleiche wie $w(x)$. @@ -40,10 +40,25 @@ Da sich die Polynome nur auf dem Intervall $[ -1,1 ]$ orthogonal verhalten, sind Beim einsetzen in die Randbedingung \ref{eq:randbedingungen}, erhält man \begin{equation} \begin{aligned} - k_a y(-1) + h_a y'(-1) &= h_a + k_a y(-1) + h_a y'(-1) &= 0 + k_b y(-1) + h_b y'(-1) &= 0 \end{aligned} -\end{equation} +\end{equation}. +Die Funktion $y(x)$ und $y'(x)$ sind in diesem Fall die Tschebyscheff Polynome (siehe \label{sub:definiton_der_tschebyscheff-Polynome}). +Es gibt zwei Arten von Tschebyscheff Polynome: die erste Art $T_n(x)$ und die zweite Art $U_n(x)$. +Jedoch beachtet man in diesem Kapitel nur die Tschebyscheff Polynome erster Art (\ref{eq:tschebyscheff-polynome}). +Die Funktion $y(x)$ wird nun mit der Funktion $T_n(x)$ ersetzt und für die Verifizierung der Randbedingung wählt man $n=2$. +Somit erhält man +\begin{equation} + \begin{aligned} + k_a T_2(-1) + h_a T_{2}'(-1) &= k_a = 0\\ + k_b T_2(1) + h_b T_{2}'(1) &= k_b = 0 +\end{aligned} +\end{equation}. +Ähnlich wie beim Beispiel der Wärmeleitung in einem homogenen Stab kann man, damit die Bedingung $|k_i|^2 + |h_i|^2\ne 0$ erfüllt ist, können beliebige $h_a \ne 0$ und $h_b \ne 0$ gewählt werden. +Somit ist erneut gezeigt, dass die Randbedingungen der Tschebyscheff-Polynome auf die Sturm-Liouville-Randbedingungen erfüllt und alle daraus resultierenden Lösungen orthogonal sind. + -- cgit v1.2.1 From b31c40a22e885205373ff1a3284935d544c99889 Mon Sep 17 00:00:00 2001 From: canuel Date: Tue, 16 Aug 2022 14:32:49 +0200 Subject: started the chapter about spherical harmonics (more specific about the derivation of them) --- buch/papers/kugel/packages.tex | 2 +- buch/papers/kugel/spherical-harmonics.tex | 387 +++++++++++++++++++++++++++++- 2 files changed, 387 insertions(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kugel/packages.tex b/buch/papers/kugel/packages.tex index 61f91ad..1c4f3e0 100644 --- a/buch/papers/kugel/packages.tex +++ b/buch/papers/kugel/packages.tex @@ -7,4 +7,4 @@ % if your paper needs special packages, add package commands as in the % following example %\usepackage{packagename} - +\usepackage{cases} diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex index 6b23ce5..c76e757 100644 --- a/buch/papers/kugel/spherical-harmonics.tex +++ b/buch/papers/kugel/spherical-harmonics.tex @@ -1,9 +1,394 @@ % vim:ts=2 sw=2 et spell: \section{Spherical Harmonics} +We finally arrived at the main section, which gives our chapter its name. The idea is to discuss spherical harmonics, their mathematical derivation and some of their properties and applications.\newline +The subsection \ref{} will be devoted to the Eigenvalue problem of the Laplace operator. Through the latter, we will derive the set of Eigenfunctions that obey the equation presented in \ref{}[TODO: reference to eigenvalue equation], which will be defined as \emph{Spherical Harmonics}. In fact, this subsection will present their mathematical derivation.\newline +In the subsection \ref{}, on the other hand, some interesting properties related to them will be discussed. Some of these will come back to help us understand in more detail why they are useful in various real-world applications, which will be presented in the section \ref{}.\newline +One specific property will be studied in more detail in the subsection \ref{}, namely the recursive property. +The last subsection is devoted to one of the most beautiful applications (In our humble opinion), namely the derivation of a Fourier-style series expansion but defined on the sphere instead of a plane.\newline +More importantly, this subsection will allow us to connect all the dots we have created with the previous sections, concluding that Fourier is just a specific case of the application of the concept of orthogonality.\newline +Our hope is that after reading this section you will appreciate the beauty and power of generalization that mathematics offers us. -\subsection{Eigenvalue Problem in Spherical Coordinates} +\subsection{Eigenvalue Problem on the Spherical surface} +\subsubsection{Unormalized Spherical Harmonics} +From the chapter \ref{}, we know that the spherical Laplacian is defined as. \begin{equation*} + \nabla^2_S := \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial}{\partial r} \right) + \frac{1}{r^2} + \left[ + \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial}{\partial\vartheta} \right) + + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2} + \right] +\end{equation*} +But we do not want to consider this algebraic monster entirely, since this includes the whole set $\mathbb{R}^3$; rather, we want to focus only on the spherical surface (as the title suggests). We can then further concretise our calculations by selecting any number for the variable $r$, so that we have a sphere and, more importantly, a spherical surface on which we can ``play''.\newline +Surely you have already heard of the unit circle, a geometric entity used extensively in many mathematical contexts. The most famous and basic among them is surely trigonometry.\newline +Extending this concept into three dimensions, we will talk about the unit sphere. This is a very famous sphere, as is the unit circle. So since we need a sphere why not use the most famous one? Thus imposing $r=1$.\newline +Now, since the variable $r$ became a constant, we can leave out all derivatives with respect to $r$, setting them to zero. Then substituting the value of $r$ for 1, we will obtain the operator we will refer to as \emph{Spherical Surface Operator}: +\begin{equation*} + \nabla^2_{\partial S} := \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial}{\partial\vartheta} \right) + + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2}. +\end{equation*} +As can be seen, for this definition, the subscript ``$\partial S$'' was used to emphasize the fact that we are on the spherical surface, which can be understood as a boundary of the sphere.\newline +Now that we have defined an operator, we can go on to calculate its eigenfunctions. As mentioned earlier, we can translate this problem at first abstract into a much more concrete problem, which has to do with the field of \emph{Partial Differential Equaitons} (PDEs). The functions we want to find are simply functions that respect the following expression: +\begin{equation}\label{kugel:eq:sph_srfc_laplace} + \nabla^2_{\partial S} f = \lambda f +\end{equation} +Which is traditionally written as follows: +\begin{equation*} + \nabla^2_{\partial S} f = -\lambda f +\end{equation*} +Perhaps the fact that we are dealing with a PDE may not be obvious at first glance, but if we extend the operator $\nabla^2_{\partial S}$ according to Eq.(\ref{kugel:eq:sph_srfc_laplace}), we will get: +\begin{equation}\label{kugel:eq:PDE_sph} + \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial f}{\partial\vartheta} \right) + + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 f}{\partial\varphi^2} + \lambda f = 0, +\end{equation} +making it emerge.\newline +All functions satisfying Eq.(\ref{kugel:eq:PDE_sph}), are called eigenfunctions. Our new goal is therefore to solve this PDE. The task seems very difficult but we can simplify it with a well-known technique, namely the \emph{separation Ansatz}. The latter consists in assuming that the function $f(\vartheta, \varphi)$ we are looking for can be factorized in the following form +\begin{equation}\label{kugel:eq:sep_ansatz_0} + f(\vartheta, \varphi) = \Theta(\vartheta)\Phi(\varphi). +\end{equation} +In short, we are saying that the effect of the two independent variables can be described using the multiplication of two functions that describe their effect separately. If we include this assumption in Eq.(\ref{kugel:eq:PDE_sph}), we have: +\begin{equation} + \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial \Theta(\vartheta)}{\partial\vartheta} \right)\Phi(\varphi) + + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 \Phi(\varphi)}{\partial\varphi^2} \Theta(\vartheta) + \lambda \Theta(\vartheta)\Phi(\varphi) = 0. \label{kugel:eq:sep_ansatz_1} +\end{equation} +Dividing Eq.(\ref{kugel:eq:sep_ansatz_1}) by $\Theta(\vartheta)\Phi(\varphi)$ and inserting an auxiliary variable $m$, which we will call the separating constant, we will have: +\begin{equation*} +\frac{1}{\Theta(\vartheta)}\sin \vartheta \frac{d}{d \vartheta} \left( \sin \vartheta \frac{d \Theta}{d \vartheta} \right) + \lambda \sin^2 \vartheta = -\frac{1}{\Phi(\varphi)} \frac{d^2\Phi(\varphi)}{d\varphi^2} = m, +\end{equation*} +which is equivalent to the following system of two \emph{Ordinary Differential Equations} (ODEs) +\begin{align} + \frac{d^2\Phi(\varphi)}{d\varphi^2} &= -m \Phi(\varphi) \label{kugel:eq:ODE_1} \\ + \sin \vartheta \frac{d}{d \vartheta} \left( \sin \vartheta \frac{d \Theta}{d \vartheta} \right) + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)\Theta(\vartheta) &= 0 \label{kugel:eq:ODE_2} +\end{align} +The solution of Eq.(\ref{kugel:eq:ODE_1}) is quite trivial. The complex exponential is obviously the function we are looking for, so we can write +\begin{equation*} + \Phi_m(\varphi) = e^{j m \varphi}, \quad m \in \mathbb{Z}. +\end{equation*} +The restriction for the separation constant $m$ arises from the fact that we require the following periodic constraint $\Phi_m(\varphi + 2\pi) = \Phi_m(\varphi)$.\newline +As for Eq.(\ref{kugel:eq:ODE_2}), the resolution will not be so straightforward. We can begin by considering the substitution $x = \cos \vartheta$. The operator $\frac{d}{d \vartheta}$ will be: +\begin{align*} + \frac{d}{d \vartheta} = \frac{dx}{d \vartheta}\frac{d}{dx} &= -\sin \vartheta \frac{d}{dx} \\ + &= -\sqrt{1-x^2} \frac{d}{dx}. +\end{align*} +Eq.(\ref{kugel:eq:ODE_2}) will then become. +\begin{align*} + \frac{-\sqrt{1-x^2}}{\sqrt{1-x^2}} \frac{d}{dx} \left( \left(\sqrt{1-x^2}\right) \left(-\sqrt{1-x^2}\right) \frac{d \Theta}{dx} \right) + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)\Theta(\vartheta) &= 0 \\ + \frac{d}{dx} \left( (1-x^2) \frac{d \Theta}{dx} \right) + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)\Theta(\vartheta) &= 0 \\ + (1-x^2)\frac{d^2 \Theta}{dx} - 2x\frac{d \Theta}{dx} + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)\Theta(\vartheta) &= 0 \\ + (1-x^2)\frac{d^2 \Theta}{dx} - 2x\frac{d \Theta}{dx} + \left( \lambda - \frac{m}{1-x^2} \right)\Theta(\vartheta) &= 0 +\end{align*} +By making two final cosmetic substitutions, namely $\Theta(\vartheta)=\Theta(\cos^{-1}x):=y(x)$ and $\lambda=n(n+1)$, we will be able to define the \emph{Associated Legendre Equation} in its standard and most familiar form +\begin{definition}{Associated Legendre Equation} + \begin{equation}\label{kugel:eq:associated_leg_eq} + (1-x^2)\frac{d^2 y}{dx} - 2x\frac{d y}{dx} + \left( n(n+1) - \frac{m}{1-x^2} \right)y(x) = 0. + \end{equation} +\end{definition} +Our new goal then became solving Eq.(\ref{kugel:eq:asssociated_leg_eq}). After that we can fit the solution into Eq.(\ref{kugel:eq:sep_ansatz_0}), obtaining $f(\vartheta, \varphi)$, the solution of the eigenvalue problem. \newline +We simplified the problem somewhat but the task still remains very difficult. We can rely on a lemma to continue but first we need to define an additional equation, namely the \emph{Legendre Equation} +\begin{definition}{Legendre equation}\newline + Setting $m=0$ in Eq.(\ref{kugel:eq:asssociated_leg_eq}), we get + \begin{equation}\label{kugel:eq:leg_eq} + (1-x^2)\frac{d^2 y}{dx} - 2x\frac{d y}{dx} + n(n+1)y(x) = 0, + \end{equation} + also known as \emph{Legendre Equation}. +\end{definition} +Now we can continue with the lemma +\begin{lemma}\label{kugel:lemma_1} + If $y_n(x)$ is a solution of Eq.(\ref{kugel:eq:leg_eq}), then the function + \begin{equation*} + y_{m,n}(x) = (1-x^2)^{\frac{m}{2}}\frac{d^m}{dx^m}y_n(x) + \end{equation*} + satisfies Eq.(\ref{kugel:eq:associated_leg_eq}) +\end{lemma} +\begin{proof} [TODO: modificare la $m$ (è già usata come costante di separazione) o forse è giusta (?)] + To begin, we can start by differentiating $m$ times Eq.\eqref{kugel:eq:leg_eq} (which is staisfied by $y(x)$), obtaining + \begin{equation}\label{eq:lagrange_mderiv} + \frac{d^m}{dx^m}\left[ (1-x^2)\frac{d^2y}{dx^2} \right] -2 \frac{d^m}{dx^m}\left[ x\frac{dy}{dx} \right] + n(n+1)\frac{d^m}{dx^m}y=0. + \end{equation} + \emph{Leibniz's theorem} says, that if we want to differentiate $m$ times a multiplication of two functions, we can use the binomial coefficients to build up a sum. This allows us to be more compact, obtaining + \begin{equation}\label{eq:leibniz} + \frac{d^m}{dx^m}[u(x)v(x)] = \sum_{i=0}^m \binom{n}{i} \frac{d^{m-i}u}{dx^{m-1}} \frac{d^{i}v}{dx^i}. + \end{equation} + Using Eq.\eqref{eq:leibniz} in Eq.\eqref{eq:lagrange_mderiv}, we have + \begin{align} + (1-x^2)\frac{d^{m+2}y}{dx^{m+2}} &+ m \frac{d}{dx}(1-x^2)\frac{d^{m+1}y}{dx^{m+1}} + \frac{m(m-1)}{2}\frac{d^{2}}{dx^{2}}(1-x^2)\frac{d^{m}y}{dx^{m}} + n(n+1)\frac{d^m{}y}{dx^{m}} \nonumber \\ + &-2\left(x\frac{d^{m+1}y}{dx^{m+1}} + m\frac{d}{dx}x\frac{d^{m}y}{dx^{m}} \right) \nonumber \\ + &= (1-x^2)\frac{d^{m+2}y}{dx^{m+2}} -2x(m+1)\frac{d^{m+1}y}{dx^{m+1}}+(n(n+1)-m(m-1)-2m)\frac{d^{m}y}{dx^{m}}=0. \label{eq:aux_3} + \end{align} + To make the notation easier to follow, a new function can be defined + \begin{equation*} + \frac{d^{m}y}{dx^{m}} := y_m. + \end{equation*} + Eq.\eqref{eq:aux_3} now becomes + \begin{equation}\label{eq:1st_subs} + (1-x^2)\frac{d^{2}y_m}{dx^{2}} -2x(m+1)\frac{dy_m}{dx}+(n(n+1)-m(m+1))y_m=0 + \end{equation} + A second function can be further defined as + \begin{equation*} + (1-x^2)^{\frac{m}{2}}\frac{d^{m}y}{dx^{m}} = (1-x^2)^{\frac{m}{2}}y_m := \hat{y}_m, + \end{equation*} + allowing to write Eq.\eqref{eq:1st_subs} as + \begin{equation}\label{eq:2st_subs} + (1-x^2)\frac{d^2}{dx^2}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] -2(m+1)x\frac{d}{dx}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] + (n(n+1)-m(m+1))\hat{y}_m(1-x^2)^{-\frac{m}{2}}=0. + \end{equation} + The goal now is to compute the two terms + \begin{align*} + \frac{d^2}{dx^2}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] &= \frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} + \frac{d\hat{y}_m}{dx}\frac{m}{2}(1-x^2)^{-\frac{m}{2}-1}2x \\ + &+ m\left( \frac{d\hat{y}_m}{dx} x (1-x^2)^{-\frac{m}{2}-1} + \hat{y}_m (1-x^2)^{-\frac{m}{2}-1} - \hat{y}_m x (-\frac{m}{2}-1)(1-x^2)^{-\frac{m}{2}} 2x\right) \\ + &= \frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} + \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-\frac{m}{2}-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-\frac{m}{2}-1}\\ + &+ m\hat{y}_m (1-x^2)^{-\frac{m}{2}-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-\frac{m}{2}-2} + \end{align*} + and + \begin{align*} + \frac{d}{dx}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] &= \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_m\frac{m}{2}(1-x^2)^{-\frac{m}{2}-1}2x \\ + &= \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_mm(1-x^2)^{-\frac{m}{2}-1}x, + \end{align*} + to use them in Eq.\eqref{eq:2st_subs}, obtaining + \begin{align*} + (1-x^2)\biggl[\frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} &+ \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-\frac{m}{2}-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-\frac{m}{2}-1} \\ + &+ m\hat{y}_m (1-x^2)^{-\frac{m}{2}-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-\frac{m}{2}-2}\biggr] \\ + &-2(m+1)x\left[ \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_mm(1-x^2)^{-\frac{m}{2}-1}x \right] \\ + &+ (n(n+1)-m(m+1))\hat{y}_m(1-x^2)^{-\frac{m}{2}}=0.\\ + \end{align*} + We can now divide by $(1-x^2)^{-\frac{m}{2}}$, obtaining + \begin{align*} + (1-x^2)\biggl[\frac{d^2\hat{y}_m}{dx^2} &+ \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-1} + m\hat{y}_m (1-x^2)^{-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-2}\biggr] \\ + &-2(m+1)x\left[ \frac{d\hat{y}_m}{dx} + \hat{y}_mm(1-x^2)^{-1}x \right] + (n(n+1)-m(m+1))\hat{y}_m\\ + &= \frac{d^2\hat{y}_m}{dx^2} + \frac{d\hat{y}_m}{dx}mx + m\frac{d\hat{y}_m}{dx}x + m\hat{y}_m + m\hat{y}_m x^2(m+2)(1-x^2)^{-1} \\ + &-2(m+1)x\left[ \frac{d\hat{y}_m}{dx} + \hat{y}_mm(1-x^2)^{-1}x \right] + (n(n+1)-m(m+1))\hat{y}_m\\ + \end{align*} + and collecting some terms + \begin{equation*} + (1-x^2)\frac{d^2\hat{y}_m}{dx^2} - 2x\frac{d\hat{y}_m}{dx} + \left( -x^2 \frac{m^2}{1-x^2} + m+n(n+1)-m(m+1)\right)\hat{y}_m=0. + \end{equation*} + Showing that + \begin{align*} + -x^2 \frac{m^2}{1-x^2} + m+n(n+1)-m(m+1) &= n(n+1)- m^2 -x^2 \frac{m^2}{1-x^2} \\ + &= n(n+1)- \frac{m}{1-x^2} + \end{align*} + implies $\hat{y}_m(x)$ being a solution of Eq.\eqref{kugel:eq:associated_leg_eq} +\end{proof} +In simpler words, if we find a solution to Eq.\eqref{kugel:eq:leg_eq}, we can extend the latter according to the Lemma \ref{kugel:lemma_1} obtaining the solution of Eq.\eqref{kugel:eq:associated_leg_eq}.\newline +We can say that we are going in the right direction, as the problem to be solved is decreasing in difficulty. We moved from having to find a solution to Eq.\eqref{kugel:eq:associated_leg_eq} to finding a solution to Eq.\eqref{kugel:eq:leg_eq}, which is much more approachable as a problem. Luckily for us, the lemma we will present below will help us extensively, which is something of an euphemism, since it will give us the solution directly. +\begin{lemma} + The polynomial function + \begin{align*} + y_n(x)&=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{(2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k}\\ + &= \frac{1}{n!2^n}\frac{d^n}{dx^n}(1-x^2)^n =: P_n(x), + \end{align*} + is a solution to the second order differential equation + \begin{equation}\label{kugel:eq:sol_leg} + (1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx} + n(n+1)y=0, \quad \forall n>0. + \end{equation} +\end{lemma} +\begin{proof} + In order to find a solution to Eq.\eqref{eq:legendre}, the following Ansatz can be performed: + \begin{equation}\label{eq:ansatz} + y(x) = \sum_{k=0}^\infty a_k x^k. + \end{equation} + Given Eq.\eqref{eq:ansatz}, then + \begin{align*} + \frac{dy}{dx} &= \sum_{k=0}^\infty k a_k x^{k-1}, \\ + \frac{d^2y}{dx^2} &= \sum_{k=0}^\infty k (k-1) a_k x^{k-2}. + \end{align*} + Eq.\eqref{eq:legendre} can be therefore written as + \begin{align} + &(1-x^2)\sum_{k=0}^\infty k (k-1) a_k x^{k-2} - 2x\sum_{k=0}^\infty k a_k x^{k-1} + n(n+1)\sum_{k=0}^\infty a_k x^k=0 \label{eq:ansatz_in_legendre} \\ + &=\sum_{k=0}^\infty k (k-1) a_k x^{k-2} - \sum_{k=0}^\infty k (k-1) a_k x^{k} - 2x\sum_{k=0}^\infty k a_k x^{k-1} + n(n+1)\sum_{k=0}^\infty a_k x^k=0. \nonumber + \end{align} + If one consider the term + \begin{equation}\label{eq:term} + \sum_{k=0}^\infty k (k-1) a_k x^{k-2}, + \end{equation} + the substitution $\tilde{k}=k-2$ yields Eq.\eqref{eq:term} to + \begin{equation*} + \sum_{\tilde{k}=-2}^\infty (\tilde{k}+2) (\tilde{k}+1) a_{\tilde{k}+2} x^{\tilde{k}}=\sum_{\tilde{k}=0}^\infty (\tilde{k}+2) (\tilde{k}+1) a_{\tilde{k}} x^{\tilde{k}}. + \end{equation*} + This means that Eq.\eqref{eq:ansatz_in_legendre} becomes + \begin{align} + &\sum_{k=0}^\infty (k+1)(k+2) a_{k+2} x^{k} - \sum_{k=0}^\infty k (k-1) a_k x^{k} - 2\sum_{k=0}^\infty k a_k x^k + n(n+1)\sum_{k=0}^\infty a_k x^k \nonumber \\ + = &\sum_{k=0}^\infty \big[ (k+1)(k+2) a_{k+2} - k (k-1) a_k - 2 k a_k + n(n+1) a_k \big] x^k \stackrel{!}{=} 0. \label{eq:condition} + \end{align} + The condition in Eq.\eqref{eq:condition} is equivalent to + \begin{equation}\label{eq:condition_2} + (k+1)(k+2) a_{k+2} - k (k-1) a_k - 2 k a_k + n(n+1) a_k = 0. + \end{equation} + We can derive a recursion formula for $a_{k+2}$ from Eq.\eqref{eq:condition_2}, which can be expressed as + \begin{equation}\label{eq:recursion} + a_{k+2}= \frac{k (k-1) - 2 k + n(n+1)}{(k+1)(k+2)}a_k = \frac{(k-n)(k+n+1)}{(k+2)(k+1)}a_k. + \end{equation} + All coefficients can be calculated using the latter. + + Following Eq.\eqref{eq:recursion}, if we want to compute $a_6$ we would have + \begin{align*} + a_{6}= -\frac{(n-4)(n+5)}{6\cdot 5}a_4 &= -\frac{(n-4)(5+n)}{6 \cdot 5} -\frac{(n-2)(n+3)}{4 \cdot 3} a_2 \\ + &= -\frac{(n-4)(n+5)}{6 \cdot 5} -\frac{(n-2)(n+3)}{4 \cdot 3} -\frac{n(n+1)}{2 \cdot 1} a_0 \\ + &= -\frac{(n+5)(n+3)(n+1)n(n-2)(n-4)}{6!} a_0. + \end{align*} + One can generalize this relation for the $i^\text{th}$ even coefficient as + \begin{equation*} + a_{2k} = (-1)^k \frac{(n+(2k-1))(n+(2k-1)-2)\hdots (n-(2k-2)+2)(n-(2k-2))}{(2k)!}a_0 + \end{equation*} + where $i=2k$. + + A similar expression can be written for the odd coefficients $a_{2k-1}$. In this case, the equation starts from $a_1$ and to find the pattern we can write the recursion for an odd coefficient, $a_7$ for example + \begin{align*} + a_{7}= -\frac{(n-5)(n+6)}{7\cdot 6}a_5 &= - \frac{(n-5)(n+6)}{7\cdot 6} -\frac{(n-3)(n+4)}{5 \cdot 4} a_3 \\ + &= - \frac{(n-5)(n+6)}{7\cdot 6} -\frac{(n-3)(n+4)}{5 \cdot 4} -\frac{(n-1)(n+2)}{3 \cdot 2} a_1 \\ + &= -\frac{(n+6)(n+4)(n+2)(n-1)(n-3)(n-5)}{7!} a_1. + \end{align*} + As before, we can generalize this equation for the $i^\text{th}$ odd coefficient + \begin{equation*} + a_{2k+1} = (-1)^k \frac{(n + 2k)(n+2k-2)\hdots(n-(2k-1)+2)(n-(2k-1))}{(2k+1)!}a_1 + \end{equation*} + where $i=2k+1$. + + Let be + \begin{align*} + y_\text{e}^K(x) &:= \sum_{k=0}^K(-1)^k \frac{(n+(2k-1))(n+(2k-1)-2)\hdots \color{red}(n-(2k-2)+2)(n-(2k-2))}{(2k)!} x^{2k}, \\ + y_\text{o}^K(x) &:= \sum_{k=0}^K(-1)^k \frac{(n + 2k)(n+2k-2)\hdots \color{blue} (n-(2k-1)+2)(n-(2k-1))}{(2k+1)!} x^{2k+1}. + \end{align*} + The solution to the Eq.\eqref{eq:legendre} can be written as + \begin{equation}\label{eq:solution} + y(x) = \lim_{K \to \infty} \left[ a_0 y_\text{e}^K(x) + a_1 y_\text{o}^K(x) \right]. + \end{equation} + + The colored parts can be analyzed separately: + \begin{itemize} + \item[\textcolor{red}{\textbullet}] Suppose that $n=n_0$ is an even number. Then the red part, for a specific value of $k=k_0$, will follow the following relation: + \begin{equation*} + n_0-(2k_0-2)=0. + \end{equation*} + From that point on, given the recursive nature of Eq.\eqref{eq:recursion}, all the subsequent coefficients will also be 0, making the sum finite. + \begin{equation*} + a_{2k}=0 \iff y_{\text{o}}^{2k}(x)=y_{\text{o}}^{2k_0}(x), \quad \forall k>k_0 + \end{equation*} + \item[\textcolor{blue}{\textbullet}] Suppose that $n=n_0$ is an odd number. Then the blue part, for a specific value of $k=k_0$, will follow the following relation + \begin{equation*} + n_0-(2k_0-1)=0. + \end{equation*} + From that point on, for the same reason as before, all the subsequent coefficients will also be 0, making the sum finite. + \begin{equation*} + a_{2k+1}=0 \iff y_{\text{o}}^{2k+1}(x)=y_{\text{o}}^{2k_0+1}(x), \quad \forall k>k_0 + \end{equation*} + \end{itemize} + + There is the possibility of expressing the solution in Eq.\eqref{eq:solution} in a more compact form, combining the two solutions $y_\text{o}^K(x)$ and $y_\text{e}^K(x)$. They are both a polynomial of maximum degree $n$, assuming $n \in \mathbb{N}$. In the case where $n$ is even, the polynomial solution + \begin{equation*} + \lim_{K\to \infty} y_\text{e}^K(x) + \end{equation*} + will be a finite sum. If instead $n$ is odd, will be + \begin{equation*} + \lim_{K\to \infty} y_\text{o}^K(x) + \end{equation*} + to be a finite sum. + + Depending on the coefficient we start with, $a_1$ or $a_0$, we will obtain the odd or even polynomial respectively. Starting with the last coefficient $a_n$ and, recursively, calculating all the others in descending order, we can express the two parts $y_\text{o}^K(x)$ and $y_\text{e}^K(x)$ with a single sum. Hence, because we start with the last coefficient, the choice concerning $a_1$ and $a_0$ will be at the end of the sum, and not at the beginning. To compact Eq.\eqref{eq:solution}, Eq.\eqref{eq:recursion} can be reconsidered to calculate the coefficient $a_{k-2}$, using $a_k$ + \begin{equation*} + a_{k-2} = -\frac{(k+2)(k+1)}{(k-n)(k+n+1)}a_k + \end{equation*} + Now the game is to find a pattern, as before. Remember that $n$ is a fixed parameter of Eq.\eqref{eq:legendre}. + \begin{align*} + a_{n-2} &= -\frac{n(n-1)}{2(2n-1)}a_n, \\ + a_{n-4} &= -\frac{(n-2)(n-3)}{4(2n-3)}a_{n-2} \\ + &= -\frac{(n-2)(n-3)}{4(2n-3)}-\frac{n(n-1)}{2(2n-1)}a_n. + \end{align*} + In general + \begin{equation}\label{eq:general_recursion} + a_{n-2k} = (-1)^k \frac{n(n-1)(n-2)(n-3) \hdots (n-2k+1)}{2\cdot4\hdots 2k(2n-1)(2n-3)\hdots(2n-2k+1)}a_n + \end{equation} + The whole solution can now be written as + \begin{align} + y(x) &= a_n x^n + a_{n-2} x^{n-2} + a_{n-4} x^{n-4} + a_{n-6} x^{n-6} + \hdots + \begin{cases} + a_1 x, \quad &\text{if } n \text{ odd} \\ + a_0, \quad &\text{if } n \text{ even} + \end{cases} \nonumber \\ + &= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} a_{n-2k}x^{n-2k} \label{eq:solution_2} + \end{align} + By considering + \begin{align} + (2n-1)(2n-3)\hdots (2n-2k+1)&=\frac{2n(2n-1)(2n-2)(2n-3)\hdots(2n-2k+1)} + {2n(2n-2)(2n-4)(2n-6)\hdots(2n-2k+2)} \nonumber \\ + &=\frac{\frac{(2n)!}{(2n-2k)!}} + {2^kn(n-1)(n-2)(n-3)\hdots(n-k+1)} \nonumber \\ + &=\frac{\frac{(2n)!}{(2n-2k)!}} + {2^k\frac{n!}{(n-k)!}}=\frac{(n-k)!(2n)!}{n!(2n-2k)!2^k} \label{eq:1_sub_recursion}, \\ + 2 \cdot 4 \hdots 2k &= 2^r 1\cdot2 \hdots r = 2^r r!\label{eq:2_sub_recursion}, \\ + n(n-1)(n-2)(n-3) \hdots (n-2k+1) &= \frac{n!}{(n-2k)!}\label{eq:3_sub_recursion}. + \end{align} + Eq.\eqref{eq:solution_2} can be rewritten as + \begin{equation}\label{eq:solution_3} + y(x)=a_n \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{n!^2(2n-2k)!}{k!(n-2k)!(n-k)!(2n)!} x^{n-2k}. + \end{equation} + Eq.\eqref{eq:solution_3} is defined for any $a_n$. By letting $a_n$ be declared as + \begin{equation*} + a_{n} := \frac{(2n)!}{2^n n!^2}, + \end{equation*} + the so called \emph{Legendre polynomial} emerges + \begin{equation}\label{eq:leg_poly} + P_n(x):=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{(2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k} + \end{equation} +\end{proof} +As can be seen, the solution is a $n$-dependent power series, traditionally denoted as $P_n(x)$. This set of polynomials are called \emph{Legendre Polynomials}, because precisely they are polynomials satisfying the Legendre equation.\newline +Now that we have a solution to Eq.\eqref{kugel:eq:leg_eq}, we can then extend Eq.\eqref{kugel:eq:sol_leg}, as stated in Lemma \ref{kugel:lemma_1}. We will then have +\begin{align*} +y_{m,n}(x) &= (1-x^2)^{\frac{m}{2}}\frac{d^m}{dx^m}P_n(x) \\ +&= \frac{1}{n!2^n}(1-x^2)^{\frac{m}{2}}\frac{d^{m+n}}{dx^{m+n}}(1-x^2)^n +\end{align*} +This set of functions are defined as \emph{Associated Legendre functions}, because similarly to before, they solve the Associated Legendre equation, defined in Eq.\eqref{kugel:eq:eq_leg}. +\begin{definition}{Associated Legendre Functions} +\begin{equation}\label{kugel:eq:associated_leg_func} +P_{m,n}(x) := \frac{1}{n!2^n}(1-x^2)^{\frac{m}{2}}\frac{d^{m+n}}{dx^{m+n}}(1-x^2)^n +\end{equation} +\end{definition} +As you may recall, previously we performed the substitution $x=\cos \vartheta$. Now we need to return to the old domain, which can be done straightforwardly: +\begin{equation*} + \Theta(\vartheta) = P_{m,n}(\cos \vartheta), +\end{equation*} +obtaining the much sought function $\Theta(\vartheta)$. \newline +So we finally reached the end of this tortuous path. Now we just need to put together all the information we have to construct $f(\vartheta, \varphi)$ in the following way: +\begin{equation}\label{kugel:eq:sph_harm_0} + f(\vartheta, \varphi) = \Theta(\vartheta)\Phi(\varphi) = P_{m,n}(\cos \vartheta)e^{jm\varphi}, \quad |m|\leq n. +\end{equation} +The constraint $|m| Date: Tue, 16 Aug 2022 15:32:01 +0200 Subject: Removed unnecessary equation indices. --- buch/papers/sturmliouville/einleitung.tex | 53 ++++++++++++++----------------- 1 file changed, 24 insertions(+), 29 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/einleitung.tex b/buch/papers/sturmliouville/einleitung.tex index 78c1800..31256eb 100644 --- a/buch/papers/sturmliouville/einleitung.tex +++ b/buch/papers/sturmliouville/einleitung.tex @@ -6,15 +6,15 @@ \section{Was ist das Sturm-Liouville-Problem\label{sturmliouville:section:teil0}} \rhead{Einleitung} Das Sturm-Liouville-Problem wurde benannt nach dem schweizerisch-französischer Mathematiker und Physiker Jacques Charles Fran\c{c}ois Sturm und dem französischer Mathematiker Joseph Liouville. -Gemeinsam haben sie in der mathematischen Physik die Sturm-Liouville-Theorie entwickelt und gilt für die Lösung von gewohnlichen Differentialgleichungen, jedoch verwendet man die Theorie öfters bei der Lösung von partiellen Differentialgleichungen. -Normalerweise betrachtet man für das Strum-Liouville-Problem eine gewöhnliche Differentialgleichung 2. Ordnung, und wenn es sich um eine partielle Differentialgleichung handelt, kann man sie mit Hilfe einiger Methoden in mehrere gewöhnliche Differentialgleichungen umwandeln, wie z. B. den Separationsansatz, die partielle Differentialgleichung mit mehreren Variablen. +Gemeinsam haben sie in der mathematischen Physik die Sturm-Liouville-Theorie entwickelt, welche für die Lösung von gewohnlichen Differentialgleichungen gilt, jedoch verwendet man die Theorie öfters bei der Lösung von partiellen Differentialgleichungen. +Normalerweise betrachtet man für das Strum-Liouville-Problem eine gewöhnliche Differentialgleichung 2. Ordnung, und wenn es sich um eine partielle Differentialgleichung handelt, kann man sie mit Hilfe einiger Methoden in mehrere gewöhnliche Differentialgleichungen umwandeln, wie zum Beispiel mit dem Separationsansatz. \begin{definition} \index{Sturm-Liouville-Gleichung}% Angenommen man hat die lineare homogene Differentialgleichung -\begin{equation} +\[ \frac{d^2y}{dx^2} + a(x)\frac{dy}{dx} + b(x)y = 0 -\end{equation} +\] und schreibt die Gleichung um in: \begin{equation} \label{eq:sturm-liouville-equation} @@ -23,7 +23,7 @@ und schreibt die Gleichung um in: , diese Gleichung wird dann Sturm-Liouville-Gleichung bezeichnet. \end{definition} -Alle homogene 2.Ordnung lineare gewöhnliche Differentialgleichungen können in die Form der Gleichung \ref{eq:sturm-liouville-equation} umgeformt werden. +Alle homogenen, linearen, gewöhnlichen, Differentialgleichungen 2.Ordnung können in die Form der Gleichung~\eqref{eq:sturm-liouville-equation} gebracht werden. Die Sturm-Liouville-Theorie besagt, dass, wenn man die Sturm-Liouville-Gleichung mit den homogenen Randbedingungen des dritten Typs\footnote{Die Randbedingung des dritten Typs, oder Robin-Randbedingungen (benannt nach dem französischen mathematischen Analytiker und angewandten Mathematiker Victor Gustave Robin), wird genannt, wenn sie einer gewöhnlichen oder partiellen Differentialgleichung auferlegt wird, so sind die Spezifikationen einer Linearkombination der Werte einer Funktion sowie die Werte ihrer Ableitung am Rande des Bereichs} \begin{equation} \begin{aligned} @@ -34,30 +34,30 @@ Die Sturm-Liouville-Theorie besagt, dass, wenn man die Sturm-Liouville-Gleichung \end{equation} kombiniert, wie schon im Kapitel \ref{sub:differentailgleichung} erwähnt, auf dem Intervall (a,b), dann bekommt man das klassische Sturm-Liouville-Problem. Wenn von der Funktion $y(x)$ die Werte $x$ des jeweiligen Randes des Definitionsbereiches anzunehmen sind, also -\begin{equation} +\[ y(a) = y(b) = 0 -\end{equation} +\] , so spricht man von einer Dirichlet-Randbedingung, und von einer Neumann-Randbedingung spricht man, wenn -\begin{equation} +\[ y'(a) = y'(b) = 0 -\end{equation} -ergibt - die Existenz und Eindeutigkeit der Lösung kann mit den zwei Randbedingungen sichergestellt werden. +\] +ist. Die Existenz und Eindeutigkeit der Lösung kann mit den zwei Randbedingungen sichergestellt werden. Lösungen die nicht Null sind, werden nicht betrachtet und diese zwei Gleichungen (\ref{eq:sturm-liouville-equation} und \ref{eq:randbedingungen}) kombiniert, nennt man Eigenfunktionen. Wenn bei der Sturm-Liouville-Gleichung \ref{eq:sturm-liouville-equation} alles konstant bleibt, aber der Wert von $\lambda$ sich ändert, erhält man eine andere Eigenfunktion, weil man eine andere gewöhnliche Differentialgleichung löst; der Parameter $\lambda$ wird als Eigenwert bezeichnet. Es ist genau das gleiche Prinzip wie bei den Matrizen, andere Eigenwerte ergeben andere Eigenvektoren. Es besteht eine Korrespondenz zwischen den Eigenwerten und den Eigenvektoren. Das gleiche gilt auch beim Sturm-Liouville-Problem, und zwar -\begin{equation} - \lambda \overset{Korrespondenz}\leftrightarrow y -\end{equation}. +\[ + \lambda \overset{Korrespondenz}\leftrightarrow y. +\] Die Theorie besagt, wenn $y_m$, $y_n$ Eigenfuktionen des Sturm-Liouville-Problems sind, die verschiedene Eigenwerte $\lambda_m$, $\lambda_n$ ($\lambda_m \neq \lambda_n$) entsprechen, so sind $y_m$, $y_n$ orthogonal zu y - dies gilt für das Intervall (a,b). Somit ergibt die Gleichung -\begin{equation} - \int_{a}^{b} w(x)y_m y_n = 0 -\end{equation}. +\[ + \int_{a}^{b} w(x)y_m y_n = 0. +\] Die Funktionen $p(x)$, $q(x)$ und $w(x)$ werden als Koeffizientenfunktionen mit ihren freien Variablen $x$ bezeichnet. Die Funktion $w(x)$ (manchmal auch $r(x)$ genannt) wird als Gewichtsfunktion oder Dichtefunktion bezeichnet. @@ -90,29 +90,29 @@ Bei einem regulären Sturm-Liouville-Problem geht es darum, ohne genaue Kenntnis \subsection{Das singuläre Sturm-Liouville-Problem\label{sub:singuläre_sturm_liouville_problem}} -Von einem singulären Sturm-Liouville-Problem spricht man, wenn die Bedingungen des regulärem Problem nicht erfüllt sind. +Von einem singulären Sturm-Liouville-Problem spricht man, wenn die Bedingungen des regulären Problems nicht erfüllt sind. \begin{definition} \label{def:singulär_sturm-liouville-problem} \index{singuläres Sturm-Liouville-Problem} -Es handelt sich um ein singuläres Sturm-Liouville-Problem, wenn: +Es handelt sich um ein singuläres Sturm-Liouville-Problem, \begin{itemize} \item wenn sein Definitionsbereich auf dem Intervall $[ \ a,b] \ $ unbeschränkt ist oder \item wenn die Koeffizienten an den Randpunkten Singularitäten haben. \end{itemize} \end{definition} -Allerdings kann nur eine der Bedingungen nicht erfüllt sein, so dass es sich bereits um ein singuläres Sturm-Liouville-Problem handelt. +Allerdings kann auch nur eine der Bedingungen nicht erfüllt sein, so dass es sich bereits um ein singuläres Sturm-Liouville-Problem handelt. \begin{beispiel} Das Randwertproblem - \begin{equation} + \[ \begin{aligned} x^2y'' + xy' + (\lambda^2x^2 - m^2)y &= 0, 0 Date: Tue, 16 Aug 2022 15:47:34 +0200 Subject: Corrected some minor mistakes in solution properties. --- buch/papers/sturmliouville/eigenschaften.tex | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex index fda8be6..87ba864 100644 --- a/buch/papers/sturmliouville/eigenschaften.tex +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -42,7 +42,7 @@ den Lösungen hervorbringt, wird der Spektralsatz benötigt. Dieser wird in der linearen Algebra oft verwendet um zu zeigen, dass eine Matrix diagonalisierbar ist, beziehungsweise dass eine Orthonormalbasis existiert. Dazu wird zunächst gezeigt, dass eine gegebene $n\times n$-Matrix $A$ aus einem -endlichdimensionalem $\mathbb{K}$-Vektorraum selbstadungiert ist, also dass +endlichdimensionalem $\mathbb{K}$-Vektorraum selbstadjungiert ist, also dass \[ \langle Av, w \rangle = @@ -67,8 +67,8 @@ Orthonormalsystem existiert. Der Spektralsatz besagt also, dass, weil $L_0$ selbstadjungiert ist, eine Orthonormalbasis aus Eigenvektoren existiert. Genauer bedeutet dies, dass alle Eigenvektoren, beziehungsweise alle Lösungen -des Sturm-Liouville-Problems orthogonal zueinander sind bezüglich dem -Skalarprodukt, in dem $L_0$ selbstadjungiert ist. +des Sturm-Liouville-Problems orthogonal zueinander sind bezüglich des +Skalarprodukts, in dem $L_0$ selbstadjungiert ist. Erfüllt also eine Differenzialgleichung die in Abschnitt~\ref{sturmliouville:section:teil0} präsentierten Eigenschaften und -- cgit v1.2.1 From 0904096277bc7944a6c0baf50f4032c2f4d9909a Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Tue, 16 Aug 2022 16:11:37 +0200 Subject: Corrected some smaller mistakes in fourier example and added authors to files. --- buch/papers/sturmliouville/eigenschaften.tex | 1 + buch/papers/sturmliouville/einleitung.tex | 1 + buch/papers/sturmliouville/tschebyscheff_beispiel.tex | 1 + buch/papers/sturmliouville/waermeleitung_beispiel.tex | 18 ++++++++++-------- 4 files changed, 13 insertions(+), 8 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex index 87ba864..85f0bf3 100644 --- a/buch/papers/sturmliouville/eigenschaften.tex +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -1,5 +1,6 @@ % % eigenschaften.tex -- Eigenschaften der Lösungen +% Author: Erik Löffler % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % diff --git a/buch/papers/sturmliouville/einleitung.tex b/buch/papers/sturmliouville/einleitung.tex index 31256eb..babc06d 100644 --- a/buch/papers/sturmliouville/einleitung.tex +++ b/buch/papers/sturmliouville/einleitung.tex @@ -1,5 +1,6 @@ % % einleitung.tex -- Beispiel-File für die Einleitung +% Author: Réda Haddouche % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % diff --git a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex index a18684f..8a99ae9 100644 --- a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex +++ b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex @@ -1,5 +1,6 @@ % % tschebyscheff_beispiel.tex +% Author: Réda Haddouche % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index b22d5f5..7a37b2b 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -1,5 +1,6 @@ % -% waermeleitung_beispiel.tex -- Beispiel Wärmeleitung in homogenem Stab. +% waermeleitung_beispiel.tex -- Beispiel Wärmeleitung in homogenem Stab. +% Author: Erik Löffler % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % @@ -17,7 +18,7 @@ die partielle Differentialgleichung \begin{equation} \label{sturmliouville:eq:example-fourier-heat-equation} \frac{\partial u}{\partial t} = - \kappa \frac{\partial^{2}u}{{\partial x}^{2}} + \kappa \frac{\partial^{2}u}{{\partial x}^{2}}, \end{equation} wobei der Stab in diesem Fall auf der X-Achse im Intervall $[0,l]$ liegt. @@ -187,7 +188,8 @@ somit auch zu orthogonalen Lösungen führen. % \subsubsection{Lösund der Differentialgleichung in x} -Als erstes wird auf die erste erste Gleichung eingegangen. +Als erstes wird auf die +Gleichung~\eqref{sturmliouville:eq:example-fourier-separated-x} eingegangen. Aufgrund der Struktur der Gleichung \[ X^{\prime \prime}(x) - \mu X(x) @@ -417,7 +419,7 @@ sein, welche Integralgrenzen zu verwenden sind. In diesem Fall haben die $\sin$ und $\cos$ Terme beispielsweise keine ganze Periode im Intervall $x \in [0, l]$ für ungerade $n$ und $m$. Um die Skalarprodukte aber korrekt zu berechnen, muss über ein ganzzahliges -Vielfaches der Periode der triginimetrischen Funktionen integriert werden. +Vielfaches der Periode der trigonometrischen Funktionen integriert werden. Dazu werden die Integralgrenzen $-l$ und $l$ verwendet und es werden ausserdem neue Funktionen $\hat{u}_c(0, x)$ für die Berechnung mit Cosinus und $\hat{u}_s(0, x)$ für die Berechnung mit Sinus angenomen, welche $u(0, t)$ @@ -487,7 +489,7 @@ nahezu alle Terme verschwinden, denn \[ \int_{-l}^{l}cos\left(\frac{m \pi}{l}x\right) dx = - 0 + 0, \] da hier über ein ganzzahliges Vielfaches der Periode integriert wird, \[ @@ -611,7 +613,7 @@ Es bleibt also noch % Lösung von T(t) % -\subsubsection{Lösund der Differentialgleichung in t} +\subsubsection{Lösung der Differentialgleichung in $t$} Zuletzt wird die zweite Gleichung der Separation~\eqref{sturmliouville:eq:example-fourier-separated-t} betrachtet. Diese wird über das charakteristische Polynom @@ -627,7 +629,7 @@ Lösung \[ T(t) = - e^{-\kappa \mu t} + e^{\kappa \mu t} \] führt und mit dem Resultat~\eqref{sturmliouville:eq:example-fourier-mu-solution} \[ @@ -637,7 +639,7 @@ führt und mit dem Resultat~\eqref{sturmliouville:eq:example-fourier-mu-solution \] ergibt. -Dieses Resultat kann nun mit allen vorhergehenden Resultaten zudammengesetzt +Dieses Resultat kann nun mit allen vorhergehenden Resultaten zusammengesetzt werden um die vollständige Lösung für das Stab-Problem zu erhalten. \subsubsection{Lösung für einen Stab mit Enden auf konstanter Temperatur} -- cgit v1.2.1 From 3bdc9c20b9a83cf1cb7f4570d99d2495576ca30d Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Tue, 16 Aug 2022 16:21:59 +0200 Subject: Removed unnecessary equation indexing in Tschebyscheff example and corrected some errors. --- .../sturmliouville/tschebyscheff_beispiel.tex | 50 ++++++++-------------- 1 file changed, 19 insertions(+), 31 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex index 8a99ae9..e86e742 100644 --- a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex +++ b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex @@ -6,12 +6,12 @@ % \subsection{Tschebyscheff-Polynome\label{sub:tschebyscheff-polynome}} -Im Kapitel \ref{sub:beispiele_sturm_liouville_problem} sind die Koeffizientenfunktionen die man braucht schon aufgeliste, und zwar mit +Im Kapitel \ref{sub:beispiele_sturm_liouville_problem} sind die Koeffizientenfunktionen die man braucht schon aufgelistet, und zwar mit \begin{align*} w(x) &= \frac{1}{\sqrt{1-x^2}} \\ p(x) &= \sqrt{1-x^2} \\ - q(x) &= 0 -\end{align*}. + q(x) &= 0. +\end{align*} Da die Sturm-Liouville-Gleichung \begin{equation} \label{eq:sturm-liouville-equation-tscheby} @@ -20,14 +20,14 @@ Da die Sturm-Liouville-Gleichung nun mit den Koeffizientenfunktionen aufgestellt werden kann, bleibt die Frage, ob es sich um ein reguläres oder singuläres Sturm-Liouville-Problem handelt. Für das reguläre Problem laut der Definition \ref{def:reguläres_sturm-liouville-problem} muss die funktion $p(x) = \sqrt{1-x^2}$, $p'(x) = -2x$, $q(x) = 0$ und $w(x) = \frac{1}{\sqrt{1-x^2}}$ stetig und reell sein - und sie sind es auch. Auf dem Intervall $(-1,1)$ sind die Tschebyscheff-Polynome erster Art mit Hilfe von Hyperbelfunktionen -\begin{equation} - T_n(x) = \cos n (\arccos x) -\end{equation}. +\[ + T_n(x) = \cos n (\arccos x). +\] Für $x>1$ und $x<-1$ sehen die Polynome wie folgt aus: -\begin{equation} +\[ T_n(x) = \left\{\begin{array}{ll} \cosh (n \arccos x), & x > 1\\ - (-1)^n \cosh (n \arccos (-x)), & x<-1 \end{array}\right. -\end{equation}, + (-1)^n \cosh (n \arccos (-x)), & x<-1 \end{array}\right., +\] jedoch ist die Orthogonalität nur auf dem Intervall $[ -1, 1]$ sichergestellt. Die nächste Bedingung beinhaltet, dass die Funktion $p(x)^{-1}$ und $w(x)>0$ sein müssen. Die Funktion @@ -39,34 +39,22 @@ ist die gleiche wie $w(x)$. Für die Verifizierung der Randbedingungen benötigt man erneut $p(x)$. Da sich die Polynome nur auf dem Intervall $[ -1,1 ]$ orthogonal verhalten, sind $a = -1$ und $b = 1$ gesetzt. Beim einsetzen in die Randbedingung \ref{eq:randbedingungen}, erhält man -\begin{equation} +\[ \begin{aligned} - k_a y(-1) + h_a y'(-1) &= 0 - k_b y(-1) + h_b y'(-1) &= 0 + k_a y(-1) + h_a y'(-1) &= 0 \\ + k_b y(-1) + h_b y'(-1) &= 0. \end{aligned} -\end{equation}. -Die Funktion $y(x)$ und $y'(x)$ sind in diesem Fall die Tschebyscheff Polynome (siehe \label{sub:definiton_der_tschebyscheff-Polynome}). +\] +Die Funktion $y(x)$ und $y'(x)$ sind in diesem Fall die Tschebyscheff Polynome (siehe \ref{sub:definiton_der_tschebyscheff-Polynome}). Es gibt zwei Arten von Tschebyscheff Polynome: die erste Art $T_n(x)$ und die zweite Art $U_n(x)$. Jedoch beachtet man in diesem Kapitel nur die Tschebyscheff Polynome erster Art (\ref{eq:tschebyscheff-polynome}). Die Funktion $y(x)$ wird nun mit der Funktion $T_n(x)$ ersetzt und für die Verifizierung der Randbedingung wählt man $n=2$. Somit erhält man -\begin{equation} +\[ \begin{aligned} k_a T_2(-1) + h_a T_{2}'(-1) &= k_a = 0\\ - k_b T_2(1) + h_b T_{2}'(1) &= k_b = 0 + k_b T_2(1) + h_b T_{2}'(1) &= k_b = 0. \end{aligned} -\end{equation}. -Ähnlich wie beim Beispiel der Wärmeleitung in einem homogenen Stab kann man, damit die Bedingung $|k_i|^2 + |h_i|^2\ne 0$ erfüllt ist, können beliebige $h_a \ne 0$ und $h_b \ne 0$ gewählt werden. -Somit ist erneut gezeigt, dass die Randbedingungen der Tschebyscheff-Polynome auf die Sturm-Liouville-Randbedingungen erfüllt und alle daraus resultierenden Lösungen orthogonal sind. - - - - - - - - - - - - +\] +Ähnlich wie beim Beispiel der Wärmeleitung in einem homogenen Stab können, damit die Bedingung $|k_i|^2 + |h_i|^2\ne 0$ erfüllt ist, beliebige $h_a \ne 0$ und $h_b \ne 0$ gewählt werden. +Somit ist erneut gezeigt, dass die Randbedingungen der Tschebyscheff-Polynome auch die Sturm-Liouville-Randbedingungen erfüllt und alle daraus resultierenden Lösungen orthogonal sind. -- cgit v1.2.1 From afb21283eeaa8178fbf1890212e177aa05a92c1b Mon Sep 17 00:00:00 2001 From: Joshua Baer Date: Tue, 16 Aug 2022 16:57:02 +0200 Subject: started e littel bit --- buch/papers/fm/00_modulation.tex | 15 +- buch/papers/fm/01_AM.tex | 52 +- buch/papers/fm/02_FM.tex | 62 +- buch/papers/fm/03_bessel.tex | 25 +- buch/papers/fm/Python animation/Bessel-FM.ipynb | 91 +-- buch/papers/fm/Python animation/Bessel-FM.py | 72 +-- buch/papers/fm/Python animation/m_t.pgf | 746 ++++++++++++++++++++++++ buch/papers/fm/Quellen/NaT_Skript_20210920.pdf | Bin 0 -> 5455101 bytes buch/papers/fm/main.tex | 17 +- 9 files changed, 945 insertions(+), 135 deletions(-) create mode 100644 buch/papers/fm/Python animation/m_t.pgf create mode 100644 buch/papers/fm/Quellen/NaT_Skript_20210920.pdf (limited to 'buch/papers') diff --git a/buch/papers/fm/00_modulation.tex b/buch/papers/fm/00_modulation.tex index e2ba39f..982d63c 100644 --- a/buch/papers/fm/00_modulation.tex +++ b/buch/papers/fm/00_modulation.tex @@ -3,11 +3,22 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % + +Durch die Modulation wird ein Nachrichtensignal \(m(t)\) auf ein Trägersignal (z.B. ein Sinus- oder Rechtecksignal) abgebildet (kombiniert). +Durch dieses Auftragen vom Nachrichtensignal \(m(t)\) kann das modulierte Signal in einem gewünschten Frequenzbereich übertragen werden. +Der ursprünglich Frequenzbereich des Nachrichtensignal \(m(t)\) erstreckt sich typischerweise von 0 Hz bis zur Bandbreite \(B_m\). +Beim Empfänger wird dann durch Demodulation das ursprüngliche Nachrichtensignal \(m(t)\) so originalgetreu wie möglich zurückgewonnen. +Beim Trägersignal \(x_c(t)\) handelt es sich um ein informationsloses Hilfssignal. +Durch die Modulation mit dem Nachrichtensignal \(m(t)\) wird es zum modulierten zu übertragenden Signal. +Für alle Erklärungen wird ein sinusförmiges Trägersignal benutzt, jedoch kann auch ein Rechtecksignal, +welches Digital einfach umzusetzten ist, +genauso als Trägersignal genutzt werden kann.\cite{fm:NAT} + \subsection{Modulationsarten\label{fm:section:modulation}} Das sinusförmige Trägersignal hat die übliche Form: \(x_c(t) = A_c \cdot \cos(\omega_c(t)+\varphi)\). -Wobei die konstanten Amplitude \(A_c\) und Phase \(\varphi\) vom Nachrichtensignal \(m(t)\) verändert wird. +Wobei die konstanten Amplitude \(A_c\) und Phase \(\varphi\) vom Nachrichtensignal \(m(t)\) verändert werden können. Der Parameter \(\omega_c\), die Trägerkreisfrequenz bzw. die Trägerfrequenz \(f_c = \frac{\omega_c}{2\pi}\), steht nicht für die modulation zur verfügung, statt dessen kann durch ihn die Frequenzachse frei gewählt werden. \newblockpunct @@ -25,6 +36,8 @@ die Phasenmodulation \(\varphi\) und dann noch die Momentankreisfrequenz \(\omeg \item PM \item FM \end{itemize} +Um modulation zu Verstehen ist es am Anschaulichst mit der AM Amplitudenmodulation, +da Phasenmodulation und Frequenzmodulation den gleichen Parameter verändert vernachlässige ich die Phasenmodulation ganz. To do: Bilder jeder Modulationsart diff --git a/buch/papers/fm/01_AM.tex b/buch/papers/fm/01_AM.tex index 21927f5..714b9a0 100644 --- a/buch/papers/fm/01_AM.tex +++ b/buch/papers/fm/01_AM.tex @@ -11,19 +11,61 @@ Nun zur Amplitudenmodulation verwenden wir das bevorzugte Trägersignal \[ x_c(t) = A_c \cdot \cos(\omega_ct). \] -Dies bringt den grossen Vorteil das, dass modulierend Signal sämtliche Anteile im Frequenzspektrum inanspruch nimmt -und das Trägersignal nur zwei komplexe Schwingungen besitzt. +Dies bringt den grossen Vorteil das, dass modulierend Signal sämtliche Anteile im Frequenzspektrum in Anspruch nimmt +und das Trägersignal nur zwei komplexe Schwingungen besitzt. Dies sieht man besonders in der Eulerischen Formel \[ x_c(t) = \frac{A_c}{2} \cdot e^{j\omega_ct}\;+\;\frac{A_c}{2} \cdot e^{-j\omega_ct}. + \label{fm:eq:AM:euler} \] Dabei ist die negative Frequenz der zweiten komplexen Schwingung zwingend erforderlich, damit in der Summe immer ein reellwertiges Trägersignal ergibt. Nun wird der Parameter \(A_c\) durch das Modulierende Signal \(m(t)\) ersetzt, wobei so \(m(t) \leqslant |1|\) normiert wurde. -\newline -\newline + +Dabei entseht wine Umhüllende kurve die unserem ursprünglichen signal \(m(t)\) entspricht. +\[ + x_c(t) = m(t) \cdot \cos(\omega_ct). +\] + +\begin{figure} + \centering + \input{papers/fm/Python animation/m_t.pgf} + \caption{modulierende Signal \(m(t)\)} + \label{fig:bessel} +\end{figure} +% TODO: +Bilder Hier beschrieib ich was AmplitudenModulation ist und mache dan den link zu Frequenzmodulation inkl Formel \[\cos( \cos x)\] so wird beschrieben das daraus eigentlich \(x_c(t) = A_c \cdot \cos(\omega_i)\) wird und somit \(x_c(t) = A_c \cdot \cos(\omega_c + \frac{d \varphi(t)}{dt})\). Da \(\sin \) abgeleitet \(\cos \) ergibt, so wird aus dem \(m(t)\) ein \( \frac{d \varphi(t)}{dt}\) in der momentan frequenz. \[ \Rightarrow \cos( \cos x) \] +\subsection{Frequenzspektrum} +Das Frequenzspektrum ist eine Darstellung von einem Signal im Frequenzbereich, das heisst man erkennt welche Frequenzen in einem Signal vorhanden sind. +Dafür muss man eine Fouriertransformation vornehmen. +Wird aus dieser Gleichung \eqref{fm:eq:AM:euler}die Fouriertransformation vorggenommen, so erhält man -\subsection{Frequenzspektrum} \ No newline at end of file +% +%Ein Ziel der Modulation besteht darin, mehrere Nachrichtensignale von verschiedenen Sendern gleichzeitig +%in verschiedenen Frequenzbereichen über den gleichen Kanal zu senden. Um dieses Frequenzmultiplexing +%störungsfrei und mit eine Vielzahl von Teilnehmern durchführen zu können, muss die spektrale Beschaffen- +%heit der modulierten Signale möglichst gut bekannt sein. +%Dank des Modulationssatzes der Fouriertransformation lässt sich das Spektrum eines gewöhnlichen AM Si- +%gnals sofort bestimmen: +%A c μ +%F +%·(M n (ω−ω c ) + M n (ω+ω c )) (5.5) +%A c ·(1+μm n (t))·cos(ω c t) ❝ s A c π (δ(ω−ω c ) + δ(ω+ω c )) + +%2 +%Das zweiseitige Spektrum des Nachrichtensignals M (ω) wird mit dem Faktor A 2 c μ gewichtet und einmal +%nach +ω c und einmal nach −ω c verschoben. Dies führt im Vergleich zum Basisbandsignal zu einer Verdop- +%pelung der Bandbreite mit je einem Seitenband links und rechts der Trägerfrequenz. Weiter beinhaltet das +%Amplitudendichtespektrum je eine Deltafunktion mit Gewicht A c π an den Stellen ±ω c , d.h. ein fester, nicht- +%modulierter Amplitudenanteil bei der eigentlichen Trägerfrequenz. +%Das Amplitudendichtespektrum ist im nachfolgenden Graphen für A c = 1 und μ = 100% dargestellt.5.3. Gewöhnliche Amplitudenmodulation +%47 +%Abbildung 5.12: Amplitudendichtespektrum von gewöhnlicher AM +%Für das Nachrichtensignal wurde in diesem Graph mit einem Keil symbolhaft ein Amplitudendichtespektrum +%|M (ω)| gewählt, bei welchem der Anteil auf der positiven und jener auf der negativen Frequenzachse visuell +%gut auseinandergehalten werden können. Ein solch geformtes Spektrum wird aber in der Praxis kaum je +%auftreten: bei periodischen Testsignalen besteht das Nachrichtensignal aus einem Linienspektrum, bei einem +%Energiesignal mit zufälligem Verlauf aus einem kontinuierlichen Spektrum, welches jedoch nicht auf diese +%einfache Art geformt sein wird \ No newline at end of file diff --git a/buch/papers/fm/02_FM.tex b/buch/papers/fm/02_FM.tex index fedfaaa..a01fb69 100644 --- a/buch/papers/fm/02_FM.tex +++ b/buch/papers/fm/02_FM.tex @@ -6,9 +6,65 @@ \section{FM \label{fm:section:teil1}} \rhead{FM} -\subsection{Frequenzspektrum} -TODO -Hier Beschreiben ich FM und FM im Frequenzspektrum. +\subsection{Frequenzmodulation} +(skript Nat ab Seite 60) +Als weiterer Parameter, um ein sinusförmiges Trägersignal \(x_c = A_c \cdot \cos(\omega_c t + \varphi)\) zu modulieren, +bietet sich neben der Amplitude \(A_c\) auch der Phasenwinkel \(\varphi\) oder die momentane Frequenzabweichung \(\frac{d\varphi}{dt}\) an. +Bei der Phasenmodulation (Englisch: phase modulation, PM) erzeugt das Nachrichtensignal \(m(t)\) eine Phasenabweichung \(\varphi(t)\) des modulierten Trägersignals im Vergleich zum nicht-modulierten Träger. Sie ist pro- +%portional zum Nachrichtensignal \(m(t)\) durch eine Skalierung mit der Phasenhubkonstanten (Englisch: phase deviation constant) +%k p [rad], +%welche die Amplitude des Nachrichtensignals auf die Phasenabweichung des +%modulierten Trägersignals abbildet: φ(t) = k p · m(t). Damit ergibt sich für das phasenmodulierte Trägersi- +%gnal: +%x PM (t) = A c · cos (ω c t + k p · m(t)) +%(5.16) +%Die modulierte Phase φ(t) verändert dabei auch die Momentanfrequenz (Englisch: instantaneous frequency) +%ω i +%, welche wie folgt berechnet wird: +%f i = 2π +%ω i (t) = ω c + +%d φ(t) +%dt +%(5.17) +%Bei der Frequenzmodulation (Englisch: frequency modulation, FM) ist die Abweichung der momentanen +%Kreisfrequenz ω i von der Trägerkreisfrequenz ω c proportional zum Nachrichtensignal m(t). Sie ergibt sich, +%indem m(t) mit der (Kreis-)Frequenzhubkonstanten (Englisch: frequency deviation constant) k f [rad/s] ska- +%liert wird: ω i (t) = ω c + k f · m(t). Diese sich zeitlich verändernde Abweichung von der Kreisfrequenz ω c +%verursacht gleichzeitig auch Schwankungen der Phase φ(t), welche wie folgt berechnet wird: +%φ(t) = +%Z t +%−∞ +%ω i (τ ) − ω c dτ = +%Somit ergibt sich für das frequenzmodulierte Trägersignal: +% +%Z t +%−∞ +%x FM (t) = A c · cos  ω c t + k f +%k f · m(t) dτ +%Z t +%−∞ +% +%m(τ ) dτ  +%(5.18) +%(5.19) +%Die Phase φ(t) hat dabei einen kontinuierlichen Verlauf, d.h. das FM-modulierte Signal x FM (t) weist keine +%Stellen auf, wo sich die Phase sprunghaft ändert. Aus diesem Grund spricht man bei frequenzmodulierten +%Signalen – speziell auch bei digitalen FM-Signalen – von einer Modulation mit kontinuierlicher Phase (Eng- +%lisch: continuous phase modulation). +%Wie aus diesen Ausführungen hervorgeht, sind Phasenmodulation und Frequenzmodulation äquivalente Mo- +%dulationsverfahren. Beide variieren sowohl die Phase φ wie auch die Momentanfrequenz ω i . Dadurch kann +%man leider nicht – wie vielleicht erhofft – je mit einem eigenen Nachrichtensignal ein gemeinsames Trägersi- +%gnal unabhängig PM- und FM-modulieren, ohne dass sich diese Modulationen für den Empfänger untrennbar +%vermischen würden. +% +%Um die mathematische Behandlung der nicht-linearen Winkelmodulation etwas zu verkürzen, ist es aufgrund +%dieser Äquivalenzen gerechtfertigt, dass PM und FM gemeinsam behandelt werden. Jeweils vor der Modu- +%lation bzw. nach der Demodulation kann dann noch eine Differentiation oder Integration durchgeführt wird, +%um von der einen Modulationsart zur anderen zu gelangen. +%\subsection{Frequenzbereich} +%Nun +%TODO +%Hier Beschreiben ich FM und FM im Frequenzspektrum. %Sed ut perspiciatis unde omnis iste natus error sit voluptatem %accusantium doloremque laudantium, totam rem aperiam, eaque ipsa %quae ab illo inventore veritatis et quasi architecto beatae vitae diff --git a/buch/papers/fm/03_bessel.tex b/buch/papers/fm/03_bessel.tex index 45f2dfd..3c2cb71 100644 --- a/buch/papers/fm/03_bessel.tex +++ b/buch/papers/fm/03_bessel.tex @@ -67,7 +67,7 @@ Mit dem \eqref{fm:eq:addth1} wird aus dem modulierten Signal = \cos(\omega_c t + \beta\sin(\omega_mt)) = - \cos(\omega_c t)\cos(\beta\sin(\omega_m t))-\sin(\omega_ct)\sin(\beta\sin(\omega_m t)). + \cos(\omega_c t)\cos(\beta\sin(\omega_m t)) - \sin(\omega_ct)\sin(\beta\sin(\omega_m t)). \label{fm:eq:start} \] %----------------------------------------------------------------------------------------------------------- @@ -89,23 +89,34 @@ mit hilfe der Besselindentität \eqref{fm:eq:besselid1} zum \end{align*} %intertext{} Funktioniert nicht. wobei mit dem Additionstheorem \eqref{fm:eq:addth2} \(A = \omega_c t\) und \(B = 2k\omega_m t \) ersetzt wurden. +Nun kann die Summe in zwei Summen \begin{align*} c(t) &= - J_0(\beta) \cdot \cos(\omega_c t) + \sum_{k=1}^\infty J_{2k}(\beta) \{ \underbrace{\cos((\omega_c - 2k \omega_m) t)} \,+\, \cos((\omega_c + 2k \omega_m) t) \} + J_0(\beta) \cdot \cos(\omega_c t) + \sum_{k=1}^\infty J_{2k}(\beta) \cos((\omega_c - 2k \omega_m) t) \,+\, \cos((\omega_c + 2k \omega_m) t) \} \\ &= - \sum_{k=-\infty}^{-1} J_{2k}(\beta) \overbrace{\cos((\omega_c +2k \omega_m) t)} - \,+\,J_0(\beta)\cdot \cos(\omega_c t+ 2\cdot0 \omega_m) + \sum_{k=\infty}^{1} J_{2k}(\beta) \underbrace{\cos((\omega_c - 2k \omega_m) t)} + \,+\,J_0(\beta)\cdot \cos(\omega_c t) \,+\, \sum_{k=1}^\infty J_{2k}(\beta)\cos((\omega_c + 2k \omega_m) t) \end{align*} -wird. -Das Minus im Ersten Term wird zur negativen Summe \(\sum_{-\infty}^{-1}\) ersetzt. -Da \(2k\) immer gerade ist, wird es durch alle negativen und positiven Ganzzahlen \(n\) ersetzt: +aufgeteilt werden. +Wenn bei der ersten Summe noch \(k\) von \(-\infty \to -1\) läuft, wird diese summe zu \(\sum_{k=-1}^{-\infty} J_{-2k}(\beta) {\cos((\omega_c + 2k \omega_m) t)} \) +Zudem kann die Besselindentität \eqref{fm:eq:besselid3} gebraucht werden. \(n \) wird mit \(2k\) ersetzt, da dies immer gerade ist so gilt: \(J_{-n}(\beta) = J_n(\beta)\) +Somit bekommt man zwei gleiche Summen +\begin{align*} + c(t) + &= + \sum_{k=-\infty}^{-1} J_{2k}(\beta) \cos((\omega_c + 2k \omega_m) t) + \,+\,J_0(\beta)\cdot \cos(\omega_c t+ 2 \cdot 0 \omega_m) + \,+\, \sum_{k=1}^\infty J_{2k}(\beta)\cos((\omega_c + 2k \omega_m) t) +\end{align*} +Diese können wir vereinfachter schreiben, \begin{align*} \sum_{n\, \text{gerade}} J_{n}(\beta) \cos((\omega_c + n \omega_m) t), \label{fm:eq:gerade} \end{align*} +da \(2k\) für alle negativen, wie positiven geraden Zahlen zählt. %---------------------------------------------------------------------------------------------------------------- \subsubsection{Sin-Teil} Nun zum zweiten Teil des Term \eqref{fm:eq:start}, den Sin-Teil diff --git a/buch/papers/fm/Python animation/Bessel-FM.ipynb b/buch/papers/fm/Python animation/Bessel-FM.ipynb index ef6300b..4074765 100644 --- a/buch/papers/fm/Python animation/Bessel-FM.ipynb +++ b/buch/papers/fm/Python animation/Bessel-FM.ipynb @@ -2,7 +2,7 @@ "cells": [ { "cell_type": "code", - "execution_count": 1, + "execution_count": 4, "metadata": {}, "outputs": [], "source": [ @@ -16,7 +16,7 @@ "# Use the pgf backend (must be set before pyplot imported)\n", "# mpl.use('pgf')\n", "\n", - "from matplotlib.widgets import Slider\n", + "\n", "def fm(beta):\n", " # Number of samplepoints\n", " N = 600\n", @@ -28,7 +28,7 @@ " #beta = 1.0\n", " y_old = np.sin(fc * 2.0*np.pi*x+beta*np.sin(fm * 2.0*np.pi*x))\n", " y = 0*x;\n", - " xf = fftfreq(N, 1 / 400)\n", + " xf = fftfreq(N, 1 / N)\n", " for k in range (-4, 4):\n", " y = sc.jv(k,beta)*np.sin((fc+k*fm) * 2.0*np.pi*x)\n", " yf = fft(y)/(fc*np.pi)\n", @@ -43,12 +43,12 @@ }, { "cell_type": "code", - "execution_count": 9, + "execution_count": 6, "metadata": {}, "outputs": [ { "data": { - "image/png": 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", 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", 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", 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" ] @@ -75,17 +75,17 @@ "# Number of samplepoints\n", "N = 800\n", "# sample spacing\n", - "T = 1.0 / 1000.0\n", + "T = 1.0 / N\n", "x = np.linspace(0.01, N*T, N)\n", "\n", "y_old = np.sin(100* 2.0*np.pi*x+1*np.sin(15* 2.0*np.pi*x))\n", "yf_old = fft(y_old)/(100*np.pi)\n", - "xf = fftfreq(N, 1 / 1000)\n", + "xf = fftfreq(N, 1 / N)\n", "plt.plot(xf, np.abs(yf_old))\n", "#plt.xlim(-150, 150)\n", "plt.show()\n", "\n", - "fm(1)" + "fm(2)" ] }, { @@ -133,56 +133,12 @@ }, { "cell_type": "code", - "execution_count": 6, - "metadata": {}, - "outputs": [ - { - "data": { - "image/png": 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", - "text/plain": [ - "
" - ] - }, - "metadata": { - "needs_background": "light" - }, - "output_type": "display_data" - } - ], - "source": [ - "from scipy import special\n", - "\n", - "def drumhead_height(n, k, distance, angle, t):\n", - " kth_zero = special.jn_zeros(n, k)[-1]\n", - " return np.cos(t) * np.cos(n*angle) * special.jn(n, distance*kth_zero)\n", - "\n", - "theta = np.r_[0:2*np.pi:50j]\n", - "radius = np.r_[0:1:50j]\n", - "x = np.array([r * np.cos(theta) for r in radius])\n", - "y = np.array([r * np.sin(theta) for r in radius])\n", - "z = np.array([drumhead_height(1, 1, r, theta, 0.5) for r in radius])\n", - "\n", - "import matplotlib.pyplot as plt\n", - "fig = plt.figure()\n", - "ax = fig.add_axes(rect=(0, 0.05, 0.95, 0.95), projection='3d')\n", - "ax.plot_surface(x, y, z, rstride=1, cstride=1, cmap='RdBu_r', vmin=-0.5, vmax=0.5)\n", - "ax.set_xlabel('X')\n", - "ax.set_ylabel('Y')\n", - "ax.set_xticks(np.arange(-1, 1.1, 0.5))\n", - "ax.set_yticks(np.arange(-1, 1.1, 0.5))\n", - "ax.set_zlabel('Z')\n", - "\n", - "plt.show()" - ] - }, - { - "cell_type": "code", - "execution_count": 98, + "execution_count": 131, "metadata": {}, "outputs": [ { "data": { - "image/png": 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", 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", 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" ] @@ -195,7 +151,7 @@ ], "source": [ "\n", - "x = np.linspace(0,0.1,1000)\n", + "x = np.linspace(0,0.1,2000)\n", "y = np.sin(100 * 2.0*np.pi*x+1.5*np.sin(30 * 2.0*np.pi*x))\n", "plt.plot(x, y, '-')\n", "plt.show()" @@ -203,22 +159,12 @@ }, { "cell_type": "code", - "execution_count": 130, + "execution_count": 12, "metadata": {}, "outputs": [ { "data": { - "text/plain": [ - "[]" - ] - }, - "execution_count": 130, - "metadata": {}, - "output_type": "execute_result" - }, - { - "data": { - "image/png": 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", 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", "text/plain": [ "
" ] @@ -237,16 +183,17 @@ "second = ratio-first\n", "odd = ratio % 2\n", "\n", - "first = int(first * length/ratio) + odd\n", - "second = int( second * length/ratio)\n", + "first = int(first * length/ratio) \n", + "second = int( second * length/ratio) + odd\n", "slop = np.array(np.append(np.zeros(first-slop) , (np.arange(slop))/slop))\n", "#steep = np.ones(int(first * length/ratio)+ odd) - np.exp(-np.arange(int(first * length/ratio) + odd)/200)\n", "steep = (np.ones(first) + slop)*0.5\n", "\n", "step = np.append(steep, np.ones(second))\n", - "m = np.sin(30 * 2.0 * np.pi * x) * step \n", + "m = np.sin(5 * 2.0 * np.pi * x) * step \n", "plt.plot(x, step, '-')\n", - "plt.plot(x, m, '-')" + "plt.plot(x, m, '-')\n", + "plt.savefig('m_t.pgf', format='pgf')" ] } ], diff --git a/buch/papers/fm/Python animation/Bessel-FM.py b/buch/papers/fm/Python animation/Bessel-FM.py index cf30e16..cb35ebd 100644 --- a/buch/papers/fm/Python animation/Bessel-FM.py +++ b/buch/papers/fm/Python animation/Bessel-FM.py @@ -4,39 +4,45 @@ from scipy.fft import fft, ifft, fftfreq import scipy.special as sc import scipy.fftpack import matplotlib.pyplot as plt -from matplotlib.widgets import Slider - -# Number of samplepoints -N = 600 -# sample spacing -T = 1.0 / 800.0 -x = np.linspace(0.01, N*T, N) -beta = 1.0 -y_old = np.sin(100.0 * 2.0*np.pi*x+beta*np.sin(50.0 * 2.0*np.pi*x)) -y = 0*x; -xf = fftfreq(N, 1 / 400) -for k in range (-5, 5): - y = sc.jv(k,beta)*np.sin((100.0+k*50) * 2.0*np.pi*x) - yf = fft(y) - plt.plot(xf, np.abs(yf)) - -axbeta =plt.axes([0.25, 0.1, 0.65, 0.03]) -beta_slider = Slider( -ax=axbeta, -label="Beta", -valmin=0.1, -valmax=3, -valinit=beta, -) - -def update(val): - line.set_ydata(fm(beta_slider.val)) - fig.canvas.draw_idle() +import matplotlib as mpl +# Use the pgf backend (must be set before pyplot imported) +mpl.use('pgf') +from matplotlib.widgets import Slider +def fm(beta): + # Number of samplepoints + N = 600 + # sample spacing + T = 1.0 / 1000.0 + fc = 100.0 + fm = 30.0 + x = np.linspace(0.01, N*T, N) + #beta = 1.0 + y_old = np.sin(fc * 2.0*np.pi*x+beta*np.sin(fm * 2.0*np.pi*x)) + y = 0*x; + xf = fftfreq(N, 1 / N) + for k in range (-4, 4): + y = sc.jv(k,beta)*np.sin((fc+k*fm) * 2.0*np.pi*x) + yf = fft(y)/(fc*np.pi) + plt.plot(xf, np.abs(yf)) + plt.xlim(-150, 150) + #plt.savefig('bessel.pgf', format='pgf') + plt.show() -beta_slider.on_changed(update) -plt.show() +fm(1) -yf_old = fft(y_old) -plt.plot(xf, np.abs(yf_old)) -plt.show() \ No newline at end of file +# Bessel-Funktion +for n in range (-2,4): + x = np.linspace(-11,11,1000) + y = sc.jv(n,x) + plt.plot(x, y, '-',label='n='+str(n)) +#plt.plot([1,1],[sc.jv(0,1),sc.jv(-1,1)],) +plt.xlim(-10,10) +plt.grid(True) +plt.ylabel('Bessel $J_n(\\beta)$') +plt.xlabel(' $ \\beta $ ') +plt.plot(x, y) +plt.legend() +#plt.show() +plt.savefig('bessel.pgf', format='pgf') +print(sc.jv(0,1)) \ No newline at end of file diff --git a/buch/papers/fm/Python animation/m_t.pgf b/buch/papers/fm/Python animation/m_t.pgf new file mode 100644 index 0000000..edcfb33 --- /dev/null +++ b/buch/papers/fm/Python animation/m_t.pgf @@ -0,0 +1,746 @@ +%% Creator: Matplotlib, PGF backend +%% +%% To include the figure in your LaTeX document, write +%% \input{.pgf} +%% +%% Make sure the required packages are loaded in your preamble +%% \usepackage{pgf} +%% +%% Also ensure that all the required font packages are loaded; for instance, +%% the lmodern package is sometimes necessary when using math font. +%% \usepackage{lmodern} +%% +%% Figures using additional raster images can only be included by \input if +%% they are in the same directory as the main LaTeX file. For loading figures +%% from other directories you can use the `import` package +%% \usepackage{import} +%% +%% and then include the figures with +%% \import{}{.pgf} +%% +%% Matplotlib used the following preamble +%% \usepackage{fontspec} +%% \setmainfont{DejaVuSerif.ttf}[Path=\detokenize{/home/joshua/.local/lib/python3.8/site-packages/matplotlib/mpl-data/fonts/ttf/}] +%% \setsansfont{DejaVuSans.ttf}[Path=\detokenize{/home/joshua/.local/lib/python3.8/site-packages/matplotlib/mpl-data/fonts/ttf/}] +%% \setmonofont{DejaVuSansMono.ttf}[Path=\detokenize{/home/joshua/.local/lib/python3.8/site-packages/matplotlib/mpl-data/fonts/ttf/}] +%% +\begingroup% +\makeatletter% +\begin{pgfpicture}% +\pgfpathrectangle{\pgfpointorigin}{\pgfqpoint{6.000000in}{4.000000in}}% +\pgfusepath{use as bounding box, clip}% +\begin{pgfscope}% +\pgfsetbuttcap% +\pgfsetmiterjoin% +\pgfsetlinewidth{0.000000pt}% +\definecolor{currentstroke}{rgb}{1.000000,1.000000,1.000000}% +\pgfsetstrokecolor{currentstroke}% 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+\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}% +\pgfsetstrokecolor{currentstroke}% +\pgfsetdash{}{0pt}% +\pgfpathmoveto{\pgfqpoint{0.750000in}{3.520000in}}% +\pgfpathlineto{\pgfqpoint{5.400000in}{3.520000in}}% +\pgfusepath{stroke}% +\end{pgfscope}% +\end{pgfpicture}% +\makeatother% +\endgroup% diff --git a/buch/papers/fm/Quellen/NaT_Skript_20210920.pdf b/buch/papers/fm/Quellen/NaT_Skript_20210920.pdf new file mode 100644 index 0000000..b9acc1f Binary files /dev/null and b/buch/papers/fm/Quellen/NaT_Skript_20210920.pdf differ diff --git a/buch/papers/fm/main.tex b/buch/papers/fm/main.tex index 731f56f..0c98427 100644 --- a/buch/papers/fm/main.tex +++ b/buch/papers/fm/main.tex @@ -14,20 +14,9 @@ Die Frequenzmodulation ist eine Modulation die man auch schon im alten Radio findet. Falls du dich an die Zeit erinnerst, konnte man zwischen \textit{FM-AM} Umschalten, dies bedeutete so viel wie: \textit{F}requenz-\textit{M}odulation und \textit{A}mplituden-\textit{M}odulation. -Durch die Modulation wird ein Nachrichtensignal \(m(t)\) auf ein Trägersignal (z.B. ein Sinus- oder Rechtecksignal) abgebildet (kombiniert). -Durch dieses Auftragen vom Nachrichtensignal \(m(t)\) kann das modulierte Signal in einem gewünschten Frequenzbereich übertragen werden. -Der ursprünglich Frequenzbereich des Nachrichtensignal \(m(t)\) erstreckt sich typischerweise von 0 Hz bis zur Bandbreite \(B_m\). -\newline -Beim Empfänger wird dann durch Demodulation das ursprüngliche Nachrichtensignal \(m(t)\) so originalgetreu wie möglich zurückgewonnen. -\newline -Beim Trägersignal \(x_c(t)\) handelt es sich um ein informationsloses Hilfssignal. -Durch die Modulation mit dem Nachrichtensignal \(m(t)\) wird es zum modulierten zu übertragenden Signal. -Für alle Erklärungen wird ein sinusförmiges Trägersignal benutzt, jedoch kann auch ein Rechtecksignal, -welches Digital einfach umzusetzten ist, -genauso als Trägersignal genutzt werden kann. -Zuerst wird erklärt was \textit{FM-AM} ist, danach wie sich diese im Frequenzspektrum verhalten. -Erst dann erklär ich dir wie die Besselfunktion mit der Frequenzmodulation( acro?) zusammenhängt. -Nun zur Modulation im nächsten Abschnitt.\cite{fm:NAT} +Um das Thema einwenig einzuschränken werde ich leider nichts über die Vertiefte, (Physikalische) zusammenhänge oder die Demodulation aufzeigen. +Dieses Kapitel soll nurdie Frequenzmodulation und ihren zusammenhang mit der Besselfunktion erklären. +Aber zuerst einmal zur Modulation selbst, wie funktioniert diese Mathematisch. \input{papers/fm/00_modulation.tex} -- cgit v1.2.1 From 1cd844f0459df9d264c5552047af320b378df8ba Mon Sep 17 00:00:00 2001 From: Nao Pross Date: Tue, 16 Aug 2022 17:16:27 +0200 Subject: kugel: Reorganize figures directory, add tikz spherical coordinates and flux --- buch/papers/kugel/figures/flux.pdf | Bin 0 -> 345665 bytes buch/papers/kugel/figures/povray/Makefile | 30 +++ buch/papers/kugel/figures/povray/curvature.maxima | 6 + buch/papers/kugel/figures/povray/curvature.pov | 139 ++++++++++ buch/papers/kugel/figures/povray/curvgraph.m | 140 ++++++++++ buch/papers/kugel/figures/povray/spherecurve.cpp | 292 +++++++++++++++++++++ buch/papers/kugel/figures/povray/spherecurve.m | 160 +++++++++++ .../papers/kugel/figures/povray/spherecurve.maxima | 13 + buch/papers/kugel/figures/povray/spherecurve.pov | 73 ++++++ .../kugel/figures/tikz/spherical-coordinates.pdf | Bin 0 -> 5824 bytes .../kugel/figures/tikz/spherical-coordinates.tex | 99 +++++++ buch/papers/kugel/images/Makefile | 30 --- buch/papers/kugel/images/curvature.maxima | 6 - buch/papers/kugel/images/curvature.pov | 139 ---------- buch/papers/kugel/images/curvgraph.m | 140 ---------- buch/papers/kugel/images/spherecurve.cpp | 292 --------------------- buch/papers/kugel/images/spherecurve.m | 160 ----------- buch/papers/kugel/images/spherecurve.maxima | 13 - buch/papers/kugel/images/spherecurve.pov | 73 ------ 19 files changed, 952 insertions(+), 853 deletions(-) create mode 100644 buch/papers/kugel/figures/flux.pdf create mode 100644 buch/papers/kugel/figures/povray/Makefile create mode 100644 buch/papers/kugel/figures/povray/curvature.maxima create mode 100644 buch/papers/kugel/figures/povray/curvature.pov create mode 100644 buch/papers/kugel/figures/povray/curvgraph.m create mode 100644 buch/papers/kugel/figures/povray/spherecurve.cpp create mode 100644 buch/papers/kugel/figures/povray/spherecurve.m create mode 100644 buch/papers/kugel/figures/povray/spherecurve.maxima create mode 100644 buch/papers/kugel/figures/povray/spherecurve.pov create mode 100644 buch/papers/kugel/figures/tikz/spherical-coordinates.pdf create mode 100644 buch/papers/kugel/figures/tikz/spherical-coordinates.tex delete mode 100644 buch/papers/kugel/images/Makefile delete mode 100644 buch/papers/kugel/images/curvature.maxima delete mode 100644 buch/papers/kugel/images/curvature.pov delete mode 100644 buch/papers/kugel/images/curvgraph.m delete mode 100644 buch/papers/kugel/images/spherecurve.cpp delete mode 100644 buch/papers/kugel/images/spherecurve.m delete mode 100644 buch/papers/kugel/images/spherecurve.maxima delete mode 100644 buch/papers/kugel/images/spherecurve.pov (limited to 'buch/papers') diff --git a/buch/papers/kugel/figures/flux.pdf b/buch/papers/kugel/figures/flux.pdf new file mode 100644 index 0000000..6a87288 Binary files /dev/null and b/buch/papers/kugel/figures/flux.pdf differ diff --git a/buch/papers/kugel/figures/povray/Makefile b/buch/papers/kugel/figures/povray/Makefile new file mode 100644 index 0000000..4226dab --- /dev/null +++ b/buch/papers/kugel/figures/povray/Makefile @@ -0,0 +1,30 @@ +# +# Makefile -- build images +# +# (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +# +all: curvature.jpg spherecurve.jpg + +curvature.inc: curvgraph.m + octave curvgraph.m + +curvature.png: curvature.pov curvature.inc + povray +A0.1 +W1920 +H1080 +Ocurvature.png curvature.pov + +curvature.jpg: curvature.png + convert curvature.png -density 300 -units PixelsPerInch curvature.jpg + +spherecurve2.inc: spherecurve.m + octave spherecurve.m + +spherecurve.png: spherecurve.pov spherecurve.inc + povray +A0.1 +W1080 +H1080 +Ospherecurve.png spherecurve.pov + +spherecurve.jpg: spherecurve.png + convert spherecurve.png -density 300 -units PixelsPerInch spherecurve.jpg + +spherecurve: spherecurve.cpp + g++ -o spherecurve -g -Wall -O spherecurve.cpp + +spherecurve.inc: spherecurve + ./spherecurve diff --git a/buch/papers/kugel/figures/povray/curvature.maxima b/buch/papers/kugel/figures/povray/curvature.maxima new file mode 100644 index 0000000..6313642 --- /dev/null +++ b/buch/papers/kugel/figures/povray/curvature.maxima @@ -0,0 +1,6 @@ + +f: exp(-r^2/sigma^2)/sigma; +laplacef: ratsimp(diff(r * diff(f,r), r) / r); +f: exp(-r^2/(2*sigma^2))/(sqrt(2)*sigma); +laplacef: ratsimp(diff(r * diff(f,r), r) / r); + diff --git a/buch/papers/kugel/figures/povray/curvature.pov b/buch/papers/kugel/figures/povray/curvature.pov new file mode 100644 index 0000000..3b15d77 --- /dev/null +++ b/buch/papers/kugel/figures/povray/curvature.pov @@ -0,0 +1,139 @@ +// +// curvature.pov +// +// (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +// + +#version 3.7; +#include "colors.inc" + +global_settings { + assumed_gamma 1 +} + +#declare imagescale = 0.09; + +camera { + location <10, 10, -40> + look_at <0, 0, 0> + right 16/9 * x * imagescale + up y * imagescale +} + +light_source { + <-10, 10, -40> color White + area_light <1,0,0> <0,0,1>, 10, 10 + adaptive 1 + jitter +} + +sky_sphere { + pigment { + color rgb<1,1,1> + } +} + +// +// draw an arrow from to with thickness with +// color +// +#macro arrow(from, to, arrowthickness, c) +#declare arrowdirection = vnormalize(to - from); +#declare arrowlength = vlength(to - from); +union { + sphere { + from, 1.1 * arrowthickness + } + cylinder { + from, + from + (arrowlength - 5 * arrowthickness) * arrowdirection, + arrowthickness + } + cone { + from + (arrowlength - 5 * arrowthickness) * arrowdirection, + 2 * arrowthickness, + to, + 0 + } + pigment { + color c + } + finish { + specular 0.9 + metallic + } +} +#end + +arrow(<-3.1,0,0>, <3.1,0,0>, 0.01, White) +arrow(<0,-1,0>, <0,1,0>, 0.01, White) +arrow(<0,0,-2.1>, <0,0,2.1>, 0.01, White) + +#include "curvature.inc" + +#declare sigma = 1; +#declare s = 1.4; +#declare N0 = 0.4; +#declare funktion = function(r) { + (exp(-r*r/(sigma*sigma)) / sigma + - + exp(-r*r/(2*sigma*sigma)) / (sqrt(2)*sigma)) / N0 +}; +#declare hypot = function(xx, yy) { sqrt(xx*xx+yy*yy) }; + +#declare Funktion = function(x,y) { funktion(hypot(x+s,y)) - funktion(hypot(x-s,y)) }; +#macro punkt(xx,yy) + +#end + +#declare griddiameter = 0.006; +union { + #declare xmin = -3; + #declare xmax = 3; + #declare ymin = -2; + #declare ymax = 2; + + + #declare xstep = 0.2; + #declare ystep = 0.02; + #declare xx = xmin; + #while (xx < xmax + xstep/2) + #declare yy = ymin; + #declare P = punkt(xx, yy); + #while (yy < ymax - ystep/2) + #declare yy = yy + ystep; + #declare Q = punkt(xx, yy); + sphere { P, griddiameter } + cylinder { P, Q, griddiameter } + #declare P = Q; + #end + sphere { P, griddiameter } + #declare xx = xx + xstep; + #end + + #declare xstep = 0.02; + #declare ystep = 0.2; + #declare yy = ymin; + #while (yy < ymax + ystep/2) + #declare xx = xmin; + #declare P = punkt(xx, yy); + #while (xx < xmax - xstep/2) + #declare xx = xx + xstep; + #declare Q = punkt(xx, yy); + sphere { P, griddiameter } + cylinder { P, Q, griddiameter } + #declare P = Q; + #end + sphere { P, griddiameter } + #declare yy = yy + ystep; + #end + + pigment { + color rgb<0.8,0.8,0.8> + } + finish { + metallic + specular 0.8 + } +} + diff --git a/buch/papers/kugel/figures/povray/curvgraph.m b/buch/papers/kugel/figures/povray/curvgraph.m new file mode 100644 index 0000000..75effd6 --- /dev/null +++ b/buch/papers/kugel/figures/povray/curvgraph.m @@ -0,0 +1,140 @@ +# +# curvature.m +# +# (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +# + +global N; +N = 10; + +global sigma2; +sigma2 = 1; + +global s; +s = 1.4; + +global cmax; +cmax = 0.9; +global cmin; +cmin = -0.9; + +global Cmax; +global Cmin; +Cmax = 0; +Cmin = 0; + +xmin = -3; +xmax = 3; +xsteps = 200; +hx = (xmax - xmin) / xsteps; + +ymin = -2; +ymax = 2; +ysteps = 200; +hy = (ymax - ymin) / ysteps; + +function retval = f0(r) + global sigma2; + retval = exp(-r^2/sigma2)/sqrt(sigma2) - exp(-r^2/(2*sigma2))/(sqrt(2*sigma2)); +end + +global N0; +N0 = f0(0) +N0 = 0.4; + +function retval = f1(x,y) + global N0; + retval = f0(hypot(x, y)) / N0; +endfunction + +function retval = f(x, y) + global s; + retval = f1(x+s, y) - f1(x-s, y); +endfunction + +function retval = curvature0(r) + global sigma2; + retval = ( + -4*(sigma2-r^2)*exp(-r^2/sigma2) + + + (2*sigma2-r^2)*exp(-r^2/(2*sigma2)) + ) / (sigma2^(5/2)); +endfunction + +function retval = curvature1(x, y) + retval = curvature0(hypot(x, y)); +endfunction + +function retval = curvature(x, y) + global s; + retval = curvature1(x+s, y) - curvature1(x-s, y); +endfunction + +function retval = farbe(x, y) + global Cmax; + global Cmin; + global cmax; + global cmin; + c = curvature(x, y); + if (c < Cmin) + Cmin = c + endif + if (c > Cmax) + Cmax = c + endif + u = (c - cmin) / (cmax - cmin); + if (u > 1) + u = 1; + endif + if (u < 0) + u = 0; + endif + color = [ u, 0.5, 1-u ]; + color = color/max(color); + color(1,4) = c/2; + retval = color; +endfunction + +function dreieck(fn, A, B, C) + fprintf(fn, "\ttriangle {\n"); + fprintf(fn, "\t <%.4f,%.4f,%.4f>,\n", A(1,1), A(1,3), A(1,2)); + fprintf(fn, "\t <%.4f,%.4f,%.4f>,\n", B(1,1), B(1,3), B(1,2)); + fprintf(fn, "\t <%.4f,%.4f,%.4f>\n", C(1,1), C(1,3), C(1,2)); + fprintf(fn, "\t}\n"); +endfunction + +function viereck(fn, punkte) + color = farbe(mean(punkte(:,1)), mean(punkte(:,2))); + fprintf(fn, " mesh {\n"); + dreieck(fn, punkte(1,:), punkte(2,:), punkte(3,:)); + dreieck(fn, punkte(2,:), punkte(3,:), punkte(4,:)); + fprintf(fn, "\tpigment { color rgb<%.4f,%.4f,%.4f> } // %.4f\n", + color(1,1), color(1,2), color(1,3), color(1,4)); + fprintf(fn, " }\n"); +endfunction + +fn = fopen("curvature.inc", "w"); +punkte = zeros(4,3); +for ix = (0:xsteps-1) + x = xmin + ix * hx; + punkte(1,1) = x; + punkte(2,1) = x; + punkte(3,1) = x + hx; + punkte(4,1) = x + hx; + for iy = (0:ysteps-1) + y = ymin + iy * hy; + punkte(1,2) = y; + punkte(2,2) = y + hy; + punkte(3,2) = y; + punkte(4,2) = y + hy; + for i = (1:4) + punkte(i,3) = f(punkte(i,1), punkte(i,2)); + endfor + viereck(fn, punkte); + end +end +#fprintf(fn, " finish { metallic specular 0.5 }\n"); +fclose(fn); + +printf("Cmax = %.4f\n", Cmax); +printf("Cmin = %.4f\n", Cmin); diff --git a/buch/papers/kugel/figures/povray/spherecurve.cpp b/buch/papers/kugel/figures/povray/spherecurve.cpp new file mode 100644 index 0000000..8ddf5e5 --- /dev/null +++ b/buch/papers/kugel/figures/povray/spherecurve.cpp @@ -0,0 +1,292 @@ +/* + * spherecurve.cpp + * + * (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule + */ +#include +#include +#include +#include +#include + +inline double sqr(double x) { return x * x; } + +/** + * \brief Class for 3d vectors (also used as colors) + */ +class vector { + double X[3]; +public: + vector() { X[0] = X[1] = X[2] = 0; } + vector(double a) { X[0] = X[1] = X[2] = a; } + vector(double x, double y, double z) { + X[0] = x; X[1] = y; X[2] = z; + } + vector(double theta, double phi) { + double s = sin(theta); + X[0] = cos(phi) * s; + X[1] = sin(phi) * s; + X[2] = cos(theta); + } + vector(const vector& other) { + for (int i = 0; i < 3; i++) { + X[i] = other.X[i]; + } + } + vector operator+(const vector& other) const { + return vector(X[0] + other.X[0], + X[1] + other.X[1], + X[2] + other.X[2]); + } + vector operator*(double l) const { + return vector(X[0] * l, X[1] * l, X[2] * l); + } + double operator*(const vector& other) const { + double s = 0; + for (int i = 0; i < 3; i++) { + s += X[i] * other.X[i]; + } + return s; + } + double norm() const { + double s = 0; + for (int i = 0; i < 3; i++) { + s += sqr(X[i]); + } + return sqrt(s); + } + vector normalize() const { + double l = norm(); + return vector(X[0]/l, X[1]/l, X[2]/l); + } + double max() const { + return std::max(X[0], std::max(X[1], X[2])); + } + double l0norm() const { + double l = 0; + for (int i = 0; i < 3; i++) { + if (fabs(X[i]) > l) { + l = fabs(X[i]); + } + } + return l; + } + vector l0normalize() const { + double l = l0norm(); + vector result(X[0]/l, X[1]/l, X[2]/l); + return result; + } + const double& operator[](int i) const { return X[i]; } + double& operator[](int i) { return X[i]; } +}; + +/** + * \brief Derived 3d vector class implementing color + * + * The constructor in this class converts a single value into a + * color on a suitable gradient. + */ +class color : public vector { +public: + static double utop; + static double ubottom; + static double green; +public: + color(double u) { + u = (u - ubottom) / (utop - ubottom); + if (u > 1) { + u = 1; + } + if (u < 0) { + u = 0; + } + u = pow(u,2); + (*this)[0] = u; + (*this)[1] = green * u * (1 - u); + (*this)[2] = 1-u; + double l = l0norm(); + for (int i = 0; i < 3; i++) { + (*this)[i] /= l; + } + } +}; + +double color::utop = 12; +double color::ubottom = -31; +double color::green = 0.5; + +/** + * \brief Surface model + * + * This class contains the definitions of the functions to plot + * and the parameters to + */ +class surfacefunction { + static vector axes[6]; + + double _a; + double _A; + + double _umin; + double _umax; +public: + double a() const { return _a; } + double A() const { return _A; } + + double umin() const { return _umin; } + double umax() const { return _umax; } + + surfacefunction(double a, double A) : _a(a), _A(A), _umin(0), _umax(0) { + } + + double f(double z) { + return A() * exp(a() * (sqr(z) - 1)); + } + + double g(double z) { + return -f(z) * 2*a() * ((2*a()*sqr(z) + (3-2*a()))*sqr(z) - 1); + } + + double F(const vector& v) { + double s = 0; + for (int i = 0; i < 6; i++) { + s += f(axes[i] * v); + } + return s / 6; + } + + double G(const vector& v) { + double s = 0; + for (int i = 0; i < 6; i++) { + s += g(axes[i] * v); + } + return s / 6; + } +protected: + color farbe(const vector& v) { + double u = G(v); + if (u < _umin) { + _umin = u; + } + if (u > _umax) { + _umax = u; + } + return color(u); + } +}; + +static double phi = (1 + sqrt(5)) / 2; +static double sl = sqrt(sqr(phi) + 1); +vector surfacefunction::axes[6] = { + vector( 0. , -1./sl, phi/sl ), + vector( 0. , 1./sl, phi/sl ), + vector( 1./sl, phi/sl, 0. ), + vector( -1./sl, phi/sl, 0. ), + vector( phi/sl, 0. , 1./sl ), + vector( -phi/sl, 0. , 1./sl ) +}; + +/** + * \brief Class to construct the plot + */ +class surface : public surfacefunction { + FILE *outfile; + + int _phisteps; + int _thetasteps; + double _hphi; + double _htheta; +public: + int phisteps() const { return _phisteps; } + int thetasteps() const { return _thetasteps; } + double hphi() const { return _hphi; } + double htheta() const { return _htheta; } + void phisteps(int s) { _phisteps = s; _hphi = 2 * M_PI / s; } + void thetasteps(int s) { _thetasteps = s; _htheta = M_PI / s; } + + surface(const std::string& filename, double a, double A) + : surfacefunction(a, A) { + outfile = fopen(filename.c_str(), "w"); + phisteps(400); + thetasteps(200); + } + + ~surface() { + fclose(outfile); + } + +private: + void triangle(const vector& v0, const vector& v1, const vector& v2) { + fprintf(outfile, " mesh {\n"); + vector c = (v0 + v1 + v2) * (1./3.); + vector color = farbe(c.normalize()); + vector V0 = v0 * (1 + F(v0)); + vector V1 = v1 * (1 + F(v1)); + vector V2 = v2 * (1 + F(v2)); + fprintf(outfile, "\ttriangle {\n"); + fprintf(outfile, "\t <%.6f,%.6f,%.6f>,\n", + V0[0], V0[2], V0[1]); + fprintf(outfile, "\t <%.6f,%.6f,%.6f>,\n", + V1[0], V1[2], V1[1]); + fprintf(outfile, "\t <%.6f,%.6f,%.6f>\n", + V2[0], V2[2], V2[1]); + fprintf(outfile, "\t}\n"); + fprintf(outfile, "\tpigment { color rgb<%.4f,%.4f,%.4f> }\n", + color[0], color[1], color[2]); + fprintf(outfile, "\tfinish { metallic specular 0.5 }\n"); + fprintf(outfile, " }\n"); + } + + void northcap() { + vector v0(0, 0, 1); + for (int i = 1; i <= phisteps(); i++) { + fprintf(outfile, " // northcap i = %d\n", i); + vector v1(htheta(), (i - 1) * hphi()); + vector v2(htheta(), i * hphi()); + triangle(v0, v1, v2); + } + } + + void southcap() { + vector v0(0, 0, -1); + for (int i = 1; i <= phisteps(); i++) { + fprintf(outfile, " // southcap i = %d\n", i); + vector v1(M_PI - htheta(), (i - 1) * hphi()); + vector v2(M_PI - htheta(), i * hphi()); + triangle(v0, v1, v2); + } + } + + void zone() { + for (int j = 1; j < thetasteps() - 1; j++) { + for (int i = 1; i <= phisteps(); i++) { + fprintf(outfile, " // zone j = %d, i = %d\n", + j, i); + vector v0( j * htheta(), (i-1) * hphi()); + vector v1((j+1) * htheta(), (i-1) * hphi()); + vector v2( j * htheta(), i * hphi()); + vector v3((j+1) * htheta(), i * hphi()); + triangle(v0, v1, v2); + triangle(v1, v2, v3); + } + } + } +public: + void draw() { + northcap(); + southcap(); + zone(); + } +}; + +/** + * \brief main function + */ +int main(int argc, char *argv[]) { + surface S("spherecurve.inc", 5, 10); + color::green = 1.0; + S.draw(); + std::cout << "umin: " << S.umin() << std::endl; + std::cout << "umax: " << S.umax() << std::endl; + return EXIT_SUCCESS; +} diff --git a/buch/papers/kugel/figures/povray/spherecurve.m b/buch/papers/kugel/figures/povray/spherecurve.m new file mode 100644 index 0000000..99d5c9a --- /dev/null +++ b/buch/papers/kugel/figures/povray/spherecurve.m @@ -0,0 +1,160 @@ +# +# spherecurve.m +# +# (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +# +global a; +a = 5; +global A; +A = 10; + +phisteps = 400; +hphi = 2 * pi / phisteps; +thetasteps = 200; +htheta = pi / thetasteps; + +function retval = f(z) + global a; + global A; + retval = A * exp(a * (z^2 - 1)); +endfunction + +function retval = g(z) + global a; + retval = -f(z) * 2 * a * (2 * a * z^4 + (3 - 2*a) * z^2 - 1); + # 2 + # - a 2 4 2 2 a z + #(%o6) - %e (4 a z + (6 a - 4 a ) z - 2 a) %e +endfunction + +phi = (1 + sqrt(5)) / 2; + +global axes; +axes = [ + 0, 0, 1, -1, phi, -phi; + 1, -1, phi, phi, 0, 0; + phi, phi, 0, 0, 1, 1; +]; +axes = axes / (sqrt(phi^2+1)); + +function retval = kugel(theta, phi) + retval = [ + cos(phi) * sin(theta); + sin(phi) * sin(theta); + cos(theta) + ]; +endfunction + +function retval = F(v) + global axes; + s = 0; + for i = (1:6) + z = axes(:,i)' * v; + s = s + f(z); + endfor + retval = s / 6; +endfunction + +function retval = F2(theta, phi) + v = kugel(theta, phi); + retval = F(v); +endfunction + +function retval = G(v) + global axes; + s = 0; + for i = (1:6) + s = s + g(axes(:,i)' * v); + endfor + retval = s / 6; +endfunction + +function retval = G2(theta, phi) + v = kugel(theta, phi); + retval = G(v); +endfunction + +function retval = cnormalize(u) + utop = 11; + ubottom = -30; + retval = (u - ubottom) / (utop - ubottom); + if (retval > 1) + retval = 1; + endif + if (retval < 0) + retval = 0; + endif +endfunction + +global umin; +umin = 0; +global umax; +umax = 0; + +function color = farbe(v) + global umin; + global umax; + u = G(v); + if (u < umin) + umin = u; + endif + if (u > umax) + umax = u; + endif + u = cnormalize(u); + color = [ u, 0.5, 1-u ]; + color = color/max(color); +endfunction + +function dreieck(fn, v0, v1, v2) + fprintf(fn, " mesh {\n"); + c = (v0 + v1 + v2) / 3; + c = c / norm(c); + color = farbe(c); + v0 = v0 * (1 + F(v0)); + v1 = v1 * (1 + F(v1)); + v2 = v2 * (1 + F(v2)); + fprintf(fn, "\ttriangle {\n"); + fprintf(fn, "\t <%.6f,%.6f,%.6f>,\n", v0(1,1), v0(3,1), v0(2,1)); + fprintf(fn, "\t <%.6f,%.6f,%.6f>,\n", v1(1,1), v1(3,1), v1(2,1)); + fprintf(fn, "\t <%.6f,%.6f,%.6f>\n", v2(1,1), v2(3,1), v2(2,1)); + fprintf(fn, "\t}\n"); + fprintf(fn, "\tpigment { color rgb<%.4f,%.4f,%.4f> }\n", + color(1,1), color(1,2), color(1,3)); + fprintf(fn, "\tfinish { metallic specular 0.5 }\n"); + fprintf(fn, " }\n"); +endfunction + +fn = fopen("spherecurve2.inc", "w"); + + for i = (1:phisteps) + # Polkappe nord + v0 = [ 0; 0; 1 ]; + v1 = kugel(htheta, (i-1) * hphi); + v2 = kugel(htheta, i * hphi); + fprintf(fn, " // i = %d\n", i); + dreieck(fn, v0, v1, v2); + + # Polkappe sued + v0 = [ 0; 0; -1 ]; + v1 = kugel(pi-htheta, (i-1) * hphi); + v2 = kugel(pi-htheta, i * hphi); + dreieck(fn, v0, v1, v2); + endfor + + for j = (1:thetasteps-2) + for i = (1:phisteps) + v0 = kugel( j * htheta, (i-1) * hphi); + v1 = kugel((j+1) * htheta, (i-1) * hphi); + v2 = kugel( j * htheta, i * hphi); + v3 = kugel((j+1) * htheta, i * hphi); + fprintf(fn, " // i = %d, j = %d\n", i, j); + dreieck(fn, v0, v1, v2); + dreieck(fn, v1, v2, v3); + endfor + endfor + +fclose(fn); + +umin +umax diff --git a/buch/papers/kugel/figures/povray/spherecurve.maxima b/buch/papers/kugel/figures/povray/spherecurve.maxima new file mode 100644 index 0000000..1e9077c --- /dev/null +++ b/buch/papers/kugel/figures/povray/spherecurve.maxima @@ -0,0 +1,13 @@ +/* + * spherecurv.maxima + * + * (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule + */ +f: exp(-a * sin(theta)^2); + +g: ratsimp(diff(sin(theta) * diff(f, theta), theta)/sin(theta)); +g: subst(z, cos(theta), g); +g: subst(sqrt(1-z^2), sin(theta), g); +ratsimp(g); + +f: ratsimp(subst(sqrt(1-z^2), sin(theta), f)); diff --git a/buch/papers/kugel/figures/povray/spherecurve.pov b/buch/papers/kugel/figures/povray/spherecurve.pov new file mode 100644 index 0000000..b1bf4b8 --- /dev/null +++ b/buch/papers/kugel/figures/povray/spherecurve.pov @@ -0,0 +1,73 @@ +// +// curvature.pov +// +// (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +// + +#version 3.7; +#include "colors.inc" + +global_settings { + assumed_gamma 1 +} + +#declare imagescale = 0.13; + +camera { + location <10, 10, -40> + look_at <0, 0, 0> + right x * imagescale + up y * imagescale +} + +light_source { + <-10, 10, -40> color White + area_light <1,0,0> <0,0,1>, 10, 10 + adaptive 1 + jitter +} + +sky_sphere { + pigment { + color rgb<1,1,1> + } +} + +// +// draw an arrow from to with thickness with +// color +// +#macro arrow(from, to, arrowthickness, c) +#declare arrowdirection = vnormalize(to - from); +#declare arrowlength = vlength(to - from); +union { + sphere { + from, 1.1 * arrowthickness + } + cylinder { + from, + from + (arrowlength - 5 * arrowthickness) * arrowdirection, + arrowthickness + } + cone { + from + (arrowlength - 5 * arrowthickness) * arrowdirection, + 2 * arrowthickness, + to, + 0 + } + pigment { + color c + } + finish { + specular 0.9 + metallic + } +} +#end + +arrow(<-2.7,0,0>, <2.7,0,0>, 0.03, White) +arrow(<0,-2.7,0>, <0,2.7,0>, 0.03, White) +arrow(<0,0,-2.7>, <0,0,2.7>, 0.03, White) + +#include "spherecurve.inc" + diff --git a/buch/papers/kugel/figures/tikz/spherical-coordinates.pdf b/buch/papers/kugel/figures/tikz/spherical-coordinates.pdf new file mode 100644 index 0000000..28f242e Binary files /dev/null and b/buch/papers/kugel/figures/tikz/spherical-coordinates.pdf differ diff --git a/buch/papers/kugel/figures/tikz/spherical-coordinates.tex b/buch/papers/kugel/figures/tikz/spherical-coordinates.tex new file mode 100644 index 0000000..3a45385 --- /dev/null +++ b/buch/papers/kugel/figures/tikz/spherical-coordinates.tex @@ -0,0 +1,99 @@ +\documentclass[tikz]{standalone} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{bm} +\usepackage{lmodern} +\usepackage{tikz-3dplot} + +\usetikzlibrary{arrows} +\usetikzlibrary{intersections} +\usetikzlibrary{math} +\usetikzlibrary{positioning} +\usetikzlibrary{arrows.meta} +\usetikzlibrary{shapes.misc} +\usetikzlibrary{calc} + +\begin{document} + +\tdplotsetmaincoords{60}{130} +\pgfmathsetmacro{\l}{2} + +\begin{tikzpicture}[ + >=latex, + tdplot_main_coords, + dot/.style = { + black, fill = black, circle, + outer sep = 0, inner sep = 0, + minimum size = .8mm + }, + round/.style = { + draw = orange, thick, circle, + minimum size = 1mm, + inner sep = 0pt, outer sep = 0pt, + }, + cross/.style = { + cross out, draw = magenta, thick, + minimum size = 1mm, + inner sep = 0pt, outer sep = 0pt + }, + ] + + % origin + \coordinate (O) at (0,0,0); + + % poles + \coordinate (NP) at (0,0,\l); + \coordinate (SP) at (0,0,-\l); + + % \draw (SP) node[dot, gray] {}; + % \draw (NP) node[dot, gray] {}; + + % gray unit circle + \tdplotdrawarc[gray]{(O)}{\l}{0}{360}{}{}; + \draw[gray, dashed] (-\l, 0, 0) to (\l, 0, 0); + \draw[gray, dashed] (0, -\l, 0) to (0, \l, 0); + + % axis + \draw[->] (O) -- ++(1.25*\l,0,0) node[left] {\(\mathbf{\hat{x}}\)}; + \draw[->] (O) -- ++(0,1.25*\l,0) node[right] {\(\mathbf{\hat{y}}\)}; + \draw[->] (O) -- ++(0,0,1.25*\l) node[above] {\(\mathbf{\hat{z}}\)}; + + % meridians + \foreach \phi in {0, 30, 60, ..., 150}{ + \tdplotsetrotatedcoords{\phi}{90}{0}; + \tdplotdrawarc[lightgray, densely dotted, tdplot_rotated_coords]{(O)}{\l}{0}{360}{}{}; + } + + % dot above and its projection + \pgfmathsetmacro{\phi}{120} + \pgfmathsetmacro{\theta}{40} + + \pgfmathsetmacro{\px}{cos(\phi)*sin(\theta)*\l} + \pgfmathsetmacro{\py}{sin(\phi)*sin(\theta)*\l} + \pgfmathsetmacro{\pz}{cos(\theta)*\l}) + + % point A + \coordinate (A) at (\px,\py,\pz); + \coordinate (Ap) at (\px,\py, 0); + + % lines + \draw[red!80!black, ->] (O) -- (A); + \draw[red!80!black, densely dashed] (O) -- (Ap) -- (A) + node[above right] {\(\mathbf{\hat{r}}\)}; + + % arcs + \tdplotdrawarc[blue!80!black, ->]{(O)}{.8\l}{0}{\phi}{}{}; + \node[below right, blue!80!black] at (.8\l,0,0) {\(\bm{\hat{\varphi}}\)}; + + \tdplotsetrotatedcoords{\phi-90}{-90}{0}; + \tdplotdrawarc[blue!80!black, ->, tdplot_rotated_coords]{(O)}{.95\l}{0}{\theta}{}{}; + \node[above right = 1mm, blue!80!black] at (0,0,.8\l) {\(\bm{\hat{\vartheta}}\)}; + + + % dots + \draw (O) node[dot] {}; + \draw (A) node[dot, fill = red!80!black] {}; + +\end{tikzpicture} +\end{document} +% vim:ts=2 sw=2 et: diff --git a/buch/papers/kugel/images/Makefile b/buch/papers/kugel/images/Makefile deleted file mode 100644 index 4226dab..0000000 --- a/buch/papers/kugel/images/Makefile +++ /dev/null @@ -1,30 +0,0 @@ -# -# Makefile -- build images -# -# (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule -# -all: curvature.jpg spherecurve.jpg - -curvature.inc: curvgraph.m - octave curvgraph.m - -curvature.png: curvature.pov curvature.inc - povray +A0.1 +W1920 +H1080 +Ocurvature.png curvature.pov - -curvature.jpg: curvature.png - convert curvature.png -density 300 -units PixelsPerInch curvature.jpg - -spherecurve2.inc: spherecurve.m - octave spherecurve.m - -spherecurve.png: spherecurve.pov spherecurve.inc - povray +A0.1 +W1080 +H1080 +Ospherecurve.png spherecurve.pov - -spherecurve.jpg: spherecurve.png - convert spherecurve.png -density 300 -units PixelsPerInch spherecurve.jpg - -spherecurve: spherecurve.cpp - g++ -o spherecurve -g -Wall -O spherecurve.cpp - -spherecurve.inc: spherecurve - ./spherecurve diff --git a/buch/papers/kugel/images/curvature.maxima b/buch/papers/kugel/images/curvature.maxima deleted file mode 100644 index 6313642..0000000 --- a/buch/papers/kugel/images/curvature.maxima +++ /dev/null @@ -1,6 +0,0 @@ - -f: exp(-r^2/sigma^2)/sigma; -laplacef: ratsimp(diff(r * diff(f,r), r) / r); -f: exp(-r^2/(2*sigma^2))/(sqrt(2)*sigma); -laplacef: ratsimp(diff(r * diff(f,r), r) / r); - diff --git a/buch/papers/kugel/images/curvature.pov b/buch/papers/kugel/images/curvature.pov deleted file mode 100644 index 3b15d77..0000000 --- a/buch/papers/kugel/images/curvature.pov +++ /dev/null @@ -1,139 +0,0 @@ -// -// curvature.pov -// -// (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule -// - -#version 3.7; -#include "colors.inc" - -global_settings { - assumed_gamma 1 -} - -#declare imagescale = 0.09; - -camera { - location <10, 10, -40> - look_at <0, 0, 0> - right 16/9 * x * imagescale - up y * imagescale -} - -light_source { - <-10, 10, -40> color White - area_light <1,0,0> <0,0,1>, 10, 10 - adaptive 1 - jitter -} - -sky_sphere { - pigment { - color rgb<1,1,1> - } -} - -// -// draw an arrow from to with thickness with -// color -// -#macro arrow(from, to, arrowthickness, c) -#declare arrowdirection = vnormalize(to - from); -#declare arrowlength = vlength(to - from); -union { - sphere { - from, 1.1 * arrowthickness - } - cylinder { - from, - from + (arrowlength - 5 * arrowthickness) * arrowdirection, - arrowthickness - } - cone { - from + (arrowlength - 5 * arrowthickness) * arrowdirection, - 2 * arrowthickness, - to, - 0 - } - pigment { - color c - } - finish { - specular 0.9 - metallic - } -} -#end - -arrow(<-3.1,0,0>, <3.1,0,0>, 0.01, White) -arrow(<0,-1,0>, <0,1,0>, 0.01, White) -arrow(<0,0,-2.1>, <0,0,2.1>, 0.01, White) - -#include "curvature.inc" - -#declare sigma = 1; -#declare s = 1.4; -#declare N0 = 0.4; -#declare funktion = function(r) { - (exp(-r*r/(sigma*sigma)) / sigma - - - exp(-r*r/(2*sigma*sigma)) / (sqrt(2)*sigma)) / N0 -}; -#declare hypot = function(xx, yy) { sqrt(xx*xx+yy*yy) }; - -#declare Funktion = function(x,y) { funktion(hypot(x+s,y)) - funktion(hypot(x-s,y)) }; -#macro punkt(xx,yy) - -#end - -#declare griddiameter = 0.006; -union { - #declare xmin = -3; - #declare xmax = 3; - #declare ymin = -2; - #declare ymax = 2; - - - #declare xstep = 0.2; - #declare ystep = 0.02; - #declare xx = xmin; - #while (xx < xmax + xstep/2) - #declare yy = ymin; - #declare P = punkt(xx, yy); - #while (yy < ymax - ystep/2) - #declare yy = yy + ystep; - #declare Q = punkt(xx, yy); - sphere { P, griddiameter } - cylinder { P, Q, griddiameter } - #declare P = Q; - #end - sphere { P, griddiameter } - #declare xx = xx + xstep; - #end - - #declare xstep = 0.02; - #declare ystep = 0.2; - #declare yy = ymin; - #while (yy < ymax + ystep/2) - #declare xx = xmin; - #declare P = punkt(xx, yy); - #while (xx < xmax - xstep/2) - #declare xx = xx + xstep; - #declare Q = punkt(xx, yy); - sphere { P, griddiameter } - cylinder { P, Q, griddiameter } - #declare P = Q; - #end - sphere { P, griddiameter } - #declare yy = yy + ystep; - #end - - pigment { - color rgb<0.8,0.8,0.8> - } - finish { - metallic - specular 0.8 - } -} - diff --git a/buch/papers/kugel/images/curvgraph.m b/buch/papers/kugel/images/curvgraph.m deleted file mode 100644 index 75effd6..0000000 --- a/buch/papers/kugel/images/curvgraph.m +++ /dev/null @@ -1,140 +0,0 @@ -# -# curvature.m -# -# (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule -# - -global N; -N = 10; - -global sigma2; -sigma2 = 1; - -global s; -s = 1.4; - -global cmax; -cmax = 0.9; -global cmin; -cmin = -0.9; - -global Cmax; -global Cmin; -Cmax = 0; -Cmin = 0; - -xmin = -3; -xmax = 3; -xsteps = 200; -hx = (xmax - xmin) / xsteps; - -ymin = -2; -ymax = 2; -ysteps = 200; -hy = (ymax - ymin) / ysteps; - -function retval = f0(r) - global sigma2; - retval = exp(-r^2/sigma2)/sqrt(sigma2) - exp(-r^2/(2*sigma2))/(sqrt(2*sigma2)); -end - -global N0; -N0 = f0(0) -N0 = 0.4; - -function retval = f1(x,y) - global N0; - retval = f0(hypot(x, y)) / N0; -endfunction - -function retval = f(x, y) - global s; - retval = f1(x+s, y) - f1(x-s, y); -endfunction - -function retval = curvature0(r) - global sigma2; - retval = ( - -4*(sigma2-r^2)*exp(-r^2/sigma2) - + - (2*sigma2-r^2)*exp(-r^2/(2*sigma2)) - ) / (sigma2^(5/2)); -endfunction - -function retval = curvature1(x, y) - retval = curvature0(hypot(x, y)); -endfunction - -function retval = curvature(x, y) - global s; - retval = curvature1(x+s, y) - curvature1(x-s, y); -endfunction - -function retval = farbe(x, y) - global Cmax; - global Cmin; - global cmax; - global cmin; - c = curvature(x, y); - if (c < Cmin) - Cmin = c - endif - if (c > Cmax) - Cmax = c - endif - u = (c - cmin) / (cmax - cmin); - if (u > 1) - u = 1; - endif - if (u < 0) - u = 0; - endif - color = [ u, 0.5, 1-u ]; - color = color/max(color); - color(1,4) = c/2; - retval = color; -endfunction - -function dreieck(fn, A, B, C) - fprintf(fn, "\ttriangle {\n"); - fprintf(fn, "\t <%.4f,%.4f,%.4f>,\n", A(1,1), A(1,3), A(1,2)); - fprintf(fn, "\t <%.4f,%.4f,%.4f>,\n", B(1,1), B(1,3), B(1,2)); - fprintf(fn, "\t <%.4f,%.4f,%.4f>\n", C(1,1), C(1,3), C(1,2)); - fprintf(fn, "\t}\n"); -endfunction - -function viereck(fn, punkte) - color = farbe(mean(punkte(:,1)), mean(punkte(:,2))); - fprintf(fn, " mesh {\n"); - dreieck(fn, punkte(1,:), punkte(2,:), punkte(3,:)); - dreieck(fn, punkte(2,:), punkte(3,:), punkte(4,:)); - fprintf(fn, "\tpigment { color rgb<%.4f,%.4f,%.4f> } // %.4f\n", - color(1,1), color(1,2), color(1,3), color(1,4)); - fprintf(fn, " }\n"); -endfunction - -fn = fopen("curvature.inc", "w"); -punkte = zeros(4,3); -for ix = (0:xsteps-1) - x = xmin + ix * hx; - punkte(1,1) = x; - punkte(2,1) = x; - punkte(3,1) = x + hx; - punkte(4,1) = x + hx; - for iy = (0:ysteps-1) - y = ymin + iy * hy; - punkte(1,2) = y; - punkte(2,2) = y + hy; - punkte(3,2) = y; - punkte(4,2) = y + hy; - for i = (1:4) - punkte(i,3) = f(punkte(i,1), punkte(i,2)); - endfor - viereck(fn, punkte); - end -end -#fprintf(fn, " finish { metallic specular 0.5 }\n"); -fclose(fn); - -printf("Cmax = %.4f\n", Cmax); -printf("Cmin = %.4f\n", Cmin); diff --git a/buch/papers/kugel/images/spherecurve.cpp b/buch/papers/kugel/images/spherecurve.cpp deleted file mode 100644 index 8ddf5e5..0000000 --- a/buch/papers/kugel/images/spherecurve.cpp +++ /dev/null @@ -1,292 +0,0 @@ -/* - * spherecurve.cpp - * - * (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule - */ -#include -#include -#include -#include -#include - -inline double sqr(double x) { return x * x; } - -/** - * \brief Class for 3d vectors (also used as colors) - */ -class vector { - double X[3]; -public: - vector() { X[0] = X[1] = X[2] = 0; } - vector(double a) { X[0] = X[1] = X[2] = a; } - vector(double x, double y, double z) { - X[0] = x; X[1] = y; X[2] = z; - } - vector(double theta, double phi) { - double s = sin(theta); - X[0] = cos(phi) * s; - X[1] = sin(phi) * s; - X[2] = cos(theta); - } - vector(const vector& other) { - for (int i = 0; i < 3; i++) { - X[i] = other.X[i]; - } - } - vector operator+(const vector& other) const { - return vector(X[0] + other.X[0], - X[1] + other.X[1], - X[2] + other.X[2]); - } - vector operator*(double l) const { - return vector(X[0] * l, X[1] * l, X[2] * l); - } - double operator*(const vector& other) const { - double s = 0; - for (int i = 0; i < 3; i++) { - s += X[i] * other.X[i]; - } - return s; - } - double norm() const { - double s = 0; - for (int i = 0; i < 3; i++) { - s += sqr(X[i]); - } - return sqrt(s); - } - vector normalize() const { - double l = norm(); - return vector(X[0]/l, X[1]/l, X[2]/l); - } - double max() const { - return std::max(X[0], std::max(X[1], X[2])); - } - double l0norm() const { - double l = 0; - for (int i = 0; i < 3; i++) { - if (fabs(X[i]) > l) { - l = fabs(X[i]); - } - } - return l; - } - vector l0normalize() const { - double l = l0norm(); - vector result(X[0]/l, X[1]/l, X[2]/l); - return result; - } - const double& operator[](int i) const { return X[i]; } - double& operator[](int i) { return X[i]; } -}; - -/** - * \brief Derived 3d vector class implementing color - * - * The constructor in this class converts a single value into a - * color on a suitable gradient. - */ -class color : public vector { -public: - static double utop; - static double ubottom; - static double green; -public: - color(double u) { - u = (u - ubottom) / (utop - ubottom); - if (u > 1) { - u = 1; - } - if (u < 0) { - u = 0; - } - u = pow(u,2); - (*this)[0] = u; - (*this)[1] = green * u * (1 - u); - (*this)[2] = 1-u; - double l = l0norm(); - for (int i = 0; i < 3; i++) { - (*this)[i] /= l; - } - } -}; - -double color::utop = 12; -double color::ubottom = -31; -double color::green = 0.5; - -/** - * \brief Surface model - * - * This class contains the definitions of the functions to plot - * and the parameters to - */ -class surfacefunction { - static vector axes[6]; - - double _a; - double _A; - - double _umin; - double _umax; -public: - double a() const { return _a; } - double A() const { return _A; } - - double umin() const { return _umin; } - double umax() const { return _umax; } - - surfacefunction(double a, double A) : _a(a), _A(A), _umin(0), _umax(0) { - } - - double f(double z) { - return A() * exp(a() * (sqr(z) - 1)); - } - - double g(double z) { - return -f(z) * 2*a() * ((2*a()*sqr(z) + (3-2*a()))*sqr(z) - 1); - } - - double F(const vector& v) { - double s = 0; - for (int i = 0; i < 6; i++) { - s += f(axes[i] * v); - } - return s / 6; - } - - double G(const vector& v) { - double s = 0; - for (int i = 0; i < 6; i++) { - s += g(axes[i] * v); - } - return s / 6; - } -protected: - color farbe(const vector& v) { - double u = G(v); - if (u < _umin) { - _umin = u; - } - if (u > _umax) { - _umax = u; - } - return color(u); - } -}; - -static double phi = (1 + sqrt(5)) / 2; -static double sl = sqrt(sqr(phi) + 1); -vector surfacefunction::axes[6] = { - vector( 0. , -1./sl, phi/sl ), - vector( 0. , 1./sl, phi/sl ), - vector( 1./sl, phi/sl, 0. ), - vector( -1./sl, phi/sl, 0. ), - vector( phi/sl, 0. , 1./sl ), - vector( -phi/sl, 0. , 1./sl ) -}; - -/** - * \brief Class to construct the plot - */ -class surface : public surfacefunction { - FILE *outfile; - - int _phisteps; - int _thetasteps; - double _hphi; - double _htheta; -public: - int phisteps() const { return _phisteps; } - int thetasteps() const { return _thetasteps; } - double hphi() const { return _hphi; } - double htheta() const { return _htheta; } - void phisteps(int s) { _phisteps = s; _hphi = 2 * M_PI / s; } - void thetasteps(int s) { _thetasteps = s; _htheta = M_PI / s; } - - surface(const std::string& filename, double a, double A) - : surfacefunction(a, A) { - outfile = fopen(filename.c_str(), "w"); - phisteps(400); - thetasteps(200); - } - - ~surface() { - fclose(outfile); - } - -private: - void triangle(const vector& v0, const vector& v1, const vector& v2) { - fprintf(outfile, " mesh {\n"); - vector c = (v0 + v1 + v2) * (1./3.); - vector color = farbe(c.normalize()); - vector V0 = v0 * (1 + F(v0)); - vector V1 = v1 * (1 + F(v1)); - vector V2 = v2 * (1 + F(v2)); - fprintf(outfile, "\ttriangle {\n"); - fprintf(outfile, "\t <%.6f,%.6f,%.6f>,\n", - V0[0], V0[2], V0[1]); - fprintf(outfile, "\t <%.6f,%.6f,%.6f>,\n", - V1[0], V1[2], V1[1]); - fprintf(outfile, "\t <%.6f,%.6f,%.6f>\n", - V2[0], V2[2], V2[1]); - fprintf(outfile, "\t}\n"); - fprintf(outfile, "\tpigment { color rgb<%.4f,%.4f,%.4f> }\n", - color[0], color[1], color[2]); - fprintf(outfile, "\tfinish { metallic specular 0.5 }\n"); - fprintf(outfile, " }\n"); - } - - void northcap() { - vector v0(0, 0, 1); - for (int i = 1; i <= phisteps(); i++) { - fprintf(outfile, " // northcap i = %d\n", i); - vector v1(htheta(), (i - 1) * hphi()); - vector v2(htheta(), i * hphi()); - triangle(v0, v1, v2); - } - } - - void southcap() { - vector v0(0, 0, -1); - for (int i = 1; i <= phisteps(); i++) { - fprintf(outfile, " // southcap i = %d\n", i); - vector v1(M_PI - htheta(), (i - 1) * hphi()); - vector v2(M_PI - htheta(), i * hphi()); - triangle(v0, v1, v2); - } - } - - void zone() { - for (int j = 1; j < thetasteps() - 1; j++) { - for (int i = 1; i <= phisteps(); i++) { - fprintf(outfile, " // zone j = %d, i = %d\n", - j, i); - vector v0( j * htheta(), (i-1) * hphi()); - vector v1((j+1) * htheta(), (i-1) * hphi()); - vector v2( j * htheta(), i * hphi()); - vector v3((j+1) * htheta(), i * hphi()); - triangle(v0, v1, v2); - triangle(v1, v2, v3); - } - } - } -public: - void draw() { - northcap(); - southcap(); - zone(); - } -}; - -/** - * \brief main function - */ -int main(int argc, char *argv[]) { - surface S("spherecurve.inc", 5, 10); - color::green = 1.0; - S.draw(); - std::cout << "umin: " << S.umin() << std::endl; - std::cout << "umax: " << S.umax() << std::endl; - return EXIT_SUCCESS; -} diff --git a/buch/papers/kugel/images/spherecurve.m b/buch/papers/kugel/images/spherecurve.m deleted file mode 100644 index 99d5c9a..0000000 --- a/buch/papers/kugel/images/spherecurve.m +++ /dev/null @@ -1,160 +0,0 @@ -# -# spherecurve.m -# -# (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule -# -global a; -a = 5; -global A; -A = 10; - -phisteps = 400; -hphi = 2 * pi / phisteps; -thetasteps = 200; -htheta = pi / thetasteps; - -function retval = f(z) - global a; - global A; - retval = A * exp(a * (z^2 - 1)); -endfunction - -function retval = g(z) - global a; - retval = -f(z) * 2 * a * (2 * a * z^4 + (3 - 2*a) * z^2 - 1); - # 2 - # - a 2 4 2 2 a z - #(%o6) - %e (4 a z + (6 a - 4 a ) z - 2 a) %e -endfunction - -phi = (1 + sqrt(5)) / 2; - -global axes; -axes = [ - 0, 0, 1, -1, phi, -phi; - 1, -1, phi, phi, 0, 0; - phi, phi, 0, 0, 1, 1; -]; -axes = axes / (sqrt(phi^2+1)); - -function retval = kugel(theta, phi) - retval = [ - cos(phi) * sin(theta); - sin(phi) * sin(theta); - cos(theta) - ]; -endfunction - -function retval = F(v) - global axes; - s = 0; - for i = (1:6) - z = axes(:,i)' * v; - s = s + f(z); - endfor - retval = s / 6; -endfunction - -function retval = F2(theta, phi) - v = kugel(theta, phi); - retval = F(v); -endfunction - -function retval = G(v) - global axes; - s = 0; - for i = (1:6) - s = s + g(axes(:,i)' * v); - endfor - retval = s / 6; -endfunction - -function retval = G2(theta, phi) - v = kugel(theta, phi); - retval = G(v); -endfunction - -function retval = cnormalize(u) - utop = 11; - ubottom = -30; - retval = (u - ubottom) / (utop - ubottom); - if (retval > 1) - retval = 1; - endif - if (retval < 0) - retval = 0; - endif -endfunction - -global umin; -umin = 0; -global umax; -umax = 0; - -function color = farbe(v) - global umin; - global umax; - u = G(v); - if (u < umin) - umin = u; - endif - if (u > umax) - umax = u; - endif - u = cnormalize(u); - color = [ u, 0.5, 1-u ]; - color = color/max(color); -endfunction - -function dreieck(fn, v0, v1, v2) - fprintf(fn, " mesh {\n"); - c = (v0 + v1 + v2) / 3; - c = c / norm(c); - color = farbe(c); - v0 = v0 * (1 + F(v0)); - v1 = v1 * (1 + F(v1)); - v2 = v2 * (1 + F(v2)); - fprintf(fn, "\ttriangle {\n"); - fprintf(fn, "\t <%.6f,%.6f,%.6f>,\n", v0(1,1), v0(3,1), v0(2,1)); - fprintf(fn, "\t <%.6f,%.6f,%.6f>,\n", v1(1,1), v1(3,1), v1(2,1)); - fprintf(fn, "\t <%.6f,%.6f,%.6f>\n", v2(1,1), v2(3,1), v2(2,1)); - fprintf(fn, "\t}\n"); - fprintf(fn, "\tpigment { color rgb<%.4f,%.4f,%.4f> }\n", - color(1,1), color(1,2), color(1,3)); - fprintf(fn, "\tfinish { metallic specular 0.5 }\n"); - fprintf(fn, " }\n"); -endfunction - -fn = fopen("spherecurve2.inc", "w"); - - for i = (1:phisteps) - # Polkappe nord - v0 = [ 0; 0; 1 ]; - v1 = kugel(htheta, (i-1) * hphi); - v2 = kugel(htheta, i * hphi); - fprintf(fn, " // i = %d\n", i); - dreieck(fn, v0, v1, v2); - - # Polkappe sued - v0 = [ 0; 0; -1 ]; - v1 = kugel(pi-htheta, (i-1) * hphi); - v2 = kugel(pi-htheta, i * hphi); - dreieck(fn, v0, v1, v2); - endfor - - for j = (1:thetasteps-2) - for i = (1:phisteps) - v0 = kugel( j * htheta, (i-1) * hphi); - v1 = kugel((j+1) * htheta, (i-1) * hphi); - v2 = kugel( j * htheta, i * hphi); - v3 = kugel((j+1) * htheta, i * hphi); - fprintf(fn, " // i = %d, j = %d\n", i, j); - dreieck(fn, v0, v1, v2); - dreieck(fn, v1, v2, v3); - endfor - endfor - -fclose(fn); - -umin -umax diff --git a/buch/papers/kugel/images/spherecurve.maxima b/buch/papers/kugel/images/spherecurve.maxima deleted file mode 100644 index 1e9077c..0000000 --- a/buch/papers/kugel/images/spherecurve.maxima +++ /dev/null @@ -1,13 +0,0 @@ -/* - * spherecurv.maxima - * - * (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule - */ -f: exp(-a * sin(theta)^2); - -g: ratsimp(diff(sin(theta) * diff(f, theta), theta)/sin(theta)); -g: subst(z, cos(theta), g); -g: subst(sqrt(1-z^2), sin(theta), g); -ratsimp(g); - -f: ratsimp(subst(sqrt(1-z^2), sin(theta), f)); diff --git a/buch/papers/kugel/images/spherecurve.pov b/buch/papers/kugel/images/spherecurve.pov deleted file mode 100644 index b1bf4b8..0000000 --- a/buch/papers/kugel/images/spherecurve.pov +++ /dev/null @@ -1,73 +0,0 @@ -// -// curvature.pov -// -// (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule -// - -#version 3.7; -#include "colors.inc" - -global_settings { - assumed_gamma 1 -} - -#declare imagescale = 0.13; - -camera { - location <10, 10, -40> - look_at <0, 0, 0> - right x * imagescale - up y * imagescale -} - -light_source { - <-10, 10, -40> color White - area_light <1,0,0> <0,0,1>, 10, 10 - adaptive 1 - jitter -} - -sky_sphere { - pigment { - color rgb<1,1,1> - } -} - -// -// draw an arrow from to with thickness with -// color -// -#macro arrow(from, to, arrowthickness, c) -#declare arrowdirection = vnormalize(to - from); -#declare arrowlength = vlength(to - from); -union { - sphere { - from, 1.1 * arrowthickness - } - cylinder { - from, - from + (arrowlength - 5 * arrowthickness) * arrowdirection, - arrowthickness - } - cone { - from + (arrowlength - 5 * arrowthickness) * arrowdirection, - 2 * arrowthickness, - to, - 0 - } - pigment { - color c - } - finish { - specular 0.9 - metallic - } -} -#end - -arrow(<-2.7,0,0>, <2.7,0,0>, 0.03, White) -arrow(<0,-2.7,0>, <0,2.7,0>, 0.03, White) -arrow(<0,0,-2.7>, <0,0,2.7>, 0.03, White) - -#include "spherecurve.inc" - -- cgit v1.2.1 From 10a72bf8d66de28f3f1b5598c37c32d29a306893 Mon Sep 17 00:00:00 2001 From: Fabian <@> Date: Tue, 16 Aug 2022 18:25:31 +0200 Subject: 3. Ueberarbeitung, Verbesserungen --- buch/papers/0f1/teil0.tex | 4 ++-- buch/papers/0f1/teil1.tex | 12 +++++------- buch/papers/0f1/teil2.tex | 20 ++++++++++++-------- buch/papers/0f1/teil3.tex | 12 ++++++------ 4 files changed, 25 insertions(+), 23 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/0f1/teil0.tex b/buch/papers/0f1/teil0.tex index 9aca368..335cf92 100644 --- a/buch/papers/0f1/teil0.tex +++ b/buch/papers/0f1/teil0.tex @@ -6,10 +6,10 @@ \section{Ausgangslage\label{0f1:section:ausgangslage}} \rhead{Ausgangslage} Die hypergeometrische Funktion $\mathstrut_0F_1$ wird in vielen Funktionen als Basisfunktion benutzt, -zum Beispiel um die Airy Funktion zu berechnen. +zum Beispiel um die Airy-Funktion zu berechnen. In der GNU Scientific Library \cite{0f1:library-gsl} ist die Funktion $\mathstrut_0F_1$ vorhanden. -Allerdings wirft die Funktion bei negativen Übergabenwerten wie zum Beispiel \verb+gsl_sf_hyperg_0F1(1, -1)+ eine Exception. +Allerdings wirft die Funktion bei negativen Übergabewerten wie zum Beispiel \verb+gsl_sf_hyperg_0F1(1, -1)+ eine Exception. Bei genauerer Untersuchung hat sich gezeigt, dass die Funktion je nach Betriebssystem funktioniert oder eben nicht. So kann die Funktion unter Windows fehlerfrei aufgerufen werden, beim Mac OS und Linux sind negative Übergabeparameter im Moment nicht möglich. Ziel dieser Arbeit war es zu evaluieren, ob es mit einfachen mathematischen Operationen möglich ist, die hypergeometrische Funktion $\mathstrut_0F_1$ zu implementieren. diff --git a/buch/papers/0f1/teil1.tex b/buch/papers/0f1/teil1.tex index c0f857d..8d00f95 100644 --- a/buch/papers/0f1/teil1.tex +++ b/buch/papers/0f1/teil1.tex @@ -6,8 +6,7 @@ \section{Mathematischer Hintergrund \label{0f1:section:mathHintergrund}} \rhead{Mathematischer Hintergrund} -Basierend auf den Herleitungen des Abschnittes \ref{buch:rekursion:section:hypergeometrische-funktion}, werden im nachfolgenden Abschnitt nochmals die Resultate -beschrieben. +Basierend auf den Herleitungen des Abschnittes \ref{buch:rekursion:section:hypergeometrische-funktion} werden im nachfolgenden Abschnitt nochmals die Resultate beschrieben. \subsection{Hypergeometrische Funktion \label{0f1:subsection:hypergeometrisch}} @@ -59,7 +58,7 @@ Angewendet auf die Funktion $\mathstrut_pF_q$ ergibt sich für $\mathstrut_0F_1$ -\subsection{Airy Funktion +\subsection{Airy-Funktion \label{0f1:subsection:airy}} Die Funktion $\operatorname{Ai}(x)$ und die verwandte Funktion $\operatorname{Bi}(x)$ werden als Airy-Funktion bezeichnet. Sie werden zur Lösung verschiedener physikalischer Probleme benutzt, wie zum Beispiel zur Lösung der Schrödinger-Gleichung \cite{0f1:wiki-airyFunktion}. @@ -70,8 +69,8 @@ Die Funktion $\operatorname{Ai}(x)$ und die verwandte Funktion $\operatorname{Bi heisst die {\em Airy-Differentialgleichung}. \end{definition} -Die Airy Funktion lässt sich auf verschiedene Arten darstellen. -Als hypergeometrische Funktion berechnet, ergibt sich wie in Abschnitt \ref{buch:differentialgleichungen:section:hypergeometrisch} hergeleitet, folgende Lösungen der Airy-Differentialgleichung zu den Anfangsbedingungen $\operatorname{Ai}(0)=1$ und $\operatorname{Ai}'(0)=0$, sowie $\operatorname{Bi}(0)=0$ und $\operatorname{Bi}'(0)=1$. +Die Airy-Funktion lässt sich auf verschiedene Arten darstellen. +Als hypergeometrische Funktion berechnet, ergeben sich wie in Abschnitt \ref{buch:differentialgleichungen:section:hypergeometrisch} hergeleitet, folgende Lösungen der Airy-Differentialgleichung zu den Anfangsbedingungen $\operatorname{Ai}(0)=1$ und $\operatorname{Ai}'(0)=0$, sowie $\operatorname{Bi}(0)=0$ und $\operatorname{Bi}'(0)=1$: \begin{align} \label{0f1:airy:hypergeometrisch:eq} @@ -96,7 +95,6 @@ x\cdot\mathstrut_0F_1\biggl( \qedhere \end{align} -Um die Stabilität der Algorithmen zu $\mathstrut_0F_1$ zu überprüfen, wird in dieser Arbeit die Airy Funktion $\operatorname{Ai}(x)$ \eqref{0f1:airy:hypergeometrisch:eq} -benutzt. +Um die Stabilität der Algorithmen zu $\mathstrut_0F_1$ zu überprüfen, wird in dieser Arbeit die Airy Funktion $\operatorname{Ai}(x)$ benutzt. diff --git a/buch/papers/0f1/teil2.tex b/buch/papers/0f1/teil2.tex index ef9f55e..9b3a586 100644 --- a/buch/papers/0f1/teil2.tex +++ b/buch/papers/0f1/teil2.tex @@ -11,7 +11,7 @@ Die Unterprogramme wurde jeweils, wie die GNU Scientific Library, in C geschrieb \subsection{Potenzreihe \label{0f1:subsection:potenzreihe}} -Die naheliegendste Lösung ist die Programmierung der Potenzreihe \eqref{0f1:umsetzung:0f1:eq}. +Die naheliegendste Lösung ist die Programmierung der Potenzreihe \begin{align} \label{0f1:umsetzung:0f1:eq} @@ -23,7 +23,7 @@ Die naheliegendste Lösung ist die Programmierung der Potenzreihe \eqref{0f1:ums \frac{1}{c} +\frac{z^1}{(c+1) \cdot 1} + \cdots - + \frac{z^{20}}{c(c+1)(c+2)\cdots(c+19) \cdot 2.4 \cdot 10^{18}} + + \frac{z^{20}}{c(c+1)(c+2)\cdots(c+19) \cdot 2.4 \cdot 10^{18}}. \end{align} \lstinputlisting[style=C,float,caption={Potenzreihe.},label={0f1:listing:potenzreihe}, firstline=59]{papers/0f1/listings/potenzreihe.c} @@ -31,15 +31,17 @@ Die naheliegendste Lösung ist die Programmierung der Potenzreihe \eqref{0f1:ums \subsection{Kettenbruch \label{0f1:subsection:kettenbruch}} Eine weitere Variante zur Berechnung von $\mathstrut_0F_1(;c;z)$ ist die Umsetzung als Kettenbruch. -Der Vorteil einer Umsetzung als Kettenbruch gegenüber der Potenzreihe, ist die schnellere Konvergenz. +Der Vorteil einer Umsetzung als Kettenbruch gegenüber der Potenzreihe ist die schnellere Konvergenz. +\subsubsection{Grundlage} Ein endlicher Kettenbruch \cite{0f1:wiki-kettenbruch} ist ein Bruch der Form \begin{equation*} -a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cdots}}} +a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cdots}}}, \end{equation*} in welchem $a_0, a_1,\dots,a_n$ und $b_1,b_2,\dots,b_n$ ganze Zahlen sind. -Nimmt man nun folgenden Gleichung \cite{0f1:wiki-fraction}: +\subsubsection{Rekursionsbeziehungen und Kettenbrüche} +Nimmt man nun folgende Gleichung \cite{0f1:wiki-fraction}: \begin{equation*} f_{i-1} - f_i = k_i z f_{i+1}, \end{equation*} @@ -48,7 +50,7 @@ Ergibt sich folgender Zusammenhang: \begin{equation*} \cfrac{f_i}{f_{i-1}} = \cfrac{1}{1+k_iz\cfrac{f_{i+1}}{f_i}} \end{equation*} - +\subsubsection{Rekursion für $\mathstrut_0F_1$} Angewendet auf die Funktion $\mathstrut_0F_1$ bedeutet dies: \begin{equation} \label{0f1:math:potenzreihe:0f1:eq} @@ -68,6 +70,7 @@ erhält man: \cfrac{\mathstrut_0F_1(;c+1;z)}{\mathstrut_0F_1(;c;z)} = \cfrac{1}{1+\cfrac{\cfrac{z}{c(c+1)}}{1+\cfrac{\cfrac{z}{(c+1)(c+2)}}{1+\cfrac{\cfrac{z}{(c+2)(c+3)}}{\cdots}}}}. \end{equation*} +\subsubsection{Algorithmus} Mit weiteren Relationen ergibt sich nach Wolfram Alpha \cite{0f1:wolfram-0f1} folgender Kettenbruch \begin{equation} \label{0f1:math:kettenbruch:0f1:eq} @@ -92,7 +95,7 @@ lässt sich zu \cfrac{A_k}{B_k} = \cfrac{b_{k+1}}{a_{k+1} + \cfrac{p}{q}} = \frac{b_{k+1} \cdot q}{a_{k+1} \cdot q + p} \end{align*} umformen. -Dies lässt sich auch durch die folgende Matrizenschreibweise ausdrücken: +Dies lässt sich auch durch die folgende Matrizenschreibweise \begin{equation*} \begin{pmatrix} A_k\\ @@ -112,6 +115,7 @@ Dies lässt sich auch durch die folgende Matrizenschreibweise ausdrücken: \end{pmatrix}. %\label{0f1:math:rekursionsformel:herleitung} \end{equation*} +ausdrücken. Wendet man dies nun auf den Kettenbruch in der Form \begin{equation*} \frac{A_k}{B_k} = a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{\cdots}{\cdots+\cfrac{b_{k-1}}{a_{k-1} + \cfrac{b_k}{a_k}}}}} @@ -166,7 +170,7 @@ Und schlussendlich kann der Näherungsbruch berechnet werden. -\subsubsection{Lösung} +\subsubsection{Algorithmus} Die Berechnung von $A_k, B_k$ gemäss \eqref{0f1:math:matrix:ende:eq} kann man auch ohne die Matrizenschreibweise \cite{0f1:kettenbrueche} aufschreiben: \begin{itemize} \item Startbedingungen: diff --git a/buch/papers/0f1/teil3.tex b/buch/papers/0f1/teil3.tex index b283b07..d7cdfe8 100644 --- a/buch/papers/0f1/teil3.tex +++ b/buch/papers/0f1/teil3.tex @@ -13,9 +13,9 @@ Ebenso kann festgestellt werden, dass je grösser der Wert $z$ in $\mathstrut_0F \subsection{Konvergenz \label{0f1:subsection:konvergenz}} -Es zeigt sich in Abbildung \ref{0f1:ausblick:plot:airy:konvergenz}, dass nach drei Iterationen ($k = 3$) die Funktionen genaue Resultate im Bereich von $-2$ bis $2$ liefert. Ebenso kann festgestellt werden, dass der Kettenbruch schneller konvergiert und im positiven Bereich sogar mit der Referenzfunktion $\operatorname{Ai}(x)$ übereinstimmt. Da die Rekursionsformel eine Abwandlung des Kettenbruches ist, verhalten sich die Funktionen in diesem Fall gleich. +Es zeigt sich in Abbildung \ref{0f1:ausblick:plot:airy:konvergenz}, dass nach drei Iterationen ($k = 3$) die Funktionen genaue Resultate im Bereich von $-2$ bis $2$ liefert. Ebenso kann festgestellt werden, dass der Kettenbruch schneller konvergiert und im positiven Bereich mit der Referenzfunktion $\operatorname{Ai}(x)$ übereinstimmt. Da die Rekursionsformel eine Abwandlung des Kettenbruches ist, verhalten sich die Funktionen in diesem Fall gleich. -Erst wenn mehrerer Iterationen gemacht werden, um die Genauigkeit zu verbessern, ist der Kettenbruch den anderen zwei Algorithmen bezüglich Konvergenz überlegen. +Erst wenn mehrerer Iterationen gerechnet werden, um die Genauigkeit zu verbessern, ist der Kettenbruch den anderen zwei Algorithmen bezüglich Konvergenz überlegen. Interessant ist auch, dass die Rekursionsformel nahezu gleich schnell wie die Potenzreihe konvergiert, aber sich danach, wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} zu beobachten ist, einschwingt. Dieses Verhalten ist auch bei grösseren $z$ zu beobachten, allerdings ist dann die Differenz zwischen dem ersten lokalen Minimum von $k$ bis zum Abbruch kleiner. Dieses Phänomen ist auf die Lösung der Rekursionsformel \eqref{0f1:math:matrix:ende:eq} zurück zu führen. Da im Gegensatz die ganz kleinen Werte nicht zu einer Konvergenz wie beim Kettenbruch führen, sondern sich noch eine Zeit lang durch die Multiplikation aufschwingen. @@ -29,7 +29,7 @@ Verändert sich der Wert von $z$ in $\mathstrut_0F_1(;c;z)$ gegen grössere posi Wohingegen die Potenzreihe (Listing \ref{0f1:listing:potenzreihe}) das Problem hat, dass je mehr Terme berechnet werden, desto schneller wächst die Fakultät im Nenner. Dies führt zu einer Bereichsüberschreitung des \verb+double+ Bereiches \cite{0f1:double}, der spätesten ab $k=167$ eintritt. Schlussendlich gibt das Unterprogramm das Resultat \verb+-nan(ind)+ zurück. Die Rekursionformel \eqref{0f1:listing:kettenbruchRekursion} liefert für sehr grosse positive Werte die genausten Ergebnisse, verglichen mit der GNU Scientific Library. Wie schon vermutet ist die Rekursionsformel, im positivem Bereich, der stabilste Algorithmus. Um die Konvergenz zu gewährleisten, muss wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, die Iterationstiefe $k$ genug gross gewählt werden. -Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Grund dafür ist die Potenz von $z$, was zum Phänomen der Auslöschung \cite{0f1:SeminarNumerik} führt. Schön zu beobachten ist dies in der Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} mit der Airy-Funktion als Test. So sind sowohl die Potenzreihe, der Kettenbruch, als auch die Rekursionsformel bis ungefähr $\frac{-15^3}{9}$ stabil. Dies macht auch Sinn, da alle Algorithmen auf der gleichen mathematischen Grundlage basieren. Danach verhält sich allerdings die Instabilität unterschiedlich. Diese programmiertechnischen Unterschiede sind auch in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} und \ref{0f1:ausblick:plot:konvergenz:negativ} festzustellen. +Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Grund dafür ist die Potenz von $z$, was zum Phänomen der Auslöschung \cite{0f1:SeminarNumerik} führt. Schön zu beobachten ist dies in der Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} mit der Airy-Funktion als Test. So sind nach Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} die Potenzreihe, der Kettenbruch, als auch die Rekursionsformel, bis ungefähr $\frac{-15^3}{9}$ stabil. Dies macht auch Sinn, da alle Algorithmen auf der gleichen mathematischen Grundlage basieren. Danach verhält sich allerdings die Instabilität unterschiedlich. Diese programmiertechnischen Unterschiede sind auch in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} und \ref{0f1:ausblick:plot:konvergenz:negativ} festzustellen. \begin{figure} \centering @@ -41,21 +41,21 @@ Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Gru \begin{figure} \centering \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzPositiv.pdf} - \caption{Konvergenz mit positivem z; Logarithmisch, vorzeichenlose dargestellte Differenz vom erwarteten Endresultat. + \caption{Konvergenz mit positivem $z$; Logarithmisch, vorzeichenlose dargestellte Differenz vom erwarteten Endresultat. \label{0f1:ausblick:plot:konvergenz:positiv}} \end{figure} \begin{figure} \centering \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzNegativ.pdf} - \caption{Konvergenz mit negativem z; Logarithmisch, vorzeichenlose dargestellte Differenz vom erwarteten Endresultat. + \caption{Konvergenz mit negativem $z$; Logarithmisch, vorzeichenlose dargestellte Differenz vom erwarteten Endresultat. \label{0f1:ausblick:plot:konvergenz:negativ}} \end{figure} \begin{figure} \centering \includegraphics[width=1\textwidth]{papers/0f1/images/stabilitaet.pdf} - \caption{Stabilität der 3 Algorithmen verglichen mit der Referenz Funktion $\operatorname{Ai}(x)$. + \caption{Stabilität der drei Algorithmen verglichen mit der Referenz Funktion $\operatorname{Ai}(x)$. \label{0f1:ausblick:plot:airy:stabilitaet}} \end{figure} -- cgit v1.2.1 From cd9bd7f2fb6e1088130c9eef5a96ea996fd14947 Mon Sep 17 00:00:00 2001 From: Fabian <@> Date: Tue, 16 Aug 2022 21:44:28 +0200 Subject: 3. Ueberarbeitung, bilder --- buch/papers/0f1/images/konvergenzNegativ.pdf | Bin 18155 -> 18226 bytes buch/papers/0f1/images/konvergenzPositiv.pdf | Bin 18581 -> 17532 bytes buch/papers/0f1/teil3.tex | 4 ++-- 3 files changed, 2 insertions(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/0f1/images/konvergenzNegativ.pdf b/buch/papers/0f1/images/konvergenzNegativ.pdf index 03b2ba1..232c964 100644 Binary files a/buch/papers/0f1/images/konvergenzNegativ.pdf and b/buch/papers/0f1/images/konvergenzNegativ.pdf differ diff --git a/buch/papers/0f1/images/konvergenzPositiv.pdf b/buch/papers/0f1/images/konvergenzPositiv.pdf index 2e45129..71b1042 100644 Binary files a/buch/papers/0f1/images/konvergenzPositiv.pdf and b/buch/papers/0f1/images/konvergenzPositiv.pdf differ diff --git a/buch/papers/0f1/teil3.tex b/buch/papers/0f1/teil3.tex index d7cdfe8..eb32c52 100644 --- a/buch/papers/0f1/teil3.tex +++ b/buch/papers/0f1/teil3.tex @@ -16,7 +16,7 @@ Ebenso kann festgestellt werden, dass je grösser der Wert $z$ in $\mathstrut_0F Es zeigt sich in Abbildung \ref{0f1:ausblick:plot:airy:konvergenz}, dass nach drei Iterationen ($k = 3$) die Funktionen genaue Resultate im Bereich von $-2$ bis $2$ liefert. Ebenso kann festgestellt werden, dass der Kettenbruch schneller konvergiert und im positiven Bereich mit der Referenzfunktion $\operatorname{Ai}(x)$ übereinstimmt. Da die Rekursionsformel eine Abwandlung des Kettenbruches ist, verhalten sich die Funktionen in diesem Fall gleich. Erst wenn mehrerer Iterationen gerechnet werden, um die Genauigkeit zu verbessern, ist der Kettenbruch den anderen zwei Algorithmen bezüglich Konvergenz überlegen. -Interessant ist auch, dass die Rekursionsformel nahezu gleich schnell wie die Potenzreihe konvergiert, aber sich danach, wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} zu beobachten ist, einschwingt. Dieses Verhalten ist auch bei grösseren $z$ zu beobachten, allerdings ist dann die Differenz zwischen dem ersten lokalen Minimum von $k$ bis zum Abbruch kleiner. +Interessant ist auch, dass die Rekursionsformel nahezu gleich schnell wie die Potenzreihe konvergiert, aber sich danach, wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} zu beobachten ist, auf. Dieses Verhalten ist auch bei grösseren $z$ zu beobachten, allerdings ist dann die Differenz zwischen dem ersten lokalen Minimum von $k$ bis zum Abbruch kleiner. Dieses Phänomen ist auf die Lösung der Rekursionsformel \eqref{0f1:math:matrix:ende:eq} zurück zu führen. Da im Gegensatz die ganz kleinen Werte nicht zu einer Konvergenz wie beim Kettenbruch führen, sondern sich noch eine Zeit lang durch die Multiplikation aufschwingen. Ist $z$ negativ wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu aufgrund des Vorzeichens zu alternierenden Termen. So steigt bei allen Algorithmen zuerst die Differenz zum erwarteten Endwert. Erst nach genügend Iterationen sind die Terme genügend klein, so dass sie das Endresultat nicht mehr signifikant beeinflussen. @@ -24,7 +24,7 @@ Auch hier konvergiert der Kettenbruch am schnellsten von allen Algorithmen. Eben \subsection{Stabilität \label{0f1:subsection:Stabilitaet}} -Verändert sich der Wert von $z$ in $\mathstrut_0F_1(;c;z)$ gegen grössere positive Werte, wie zum Beispiel $c = 800$ liefert die Kettenbruch-Funktion (Listing \ref{0f1:listing:kettenbruchIterativ}) \verb+inf+ zurück. Dies könnte durch ein Abbruchkriterien abgefangen werden. Allerdings würde das, bei grossen Werten zulasten der Genauigkeit gehen. Trotzdem könnte, je nach Anwendung, auf ein paar Nachkommastellen verzichtet werden. +Verändert sich der Wert von $z$ in $\mathstrut_0F_1(;c;z)$ gegen grössere positive Werte, wie zum Beispiel $c = 800$ liefert die Kettenbruch-Funktion (Listing \ref{0f1:listing:kettenbruchIterativ}) \verb+inf+ zurück. Dies könnte durch ein Abbruchkriterien abgefangen werden. Allerdings würde das bei grossen Werten zulasten der Genauigkeit gehen. Trotzdem könnte, je nach Anwendung, auf ein paar Nachkommastellen verzichtet werden. Wohingegen die Potenzreihe (Listing \ref{0f1:listing:potenzreihe}) das Problem hat, dass je mehr Terme berechnet werden, desto schneller wächst die Fakultät im Nenner. Dies führt zu einer Bereichsüberschreitung des \verb+double+ Bereiches \cite{0f1:double}, der spätesten ab $k=167$ eintritt. Schlussendlich gibt das Unterprogramm das Resultat \verb+-nan(ind)+ zurück. Die Rekursionformel \eqref{0f1:listing:kettenbruchRekursion} liefert für sehr grosse positive Werte die genausten Ergebnisse, verglichen mit der GNU Scientific Library. Wie schon vermutet ist die Rekursionsformel, im positivem Bereich, der stabilste Algorithmus. Um die Konvergenz zu gewährleisten, muss wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, die Iterationstiefe $k$ genug gross gewählt werden. -- cgit v1.2.1 From 7d969250f8860f407255091a61b0b441b172c524 Mon Sep 17 00:00:00 2001 From: Fabian <@> Date: Tue, 16 Aug 2022 22:53:23 +0200 Subject: 3.Ueberarbeitung, bilder2 --- buch/papers/0f1/images/konvergenzNegativ.pdf | Bin 18226 -> 18524 bytes buch/papers/0f1/images/konvergenzPositiv.pdf | Bin 17532 -> 18253 bytes buch/papers/0f1/teil3.tex | 12 +++++------- 3 files changed, 5 insertions(+), 7 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/0f1/images/konvergenzNegativ.pdf b/buch/papers/0f1/images/konvergenzNegativ.pdf index 232c964..07d2a44 100644 Binary files a/buch/papers/0f1/images/konvergenzNegativ.pdf and b/buch/papers/0f1/images/konvergenzNegativ.pdf differ diff --git a/buch/papers/0f1/images/konvergenzPositiv.pdf b/buch/papers/0f1/images/konvergenzPositiv.pdf index 71b1042..8e1e7e4 100644 Binary files a/buch/papers/0f1/images/konvergenzPositiv.pdf and b/buch/papers/0f1/images/konvergenzPositiv.pdf differ diff --git a/buch/papers/0f1/teil3.tex b/buch/papers/0f1/teil3.tex index eb32c52..b6c0f4f 100644 --- a/buch/papers/0f1/teil3.tex +++ b/buch/papers/0f1/teil3.tex @@ -15,12 +15,10 @@ Ebenso kann festgestellt werden, dass je grösser der Wert $z$ in $\mathstrut_0F \label{0f1:subsection:konvergenz}} Es zeigt sich in Abbildung \ref{0f1:ausblick:plot:airy:konvergenz}, dass nach drei Iterationen ($k = 3$) die Funktionen genaue Resultate im Bereich von $-2$ bis $2$ liefert. Ebenso kann festgestellt werden, dass der Kettenbruch schneller konvergiert und im positiven Bereich mit der Referenzfunktion $\operatorname{Ai}(x)$ übereinstimmt. Da die Rekursionsformel eine Abwandlung des Kettenbruches ist, verhalten sich die Funktionen in diesem Fall gleich. -Erst wenn mehrerer Iterationen gerechnet werden, um die Genauigkeit zu verbessern, ist der Kettenbruch den anderen zwei Algorithmen bezüglich Konvergenz überlegen. -Interessant ist auch, dass die Rekursionsformel nahezu gleich schnell wie die Potenzreihe konvergiert, aber sich danach, wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} zu beobachten ist, auf. Dieses Verhalten ist auch bei grösseren $z$ zu beobachten, allerdings ist dann die Differenz zwischen dem ersten lokalen Minimum von $k$ bis zum Abbruch kleiner. -Dieses Phänomen ist auf die Lösung der Rekursionsformel \eqref{0f1:math:matrix:ende:eq} zurück zu führen. Da im Gegensatz die ganz kleinen Werte nicht zu einer Konvergenz wie beim Kettenbruch führen, sondern sich noch eine Zeit lang durch die Multiplikation aufschwingen. +Erst wenn mehrerer Iterationen gerechnet werden, ist wie Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, der Kettenbruch den anderen zwei Algorithmen bezüglich Konvergenz überlegen. Allerdings muss beachtet werden, dass die Rekursionsformel zwar erst nach 35 Approximationen gänzlich konvergiert, nach 27 Iterationen sich nicht mehr gross verändert. + +Ist $z$ negativ wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu aufgrund des Vorzeichens zu alternierenden Termen. So steigt bei allen Algorithmen zuerst die Differenz zum erwarteten Endwert. Erst nach genügend Iterationen sind die Terme so klein, dass sie das Endresultat nicht mehr signifikant beeinflussen. Während die Potenzreihe zusammen mit dem Kettenbruch nach 34 Approximationen konvergiert, braucht die Rekursionsformel noch zwei Iterationen mehr. Wohingegen die Rekursionsformel der genauste Algorithmus im negativen Bereich ist. Da der Computer mit einer relativen Genauigkeit von $10^{-15}$ rechnet, ist dies das Maximum an Präzision, dass erreicht werden kann. -Ist $z$ negativ wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu aufgrund des Vorzeichens zu alternierenden Termen. So steigt bei allen Algorithmen zuerst die Differenz zum erwarteten Endwert. Erst nach genügend Iterationen sind die Terme genügend klein, so dass sie das Endresultat nicht mehr signifikant beeinflussen. -Auch hier konvergiert der Kettenbruch am schnellsten von allen Algorithmen. Ebenso bricht die Rekursionsformel nahezu gleichzeitig mit der Potenzreihe ab. \subsection{Stabilität \label{0f1:subsection:Stabilitaet}} @@ -41,14 +39,14 @@ Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Gru \begin{figure} \centering \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzPositiv.pdf} - \caption{Konvergenz mit positivem $z$; Logarithmisch, vorzeichenlose dargestellte Differenz vom erwarteten Endresultat. + \caption{Konvergenz mit positivem $z$; Logarithmisch dargestellte absoluter Fehler. \label{0f1:ausblick:plot:konvergenz:positiv}} \end{figure} \begin{figure} \centering \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzNegativ.pdf} - \caption{Konvergenz mit negativem $z$; Logarithmisch, vorzeichenlose dargestellte Differenz vom erwarteten Endresultat. + \caption{Konvergenz mit negativem $z$; Logarithmisch dargestellte absoluter Fehler. \label{0f1:ausblick:plot:konvergenz:negativ}} \end{figure} -- cgit v1.2.1 From 4a9a4ca761db0007138a778a87c652505570b071 Mon Sep 17 00:00:00 2001 From: Nao Pross Date: Tue, 16 Aug 2022 23:39:32 +0200 Subject: kugel: Update figures makefile --- buch/papers/kugel/Makefile | 3 ++- .../kugel/figures/tikz/spherical-coordinates.pdf | Bin 5824 -> 40319 bytes 2 files changed, 2 insertions(+), 1 deletion(-) (limited to 'buch/papers') diff --git a/buch/papers/kugel/Makefile b/buch/papers/kugel/Makefile index f798a55..995206b 100644 --- a/buch/papers/kugel/Makefile +++ b/buch/papers/kugel/Makefile @@ -5,5 +5,6 @@ # images: - @echo "no images to be created in kugel" + $(MAKE) -C ./figures/povray/ + $(MAKE) -C ./figures/tikz/ diff --git a/buch/papers/kugel/figures/tikz/spherical-coordinates.pdf b/buch/papers/kugel/figures/tikz/spherical-coordinates.pdf index 28f242e..1bff016 100644 Binary files a/buch/papers/kugel/figures/tikz/spherical-coordinates.pdf and b/buch/papers/kugel/figures/tikz/spherical-coordinates.pdf differ -- cgit v1.2.1 From c4cf68ac67f7fbadaacae64597ae713a6879f944 Mon Sep 17 00:00:00 2001 From: Nao Pross Date: Tue, 16 Aug 2022 23:39:59 +0200 Subject: kugel: Comment out preliminaries, review manu's work until legendre --- buch/papers/kugel/main.tex | 2 +- buch/papers/kugel/packages.tex | 5 + buch/papers/kugel/spherical-harmonics.tex | 229 +++++++++++++++++++++++------- 3 files changed, 180 insertions(+), 56 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kugel/main.tex b/buch/papers/kugel/main.tex index 98d9cb2..a281cae 100644 --- a/buch/papers/kugel/main.tex +++ b/buch/papers/kugel/main.tex @@ -11,7 +11,7 @@ \chapterauthor{Manuel Cattaneo, Naoki Pross} \input{papers/kugel/introduction} -\input{papers/kugel/preliminaries} +% \input{papers/kugel/preliminaries} \input{papers/kugel/spherical-harmonics} \input{papers/kugel/applications} diff --git a/buch/papers/kugel/packages.tex b/buch/papers/kugel/packages.tex index 1c4f3e0..b0e1f61 100644 --- a/buch/papers/kugel/packages.tex +++ b/buch/papers/kugel/packages.tex @@ -8,3 +8,8 @@ % following example %\usepackage{packagename} \usepackage{cases} + +\newcommand{\kugeltodo}[1]{\textcolor{red!70!black}{\texttt{[TODO: #1]}}} + +\DeclareMathOperator{\sphlaplacian}{\nabla^2_{\mathit{S}}} +\DeclareMathOperator{\surflaplacian}{\nabla^2_{\partial \mathit{S}}} diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex index c76e757..70657c9 100644 --- a/buch/papers/kugel/spherical-harmonics.tex +++ b/buch/papers/kugel/spherical-harmonics.tex @@ -1,70 +1,189 @@ -% vim:ts=2 sw=2 et spell: +% vim:ts=2 sw=2 et spell tw=80: \section{Spherical Harmonics} -We finally arrived at the main section, which gives our chapter its name. The idea is to discuss spherical harmonics, their mathematical derivation and some of their properties and applications.\newline -The subsection \ref{} will be devoted to the Eigenvalue problem of the Laplace operator. Through the latter, we will derive the set of Eigenfunctions that obey the equation presented in \ref{}[TODO: reference to eigenvalue equation], which will be defined as \emph{Spherical Harmonics}. In fact, this subsection will present their mathematical derivation.\newline -In the subsection \ref{}, on the other hand, some interesting properties related to them will be discussed. Some of these will come back to help us understand in more detail why they are useful in various real-world applications, which will be presented in the section \ref{}.\newline -One specific property will be studied in more detail in the subsection \ref{}, namely the recursive property. -The last subsection is devoted to one of the most beautiful applications (In our humble opinion), namely the derivation of a Fourier-style series expansion but defined on the sphere instead of a plane.\newline -More importantly, this subsection will allow us to connect all the dots we have created with the previous sections, concluding that Fourier is just a specific case of the application of the concept of orthogonality.\newline -Our hope is that after reading this section you will appreciate the beauty and power of generalization that mathematics offers us. -\subsection{Eigenvalue Problem on the Spherical surface} -\subsubsection{Unormalized Spherical Harmonics} -From the chapter \ref{}, we know that the spherical Laplacian is defined as. \begin{equation*} - \nabla^2_S := \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial}{\partial r} \right) + \frac{1}{r^2} - \left[ - \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial}{\partial\vartheta} \right) - + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2} - \right] -\end{equation*} -But we do not want to consider this algebraic monster entirely, since this includes the whole set $\mathbb{R}^3$; rather, we want to focus only on the spherical surface (as the title suggests). We can then further concretise our calculations by selecting any number for the variable $r$, so that we have a sphere and, more importantly, a spherical surface on which we can ``play''.\newline -Surely you have already heard of the unit circle, a geometric entity used extensively in many mathematical contexts. The most famous and basic among them is surely trigonometry.\newline -Extending this concept into three dimensions, we will talk about the unit sphere. This is a very famous sphere, as is the unit circle. So since we need a sphere why not use the most famous one? Thus imposing $r=1$.\newline -Now, since the variable $r$ became a constant, we can leave out all derivatives with respect to $r$, setting them to zero. Then substituting the value of $r$ for 1, we will obtain the operator we will refer to as \emph{Spherical Surface Operator}: +\if 0 +\kugeltodo{Rewrite this section if the preliminaries become an addendum} +We finally arrived at the main section, which gives our chapter its name. The +idea is to discuss spherical harmonics, their mathematical derivation and some +of their properties and applications. + +The subsection \ref{} \kugeltodo{Fix references} will be devoted to the +Eigenvalue problem of the Laplace operator. Through the latter we will derive +the set of Eigenfunctions that obey the equation presented in \ref{} +\kugeltodo{reference to eigenvalue equation}, which will be defined as +\emph{Spherical Harmonics}. In fact, this subsection will present their +mathematical derivation. + +In the subsection \ref{}, on the other hand, some interesting properties +related to them will be discussed. Some of these will come back to help us +understand in more detail why they are useful in various real-world +applications, which will be presented in the section \ref{}. + +One specific property will be studied in more detail in the subsection \ref{}, +namely the recursive property. The last subsection is devoted to one of the +most beautiful applications (In our humble opinion), namely the derivation of a +Fourier-style series expansion but defined on the sphere instead of a plane. +More importantly, this subsection will allow us to connect all the dots we have +created with the previous sections, concluding that Fourier is just a specific +case of the application of the concept of orthogonality. Our hope is that after +reading this section you will appreciate the beauty and power of generalization +that mathematics offers us. +\fi + +\subsection{Eigenvalue Problem} + +\begin{figure} + \centering + \includegraphics{papers/kugel/figures/tikz/spherical-coordinates} + \caption{ + Spherical coordinate system. Space is described with the free variables $r + \in \mathbb{R}_0^+$, $\vartheta \in [0; \pi]$ and $\varphi \in [0; 2\pi)$. + \label{kugel:fig:spherical-coordinates} + } +\end{figure} + +From Section \ref{buch:pde:section:kugel}, we know that the spherical Laplacian +in the spherical coordinate system (shown in Figure +\ref{kugel:fig:spherical-coordinates}) is is defined as \begin{equation*} - \nabla^2_{\partial S} := \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial}{\partial\vartheta} \right) - + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2}. + \sphlaplacian := + \frac{1}{r^2} \frac{\partial}{\partial r} \left( + r^2 \frac{\partial}{\partial r} + \right) + + \frac{1}{r^2} \left[ + \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( + \sin\vartheta \frac{\partial}{\partial\vartheta} + \right) + + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2} + \right]. \end{equation*} -As can be seen, for this definition, the subscript ``$\partial S$'' was used to emphasize the fact that we are on the spherical surface, which can be understood as a boundary of the sphere.\newline -Now that we have defined an operator, we can go on to calculate its eigenfunctions. As mentioned earlier, we can translate this problem at first abstract into a much more concrete problem, which has to do with the field of \emph{Partial Differential Equaitons} (PDEs). The functions we want to find are simply functions that respect the following expression: -\begin{equation}\label{kugel:eq:sph_srfc_laplace} - \nabla^2_{\partial S} f = \lambda f +But we will not consider this algebraic monstrosity in its entirety. As the +title suggests, we will only care about the \emph{surface} of the sphere. This +is for many reasons, but mainly to simplify reduce the already broad scope of +this text. Concretely, we will always work on the unit sphere, which just means +that we set $r = 1$ and keep only $\vartheta$ and $\varphi$ as free variables. +Now, since the variable $r$ became a constant, we can leave out all derivatives +with respect to $r$ and substitute all $r$'s with 1's to obtain a new operator +that deserves its own name. + +\begin{definition}[Surface spherical Laplacian] + \label{kugel:def:surface-laplacian} + The operator + \begin{equation*} + \surflaplacian := + \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( + \sin\vartheta \frac{\partial}{\partial\vartheta} + \right) + + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2}, + \end{equation*} + is called the surface spherical Laplacian. +\end{definition} + +In the definition, the subscript ``$\partial S$'' was used to emphasize the +fact that we are on the spherical surface, which can be understood as being the +boundary of the sphere. But what does it actually do? To get an intuition, +first of all, notice the fact that $\surflaplacian$ have second derivatives, +which means that this a measure of \emph{curvature}; But curvature of what? To +get an even stronger intuition we will go into geometry, were curvature can be +grasped very well visually. Consider figure \ref{kugel:fig:curvature} where the +curvature is shown using colors. First we have the curvature of a curve in 1D, +then the curvature of a surface (2D), and finally the curvature of a function on +the surface of the unit sphere. + +\begin{figure} + \centering + \includegraphics[width=.3\linewidth]{papers/kugel/figures/tikz/curvature-1d} + \hskip 5mm + \includegraphics[width=.3\linewidth]{papers/kugel/figures/povray/curvature} + \hskip 5mm + \includegraphics[width=.3\linewidth]{papers/kugel/figures/povray/spherecurve} + \caption{ + \kugeltodo{Fix alignment / size, add caption. Would be nice to match colors.} + \label{kugel:fig:curvature} + } +\end{figure} + +Now that we have defined an operator, we can go and study its eigenfunctions, +which means that we would like to find the functions $f(\vartheta, \varphi)$ +that satisfy the equation +\begin{equation} \label{kuvel:eqn:eigen} + \surflaplacian f = -\lambda f. \end{equation} -Which is traditionally written as follows: -\begin{equation*} - \nabla^2_{\partial S} f = -\lambda f -\end{equation*} -Perhaps the fact that we are dealing with a PDE may not be obvious at first glance, but if we extend the operator $\nabla^2_{\partial S}$ according to Eq.(\ref{kugel:eq:sph_srfc_laplace}), we will get: -\begin{equation}\label{kugel:eq:PDE_sph} - \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial f}{\partial\vartheta} \right) - + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 f}{\partial\varphi^2} + \lambda f = 0, +Perhaps it may not be obvious at first glance, but we are in fact dealing with a +partial differential equation (PDE). If we unpack the notation of the operator +$\nabla^2_{\partial S}$ according to definition +\ref{kugel:def:surface-laplacian}, we get: +\begin{equation} \label{kugel:eqn:eigen-pde} + \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( + \sin\vartheta \frac{\partial f}{\partial\vartheta} + \right) + + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 f}{\partial\varphi^2} + + \lambda f = 0. \end{equation} -making it emerge.\newline -All functions satisfying Eq.(\ref{kugel:eq:PDE_sph}), are called eigenfunctions. Our new goal is therefore to solve this PDE. The task seems very difficult but we can simplify it with a well-known technique, namely the \emph{separation Ansatz}. The latter consists in assuming that the function $f(\vartheta, \varphi)$ we are looking for can be factorized in the following form -\begin{equation}\label{kugel:eq:sep_ansatz_0} +Since all functions satisfying \eqref{kugel:eqn:eigen-pde} are the +\emph{eigenfunctions} of $\surflaplacian$, our new goal is to solve this PDE. +The task may seem very difficult but we can simplify it with a well-known +technique: \emph{the separation Ansatz}. It consists in assuming that the +function $f(\vartheta, \varphi)$ can be factorized in the following form: +\begin{equation} \label{kugel:eqn:sep-ansatz:0} f(\vartheta, \varphi) = \Theta(\vartheta)\Phi(\varphi). \end{equation} -In short, we are saying that the effect of the two independent variables can be described using the multiplication of two functions that describe their effect separately. If we include this assumption in Eq.(\ref{kugel:eq:PDE_sph}), we have: -\begin{equation} - \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial \Theta(\vartheta)}{\partial\vartheta} \right)\Phi(\varphi) - + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 \Phi(\varphi)}{\partial\varphi^2} \Theta(\vartheta) + \lambda \Theta(\vartheta)\Phi(\varphi) = 0. \label{kugel:eq:sep_ansatz_1} +In other words, we are saying that the effect of the two independent variables +can be described using the multiplication of two functions that describe their +effect separately. This separation process was already presented in section +\ref{buch:pde:section:kugel}, but we will briefly rehearse it here for +convenience. If we substitute this assumption in +\eqref{kugel:eqn:eigen-pde}, we have: +\begin{equation} \label{kugel:eqn:sep-ansatz:1} + \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( + \sin\vartheta \frac{\partial \Theta(\vartheta)}{\partial\vartheta} + \right) \Phi(\varphi) + + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 \Phi(\varphi)}{\partial\varphi^2} + \Theta(\vartheta) + + \lambda \Theta(\vartheta)\Phi(\varphi) = 0. \end{equation} -Dividing Eq.(\ref{kugel:eq:sep_ansatz_1}) by $\Theta(\vartheta)\Phi(\varphi)$ and inserting an auxiliary variable $m$, which we will call the separating constant, we will have: -\begin{equation*} -\frac{1}{\Theta(\vartheta)}\sin \vartheta \frac{d}{d \vartheta} \left( \sin \vartheta \frac{d \Theta}{d \vartheta} \right) + \lambda \sin^2 \vartheta = -\frac{1}{\Phi(\varphi)} \frac{d^2\Phi(\varphi)}{d\varphi^2} = m, -\end{equation*} -which is equivalent to the following system of two \emph{Ordinary Differential Equations} (ODEs) -\begin{align} - \frac{d^2\Phi(\varphi)}{d\varphi^2} &= -m \Phi(\varphi) \label{kugel:eq:ODE_1} \\ - \sin \vartheta \frac{d}{d \vartheta} \left( \sin \vartheta \frac{d \Theta}{d \vartheta} \right) + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)\Theta(\vartheta) &= 0 \label{kugel:eq:ODE_2} -\end{align} -The solution of Eq.(\ref{kugel:eq:ODE_1}) is quite trivial. The complex exponential is obviously the function we are looking for, so we can write +Dividing by $\Theta(\vartheta)\Phi(\varphi)$ and introducing an auxiliary +variable $m$, the separation constant, yields: \begin{equation*} - \Phi_m(\varphi) = e^{j m \varphi}, \quad m \in \mathbb{Z}. + \frac{1}{\Theta(\vartheta)}\sin \vartheta \frac{d}{d \vartheta} \left( + \sin \vartheta \frac{d \Theta}{d \vartheta} + \right) + + \lambda \sin^2 \vartheta + = -\frac{1}{\Phi(\varphi)} \frac{d^2\Phi(\varphi)}{d\varphi^2} + = m, \end{equation*} -The restriction for the separation constant $m$ arises from the fact that we require the following periodic constraint $\Phi_m(\varphi + 2\pi) = \Phi_m(\varphi)$.\newline -As for Eq.(\ref{kugel:eq:ODE_2}), the resolution will not be so straightforward. We can begin by considering the substitution $x = \cos \vartheta$. The operator $\frac{d}{d \vartheta}$ will be: +which is equivalent to the following system of 2 first order differential +equations (ODEs): +\begin{subequations} + \begin{gather} + \frac{d^2\Phi(\varphi)}{d\varphi^2} = -m \Phi(\varphi), + \label{kugel:eqn:ode-phi} \\ + \sin \vartheta \frac{d}{d \vartheta} \left( + \sin \vartheta \frac{d \Theta}{d \vartheta} + \right) + + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right) + \Theta(\vartheta) = 0 + \label{kugel:eqn:ode-theta}. + \end{gather} +\end{subequations} +The solution of \eqref{kugel:eqn:ode-phi} is easy to find: The complex +exponential is obviously the function we are looking for. So we can directly +write the solutions +\begin{equation} \label{kugel:eqn:ode-phi-sol} + \Phi(\varphi) = e^{i m \varphi}, \quad m \in \mathbb{Z}. +\end{equation} +The restriction that the separation constant $m$ needs to be an integer arises +from the fact that we require a $2\pi$-periodicity in $\varphi$ since +$\Phi(\varphi + 2\pi) = \Phi(\varphi)$. Unfortunately, solving +\eqref{kugel:eqn:ode-theta} is not so straightforward. Actually it is quite +difficult, and the process is so involved that it will require a dedicated +section of its own. + +\subsection{Legendre Functions} + +To solve \eqref{kugel:eqn:ode-theta} +We can begin by considering the substitution $x = \cos \vartheta$. The operator $\frac{d}{d \vartheta}$ will be: \begin{align*} \frac{d}{d \vartheta} = \frac{dx}{d \vartheta}\frac{d}{dx} &= -\sin \vartheta \frac{d}{dx} \\ &= -\sqrt{1-x^2} \frac{d}{dx}. -- cgit v1.2.1 From 4a97506a4759a46f3263aee2c46d684aed0fb104 Mon Sep 17 00:00:00 2001 From: Fabian <@> Date: Wed, 17 Aug 2022 01:35:28 +0200 Subject: 3. Ueberarbeitung, done --- buch/papers/0f1/teil2.tex | 35 ++++++++++++++++++++++++++--------- buch/papers/0f1/teil3.tex | 8 ++++---- 2 files changed, 30 insertions(+), 13 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/0f1/teil2.tex b/buch/papers/0f1/teil2.tex index 9b3a586..64f8d83 100644 --- a/buch/papers/0f1/teil2.tex +++ b/buch/papers/0f1/teil2.tex @@ -41,37 +41,54 @@ a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cdots}}}, in welchem $a_0, a_1,\dots,a_n$ und $b_1,b_2,\dots,b_n$ ganze Zahlen sind. \subsubsection{Rekursionsbeziehungen und Kettenbrüche} -Nimmt man nun folgende Gleichung \cite{0f1:wiki-fraction}: +Will man einen Kettenbruch für das Verhältnis $\frac{f_i(z)}{f_{i-1}(z)}$ finden, braucht man dazu eine Relation der analytischer Funktion $f_i(z)$. +Nimmt man die Gleichung \cite{0f1:wiki-fraction}: \begin{equation*} f_{i-1} - f_i = k_i z f_{i+1}, \end{equation*} wo $f_i$ analytische Funktionen sind und $i > 0$ ist, sowie $k_i$ konstant. Ergibt sich folgender Zusammenhang: \begin{equation*} - \cfrac{f_i}{f_{i-1}} = \cfrac{1}{1+k_iz\cfrac{f_{i+1}}{f_i}} + \cfrac{f_i}{f_{i-1}} = \cfrac{1}{1+k_iz\cfrac{f_{i+1}}{f_i}}. \end{equation*} +Geht man einen Schritt weiter und nimmt für $g_i = \frac{f_i}{f_{i-1}}$ an, kommt man zur Formel +\begin{equation*} + g_i = \cfrac{1}{1+k_izg_{i+1}}. +\end{equation*} +Setzt man dies nun für $g_1$ in den Bruch ein, ergibt sich folgendes: +\begin{equation*} + g_1 = \cfrac{f_1}{f_0} = \cfrac{1}{1+k_izg_2} = \cfrac{1}{1+\cfrac{k_1z}{1+k_2zg_3}} = \cdots +\end{equation*} +Repetiert man dies unendlich, erhält man einen Kettenbruch in der Form: +\begin{equation} + \label{0f1:math:rekursion:eq} + \cfrac{f_1}{f_0} = \cfrac{1}{1+\cfrac{k_1z}{1+\cfrac{k_2z}{1+\cfrac{k_3z}{\cdots}}}}. +\end{equation} + \subsubsection{Rekursion für $\mathstrut_0F_1$} -Angewendet auf die Funktion $\mathstrut_0F_1$ bedeutet dies: +Angewendet auf die Potenzreihe \begin{equation} \label{0f1:math:potenzreihe:0f1:eq} \mathstrut_0F_1(;c;z) = 1 + \frac{z}{c\cdot1!} + \frac{z^2}{c(c+1)\cdot2!} + \frac{z^3}{c(c+1)(c+2)\cdot3!} + \cdots \end{equation} -Durch Substitution kann bewiesen werden, dass die nachfolgende Formel eine Relation zur obigen Potenzreihe \eqref{0f1:math:potenzreihe:0f1:eq} ist: +kann durch Substitution bewiesen werden, dass \begin{equation*} - \mathstrut_0F_1(;c-1;z) - \mathstrut_0F_1(;c;z) = \frac{z}{c(c-1)} \cdot \mathstrut_0F_1(;c+1;z). + \mathstrut_0F_1(;c-1;z) - \mathstrut_0F_1(;c;z) = \frac{z}{c(c-1)} \cdot \mathstrut_0F_1(;c+1;z) \end{equation*} +eine Relation dazu ist. Wenn man für $f_i$ und $k_i$ folgende Annahme trifft: \begin{align*} - f_i =& \mathstrut_0F_1(;c+1;z)\\ - k_i =& \frac{1}{(c+1)(c+i-1)} + f_i =& \mathstrut_0F_1(;c+i;z)\\ + k_i =& \frac{1}{(c+i)(c+i-1)} \end{align*} -erhält man: +und in die Formel \eqref{0f1:math:rekursion:eq} einsetzt, erhält man: \begin{equation*} \cfrac{\mathstrut_0F_1(;c+1;z)}{\mathstrut_0F_1(;c;z)} = \cfrac{1}{1+\cfrac{\cfrac{z}{c(c+1)}}{1+\cfrac{\cfrac{z}{(c+1)(c+2)}}{1+\cfrac{\cfrac{z}{(c+2)(c+3)}}{\cdots}}}}. \end{equation*} \subsubsection{Algorithmus} -Mit weiteren Relationen ergibt sich nach Wolfram Alpha \cite{0f1:wolfram-0f1} folgender Kettenbruch +Da mit obigen Formeln nur ein Verhältnis zwischen $ \frac{\mathstrut_0F_1(;c+1;z)}{\mathstrut_0F_1(;c;z)}$ berechnet wurde, braucht es weitere Relationen um $\mathstrut_0F_1(;c;z)$ zu erhalten. +So ergeben ähnliche Relationen nach Wolfram Alpha \cite{0f1:wolfram-0f1} folgender Kettenbruch \begin{equation} \label{0f1:math:kettenbruch:0f1:eq} \mathstrut_0F_1(;c;z) = 1 + \cfrac{\cfrac{z}{c}}{1+\cfrac{-\cfrac{z}{2(c+1)}}{1+\cfrac{z}{2(c+1)}+\cfrac{-\cfrac{z}{3(c+2)}}{1+\cfrac{z}{5(c+4)} + \cdots}}}, diff --git a/buch/papers/0f1/teil3.tex b/buch/papers/0f1/teil3.tex index b6c0f4f..2afc34b 100644 --- a/buch/papers/0f1/teil3.tex +++ b/buch/papers/0f1/teil3.tex @@ -15,9 +15,9 @@ Ebenso kann festgestellt werden, dass je grösser der Wert $z$ in $\mathstrut_0F \label{0f1:subsection:konvergenz}} Es zeigt sich in Abbildung \ref{0f1:ausblick:plot:airy:konvergenz}, dass nach drei Iterationen ($k = 3$) die Funktionen genaue Resultate im Bereich von $-2$ bis $2$ liefert. Ebenso kann festgestellt werden, dass der Kettenbruch schneller konvergiert und im positiven Bereich mit der Referenzfunktion $\operatorname{Ai}(x)$ übereinstimmt. Da die Rekursionsformel eine Abwandlung des Kettenbruches ist, verhalten sich die Funktionen in diesem Fall gleich. -Erst wenn mehrerer Iterationen gerechnet werden, ist wie Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, der Kettenbruch den anderen zwei Algorithmen bezüglich Konvergenz überlegen. Allerdings muss beachtet werden, dass die Rekursionsformel zwar erst nach 35 Approximationen gänzlich konvergiert, nach 27 Iterationen sich nicht mehr gross verändert. +Erst wenn mehrerer Iterationen gerechnet werden, ist wie Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, der Kettenbruch den anderen zwei Algorithmen bezüglich Konvergenz überlegen. Bei der Rekursionsformel muss beachtet werden, dass sie zwar erst nach 35 Approximationen gänzlich konvergiert, allerdings nach 27 Iterationen sich nicht mehr gross verändert. -Ist $z$ negativ wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu aufgrund des Vorzeichens zu alternierenden Termen. So steigt bei allen Algorithmen zuerst die Differenz zum erwarteten Endwert. Erst nach genügend Iterationen sind die Terme so klein, dass sie das Endresultat nicht mehr signifikant beeinflussen. Während die Potenzreihe zusammen mit dem Kettenbruch nach 34 Approximationen konvergiert, braucht die Rekursionsformel noch zwei Iterationen mehr. Wohingegen die Rekursionsformel der genauste Algorithmus im negativen Bereich ist. Da der Computer mit einer relativen Genauigkeit von $10^{-15}$ rechnet, ist dies das Maximum an Präzision, dass erreicht werden kann. +Ist $z$ negativ wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu aufgrund des Vorzeichens zu alternierenden Termen. So steigt bei allen Algorithmen zuerst die Differenz zum erwarteten Endwert. Erst nach genügend Iterationen sind die Terme so klein, dass sie das Endresultat nicht mehr signifikant beeinflussen. Während die Potenzreihe zusammen mit dem Kettenbruch nach 34 Approximationen konvergiert, braucht die Rekursionsformel noch zwei Iterationen mehr. \subsection{Stabilität @@ -39,14 +39,14 @@ Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Gru \begin{figure} \centering \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzPositiv.pdf} - \caption{Konvergenz mit positivem $z$; Logarithmisch dargestellte absoluter Fehler. + \caption{Konvergenz mit positivem $z$; Logarithmisch dargestellter absoluter Fehler. \label{0f1:ausblick:plot:konvergenz:positiv}} \end{figure} \begin{figure} \centering \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzNegativ.pdf} - \caption{Konvergenz mit negativem $z$; Logarithmisch dargestellte absoluter Fehler. + \caption{Konvergenz mit negativem $z$; Logarithmisch dargestellter absoluter Fehler. \label{0f1:ausblick:plot:konvergenz:negativ}} \end{figure} -- cgit v1.2.1 From 46f7c56ca3b5cb512ed2b82beefeb2057af0d8cf Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Wed, 17 Aug 2022 13:30:40 +0200 Subject: Corrected formatting errors in fourier example. --- buch/papers/sturmliouville/waermeleitung_beispiel.tex | 16 ++++++++-------- 1 file changed, 8 insertions(+), 8 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 7a37b2b..a72c562 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -20,7 +20,7 @@ die partielle Differentialgleichung \frac{\partial u}{\partial t} = \kappa \frac{\partial^{2}u}{{\partial x}^{2}}, \end{equation} -wobei der Stab in diesem Fall auf der X-Achse im Intervall $[0,l]$ liegt. +wobei der Stab in diesem Fall auf der $X$-Achse im Intervall $[0,l]$ liegt. Da diese Differentialgleichung das Problem allgemein für einen homogenen Stab beschreibt, werden zusätzliche Bedingungen benötigt, um beispielsweise @@ -35,7 +35,7 @@ Tempreatur gehalten werden. Die Enden des Stabes auf konstanter Temperatur zu halten bedeutet, dass die Lösungsfunktion $u(t,x)$ bei $x = 0$ und $x = l$ nur die vorgegebene Temperatur zurückgeben darf. Diese wird einfachheitshalber als $0$ angenomen. -Es folgen nun +Es folgt nun \begin{equation} \label{sturmliouville:eq:example-fourier-boundary-condition-ends-constant} u(t,0) @@ -52,7 +52,7 @@ als Randbedingungen. \subsubsection{Randbedingungen für Stab mit isolierten Enden} -Bei isolierten Enden des Stabes können belibige Temperaturen für $x = 0$ und +Bei isolierten Enden des Stabes können beliebige Temperaturen für $x = 0$ und $x = l$ auftreten. In diesem Fall ist es nicht erlaubt, dass Wärme vom Stab an die Umgebung oder von der Umgebung an den Stab abgegeben wird. @@ -187,7 +187,7 @@ somit auch zu orthogonalen Lösungen führen. % Lösung von X(x), Teil mu % -\subsubsection{Lösund der Differentialgleichung in x} +\subsubsection{Lösund der Differentialgleichung in $x$} Als erstes wird auf die Gleichung~\eqref{sturmliouville:eq:example-fourier-separated-x} eingegangen. Aufgrund der Struktur der Gleichung @@ -473,7 +473,7 @@ berechnet: \\ 2\int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx =& - a_0 \int_{-l}^{l}cos\left(\frac{m \pi}{l}x\right) dx + a_0 \int_{-l}^{l}\cos\left(\frac{m \pi}{l}x\right) dx + \sum_{n = 1}^{\infty}\left[a_n\int_{-l}^{l}\cos\left(\frac{n\pi}{l}x\right) \cos\left(\frac{m \pi}{l}x\right)dx\right] @@ -487,7 +487,7 @@ berechnet: Betrachtet man nun die Summanden auf der rechten Seite stellt man fest, dass nahezu alle Terme verschwinden, denn \[ - \int_{-l}^{l}cos\left(\frac{m \pi}{l}x\right) dx + \int_{-l}^{l}\cos\left(\frac{m \pi}{l}x\right) dx = 0, \] @@ -528,10 +528,10 @@ mit $u = \frac{m \pi}{l}x$ substituiert wird: \frac{\sin\left(2u\right)}{4}\right]_{u=-m\pi}^{m\pi} \\ &= - a_m\frac{l}{m\pi}\left(\frac{m\pi}{2} + + a_m\frac{l}{m\pi}\biggl(\frac{m\pi}{2} + \underbrace{\frac{\sin\left(2m\pi\right)}{4}}_{\displaystyle = 0} - \frac{-m\pi}{2} - - \underbrace{\frac{\sin\left(-2m\pi\right)}{4}}_{\displaystyle = 0}\right) + \underbrace{\frac{\sin\left(-2m\pi\right)}{4}}_{\displaystyle = 0}\biggr) \\ &= a_m l -- cgit v1.2.1 From c06cb44760c3258ab6d5d30451283371b4881346 Mon Sep 17 00:00:00 2001 From: Nao Pross Date: Wed, 17 Aug 2022 14:40:44 +0200 Subject: kugel: Add reference --- buch/papers/kugel/references.bib | 11 +++++++++++ 1 file changed, 11 insertions(+) (limited to 'buch/papers') diff --git a/buch/papers/kugel/references.bib b/buch/papers/kugel/references.bib index b74c5cd..e5d6452 100644 --- a/buch/papers/kugel/references.bib +++ b/buch/papers/kugel/references.bib @@ -192,4 +192,15 @@ Created by Henry Reich}, urldate = {2022-08-01}, date = {2022}, file = {Metric Spaces\: Completeness:/Users/npross/Zotero/storage/5JYEE8NF/completeness.html:text/html}, +} + +@book{bell_special_2004, + location = {Mineola, {NY}}, + title = {Special functions for scientists and engineers}, + isbn = {978-0-486-43521-3}, + series = {Dover books on mathematics}, + pagetotal = {247}, + publisher = {Dover Publ}, + author = {Bell, William Wallace}, + date = {2004}, } \ No newline at end of file -- cgit v1.2.1 From 030aa1f0d5bb3020c909ff7cedd102ea5ff69927 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Erik=20L=C3=B6ffler?= Date: Wed, 17 Aug 2022 15:47:34 +0200 Subject: Revised solution properties section. --- buch/papers/sturmliouville/eigenschaften.tex | 32 ++++++++++++++++------------ 1 file changed, 18 insertions(+), 14 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex index 85f0bf3..bef8a39 100644 --- a/buch/papers/sturmliouville/eigenschaften.tex +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -37,31 +37,35 @@ für die Lösungen des Sturm-Liouville-Problems zur Folge hat. \subsubsection{Exkurs zum Spektralsatz} -Um zu verstehen was für Eigenschaften der selbstadjungierte Operator $L_0$ in +Um zu verstehen welche Eigenschaften der selbstadjungierte Operator $L_0$ in den Lösungen hervorbringt, wird der Spektralsatz benötigt. Dieser wird in der linearen Algebra oft verwendet um zu zeigen, dass eine Matrix diagonalisierbar ist, beziehungsweise dass eine Orthonormalbasis existiert. -Dazu wird zunächst gezeigt, dass eine gegebene $n\times n$-Matrix $A$ aus einem -endlichdimensionalem $\mathbb{K}$-Vektorraum selbstadjungiert ist, also dass + +Im Fall einer gegebenen $n\times n$-Matrix $A$ mit reellen Einträgen wird dazu +zunächst gezeigt, dass $A$ selbstadjungiert ist, also dass \[ \langle Av, w \rangle = \langle v, Aw \rangle \] -für $ v, w \in \mathbb{K}^n$ gilt. -Ist dies der Fall, folgt direkt, dass $A$ auch normal ist. -Dann wird die Aussage des Spektralsatzes -\cite{sturmliouville:spektralsatz-wiki} verwended, welche besagt, dass für -Endomorphismen genau dann eine Orthonormalbasis aus Eigenvektoren existiert, -wenn sie normal sind und nur Eigenwerte aus $\mathbb{K}$ besitzten. +für $ v, w \in \mathbb{R}^n$ gilt. +Ist dies der Fall, kann die Aussage des Spektralsatzes +\cite{sturmliouville:spektralsatz-wiki} verwended werden. +Daraus folgt dann, dass eine Orthonormalbasis aus Eigenvektoren existiert, +wenn $A$ nur Eigenwerte aus $\mathbb{R}$ besitzt. Dies ist allerdings nicht die Einzige Version des Spektralsatzes. -Unter anderen gibt es den Spektralsatz für kompakte Operatoren -\cite{sturmliouville:spektralsatz-wiki}. -Dieser besagt, dass wenn ein linearer kompakter Operator in -$\mathbb{R}$ selbstadjungiert ist, ein (eventuell endliches) -Orthonormalsystem existiert. +Unter anderen gibt es den Spektralsatz für kompakte Operatoren +\cite{sturmliouville:spektralsatz-wiki}, welcher für das +Sturm-Liouville-Problem von Bedeutung ist. +Welche Voraussetzungen erfüllt sein müssen, um diese Version des +Satzes verwenden zu können, wird hier aber nicht diskutiert und kann bei den +Beispielen in diesem Kapitel als gegeben betrachtet werden. +Grundsätzlich ist die Aussage in dieser Version dieselbe, wie bei den Matrizen, +also dass für ein Operator eine Orthonormalbasis aus Eigenvektoren existiert, +falls er selbstadjungiert ist. \subsubsection{Anwendung des Spektralsatzes auf $L_0$} -- cgit v1.2.1 From 88cc0ae20e53d67b671f0713a7921c18736bdf3e Mon Sep 17 00:00:00 2001 From: Nao Pross Date: Wed, 17 Aug 2022 16:12:40 +0200 Subject: kugel: fix figures makefile, add curvature-1d --- buch/papers/kugel/figures/tikz/Makefile | 12 + buch/papers/kugel/figures/tikz/curvature-1d.dat | 500 ++++++++++++++++++++++++ buch/papers/kugel/figures/tikz/curvature-1d.pdf | Bin 0 -> 15387 bytes buch/papers/kugel/figures/tikz/curvature-1d.py | 32 ++ buch/papers/kugel/figures/tikz/curvature-1d.tex | 21 + 5 files changed, 565 insertions(+) create mode 100644 buch/papers/kugel/figures/tikz/Makefile create mode 100644 buch/papers/kugel/figures/tikz/curvature-1d.dat create mode 100644 buch/papers/kugel/figures/tikz/curvature-1d.pdf create mode 100644 buch/papers/kugel/figures/tikz/curvature-1d.py create mode 100644 buch/papers/kugel/figures/tikz/curvature-1d.tex (limited to 'buch/papers') diff --git a/buch/papers/kugel/figures/tikz/Makefile b/buch/papers/kugel/figures/tikz/Makefile new file mode 100644 index 0000000..4ec4e5a --- /dev/null +++ b/buch/papers/kugel/figures/tikz/Makefile @@ -0,0 +1,12 @@ +FIGURES := spherical-coordinates.pdf curvature-1d.pdf + +all: $(FIGURES) + +%.pdf: %.tex + pdflatex $< + +curvature-1d.pdf: curvature-1d.tex curvature-1d.dat + pdflatex curvature-1d.tex + +curvature-1d.dat: curvature-1d.py + python3 $< diff --git a/buch/papers/kugel/figures/tikz/curvature-1d.dat b/buch/papers/kugel/figures/tikz/curvature-1d.dat new file mode 100644 index 0000000..6622398 --- /dev/null +++ b/buch/papers/kugel/figures/tikz/curvature-1d.dat @@ -0,0 +1,500 @@ +0.000000000000000000e+00 1.000000000000000000e+00 5.000007286987066095e+02 +2.004008016032064049e-02 1.025056790958151831e+00 4.899813296957295279e+02 +4.008016032064128098e-02 1.050121598724190752e+00 4.799659543969494848e+02 +6.012024048096192147e-02 1.075186379145250948e+00 4.699586248906784931e+02 +8.016032064128256196e-02 1.100243094530689136e+00 4.599633600341078932e+02 +1.002004008016031955e-01 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4.955978889110630448e+00 7.720120875228209343e+02 diff --git a/buch/papers/kugel/figures/tikz/curvature-1d.pdf b/buch/papers/kugel/figures/tikz/curvature-1d.pdf new file mode 100644 index 0000000..6425af6 Binary files /dev/null and b/buch/papers/kugel/figures/tikz/curvature-1d.pdf differ diff --git a/buch/papers/kugel/figures/tikz/curvature-1d.py b/buch/papers/kugel/figures/tikz/curvature-1d.py new file mode 100644 index 0000000..4710fc8 --- /dev/null +++ b/buch/papers/kugel/figures/tikz/curvature-1d.py @@ -0,0 +1,32 @@ +import numpy as np +import matplotlib.pyplot as plt + + +@np.vectorize +def fn(x): + return (x ** 2) * 2 / 100 + (1 + x / 4) + np.sin(x) + +@np.vectorize +def ddfn(x): + return 2 * 5 / 100 - np.sin(x) + +x = np.linspace(0, 10, 500) +y = fn(x) +ddy = ddfn(x) + +cmap = ddy - np.min(ddy) +cmap = cmap * 1000 / np.max(cmap) + +plt.plot(x, y) +plt.plot(x, ddy) +# plt.plot(x, cmap) + +plt.show() + +fname = "curvature-1d.dat" +np.savetxt(fname, np.array([x, y, cmap]).T, delimiter=" ") + +# with open(fname, "w") as f: +# # f.write("x y cmap\n") +# for xv, yv, cv in zip(x, y, cmap): +# f.write(f"{xv} {yv} {cv}\n") diff --git a/buch/papers/kugel/figures/tikz/curvature-1d.tex b/buch/papers/kugel/figures/tikz/curvature-1d.tex new file mode 100644 index 0000000..6983fb0 --- /dev/null +++ b/buch/papers/kugel/figures/tikz/curvature-1d.tex @@ -0,0 +1,21 @@ +% vim:ts=2 sw=2 et: +\documentclass[tikz, border=5mm]{standalone} +\usepackage{pgfplots} + +\begin{document} +\begin{tikzpicture} + \begin{axis}[ + clip = false, + width = 8cm, height = 6cm, + xtick = \empty, ytick = \empty, + colormap name = viridis, + axis lines = middle, + axis line style = {ultra thick, -latex} + ] + \addplot+[ + smooth, mark=none, line width = 3pt, mesh, + point meta=explicit, + ] file {curvature-1d.dat}; + \end{axis} +\end{tikzpicture} +\end{document} -- cgit v1.2.1 From 7e51ae842c61ba338aec179d71fab2d041ebe8c5 Mon Sep 17 00:00:00 2001 From: Nao Pross Date: Wed, 17 Aug 2022 16:29:41 +0200 Subject: kugel: Review manu's text, improve legendre functions --- buch/papers/kugel/main.tex | 1 + buch/papers/kugel/proofs.tex | 245 +++++++++++++++++ buch/papers/kugel/spherical-harmonics.tex | 424 ++++++++++-------------------- 3 files changed, 385 insertions(+), 285 deletions(-) create mode 100644 buch/papers/kugel/proofs.tex (limited to 'buch/papers') diff --git a/buch/papers/kugel/main.tex b/buch/papers/kugel/main.tex index a281cae..ad19178 100644 --- a/buch/papers/kugel/main.tex +++ b/buch/papers/kugel/main.tex @@ -14,6 +14,7 @@ % \input{papers/kugel/preliminaries} \input{papers/kugel/spherical-harmonics} \input{papers/kugel/applications} +\input{papers/kugel/proofs} \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/kugel/proofs.tex b/buch/papers/kugel/proofs.tex new file mode 100644 index 0000000..143caa8 --- /dev/null +++ b/buch/papers/kugel/proofs.tex @@ -0,0 +1,245 @@ +% vim:ts=2 sw=2 et spell tw=80: +\section{Proofs} + +\subsection{Legendre Functions} \label{kugel:sec:proofs:legendre} + +\kugeltodo{Fix theorem numbers to match, review text.} + +\begin{lemma} + The polynomial function + \begin{align*} + y_n(x)&=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{(2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k}\\ + &= \frac{1}{n!2^n}\frac{d^n}{dx^n}(1-x^2)^n =: P_n(x), + \end{align*} + is a solution to the second order differential equation + \begin{equation}\label{kugel:eq:sol_leg} + (1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx} + n(n+1)y=0, \quad \forall n>0. + \end{equation} +\end{lemma} +\begin{proof} + In order to find a solution to Eq.\eqref{eq:legendre}, the following Ansatz can be performed: + \begin{equation}\label{eq:ansatz} + y(x) = \sum_{k=0}^\infty a_k x^k. + \end{equation} + Given Eq.\eqref{eq:ansatz}, then + \begin{align*} + \frac{dy}{dx} &= \sum_{k=0}^\infty k a_k x^{k-1}, \\ + \frac{d^2y}{dx^2} &= \sum_{k=0}^\infty k (k-1) a_k x^{k-2}. + \end{align*} + Eq.\eqref{eq:legendre} can be therefore written as + \begin{align} + &(1-x^2)\sum_{k=0}^\infty k (k-1) a_k x^{k-2} - 2x\sum_{k=0}^\infty k a_k x^{k-1} + n(n+1)\sum_{k=0}^\infty a_k x^k=0 \label{eq:ansatz_in_legendre} \\ + &=\sum_{k=0}^\infty k (k-1) a_k x^{k-2} - \sum_{k=0}^\infty k (k-1) a_k x^{k} - 2x\sum_{k=0}^\infty k a_k x^{k-1} + n(n+1)\sum_{k=0}^\infty a_k x^k=0. \nonumber + \end{align} + If one consider the term + \begin{equation}\label{eq:term} + \sum_{k=0}^\infty k (k-1) a_k x^{k-2}, + \end{equation} + the substitution $\tilde{k}=k-2$ yields Eq.\eqref{eq:term} to + \begin{equation*} + \sum_{\tilde{k}=-2}^\infty (\tilde{k}+2) (\tilde{k}+1) a_{\tilde{k}+2} x^{\tilde{k}}=\sum_{\tilde{k}=0}^\infty (\tilde{k}+2) (\tilde{k}+1) a_{\tilde{k}} x^{\tilde{k}}. + \end{equation*} + This means that Eq.\eqref{eq:ansatz_in_legendre} becomes + \begin{align} + &\sum_{k=0}^\infty (k+1)(k+2) a_{k+2} x^{k} - \sum_{k=0}^\infty k (k-1) a_k x^{k} - 2\sum_{k=0}^\infty k a_k x^k + n(n+1)\sum_{k=0}^\infty a_k x^k \nonumber \\ + = &\sum_{k=0}^\infty \big[ (k+1)(k+2) a_{k+2} - k (k-1) a_k - 2 k a_k + n(n+1) a_k \big] x^k \stackrel{!}{=} 0. \label{eq:condition} + \end{align} + The condition in Eq.\eqref{eq:condition} is equivalent to + \begin{equation}\label{eq:condition_2} + (k+1)(k+2) a_{k+2} - k (k-1) a_k - 2 k a_k + n(n+1) a_k = 0. + \end{equation} + We can derive a recursion formula for $a_{k+2}$ from Eq.\eqref{eq:condition_2}, which can be expressed as + \begin{equation}\label{eq:recursion} + a_{k+2}= \frac{k (k-1) - 2 k + n(n+1)}{(k+1)(k+2)}a_k = \frac{(k-n)(k+n+1)}{(k+2)(k+1)}a_k. + \end{equation} + All coefficients can be calculated using the latter. + + Following Eq.\eqref{eq:recursion}, if we want to compute $a_6$ we would have + \begin{align*} + a_{6}= -\frac{(n-4)(n+5)}{6\cdot 5}a_4 &= -\frac{(n-4)(5+n)}{6 \cdot 5} -\frac{(n-2)(n+3)}{4 \cdot 3} a_2 \\ + &= -\frac{(n-4)(n+5)}{6 \cdot 5} -\frac{(n-2)(n+3)}{4 \cdot 3} -\frac{n(n+1)}{2 \cdot 1} a_0 \\ + &= -\frac{(n+5)(n+3)(n+1)n(n-2)(n-4)}{6!} a_0. + \end{align*} + One can generalize this relation for the $i^\text{th}$ even coefficient as + \begin{equation*} + a_{2k} = (-1)^k \frac{(n+(2k-1))(n+(2k-1)-2)\hdots (n-(2k-2)+2)(n-(2k-2))}{(2k)!}a_0 + \end{equation*} + where $i=2k$. + + A similar expression can be written for the odd coefficients $a_{2k-1}$. In this case, the equation starts from $a_1$ and to find the pattern we can write the recursion for an odd coefficient, $a_7$ for example + \begin{align*} + a_{7}= -\frac{(n-5)(n+6)}{7\cdot 6}a_5 &= - \frac{(n-5)(n+6)}{7\cdot 6} -\frac{(n-3)(n+4)}{5 \cdot 4} a_3 \\ + &= - \frac{(n-5)(n+6)}{7\cdot 6} -\frac{(n-3)(n+4)}{5 \cdot 4} -\frac{(n-1)(n+2)}{3 \cdot 2} a_1 \\ + &= -\frac{(n+6)(n+4)(n+2)(n-1)(n-3)(n-5)}{7!} a_1. + \end{align*} + As before, we can generalize this equation for the $i^\text{th}$ odd coefficient + \begin{equation*} + a_{2k+1} = (-1)^k \frac{(n + 2k)(n+2k-2)\hdots(n-(2k-1)+2)(n-(2k-1))}{(2k+1)!}a_1 + \end{equation*} + where $i=2k+1$. + + Let be + \begin{align*} + y_\text{e}^K(x) &:= \sum_{k=0}^K(-1)^k \frac{(n+(2k-1))(n+(2k-1)-2)\hdots \color{red}(n-(2k-2)+2)(n-(2k-2))}{(2k)!} x^{2k}, \\ + y_\text{o}^K(x) &:= \sum_{k=0}^K(-1)^k \frac{(n + 2k)(n+2k-2)\hdots \color{blue} (n-(2k-1)+2)(n-(2k-1))}{(2k+1)!} x^{2k+1}. + \end{align*} + The solution to the Eq.\eqref{eq:legendre} can be written as + \begin{equation}\label{eq:solution} + y(x) = \lim_{K \to \infty} \left[ a_0 y_\text{e}^K(x) + a_1 y_\text{o}^K(x) \right]. + \end{equation} + + The colored parts can be analyzed separately: + \begin{itemize} + \item[\textcolor{red}{\textbullet}] Suppose that $n=n_0$ is an even number. Then the red part, for a specific value of $k=k_0$, will follow the following relation: + \begin{equation*} + n_0-(2k_0-2)=0. + \end{equation*} + From that point on, given the recursive nature of Eq.\eqref{eq:recursion}, all the subsequent coefficients will also be 0, making the sum finite. + \begin{equation*} + a_{2k}=0 \iff y_{\text{o}}^{2k}(x)=y_{\text{o}}^{2k_0}(x), \quad \forall k>k_0 + \end{equation*} + \item[\textcolor{blue}{\textbullet}] Suppose that $n=n_0$ is an odd number. Then the blue part, for a specific value of $k=k_0$, will follow the following relation + \begin{equation*} + n_0-(2k_0-1)=0. + \end{equation*} + From that point on, for the same reason as before, all the subsequent coefficients will also be 0, making the sum finite. + \begin{equation*} + a_{2k+1}=0 \iff y_{\text{o}}^{2k+1}(x)=y_{\text{o}}^{2k_0+1}(x), \quad \forall k>k_0 + \end{equation*} + \end{itemize} + + There is the possibility of expressing the solution in Eq.\eqref{eq:solution} in a more compact form, combining the two solutions $y_\text{o}^K(x)$ and $y_\text{e}^K(x)$. They are both a polynomial of maximum degree $n$, assuming $n \in \mathbb{N}$. In the case where $n$ is even, the polynomial solution + \begin{equation*} + \lim_{K\to \infty} y_\text{e}^K(x) + \end{equation*} + will be a finite sum. If instead $n$ is odd, will be + \begin{equation*} + \lim_{K\to \infty} y_\text{o}^K(x) + \end{equation*} + to be a finite sum. + + Depending on the coefficient we start with, $a_1$ or $a_0$, we will obtain the odd or even polynomial respectively. Starting with the last coefficient $a_n$ and, recursively, calculating all the others in descending order, we can express the two parts $y_\text{o}^K(x)$ and $y_\text{e}^K(x)$ with a single sum. Hence, because we start with the last coefficient, the choice concerning $a_1$ and $a_0$ will be at the end of the sum, and not at the beginning. To compact Eq.\eqref{eq:solution}, Eq.\eqref{eq:recursion} can be reconsidered to calculate the coefficient $a_{k-2}$, using $a_k$ + \begin{equation*} + a_{k-2} = -\frac{(k+2)(k+1)}{(k-n)(k+n+1)}a_k + \end{equation*} + Now the game is to find a pattern, as before. Remember that $n$ is a fixed parameter of Eq.\eqref{eq:legendre}. + \begin{align*} + a_{n-2} &= -\frac{n(n-1)}{2(2n-1)}a_n, \\ + a_{n-4} &= -\frac{(n-2)(n-3)}{4(2n-3)}a_{n-2} \\ + &= -\frac{(n-2)(n-3)}{4(2n-3)}-\frac{n(n-1)}{2(2n-1)}a_n. + \end{align*} + In general + \begin{equation}\label{eq:general_recursion} + a_{n-2k} = (-1)^k \frac{n(n-1)(n-2)(n-3) \hdots (n-2k+1)}{2\cdot4\hdots 2k(2n-1)(2n-3)\hdots(2n-2k+1)}a_n + \end{equation} + The whole solution can now be written as + \begin{align} + y(x) &= a_n x^n + a_{n-2} x^{n-2} + a_{n-4} x^{n-4} + a_{n-6} x^{n-6} + \hdots + \begin{cases} + a_1 x, \quad &\text{if } n \text{ odd} \\ + a_0, \quad &\text{if } n \text{ even} + \end{cases} \nonumber \\ + &= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} a_{n-2k}x^{n-2k} \label{eq:solution_2} + \end{align} + By considering + \begin{align} + (2n-1)(2n-3)\hdots (2n-2k+1)&=\frac{2n(2n-1)(2n-2)(2n-3)\hdots(2n-2k+1)} + {2n(2n-2)(2n-4)(2n-6)\hdots(2n-2k+2)} \nonumber \\ + &=\frac{\frac{(2n)!}{(2n-2k)!}} + {2^kn(n-1)(n-2)(n-3)\hdots(n-k+1)} \nonumber \\ + &=\frac{\frac{(2n)!}{(2n-2k)!}} + {2^k\frac{n!}{(n-k)!}}=\frac{(n-k)!(2n)!}{n!(2n-2k)!2^k} \label{eq:1_sub_recursion}, \\ + 2 \cdot 4 \hdots 2k &= 2^r 1\cdot2 \hdots r = 2^r r!\label{eq:2_sub_recursion}, \\ + n(n-1)(n-2)(n-3) \hdots (n-2k+1) &= \frac{n!}{(n-2k)!}\label{eq:3_sub_recursion}. + \end{align} + Eq.\eqref{eq:solution_2} can be rewritten as + \begin{equation}\label{eq:solution_3} + y(x)=a_n \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{n!^2(2n-2k)!}{k!(n-2k)!(n-k)!(2n)!} x^{n-2k}. + \end{equation} + Eq.\eqref{eq:solution_3} is defined for any $a_n$. By letting $a_n$ be declared as + \begin{equation*} + a_{n} := \frac{(2n)!}{2^n n!^2}, + \end{equation*} + the so called \emph{Legendre polynomial} emerges + \begin{equation}\label{eq:leg_poly} + P_n(x):=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{(2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k} + \end{equation} +\end{proof} + + +\begin{lemma} + If $Z_n(z)$ is a solution of the Legendre equation \eqref{kugel:eqn:legendre}, + then + \begin{equation*} + P^m_n(z) = (1 - z^2)^{m/2} \frac{d^m}{dz^m}Z_n(z) + \end{equation*} + solves the associated Legendre equation \eqref{kugel:eqn:associated-legendre}. +\end{lemma} +% \begin{proof} [TODO: modificare la $m$ (è già usata come costante di separazione) o forse è giusta (?)] +\begin{proof} + To begin, we can start by differentiating $m$ times Eq.\eqref{kugel:eq:leg_eq} (which is staisfied by $y(x)$), obtaining + \begin{equation}\label{eq:lagrange_mderiv} + \frac{d^m}{dx^m}\left[ (1-x^2)\frac{d^2y}{dx^2} \right] -2 \frac{d^m}{dx^m}\left[ x\frac{dy}{dx} \right] + n(n+1)\frac{d^m}{dx^m}y=0. + \end{equation} + \emph{Leibniz's theorem} says, that if we want to differentiate $m$ times a multiplication of two functions, we can use the binomial coefficients to build up a sum. This allows us to be more compact, obtaining + \begin{equation}\label{eq:leibniz} + \frac{d^m}{dx^m}[u(x)v(x)] = \sum_{i=0}^m \binom{n}{i} \frac{d^{m-i}u}{dx^{m-1}} \frac{d^{i}v}{dx^i}. + \end{equation} + Using Eq.\eqref{eq:leibniz} in Eq.\eqref{eq:lagrange_mderiv}, we have + \begin{align} + (1-x^2)\frac{d^{m+2}y}{dx^{m+2}} &+ m \frac{d}{dx}(1-x^2)\frac{d^{m+1}y}{dx^{m+1}} + \frac{m(m-1)}{2}\frac{d^{2}}{dx^{2}}(1-x^2)\frac{d^{m}y}{dx^{m}} + n(n+1)\frac{d^m{}y}{dx^{m}} \nonumber \\ + &-2\left(x\frac{d^{m+1}y}{dx^{m+1}} + m\frac{d}{dx}x\frac{d^{m}y}{dx^{m}} \right) \nonumber \\ + &= (1-x^2)\frac{d^{m+2}y}{dx^{m+2}} -2x(m+1)\frac{d^{m+1}y}{dx^{m+1}}+(n(n+1)-m(m-1)-2m)\frac{d^{m}y}{dx^{m}}=0. \label{eq:aux_3} + \end{align} + To make the notation easier to follow, a new function can be defined + \begin{equation*} + \frac{d^{m}y}{dx^{m}} := y_m. + \end{equation*} + Eq.\eqref{eq:aux_3} now becomes + \begin{equation}\label{eq:1st_subs} + (1-x^2)\frac{d^{2}y_m}{dx^{2}} -2x(m+1)\frac{dy_m}{dx}+(n(n+1)-m(m+1))y_m=0 + \end{equation} + A second function can be further defined as + \begin{equation*} + (1-x^2)^{\frac{m}{2}}\frac{d^{m}y}{dx^{m}} = (1-x^2)^{\frac{m}{2}}y_m := \hat{y}_m, + \end{equation*} + allowing to write Eq.\eqref{eq:1st_subs} as + \begin{equation}\label{eq:2st_subs} + (1-x^2)\frac{d^2}{dx^2}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] -2(m+1)x\frac{d}{dx}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] + (n(n+1)-m(m+1))\hat{y}_m(1-x^2)^{-\frac{m}{2}}=0. + \end{equation} + The goal now is to compute the two terms + \begin{align*} + \frac{d^2}{dx^2}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] &= \frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} + \frac{d\hat{y}_m}{dx}\frac{m}{2}(1-x^2)^{-\frac{m}{2}-1}2x \\ + &+ m\left( \frac{d\hat{y}_m}{dx} x (1-x^2)^{-\frac{m}{2}-1} + \hat{y}_m (1-x^2)^{-\frac{m}{2}-1} - \hat{y}_m x (-\frac{m}{2}-1)(1-x^2)^{-\frac{m}{2}} 2x\right) \\ + &= \frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} + \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-\frac{m}{2}-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-\frac{m}{2}-1}\\ + &+ m\hat{y}_m (1-x^2)^{-\frac{m}{2}-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-\frac{m}{2}-2} + \end{align*} + and + \begin{align*} + \frac{d}{dx}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] &= \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_m\frac{m}{2}(1-x^2)^{-\frac{m}{2}-1}2x \\ + &= \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_mm(1-x^2)^{-\frac{m}{2}-1}x, + \end{align*} + to use them in Eq.\eqref{eq:2st_subs}, obtaining + \begin{align*} + (1-x^2)\biggl[\frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} &+ \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-\frac{m}{2}-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-\frac{m}{2}-1} \\ + &+ m\hat{y}_m (1-x^2)^{-\frac{m}{2}-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-\frac{m}{2}-2}\biggr] \\ + &-2(m+1)x\left[ \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_mm(1-x^2)^{-\frac{m}{2}-1}x \right] \\ + &+ (n(n+1)-m(m+1))\hat{y}_m(1-x^2)^{-\frac{m}{2}}=0.\\ + \end{align*} + We can now divide by $(1-x^2)^{-\frac{m}{2}}$, obtaining + \begin{align*} + (1-x^2)\biggl[\frac{d^2\hat{y}_m}{dx^2} &+ \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-1} + m\hat{y}_m (1-x^2)^{-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-2}\biggr] \\ + &-2(m+1)x\left[ \frac{d\hat{y}_m}{dx} + \hat{y}_mm(1-x^2)^{-1}x \right] + (n(n+1)-m(m+1))\hat{y}_m\\ + &= \frac{d^2\hat{y}_m}{dx^2} + \frac{d\hat{y}_m}{dx}mx + m\frac{d\hat{y}_m}{dx}x + m\hat{y}_m + m\hat{y}_m x^2(m+2)(1-x^2)^{-1} \\ + &-2(m+1)x\left[ \frac{d\hat{y}_m}{dx} + \hat{y}_mm(1-x^2)^{-1}x \right] + (n(n+1)-m(m+1))\hat{y}_m\\ + \end{align*} + and collecting some terms + \begin{equation*} + (1-x^2)\frac{d^2\hat{y}_m}{dx^2} - 2x\frac{d\hat{y}_m}{dx} + \left( -x^2 \frac{m^2}{1-x^2} + m+n(n+1)-m(m+1)\right)\hat{y}_m=0. + \end{equation*} + Showing that + \begin{align*} + -x^2 \frac{m^2}{1-x^2} + m+n(n+1)-m(m+1) &= n(n+1)- m^2 -x^2 \frac{m^2}{1-x^2} \\ + &= n(n+1)- \frac{m}{1-x^2} + \end{align*} + implies $\hat{y}_m(x)$ being a solution of Eq.\eqref{kugel:eq:associated_leg_eq} +\end{proof} diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex index 70657c9..5645941 100644 --- a/buch/papers/kugel/spherical-harmonics.tex +++ b/buch/papers/kugel/spherical-harmonics.tex @@ -1,6 +1,6 @@ % vim:ts=2 sw=2 et spell tw=80: -\section{Spherical Harmonics} +\section{Construction of the Spherical Harmonics} \if 0 \kugeltodo{Rewrite this section if the preliminaries become an addendum} @@ -111,7 +111,7 @@ that satisfy the equation \surflaplacian f = -\lambda f. \end{equation} Perhaps it may not be obvious at first glance, but we are in fact dealing with a -partial differential equation (PDE). If we unpack the notation of the operator +partial differential equation (PDE) \kugeltodo{Boundary conditions?}. If we unpack the notation of the operator $\nabla^2_{\partial S}$ according to definition \ref{kugel:def:surface-laplacian}, we get: \begin{equation} \label{kugel:eqn:eigen-pde} @@ -126,7 +126,7 @@ Since all functions satisfying \eqref{kugel:eqn:eigen-pde} are the The task may seem very difficult but we can simplify it with a well-known technique: \emph{the separation Ansatz}. It consists in assuming that the function $f(\vartheta, \varphi)$ can be factorized in the following form: -\begin{equation} \label{kugel:eqn:sep-ansatz:0} +\begin{equation} f(\vartheta, \varphi) = \Theta(\vartheta)\Phi(\varphi). \end{equation} In other words, we are saying that the effect of the two independent variables @@ -135,34 +135,34 @@ effect separately. This separation process was already presented in section \ref{buch:pde:section:kugel}, but we will briefly rehearse it here for convenience. If we substitute this assumption in \eqref{kugel:eqn:eigen-pde}, we have: -\begin{equation} \label{kugel:eqn:sep-ansatz:1} +\begin{equation*} \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial \Theta(\vartheta)}{\partial\vartheta} \right) \Phi(\varphi) + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 \Phi(\varphi)}{\partial\varphi^2} \Theta(\vartheta) + \lambda \Theta(\vartheta)\Phi(\varphi) = 0. -\end{equation} +\end{equation*} Dividing by $\Theta(\vartheta)\Phi(\varphi)$ and introducing an auxiliary -variable $m$, the separation constant, yields: +variable $m^2$, the separation constant, yields: \begin{equation*} \frac{1}{\Theta(\vartheta)}\sin \vartheta \frac{d}{d \vartheta} \left( \sin \vartheta \frac{d \Theta}{d \vartheta} \right) + \lambda \sin^2 \vartheta = -\frac{1}{\Phi(\varphi)} \frac{d^2\Phi(\varphi)}{d\varphi^2} - = m, + = m^2, \end{equation*} which is equivalent to the following system of 2 first order differential equations (ODEs): \begin{subequations} \begin{gather} - \frac{d^2\Phi(\varphi)}{d\varphi^2} = -m \Phi(\varphi), + \frac{d^2\Phi(\varphi)}{d\varphi^2} = -m^2 \Phi(\varphi), \label{kugel:eqn:ode-phi} \\ \sin \vartheta \frac{d}{d \vartheta} \left( \sin \vartheta \frac{d \Theta}{d \vartheta} \right) - + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right) + + \left( \lambda - \frac{m^2}{\sin^2 \vartheta} \right) \Theta(\vartheta) = 0 \label{kugel:eqn:ode-theta}. \end{gather} @@ -174,291 +174,141 @@ write the solutions \Phi(\varphi) = e^{i m \varphi}, \quad m \in \mathbb{Z}. \end{equation} The restriction that the separation constant $m$ needs to be an integer arises -from the fact that we require a $2\pi$-periodicity in $\varphi$ since -$\Phi(\varphi + 2\pi) = \Phi(\varphi)$. Unfortunately, solving -\eqref{kugel:eqn:ode-theta} is not so straightforward. Actually it is quite -difficult, and the process is so involved that it will require a dedicated -section of its own. +from the fact that we require a $2\pi$-periodicity in $\varphi$ since the +coordinate systems requires that $\Phi(\varphi + 2\pi) = \Phi(\varphi)$. +Unfortunately, solving \eqref{kugel:eqn:ode-theta} is as straightforward, +actually, it is quite difficult, and the process is so involved that it will +require a dedicated section of its own. \subsection{Legendre Functions} -To solve \eqref{kugel:eqn:ode-theta} -We can begin by considering the substitution $x = \cos \vartheta$. The operator $\frac{d}{d \vartheta}$ will be: -\begin{align*} - \frac{d}{d \vartheta} = \frac{dx}{d \vartheta}\frac{d}{dx} &= -\sin \vartheta \frac{d}{dx} \\ - &= -\sqrt{1-x^2} \frac{d}{dx}. -\end{align*} -Eq.(\ref{kugel:eq:ODE_2}) will then become. +To solve \eqref{kugel:eqn:ode-theta} we start with the substitution $z = \cos +\vartheta$ \kugeltodo{Explain geometric origin with picture}. The operator +$\frac{d}{d \vartheta}$ becomes +\begin{equation*} + \frac{d}{d \vartheta} + = \frac{dz}{d \vartheta}\frac{d}{dz} + = -\sin \vartheta \frac{d}{dz} + = -\sqrt{1-z^2} \frac{d}{dz}, +\end{equation*} +since $\sin \vartheta = \sqrt{1 - \cos^2 \vartheta} = \sqrt{1 - z^2}$, and +then \eqref{kugel:eqn:ode-theta} becomes \begin{align*} - \frac{-\sqrt{1-x^2}}{\sqrt{1-x^2}} \frac{d}{dx} \left( \left(\sqrt{1-x^2}\right) \left(-\sqrt{1-x^2}\right) \frac{d \Theta}{dx} \right) + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)\Theta(\vartheta) &= 0 \\ - \frac{d}{dx} \left( (1-x^2) \frac{d \Theta}{dx} \right) + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)\Theta(\vartheta) &= 0 \\ - (1-x^2)\frac{d^2 \Theta}{dx} - 2x\frac{d \Theta}{dx} + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)\Theta(\vartheta) &= 0 \\ - (1-x^2)\frac{d^2 \Theta}{dx} - 2x\frac{d \Theta}{dx} + \left( \lambda - \frac{m}{1-x^2} \right)\Theta(\vartheta) &= 0 + \frac{-\sqrt{1-z^2}}{\sqrt{1-z^2}} \frac{d}{dz} \left[ + \left(\sqrt{1-z^2}\right) \left(-\sqrt{1-z^2}\right) \frac{d \Theta}{dz} + \right] + + \left( \lambda - \frac{m^2}{1 - z^2} \right)\Theta(\vartheta) &= 0, + \\ + \frac{d}{dz} \left[ (1-z^2) \frac{d \Theta}{dz} \right] + + \left( \lambda - \frac{m^2}{1 - z^2} \right)\Theta(\vartheta) &= 0, + \\ + (1-z^2)\frac{d^2 \Theta}{dz} - 2z\frac{d \Theta}{dz} + + \left( \lambda - \frac{m^2}{1 - z^2} \right)\Theta(\vartheta) &= 0. \end{align*} -By making two final cosmetic substitutions, namely $\Theta(\vartheta)=\Theta(\cos^{-1}x):=y(x)$ and $\lambda=n(n+1)$, we will be able to define the \emph{Associated Legendre Equation} in its standard and most familiar form -\begin{definition}{Associated Legendre Equation} - \begin{equation}\label{kugel:eq:associated_leg_eq} - (1-x^2)\frac{d^2 y}{dx} - 2x\frac{d y}{dx} + \left( n(n+1) - \frac{m}{1-x^2} \right)y(x) = 0. - \end{equation} -\end{definition} -Our new goal then became solving Eq.(\ref{kugel:eq:asssociated_leg_eq}). After that we can fit the solution into Eq.(\ref{kugel:eq:sep_ansatz_0}), obtaining $f(\vartheta, \varphi)$, the solution of the eigenvalue problem. \newline -We simplified the problem somewhat but the task still remains very difficult. We can rely on a lemma to continue but first we need to define an additional equation, namely the \emph{Legendre Equation} -\begin{definition}{Legendre equation}\newline - Setting $m=0$ in Eq.(\ref{kugel:eq:asssociated_leg_eq}), we get - \begin{equation}\label{kugel:eq:leg_eq} - (1-x^2)\frac{d^2 y}{dx} - 2x\frac{d y}{dx} + n(n+1)y(x) = 0, - \end{equation} - also known as \emph{Legendre Equation}. -\end{definition} -Now we can continue with the lemma -\begin{lemma}\label{kugel:lemma_1} - If $y_n(x)$ is a solution of Eq.(\ref{kugel:eq:leg_eq}), then the function - \begin{equation*} - y_{m,n}(x) = (1-x^2)^{\frac{m}{2}}\frac{d^m}{dx^m}y_n(x) - \end{equation*} - satisfies Eq.(\ref{kugel:eq:associated_leg_eq}) +By making two final cosmetic substitutions, namely $Z(z) = \Theta(\cos^{-1}z)$ +and $\lambda = n(n+1)$, we obtain what is known in the literature as the +\emph{associated Legendre equation of order $m$}: +\nocite{olver_introduction_2013} +\begin{equation} \label{kugel:eqn:associated-legendre} + (1 - z^2)\frac{d^2 Z}{dz} + - 2z\frac{d Z}{dz} + + \left( n(n + 1) - \frac{m^2}{1 - z^2} \right) Z(z) = 0, + \quad + z \in [-1; 1], m \in \mathbb{Z}. +\end{equation} + +Our new goal has therefore become to solve +\eqref{kugel:eqn:associated-legendre}, since if we find a solution for $Z(z)$ we +can perform the substitution backwards and get back to our eigenvalue problem. +However, the associated Legendre equation is not any easier, so to attack the +problem we will look for the solutions in the easier special case when $m = 0$. +This reduces the problem because it removes the double pole, which is always +tricky to deal with. In fact, the reduced problem when $m = 0$ is known as the +\emph{Legendre equation}: +\begin{equation} \label{kugel:eqn:legendre} + (1 - z^2)\frac{d^2 Z}{dz} + - 2z\frac{d Z}{dz} + + n(n + 1) Z(z) = 0, + \quad + z \in [-1; 1]. +\end{equation} + +The Legendre equation is a second order differential equation, and therefore it +has 2 independent solutions, which are known as \emph{Legendre functions} of the +first and second kind. For the scope of this text we will only derive a special +case of the former that is known known as the \emph{Legendre polynomials}, since +we only need a solution between $-1$ and $1$. + +\begin{lemma}[Legendre polynomials] + \label{kugel:lem:legendre-poly} + The polynomial function + \[ + P_n(z) = \sum^{\lfloor n/2 \rfloor}_{k=0} + \frac{(-1)^k}{2^n s^k!} \frac{(2n - 2k)!}{(n - k)! (n-2k)!} z^{n - 2k} + \] + is the only finite solution of the Legendre equation + \eqref{kugel:eqn:legendre} when $n \in \mathbb{Z}$ and $z \in [-1; 1]$. \end{lemma} -\begin{proof} [TODO: modificare la $m$ (è già usata come costante di separazione) o forse è giusta (?)] - To begin, we can start by differentiating $m$ times Eq.\eqref{kugel:eq:leg_eq} (which is staisfied by $y(x)$), obtaining - \begin{equation}\label{eq:lagrange_mderiv} - \frac{d^m}{dx^m}\left[ (1-x^2)\frac{d^2y}{dx^2} \right] -2 \frac{d^m}{dx^m}\left[ x\frac{dy}{dx} \right] + n(n+1)\frac{d^m}{dx^m}y=0. - \end{equation} - \emph{Leibniz's theorem} says, that if we want to differentiate $m$ times a multiplication of two functions, we can use the binomial coefficients to build up a sum. This allows us to be more compact, obtaining - \begin{equation}\label{eq:leibniz} - \frac{d^m}{dx^m}[u(x)v(x)] = \sum_{i=0}^m \binom{n}{i} \frac{d^{m-i}u}{dx^{m-1}} \frac{d^{i}v}{dx^i}. - \end{equation} - Using Eq.\eqref{eq:leibniz} in Eq.\eqref{eq:lagrange_mderiv}, we have - \begin{align} - (1-x^2)\frac{d^{m+2}y}{dx^{m+2}} &+ m \frac{d}{dx}(1-x^2)\frac{d^{m+1}y}{dx^{m+1}} + \frac{m(m-1)}{2}\frac{d^{2}}{dx^{2}}(1-x^2)\frac{d^{m}y}{dx^{m}} + n(n+1)\frac{d^m{}y}{dx^{m}} \nonumber \\ - &-2\left(x\frac{d^{m+1}y}{dx^{m+1}} + m\frac{d}{dx}x\frac{d^{m}y}{dx^{m}} \right) \nonumber \\ - &= (1-x^2)\frac{d^{m+2}y}{dx^{m+2}} -2x(m+1)\frac{d^{m+1}y}{dx^{m+1}}+(n(n+1)-m(m-1)-2m)\frac{d^{m}y}{dx^{m}}=0. \label{eq:aux_3} - \end{align} - To make the notation easier to follow, a new function can be defined - \begin{equation*} - \frac{d^{m}y}{dx^{m}} := y_m. - \end{equation*} - Eq.\eqref{eq:aux_3} now becomes - \begin{equation}\label{eq:1st_subs} - (1-x^2)\frac{d^{2}y_m}{dx^{2}} -2x(m+1)\frac{dy_m}{dx}+(n(n+1)-m(m+1))y_m=0 - \end{equation} - A second function can be further defined as - \begin{equation*} - (1-x^2)^{\frac{m}{2}}\frac{d^{m}y}{dx^{m}} = (1-x^2)^{\frac{m}{2}}y_m := \hat{y}_m, - \end{equation*} - allowing to write Eq.\eqref{eq:1st_subs} as - \begin{equation}\label{eq:2st_subs} - (1-x^2)\frac{d^2}{dx^2}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] -2(m+1)x\frac{d}{dx}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] + (n(n+1)-m(m+1))\hat{y}_m(1-x^2)^{-\frac{m}{2}}=0. - \end{equation} - The goal now is to compute the two terms - \begin{align*} - \frac{d^2}{dx^2}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] &= \frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} + \frac{d\hat{y}_m}{dx}\frac{m}{2}(1-x^2)^{-\frac{m}{2}-1}2x \\ - &+ m\left( \frac{d\hat{y}_m}{dx} x (1-x^2)^{-\frac{m}{2}-1} + \hat{y}_m (1-x^2)^{-\frac{m}{2}-1} - \hat{y}_m x (-\frac{m}{2}-1)(1-x^2)^{-\frac{m}{2}} 2x\right) \\ - &= \frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} + \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-\frac{m}{2}-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-\frac{m}{2}-1}\\ - &+ m\hat{y}_m (1-x^2)^{-\frac{m}{2}-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-\frac{m}{2}-2} - \end{align*} - and - \begin{align*} - \frac{d}{dx}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] &= \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_m\frac{m}{2}(1-x^2)^{-\frac{m}{2}-1}2x \\ - &= \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_mm(1-x^2)^{-\frac{m}{2}-1}x, - \end{align*} - to use them in Eq.\eqref{eq:2st_subs}, obtaining - \begin{align*} - (1-x^2)\biggl[\frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} &+ \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-\frac{m}{2}-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-\frac{m}{2}-1} \\ - &+ m\hat{y}_m (1-x^2)^{-\frac{m}{2}-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-\frac{m}{2}-2}\biggr] \\ - &-2(m+1)x\left[ \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_mm(1-x^2)^{-\frac{m}{2}-1}x \right] \\ - &+ (n(n+1)-m(m+1))\hat{y}_m(1-x^2)^{-\frac{m}{2}}=0.\\ - \end{align*} - We can now divide by $(1-x^2)^{-\frac{m}{2}}$, obtaining - \begin{align*} - (1-x^2)\biggl[\frac{d^2\hat{y}_m}{dx^2} &+ \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-1} + m\hat{y}_m (1-x^2)^{-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-2}\biggr] \\ - &-2(m+1)x\left[ \frac{d\hat{y}_m}{dx} + \hat{y}_mm(1-x^2)^{-1}x \right] + (n(n+1)-m(m+1))\hat{y}_m\\ - &= \frac{d^2\hat{y}_m}{dx^2} + \frac{d\hat{y}_m}{dx}mx + m\frac{d\hat{y}_m}{dx}x + m\hat{y}_m + m\hat{y}_m x^2(m+2)(1-x^2)^{-1} \\ - &-2(m+1)x\left[ \frac{d\hat{y}_m}{dx} + \hat{y}_mm(1-x^2)^{-1}x \right] + (n(n+1)-m(m+1))\hat{y}_m\\ - \end{align*} - and collecting some terms - \begin{equation*} - (1-x^2)\frac{d^2\hat{y}_m}{dx^2} - 2x\frac{d\hat{y}_m}{dx} + \left( -x^2 \frac{m^2}{1-x^2} + m+n(n+1)-m(m+1)\right)\hat{y}_m=0. - \end{equation*} - Showing that - \begin{align*} - -x^2 \frac{m^2}{1-x^2} + m+n(n+1)-m(m+1) &= n(n+1)- m^2 -x^2 \frac{m^2}{1-x^2} \\ - &= n(n+1)- \frac{m}{1-x^2} - \end{align*} - implies $\hat{y}_m(x)$ being a solution of Eq.\eqref{kugel:eq:associated_leg_eq} +\begin{proof} + This results is derived in section \ref{kugel:sec:proofs:legendre}. \end{proof} -In simpler words, if we find a solution to Eq.\eqref{kugel:eq:leg_eq}, we can extend the latter according to the Lemma \ref{kugel:lemma_1} obtaining the solution of Eq.\eqref{kugel:eq:associated_leg_eq}.\newline -We can say that we are going in the right direction, as the problem to be solved is decreasing in difficulty. We moved from having to find a solution to Eq.\eqref{kugel:eq:associated_leg_eq} to finding a solution to Eq.\eqref{kugel:eq:leg_eq}, which is much more approachable as a problem. Luckily for us, the lemma we will present below will help us extensively, which is something of an euphemism, since it will give us the solution directly. -\begin{lemma} - The polynomial function - \begin{align*} - y_n(x)&=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{(2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k}\\ - &= \frac{1}{n!2^n}\frac{d^n}{dx^n}(1-x^2)^n =: P_n(x), - \end{align*} - is a solution to the second order differential equation - \begin{equation}\label{kugel:eq:sol_leg} - (1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx} + n(n+1)y=0, \quad \forall n>0. - \end{equation} + +Since the Legendre \emph{polynomials} are indeed polynomials, they can also be +expressed using the hypergeometric functions described in section +\ref{buch:rekursion:section:hypergeometrische-funktion}, so in fact +\begin{equation} + P_n(z) = {}_2F_1 \left( \begin{matrix} + n + 1, & -n \\ \multicolumn{2}{c}{1} + \end{matrix} ; \frac{1 - z}{2} \right). +\end{equation} +Further, there are a few more interesting but not very relevant forms to write +$P_n(z)$ such as \emph{Rodrigues' formula} and \emph{Laplace's integral +representation} which are +\begin{equation*} + P_n(z) = \frac{1}{2^n} \frac{d^n}{dz^n} (x^2 - 1)^n, + \qquad \text{and} \qquad + P_n(z) = \frac{1}{\pi} \int_0^\pi \left( + z + \cos\vartheta \sqrt{z^2 - 1} + \right) \, d\vartheta +\end{equation*} +respectively, both of which we will not prove (see chapter 3 of +\cite{bell_special_2004} for a proof). Now that we have a solution for the +Legendre equation, we can make use of the following lemma patch the solutions +such that they also become solutions of the associated Legendre equation +\eqref{kugel:eqn:associated-legendre}. + +\begin{lemma} \label{kugel:lem:extend-legendre} + If $Z_n(z)$ is a solution of the Legendre equation \eqref{kugel:eqn:legendre}, + then + \begin{equation*} + Z^m_n(z) = (1 - z^2)^{m/2} \frac{d^m}{dz^m}Z_n(z) + \end{equation*} + solves the associated Legendre equation \eqref{kugel:eqn:associated-legendre}. + \nocite{bell_special_2004} \end{lemma} \begin{proof} - In order to find a solution to Eq.\eqref{eq:legendre}, the following Ansatz can be performed: - \begin{equation}\label{eq:ansatz} - y(x) = \sum_{k=0}^\infty a_k x^k. - \end{equation} - Given Eq.\eqref{eq:ansatz}, then - \begin{align*} - \frac{dy}{dx} &= \sum_{k=0}^\infty k a_k x^{k-1}, \\ - \frac{d^2y}{dx^2} &= \sum_{k=0}^\infty k (k-1) a_k x^{k-2}. - \end{align*} - Eq.\eqref{eq:legendre} can be therefore written as - \begin{align} - &(1-x^2)\sum_{k=0}^\infty k (k-1) a_k x^{k-2} - 2x\sum_{k=0}^\infty k a_k x^{k-1} + n(n+1)\sum_{k=0}^\infty a_k x^k=0 \label{eq:ansatz_in_legendre} \\ - &=\sum_{k=0}^\infty k (k-1) a_k x^{k-2} - \sum_{k=0}^\infty k (k-1) a_k x^{k} - 2x\sum_{k=0}^\infty k a_k x^{k-1} + n(n+1)\sum_{k=0}^\infty a_k x^k=0. \nonumber - \end{align} - If one consider the term - \begin{equation}\label{eq:term} - \sum_{k=0}^\infty k (k-1) a_k x^{k-2}, - \end{equation} - the substitution $\tilde{k}=k-2$ yields Eq.\eqref{eq:term} to - \begin{equation*} - \sum_{\tilde{k}=-2}^\infty (\tilde{k}+2) (\tilde{k}+1) a_{\tilde{k}+2} x^{\tilde{k}}=\sum_{\tilde{k}=0}^\infty (\tilde{k}+2) (\tilde{k}+1) a_{\tilde{k}} x^{\tilde{k}}. - \end{equation*} - This means that Eq.\eqref{eq:ansatz_in_legendre} becomes - \begin{align} - &\sum_{k=0}^\infty (k+1)(k+2) a_{k+2} x^{k} - \sum_{k=0}^\infty k (k-1) a_k x^{k} - 2\sum_{k=0}^\infty k a_k x^k + n(n+1)\sum_{k=0}^\infty a_k x^k \nonumber \\ - = &\sum_{k=0}^\infty \big[ (k+1)(k+2) a_{k+2} - k (k-1) a_k - 2 k a_k + n(n+1) a_k \big] x^k \stackrel{!}{=} 0. \label{eq:condition} - \end{align} - The condition in Eq.\eqref{eq:condition} is equivalent to - \begin{equation}\label{eq:condition_2} - (k+1)(k+2) a_{k+2} - k (k-1) a_k - 2 k a_k + n(n+1) a_k = 0. - \end{equation} - We can derive a recursion formula for $a_{k+2}$ from Eq.\eqref{eq:condition_2}, which can be expressed as - \begin{equation}\label{eq:recursion} - a_{k+2}= \frac{k (k-1) - 2 k + n(n+1)}{(k+1)(k+2)}a_k = \frac{(k-n)(k+n+1)}{(k+2)(k+1)}a_k. - \end{equation} - All coefficients can be calculated using the latter. - - Following Eq.\eqref{eq:recursion}, if we want to compute $a_6$ we would have - \begin{align*} - a_{6}= -\frac{(n-4)(n+5)}{6\cdot 5}a_4 &= -\frac{(n-4)(5+n)}{6 \cdot 5} -\frac{(n-2)(n+3)}{4 \cdot 3} a_2 \\ - &= -\frac{(n-4)(n+5)}{6 \cdot 5} -\frac{(n-2)(n+3)}{4 \cdot 3} -\frac{n(n+1)}{2 \cdot 1} a_0 \\ - &= -\frac{(n+5)(n+3)(n+1)n(n-2)(n-4)}{6!} a_0. - \end{align*} - One can generalize this relation for the $i^\text{th}$ even coefficient as - \begin{equation*} - a_{2k} = (-1)^k \frac{(n+(2k-1))(n+(2k-1)-2)\hdots (n-(2k-2)+2)(n-(2k-2))}{(2k)!}a_0 - \end{equation*} - where $i=2k$. - - A similar expression can be written for the odd coefficients $a_{2k-1}$. In this case, the equation starts from $a_1$ and to find the pattern we can write the recursion for an odd coefficient, $a_7$ for example - \begin{align*} - a_{7}= -\frac{(n-5)(n+6)}{7\cdot 6}a_5 &= - \frac{(n-5)(n+6)}{7\cdot 6} -\frac{(n-3)(n+4)}{5 \cdot 4} a_3 \\ - &= - \frac{(n-5)(n+6)}{7\cdot 6} -\frac{(n-3)(n+4)}{5 \cdot 4} -\frac{(n-1)(n+2)}{3 \cdot 2} a_1 \\ - &= -\frac{(n+6)(n+4)(n+2)(n-1)(n-3)(n-5)}{7!} a_1. - \end{align*} - As before, we can generalize this equation for the $i^\text{th}$ odd coefficient - \begin{equation*} - a_{2k+1} = (-1)^k \frac{(n + 2k)(n+2k-2)\hdots(n-(2k-1)+2)(n-(2k-1))}{(2k+1)!}a_1 - \end{equation*} - where $i=2k+1$. - - Let be - \begin{align*} - y_\text{e}^K(x) &:= \sum_{k=0}^K(-1)^k \frac{(n+(2k-1))(n+(2k-1)-2)\hdots \color{red}(n-(2k-2)+2)(n-(2k-2))}{(2k)!} x^{2k}, \\ - y_\text{o}^K(x) &:= \sum_{k=0}^K(-1)^k \frac{(n + 2k)(n+2k-2)\hdots \color{blue} (n-(2k-1)+2)(n-(2k-1))}{(2k+1)!} x^{2k+1}. - \end{align*} - The solution to the Eq.\eqref{eq:legendre} can be written as - \begin{equation}\label{eq:solution} - y(x) = \lim_{K \to \infty} \left[ a_0 y_\text{e}^K(x) + a_1 y_\text{o}^K(x) \right]. - \end{equation} - - The colored parts can be analyzed separately: - \begin{itemize} - \item[\textcolor{red}{\textbullet}] Suppose that $n=n_0$ is an even number. Then the red part, for a specific value of $k=k_0$, will follow the following relation: - \begin{equation*} - n_0-(2k_0-2)=0. - \end{equation*} - From that point on, given the recursive nature of Eq.\eqref{eq:recursion}, all the subsequent coefficients will also be 0, making the sum finite. - \begin{equation*} - a_{2k}=0 \iff y_{\text{o}}^{2k}(x)=y_{\text{o}}^{2k_0}(x), \quad \forall k>k_0 - \end{equation*} - \item[\textcolor{blue}{\textbullet}] Suppose that $n=n_0$ is an odd number. Then the blue part, for a specific value of $k=k_0$, will follow the following relation - \begin{equation*} - n_0-(2k_0-1)=0. - \end{equation*} - From that point on, for the same reason as before, all the subsequent coefficients will also be 0, making the sum finite. - \begin{equation*} - a_{2k+1}=0 \iff y_{\text{o}}^{2k+1}(x)=y_{\text{o}}^{2k_0+1}(x), \quad \forall k>k_0 - \end{equation*} - \end{itemize} - - There is the possibility of expressing the solution in Eq.\eqref{eq:solution} in a more compact form, combining the two solutions $y_\text{o}^K(x)$ and $y_\text{e}^K(x)$. They are both a polynomial of maximum degree $n$, assuming $n \in \mathbb{N}$. In the case where $n$ is even, the polynomial solution - \begin{equation*} - \lim_{K\to \infty} y_\text{e}^K(x) - \end{equation*} - will be a finite sum. If instead $n$ is odd, will be - \begin{equation*} - \lim_{K\to \infty} y_\text{o}^K(x) - \end{equation*} - to be a finite sum. - - Depending on the coefficient we start with, $a_1$ or $a_0$, we will obtain the odd or even polynomial respectively. Starting with the last coefficient $a_n$ and, recursively, calculating all the others in descending order, we can express the two parts $y_\text{o}^K(x)$ and $y_\text{e}^K(x)$ with a single sum. Hence, because we start with the last coefficient, the choice concerning $a_1$ and $a_0$ will be at the end of the sum, and not at the beginning. To compact Eq.\eqref{eq:solution}, Eq.\eqref{eq:recursion} can be reconsidered to calculate the coefficient $a_{k-2}$, using $a_k$ - \begin{equation*} - a_{k-2} = -\frac{(k+2)(k+1)}{(k-n)(k+n+1)}a_k - \end{equation*} - Now the game is to find a pattern, as before. Remember that $n$ is a fixed parameter of Eq.\eqref{eq:legendre}. - \begin{align*} - a_{n-2} &= -\frac{n(n-1)}{2(2n-1)}a_n, \\ - a_{n-4} &= -\frac{(n-2)(n-3)}{4(2n-3)}a_{n-2} \\ - &= -\frac{(n-2)(n-3)}{4(2n-3)}-\frac{n(n-1)}{2(2n-1)}a_n. - \end{align*} - In general - \begin{equation}\label{eq:general_recursion} - a_{n-2k} = (-1)^k \frac{n(n-1)(n-2)(n-3) \hdots (n-2k+1)}{2\cdot4\hdots 2k(2n-1)(2n-3)\hdots(2n-2k+1)}a_n - \end{equation} - The whole solution can now be written as - \begin{align} - y(x) &= a_n x^n + a_{n-2} x^{n-2} + a_{n-4} x^{n-4} + a_{n-6} x^{n-6} + \hdots + \begin{cases} - a_1 x, \quad &\text{if } n \text{ odd} \\ - a_0, \quad &\text{if } n \text{ even} - \end{cases} \nonumber \\ - &= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} a_{n-2k}x^{n-2k} \label{eq:solution_2} - \end{align} - By considering - \begin{align} - (2n-1)(2n-3)\hdots (2n-2k+1)&=\frac{2n(2n-1)(2n-2)(2n-3)\hdots(2n-2k+1)} - {2n(2n-2)(2n-4)(2n-6)\hdots(2n-2k+2)} \nonumber \\ - &=\frac{\frac{(2n)!}{(2n-2k)!}} - {2^kn(n-1)(n-2)(n-3)\hdots(n-k+1)} \nonumber \\ - &=\frac{\frac{(2n)!}{(2n-2k)!}} - {2^k\frac{n!}{(n-k)!}}=\frac{(n-k)!(2n)!}{n!(2n-2k)!2^k} \label{eq:1_sub_recursion}, \\ - 2 \cdot 4 \hdots 2k &= 2^r 1\cdot2 \hdots r = 2^r r!\label{eq:2_sub_recursion}, \\ - n(n-1)(n-2)(n-3) \hdots (n-2k+1) &= \frac{n!}{(n-2k)!}\label{eq:3_sub_recursion}. - \end{align} - Eq.\eqref{eq:solution_2} can be rewritten as - \begin{equation}\label{eq:solution_3} - y(x)=a_n \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{n!^2(2n-2k)!}{k!(n-2k)!(n-k)!(2n)!} x^{n-2k}. - \end{equation} - Eq.\eqref{eq:solution_3} is defined for any $a_n$. By letting $a_n$ be declared as - \begin{equation*} - a_{n} := \frac{(2n)!}{2^n n!^2}, - \end{equation*} - the so called \emph{Legendre polynomial} emerges - \begin{equation}\label{eq:leg_poly} - P_n(x):=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{(2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k} - \end{equation} + See section \ref{kugel:sec:proofs:legendre}. \end{proof} -As can be seen, the solution is a $n$-dependent power series, traditionally denoted as $P_n(x)$. This set of polynomials are called \emph{Legendre Polynomials}, because precisely they are polynomials satisfying the Legendre equation.\newline -Now that we have a solution to Eq.\eqref{kugel:eq:leg_eq}, we can then extend Eq.\eqref{kugel:eq:sol_leg}, as stated in Lemma \ref{kugel:lemma_1}. We will then have -\begin{align*} -y_{m,n}(x) &= (1-x^2)^{\frac{m}{2}}\frac{d^m}{dx^m}P_n(x) \\ -&= \frac{1}{n!2^n}(1-x^2)^{\frac{m}{2}}\frac{d^{m+n}}{dx^{m+n}}(1-x^2)^n -\end{align*} -This set of functions are defined as \emph{Associated Legendre functions}, because similarly to before, they solve the Associated Legendre equation, defined in Eq.\eqref{kugel:eq:eq_leg}. -\begin{definition}{Associated Legendre Functions} -\begin{equation}\label{kugel:eq:associated_leg_func} -P_{m,n}(x) := \frac{1}{n!2^n}(1-x^2)^{\frac{m}{2}}\frac{d^{m+n}}{dx^{m+n}}(1-x^2)^n -\end{equation} + +What is happening in lemma \ref{kugel:lem:extend-legendre}, is that we are +essentially inserting a square root function in the solution in order to be able +to reach the parts of the domain near the poles at $\pm 1$ of the associated +Legendre equation, which is not possible only using power series +\kugeltodo{Reference book theory on extended power series method.}. Now, since +we have a solution in our domain, namely $P_n(z)$, we can insert it in the lemma +obtain the \emph{associated Legendre functions}. + +\begin{definition}[Ferrers or Associated Legendre functions] + The functions + \begin{equation}\label{kugel:eq:associated_leg_func} + P^m_n (z) = \frac{1}{n!2^n}(1-z^2)^{\frac{m}{2}}\frac{d^{m}}{dz^{m}} P_n(z) + = \frac{1}{n!2^n}(1-z^2)^{\frac{m}{2}}\frac{d^{m+n}}{dz^{m+n}}(1-z^2)^n + \end{equation} + are known as Ferrers or associated Legendre functions. \end{definition} + +\subsection{Spherical Harmonics} + As you may recall, previously we performed the substitution $x=\cos \vartheta$. Now we need to return to the old domain, which can be done straightforwardly: \begin{equation*} \Theta(\vartheta) = P_{m,n}(\cos \vartheta), @@ -512,6 +362,10 @@ Ora, visto che la soluzione dell'eigenfunction problem è formata dalla moltipli \subsection{Recurrence Relations} -\section{Series Expansions in \(C(S^2)\)} +\section{Series Expansions in $C(S^2)$} -\nocite{olver_introduction_2013} +\subsection{Orthogonality of $P_n$, $P^m_n$ and $Y^m_n$} + +\subsection{Series Expansion} + +\subsection{Fourier on $S^2$} -- cgit v1.2.1 From d0c30778c51d0940b93b488183f50ec8aa5fa0f0 Mon Sep 17 00:00:00 2001 From: Nao Pross Date: Wed, 17 Aug 2022 16:30:46 +0200 Subject: kugel: Add povray images --- buch/papers/kugel/figures/povray/curvature.jpg | Bin 0 -> 265649 bytes buch/papers/kugel/figures/povray/curvature.png | Bin 0 -> 590402 bytes buch/papers/kugel/figures/povray/spherecurve.jpg | Bin 0 -> 171287 bytes buch/papers/kugel/figures/povray/spherecurve.png | Bin 0 -> 423490 bytes 4 files changed, 0 insertions(+), 0 deletions(-) create mode 100644 buch/papers/kugel/figures/povray/curvature.jpg create mode 100644 buch/papers/kugel/figures/povray/curvature.png create mode 100644 buch/papers/kugel/figures/povray/spherecurve.jpg create mode 100644 buch/papers/kugel/figures/povray/spherecurve.png (limited to 'buch/papers') diff --git a/buch/papers/kugel/figures/povray/curvature.jpg b/buch/papers/kugel/figures/povray/curvature.jpg new file mode 100644 index 0000000..6448966 Binary files /dev/null and b/buch/papers/kugel/figures/povray/curvature.jpg differ diff --git a/buch/papers/kugel/figures/povray/curvature.png b/buch/papers/kugel/figures/povray/curvature.png new file mode 100644 index 0000000..20268f2 Binary files /dev/null and b/buch/papers/kugel/figures/povray/curvature.png differ diff --git a/buch/papers/kugel/figures/povray/spherecurve.jpg b/buch/papers/kugel/figures/povray/spherecurve.jpg new file mode 100644 index 0000000..cd2e7c8 Binary files /dev/null and b/buch/papers/kugel/figures/povray/spherecurve.jpg differ diff --git a/buch/papers/kugel/figures/povray/spherecurve.png b/buch/papers/kugel/figures/povray/spherecurve.png new file mode 100644 index 0000000..ff24371 Binary files /dev/null and b/buch/papers/kugel/figures/povray/spherecurve.png differ -- cgit v1.2.1 From 494636b6d00b0697bda4c5840a3666b0867f22e8 Mon Sep 17 00:00:00 2001 From: Nao Pross Date: Wed, 17 Aug 2022 17:18:13 +0200 Subject: kugel: Minor changes --- buch/papers/kugel/main.tex | 2 +- buch/papers/kugel/packages.tex | 5 ++ buch/papers/kugel/preliminaries.tex | 8 +-- buch/papers/kugel/spherical-harmonics.tex | 87 ++++++++++++++++++++++--------- 4 files changed, 73 insertions(+), 29 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/kugel/main.tex b/buch/papers/kugel/main.tex index ad19178..d063f87 100644 --- a/buch/papers/kugel/main.tex +++ b/buch/papers/kugel/main.tex @@ -11,7 +11,7 @@ \chapterauthor{Manuel Cattaneo, Naoki Pross} \input{papers/kugel/introduction} -% \input{papers/kugel/preliminaries} +\input{papers/kugel/preliminaries} \input{papers/kugel/spherical-harmonics} \input{papers/kugel/applications} \input{papers/kugel/proofs} diff --git a/buch/papers/kugel/packages.tex b/buch/papers/kugel/packages.tex index b0e1f61..ead7653 100644 --- a/buch/papers/kugel/packages.tex +++ b/buch/papers/kugel/packages.tex @@ -1,3 +1,4 @@ +% vim:ts=2 sw=2 et: % % packages.tex -- packages required by the paper kugel % @@ -10,6 +11,10 @@ \usepackage{cases} \newcommand{\kugeltodo}[1]{\textcolor{red!70!black}{\texttt{[TODO: #1]}}} +\newcommand{\kugelplaceholderfig}[2]{ \begin{tikzpicture}% + \fill[lightgray!20] (0, 0) rectangle (#1, #2);% + \node[gray, anchor = center] at ({#1 / 2}, {#2 / 2}) {\Huge \ttfamily \bfseries TODO}; + \end{tikzpicture}} \DeclareMathOperator{\sphlaplacian}{\nabla^2_{\mathit{S}}} \DeclareMathOperator{\surflaplacian}{\nabla^2_{\partial \mathit{S}}} diff --git a/buch/papers/kugel/preliminaries.tex b/buch/papers/kugel/preliminaries.tex index 03cd421..e48abe4 100644 --- a/buch/papers/kugel/preliminaries.tex +++ b/buch/papers/kugel/preliminaries.tex @@ -44,23 +44,23 @@ numbers \(\mathbb{R}\). \) \end{definition} -\texttt{TODO: Text here.} +\kugeltodo{Text here.} \begin{definition}[Span] \end{definition} -\texttt{TODO: Text here.} +\kugeltodo{Text here.} \begin{definition}[Linear independence] \end{definition} -\texttt{TODO: Text here.} +\kugeltodo{Text here.} \begin{definition}[Basis] \end{definition} -\texttt{TODO: Text here.} +\kugeltodo{Text here.} \begin{definition}[Inner product] \label{kugel:def:inner-product} \nocite{axler_linear_2014} diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex index 5645941..2ded50b 100644 --- a/buch/papers/kugel/spherical-harmonics.tex +++ b/buch/papers/kugel/spherical-harmonics.tex @@ -2,8 +2,8 @@ \section{Construction of the Spherical Harmonics} -\if 0 -\kugeltodo{Rewrite this section if the preliminaries become an addendum} +\kugeltodo{Review text, or rewrite if preliminaries becomes an addendum} + We finally arrived at the main section, which gives our chapter its name. The idea is to discuss spherical harmonics, their mathematical derivation and some of their properties and applications. @@ -29,9 +29,9 @@ created with the previous sections, concluding that Fourier is just a specific case of the application of the concept of orthogonality. Our hope is that after reading this section you will appreciate the beauty and power of generalization that mathematics offers us. -\fi \subsection{Eigenvalue Problem} +\label{kugel:sec:construction:eigenvalue} \begin{figure} \centering @@ -111,8 +111,9 @@ that satisfy the equation \surflaplacian f = -\lambda f. \end{equation} Perhaps it may not be obvious at first glance, but we are in fact dealing with a -partial differential equation (PDE) \kugeltodo{Boundary conditions?}. If we unpack the notation of the operator -$\nabla^2_{\partial S}$ according to definition +partial differential equation (PDE) \kugeltodo{Boundary conditions?}. If we +unpack the notation of the operator $\nabla^2_{\partial S}$ according to +definition \ref{kugel:def:surface-laplacian}, we get: \begin{equation} \label{kugel:eqn:eigen-pde} \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( @@ -139,7 +140,8 @@ convenience. If we substitute this assumption in \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial \Theta(\vartheta)}{\partial\vartheta} \right) \Phi(\varphi) - + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 \Phi(\varphi)}{\partial\varphi^2} + + \frac{1}{\sin^2 \vartheta} + \frac{\partial^2 \Phi(\varphi)}{\partial\varphi^2} \Theta(\vartheta) + \lambda \Theta(\vartheta)\Phi(\varphi) = 0. \end{equation*} @@ -182,6 +184,14 @@ require a dedicated section of its own. \subsection{Legendre Functions} +\begin{figure} + \centering + \kugelplaceholderfig{.8\textwidth}{5cm} + \caption{ + \kugeltodo{Why $z = \cos \vartheta$.} + } +\end{figure} + To solve \eqref{kugel:eqn:ode-theta} we start with the substitution $z = \cos \vartheta$ \kugeltodo{Explain geometric origin with picture}. The operator $\frac{d}{d \vartheta}$ becomes @@ -298,26 +308,19 @@ Legendre equation, which is not possible only using power series we have a solution in our domain, namely $P_n(z)$, we can insert it in the lemma obtain the \emph{associated Legendre functions}. -\begin{definition}[Ferrers or Associated Legendre functions] +\begin{definition}[Ferrers or associated Legendre functions] + \label{kugel:def:ferrers-functions} The functions - \begin{equation}\label{kugel:eq:associated_leg_func} + \begin{equation} P^m_n (z) = \frac{1}{n!2^n}(1-z^2)^{\frac{m}{2}}\frac{d^{m}}{dz^{m}} P_n(z) = \frac{1}{n!2^n}(1-z^2)^{\frac{m}{2}}\frac{d^{m+n}}{dz^{m+n}}(1-z^2)^n \end{equation} are known as Ferrers or associated Legendre functions. \end{definition} -\subsection{Spherical Harmonics} +\kugeltodo{Discuss $|m| \leq n$.} -As you may recall, previously we performed the substitution $x=\cos \vartheta$. Now we need to return to the old domain, which can be done straightforwardly: -\begin{equation*} - \Theta(\vartheta) = P_{m,n}(\cos \vartheta), -\end{equation*} -obtaining the much sought function $\Theta(\vartheta)$. \newline -So we finally reached the end of this tortuous path. Now we just need to put together all the information we have to construct $f(\vartheta, \varphi)$ in the following way: -\begin{equation}\label{kugel:eq:sph_harm_0} - f(\vartheta, \varphi) = \Theta(\vartheta)\Phi(\varphi) = P_{m,n}(\cos \vartheta)e^{jm\varphi}, \quad |m|\leq n. -\end{equation} +\if 0 The constraint $|m| Date: Wed, 17 Aug 2022 21:38:44 +0200 Subject: letzte Korrektur --- buch/papers/0f1/teil2.tex | 18 +++++++++--------- buch/papers/0f1/teil3.tex | 4 ++-- 2 files changed, 11 insertions(+), 11 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/0f1/teil2.tex b/buch/papers/0f1/teil2.tex index 64f8d83..fdcb0fc 100644 --- a/buch/papers/0f1/teil2.tex +++ b/buch/papers/0f1/teil2.tex @@ -41,13 +41,13 @@ a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cdots}}}, in welchem $a_0, a_1,\dots,a_n$ und $b_1,b_2,\dots,b_n$ ganze Zahlen sind. \subsubsection{Rekursionsbeziehungen und Kettenbrüche} -Will man einen Kettenbruch für das Verhältnis $\frac{f_i(z)}{f_{i-1}(z)}$ finden, braucht man dazu eine Relation der analytischer Funktion $f_i(z)$. -Nimmt man die Gleichung \cite{0f1:wiki-fraction}: +Wenn es eine Relation analytischer Funktion $f_i(z)$ hat, dann gibt es einen Kettenbruch für das Verhältnis $\frac{f_i(z)}{f_{i-1}(z)}$ \cite{0f1:wiki-fraction}. +Nimmt man die Gleichung \begin{equation*} f_{i-1} - f_i = k_i z f_{i+1}, \end{equation*} wo $f_i$ analytische Funktionen sind und $i > 0$ ist, sowie $k_i$ konstant. -Ergibt sich folgender Zusammenhang: +Ergibt sich der Zusammenhang \begin{equation*} \cfrac{f_i}{f_{i-1}} = \cfrac{1}{1+k_iz\cfrac{f_{i+1}}{f_i}}. \end{equation*} @@ -55,7 +55,7 @@ Geht man einen Schritt weiter und nimmt für $g_i = \frac{f_i}{f_{i-1}}$ an, kom \begin{equation*} g_i = \cfrac{1}{1+k_izg_{i+1}}. \end{equation*} -Setzt man dies nun für $g_1$ in den Bruch ein, ergibt sich folgendes: +Setzt man dies nun für $g_1$ in den Bruch ein, ergibt sich \begin{equation*} g_1 = \cfrac{f_1}{f_0} = \cfrac{1}{1+k_izg_2} = \cfrac{1}{1+\cfrac{k_1z}{1+k_2zg_3}} = \cdots \end{equation*} @@ -76,19 +76,19 @@ kann durch Substitution bewiesen werden, dass \mathstrut_0F_1(;c-1;z) - \mathstrut_0F_1(;c;z) = \frac{z}{c(c-1)} \cdot \mathstrut_0F_1(;c+1;z) \end{equation*} eine Relation dazu ist. -Wenn man für $f_i$ und $k_i$ folgende Annahme trifft: +Wenn man für $f_i$ und $k_i$ die Annahme \begin{align*} f_i =& \mathstrut_0F_1(;c+i;z)\\ k_i =& \frac{1}{(c+i)(c+i-1)} \end{align*} -und in die Formel \eqref{0f1:math:rekursion:eq} einsetzt, erhält man: +trifft und in die Formel \eqref{0f1:math:rekursion:eq} einsetzt, erhält man: \begin{equation*} \cfrac{\mathstrut_0F_1(;c+1;z)}{\mathstrut_0F_1(;c;z)} = \cfrac{1}{1+\cfrac{\cfrac{z}{c(c+1)}}{1+\cfrac{\cfrac{z}{(c+1)(c+2)}}{1+\cfrac{\cfrac{z}{(c+2)(c+3)}}{\cdots}}}}. \end{equation*} \subsubsection{Algorithmus} Da mit obigen Formeln nur ein Verhältnis zwischen $ \frac{\mathstrut_0F_1(;c+1;z)}{\mathstrut_0F_1(;c;z)}$ berechnet wurde, braucht es weitere Relationen um $\mathstrut_0F_1(;c;z)$ zu erhalten. -So ergeben ähnliche Relationen nach Wolfram Alpha \cite{0f1:wolfram-0f1} folgender Kettenbruch +So ergeben ähnliche Relationen nach Wolfram Alpha \cite{0f1:wolfram-0f1} den Kettenbruch \begin{equation} \label{0f1:math:kettenbruch:0f1:eq} \mathstrut_0F_1(;c;z) = 1 + \cfrac{\cfrac{z}{c}}{1+\cfrac{-\cfrac{z}{2(c+1)}}{1+\cfrac{z}{2(c+1)}+\cfrac{-\cfrac{z}{3(c+2)}}{1+\cfrac{z}{5(c+4)} + \cdots}}}, @@ -112,7 +112,7 @@ lässt sich zu \cfrac{A_k}{B_k} = \cfrac{b_{k+1}}{a_{k+1} + \cfrac{p}{q}} = \frac{b_{k+1} \cdot q}{a_{k+1} \cdot q + p} \end{align*} umformen. -Dies lässt sich auch durch die folgende Matrizenschreibweise +Dies lässt sich auch durch die Matrizenschreibweise \begin{equation*} \begin{pmatrix} A_k\\ @@ -137,7 +137,7 @@ Wendet man dies nun auf den Kettenbruch in der Form \begin{equation*} \frac{A_k}{B_k} = a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{\cdots}{\cdots+\cfrac{b_{k-1}}{a_{k-1} + \cfrac{b_k}{a_k}}}}} \end{equation*} -an, ergibt sich folgende Matrixdarstellungen: +an, ergibt sich die Matrixdarstellungen: \begin{align*} \begin{pmatrix} diff --git a/buch/papers/0f1/teil3.tex b/buch/papers/0f1/teil3.tex index 2afc34b..147668a 100644 --- a/buch/papers/0f1/teil3.tex +++ b/buch/papers/0f1/teil3.tex @@ -15,9 +15,9 @@ Ebenso kann festgestellt werden, dass je grösser der Wert $z$ in $\mathstrut_0F \label{0f1:subsection:konvergenz}} Es zeigt sich in Abbildung \ref{0f1:ausblick:plot:airy:konvergenz}, dass nach drei Iterationen ($k = 3$) die Funktionen genaue Resultate im Bereich von $-2$ bis $2$ liefert. Ebenso kann festgestellt werden, dass der Kettenbruch schneller konvergiert und im positiven Bereich mit der Referenzfunktion $\operatorname{Ai}(x)$ übereinstimmt. Da die Rekursionsformel eine Abwandlung des Kettenbruches ist, verhalten sich die Funktionen in diesem Fall gleich. -Erst wenn mehrerer Iterationen gerechnet werden, ist wie Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, der Kettenbruch den anderen zwei Algorithmen bezüglich Konvergenz überlegen. Bei der Rekursionsformel muss beachtet werden, dass sie zwar erst nach 35 Approximationen gänzlich konvergiert, allerdings nach 27 Iterationen sich nicht mehr gross verändert. +Erst wenn mehrerer Iterationen gerechnet werden, ist wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, der Kettenbruch den anderen zwei Algorithmen bezüglich Konvergenz überlegen. Bei der Rekursionsformel muss beachtet werden, dass sie zwar erst nach 35 Approximationen gänzlich konvergiert, allerdings nach 27 Iterationen sich nicht mehr gross verändert. -Ist $z$ negativ wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu aufgrund des Vorzeichens zu alternierenden Termen. So steigt bei allen Algorithmen zuerst die Differenz zum erwarteten Endwert. Erst nach genügend Iterationen sind die Terme so klein, dass sie das Endresultat nicht mehr signifikant beeinflussen. Während die Potenzreihe zusammen mit dem Kettenbruch nach 34 Approximationen konvergiert, braucht die Rekursionsformel noch zwei Iterationen mehr. +Ist $z$ negativ, wie in Abbildung \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies aufgrund des Vorzeichens zu alternierenden Termen. So steigt bei allen Algorithmen zuerst die Differenz zum erwarteten Endwert. Erst nach genügend Iterationen sind die Terme so klein, dass sie das Endresultat nicht mehr signifikant beeinflussen. Während die Potenzreihe zusammen mit dem Kettenbruch nach 34 Approximationen konvergiert, braucht die Rekursionsformel noch zwei Iterationen mehr. \subsection{Stabilität -- cgit v1.2.1 From 8faafd84edbd5dc53a693513d970fe5ab67d8b5c Mon Sep 17 00:00:00 2001 From: Nao Pross Date: Wed, 17 Aug 2022 23:20:35 +0200 Subject: Tim ist kein Zeichner --- buch/papers/kreismembran/Makefile | 4 +- buch/papers/kreismembran/images/TikzSaite.pdf | Bin 0 -> 17625 bytes buch/papers/kreismembran/images/TikzSaite.tex | 57 ++++++++++++++++++++++++++ buch/papers/kreismembran/teil0.tex | 3 +- 4 files changed, 61 insertions(+), 3 deletions(-) create mode 100644 buch/papers/kreismembran/images/TikzSaite.pdf create mode 100644 buch/papers/kreismembran/images/TikzSaite.tex (limited to 'buch/papers') diff --git a/buch/papers/kreismembran/Makefile b/buch/papers/kreismembran/Makefile index ce3c89f..a13f2cf 100644 --- a/buch/papers/kreismembran/Makefile +++ b/buch/papers/kreismembran/Makefile @@ -4,6 +4,6 @@ # (c) 2020 Prof Dr Andreas Mueller # -images: - @echo "no images to be created in kreismembran" +images/TikzSaite.pdf: images/TikzSaite.tex + cd images && pdflatex TikzSaite.tex diff --git a/buch/papers/kreismembran/images/TikzSaite.pdf b/buch/papers/kreismembran/images/TikzSaite.pdf new file mode 100644 index 0000000..f95ceb9 Binary files /dev/null and b/buch/papers/kreismembran/images/TikzSaite.pdf differ diff --git a/buch/papers/kreismembran/images/TikzSaite.tex b/buch/papers/kreismembran/images/TikzSaite.tex new file mode 100644 index 0000000..bf3d8f6 --- /dev/null +++ b/buch/papers/kreismembran/images/TikzSaite.tex @@ -0,0 +1,57 @@ +% vim: ts=2 sw=2 et : +\documentclass[tikz, border=2mm]{standalone} + +\usepackage{times} +\usepackage{txfonts} + +\begin{document} + \begin{tikzpicture}[ + axis/.style = {very thick, -latex}, + axis tick/.style = { + draw, draw = black, fill = black, rectangle, + inner sep = 0pt, + minimum height = 2mm, + minimum width = 1pt, + }, + string/.style = { + ultra thick, draw = black, + }, + string end/.style = { + string, circle, fill = gray, + inner sep = 0pt, minimum size = 1mm, + }, + force/.style = { + very thick, draw = gray, -latex, + }, + ] + + % axes + \draw[axis] (0, 0) -- (8cm, 0) node[right] {$x$}; + \draw[axis] (0, 0) -- (0, 5cm) node[above] {$u(x, t)$}; + + % axes ticks + \node[axis tick, label = {-90:$x_0$}] at (2cm, 0) {}; + \node[axis tick, label = {-90:$x_0 + dx$}] at (6cm, 0) {}; + + % string + \coordinate (A) at (2cm, 2cm); + \coordinate (B) at (6cm, 4cm); + + \draw[string] (A) to[out = 40, in = 200] (B); + + \draw[force] (A) -- ++(220:15mm) node[gray, below right] {$T_1$}; + \draw[force] (B) -- ++(20:15mm) node[gray, above left] {$T_2$}; + + \draw[dashed, gray, thick] (A) -- ++(-15mm, 0); + \draw[gray, thick] (A) ++ (-7mm,0) arc (180:220:7mm) + node[midway, left] {$\alpha$}; + + \draw[dashed, gray, thick] (B) -- ++(15mm, 0); + \draw[gray, thick] (B) ++ (7mm,0) arc (0:20:7mm) + node[pos = 0, below] {$\beta$}; + + \node[string end, label={110:$P_1$}] at (A) {}; + \node[string end, label={110:$P_2$}] at (B) {}; + + \end{tikzpicture} +\end{document} diff --git a/buch/papers/kreismembran/teil0.tex b/buch/papers/kreismembran/teil0.tex index 27c6f0f..e962aab 100644 --- a/buch/papers/kreismembran/teil0.tex +++ b/buch/papers/kreismembran/teil0.tex @@ -42,7 +42,8 @@ Die Verbindung zwischen Membran und Saite ist intuitiv ersichtlich, stellt man s \begin{figure} \begin{center} - \includegraphics[width=5cm,angle=-90]{papers/kreismembran/images/Saite.pdf} + % \includegraphics[width=5cm,angle=-90]{papers/kreismembran/images/Saite.pdf} + \includegraphics[]{papers/kreismembran/images/TikzSaite.pdf} \caption{Infinitesimales Stück einer Saite} \label{kreismembran:im:Saite} \end{center} -- cgit v1.2.1 From 3d0b6bf8410b37fd6d68a83ef08c6794cfdad8cd Mon Sep 17 00:00:00 2001 From: haddoucher Date: Thu, 18 Aug 2022 11:49:37 +0200 Subject: Korrektur Einleitung Alles korrigiert --- buch/papers/sturmliouville/einleitung.tex | 53 +++++++++++++++++-------------- 1 file changed, 29 insertions(+), 24 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/einleitung.tex b/buch/papers/sturmliouville/einleitung.tex index 78c1800..163f033 100644 --- a/buch/papers/sturmliouville/einleitung.tex +++ b/buch/papers/sturmliouville/einleitung.tex @@ -5,25 +5,36 @@ % \section{Was ist das Sturm-Liouville-Problem\label{sturmliouville:section:teil0}} \rhead{Einleitung} -Das Sturm-Liouville-Problem wurde benannt nach dem schweizerisch-französischer Mathematiker und Physiker Jacques Charles Fran\c{c}ois Sturm und dem französischer Mathematiker Joseph Liouville. -Gemeinsam haben sie in der mathematischen Physik die Sturm-Liouville-Theorie entwickelt und gilt für die Lösung von gewohnlichen Differentialgleichungen, jedoch verwendet man die Theorie öfters bei der Lösung von partiellen Differentialgleichungen. -Normalerweise betrachtet man für das Strum-Liouville-Problem eine gewöhnliche Differentialgleichung 2. Ordnung, und wenn es sich um eine partielle Differentialgleichung handelt, kann man sie mit Hilfe einiger Methoden in mehrere gewöhnliche Differentialgleichungen umwandeln, wie z. B. den Separationsansatz, die partielle Differentialgleichung mit mehreren Variablen. +Das Sturm-Liouville-Problem wurde benannt nach dem schweizerisch-französischen Mathematiker und Physiker Jacques Charles Fran\c{c}ois Sturm und dem französischen Mathematiker Joseph Liouville. +Gemeinsam haben sie in der mathematischen Physik die Sturm-Liouville-Theorie entwickelt und gilt für die Lösung von gewöhnlichen Differentialgleichungen, jedoch verwendet man die Theorie öfters bei der Lösung von partiellen Differentialgleichungen. +Normalerweise betrachtet man für das Strum-Liouville-Problem eine gewöhnliche Differentialgleichung 2. Ordnung, und wenn es sich um eine partielle Differentialgleichung handelt, kann man sie in mehrere gewöhnliche Differentialgleichungen umwandeln. Wie z. B. den Separationsansatz, die partielle Differentialgleichung mit mehreren Variablen. \begin{definition} \index{Sturm-Liouville-Gleichung}% -Angenommen man hat die lineare homogene Differentialgleichung +Wenn die lineare homogene Differentialgleichung \begin{equation} \frac{d^2y}{dx^2} + a(x)\frac{dy}{dx} + b(x)y = 0 \end{equation} -und schreibt die Gleichung um in: +als \begin{equation} \label{eq:sturm-liouville-equation} \frac{d}{dx}\lbrack p(x) \frac{dy}{dx} \rbrack + \lbrack q(x) + \lambda w(x) \rbrack y = 0 \end{equation} -, diese Gleichung wird dann Sturm-Liouville-Gleichung bezeichnet. +geschrieben werden kann, dann wird diese Gleichung als Sturm-Liouville-Gleichung bezeichnet. \end{definition} +Alle homogene 2. Ordnung lineare gewöhnliche Differentialgleichungen können in die Form der Gleichung \ref{eq:sturm-liouville-equation} umgeformt werden. + +\subsection{Randbedingungen\label{sub:was-ist-das-slp-randbedingungen}} +Wenn von der Funktion $y(x)$ die Werte $x$ des jeweiligen Randes des Definitionsbereiches anzunehmen sind, also +\begin{equation} + y(a) = y(b) = 0 +\end{equation} +, so spricht man von einer Dirichlet-Randbedingung\footnote{Die Dirichlet-Randbedingung oder auch Randbedingung des ersten Typs genannt ist nach dem deutschen Mathematiker Peter Gstav Lejeune Dirichlet benannt. Sie findet Anwendung auf gewöhnliche oder patielle Differentialgleichungen und gibt mit der Bedingung die Werte an, die für die abgeleitete Lösung innerhalb der Domänengrenze gelten.}, und von einer Neumann-Randbedingung\footnote{Die Neumann-Randbedingung oder auch Randbedingung des zweiten Typs genannt, ist nach dem deutschen Mathematiker Carl Neumann benannt. Sie legt die Werte fest, die eine Lösung entlang der Domänengrenze annehmen muss, wenn eine gewöhnliche oder partielle Differentialgleichung gestellt wird.} spricht man, wenn +\begin{equation} + y'(a) = y'(b) = 0 +\end{equation} +ergibt. -Alle homogene 2.Ordnung lineare gewöhnliche Differentialgleichungen können in die Form der Gleichung \ref{eq:sturm-liouville-equation} umgeformt werden. Die Sturm-Liouville-Theorie besagt, dass, wenn man die Sturm-Liouville-Gleichung mit den homogenen Randbedingungen des dritten Typs\footnote{Die Randbedingung des dritten Typs, oder Robin-Randbedingungen (benannt nach dem französischen mathematischen Analytiker und angewandten Mathematiker Victor Gustave Robin), wird genannt, wenn sie einer gewöhnlichen oder partiellen Differentialgleichung auferlegt wird, so sind die Spezifikationen einer Linearkombination der Werte einer Funktion sowie die Werte ihrer Ableitung am Rande des Bereichs} \begin{equation} \begin{aligned} @@ -32,17 +43,10 @@ Die Sturm-Liouville-Theorie besagt, dass, wenn man die Sturm-Liouville-Gleichung k_b y(b) + h_b p(b) y'(b) &= 0 \end{aligned} \end{equation} -kombiniert, wie schon im Kapitel \ref{sub:differentailgleichung} erwähnt, auf dem Intervall (a,b), dann bekommt man das klassische Sturm-Liouville-Problem. -Wenn von der Funktion $y(x)$ die Werte $x$ des jeweiligen Randes des Definitionsbereiches anzunehmen sind, also -\begin{equation} - y(a) = y(b) = 0 -\end{equation} -, so spricht man von einer Dirichlet-Randbedingung, und von einer Neumann-Randbedingung spricht man, wenn -\begin{equation} - y'(a) = y'(b) = 0 -\end{equation} -ergibt - die Existenz und Eindeutigkeit der Lösung kann mit den zwei Randbedingungen sichergestellt werden. -Lösungen die nicht Null sind, werden nicht betrachtet und diese zwei Gleichungen (\ref{eq:sturm-liouville-equation} und \ref{eq:randbedingungen}) kombiniert, nennt man Eigenfunktionen. +kombiniert, dann bekommt man das klassische Sturm-Liouville-Problem. + +\subsection{Eigenwertproblem} +Die Gleichungen \ref{eq:sturm-liouville-equation} hat die Form eines Eigenwertproblems Wenn bei der Sturm-Liouville-Gleichung \ref{eq:sturm-liouville-equation} alles konstant bleibt, aber der Wert von $\lambda$ sich ändert, erhält man eine andere Eigenfunktion, weil man eine andere gewöhnliche Differentialgleichung löst; der Parameter $\lambda$ wird als Eigenwert bezeichnet. Es ist genau das gleiche Prinzip wie bei den Matrizen, andere Eigenwerte ergeben andere Eigenvektoren. @@ -59,6 +63,7 @@ Somit ergibt die Gleichung \int_{a}^{b} w(x)y_m y_n = 0 \end{equation}. +\subsection{Koeffizientenfunktionen} Die Funktionen $p(x)$, $q(x)$ und $w(x)$ werden als Koeffizientenfunktionen mit ihren freien Variablen $x$ bezeichnet. Die Funktion $w(x)$ (manchmal auch $r(x)$ genannt) wird als Gewichtsfunktion oder Dichtefunktion bezeichnet. Es gibt zwei verschiedene Sturm-Liouville-Probleme: das reguläre Sturm-Liouville-Problem und das singuläre Sturm-Liouville-Problem. @@ -76,12 +81,12 @@ Damit es sich um ein reguläres Sturm-Liouville-Problem handelt, müssen einige Die Bedingungen für ein reguläres Sturm-Liouville-Problem sind: \begin{itemize} \item Die Funktionen $p(x), p'(x), q(x)$ und $w(x)$ müssen stetig und reell sein. - \item sowie müssen in einem Endlichen Intervall $[ \ a,b] \ $ integrierbar sein. - \item $p(x)^{-1}$ und $w(x)$ sind $>0$. + \item sowie müssen in einem endlichen Intervall $[a,b]$ integrierbar sein. + \item $p(x)$ und $w(x)$ sind $>0$. \item Es gelten die Randbedingungen \ref{eq:randbedingungen}, wobei $|k_i|^2 + |h_i|^2\ne 0$ mit $i=a,b$. \end{itemize} \end{definition} -Bei einem regulären Sturm-Liouville-Problem geht es darum, ohne genaue Kenntnis der Eigenfunktionen diese dennoch beschreiben zu können. +Bei einem regulären Sturm-Liouville-Problem geht es darum, wichtige Eigenschaften der Eigenfunktionen beschreiben zu können, ohne sie genau zu kennen. % @@ -111,7 +116,7 @@ Allerdings kann nur eine der Bedingungen nicht erfüllt sein, so dass es sich be \end{aligned} \end{equation} ist kein reguläres Sturm-Liouville-Problem. - Weil wenn man die Gleichung in die Sturm-Liouville Form umformen, dann ergeben die Koeffizientenfunktionen $p(x) = w(x) = x$ und $q(x) = -m^2/x$. + Wenn man die Gleichung in die Sturm-Liouville Form umformen, dann ergeben die Koeffizientenfunktionen $p(x) = w(x) = x$ und $q(x) = -m^2/x$. Schaut man jetzt die Bedingungen im Kapitel \ref{sub:reguläre_sturm_liouville_problem} an und vergleicht diese unseren Koeffizientenfunktionen, so erkennt man einige Probleme: \begin{itemize} \item $p(x)$ und $w(x)$ sind nicht positiv, wenn $x = 0$ ist. @@ -121,9 +126,9 @@ Allerdings kann nur eine der Bedingungen nicht erfüllt sein, so dass es sich be \end{beispiel} Verwendet man das reguläre Sturm-Liouville-Problem, obwohl eine oder beide Bedingungen nicht erfüllt sind, dann ist es schwierig zu sagen, ob die Lösung fundierte Ergebnisse hat. -Es ist schwierig, bestehende Kriterien anzuwenden, da die Formulierungen z.B. in der Lösungsfunktion liegen. +Es ist schwierig, Kriterien anzuwenden, da die Formulierungen z. B. in der Lösungsfunktion liegen. Das Spektrum besteht im singulärem Problem nicht mehr nur aus Eigenwerte, sondern kann auch einen stetigen Anteil enthalten. -Ähnlich wie bei der Fourier-Reihe gegenüber der Fourier-Transformation gibt es immer noch eine zugehörige Eigenfunktionsentwicklung, und zwar die Integraltransformation sowie gibt es weiterhin eine verallgemeinerte Eigenfunktionen. +Ähnlich wie bei der Fourier-Reihe gegenüber der Fourier-Transformation gibt es immer noch eine zugehörige Eigenfunktionsentwicklung, und zwar die Integraltransformation sowie gibt es weiterhin verallgemeinerte Eigenfunktionen. -- cgit v1.2.1 From 354d497301c69137fd00566b42868370d2bd46a3 Mon Sep 17 00:00:00 2001 From: haddoucher Date: Thu, 18 Aug 2022 11:57:07 +0200 Subject: einleitung fertig korrigiert --- buch/papers/sturmliouville/einleitung.tex | 3 +-- 1 file changed, 1 insertion(+), 2 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/einleitung.tex b/buch/papers/sturmliouville/einleitung.tex index 163f033..700ea1d 100644 --- a/buch/papers/sturmliouville/einleitung.tex +++ b/buch/papers/sturmliouville/einleitung.tex @@ -125,9 +125,8 @@ Allerdings kann nur eine der Bedingungen nicht erfüllt sein, so dass es sich be \end{itemize} \end{beispiel} -Verwendet man das reguläre Sturm-Liouville-Problem, obwohl eine oder beide Bedingungen nicht erfüllt sind, dann ist es schwierig zu sagen, ob die Lösung fundierte Ergebnisse hat. +Verwendet man das reguläre Sturm-Liouville-Problem, obwohl eine oder beide Bedingungen nicht erfüllt sind, dann ist es schwierig zu sagen, ob die Lösung eindeutige Ergebnisse hat. Es ist schwierig, Kriterien anzuwenden, da die Formulierungen z. B. in der Lösungsfunktion liegen. -Das Spektrum besteht im singulärem Problem nicht mehr nur aus Eigenwerte, sondern kann auch einen stetigen Anteil enthalten. Ähnlich wie bei der Fourier-Reihe gegenüber der Fourier-Transformation gibt es immer noch eine zugehörige Eigenfunktionsentwicklung, und zwar die Integraltransformation sowie gibt es weiterhin verallgemeinerte Eigenfunktionen. -- cgit v1.2.1 From b3611aa8b6f2c56c8940c18c582de0fd3dd205f2 Mon Sep 17 00:00:00 2001 From: haddoucher Date: Thu, 18 Aug 2022 12:31:12 +0200 Subject: tschebyscheff kapitel fertig geschrieben --- buch/papers/sturmliouville/einleitung.tex | 13 ++++----- .../sturmliouville/tschebyscheff_beispiel.tex | 31 +++++++++++++++------- 2 files changed, 28 insertions(+), 16 deletions(-) (limited to 'buch/papers') diff --git a/buch/papers/sturmliouville/einleitung.tex b/buch/papers/sturmliouville/einleitung.tex index 700ea1d..d497622 100644 --- a/buch/papers/sturmliouville/einleitung.tex +++ b/buch/papers/sturmliouville/einleitung.tex @@ -27,9 +27,9 @@ Alle homogene 2. Ordnung lineare gewöhnliche Differentialgleichungen können in \subsection{Randbedingungen\label{sub:was-ist-das-slp-randbedingungen}} Wenn von der Funktion $y(x)$ die Werte $x$ des jeweiligen Randes des Definitionsbereiches anzunehmen sind, also \begin{equation} - y(a) = y(b) = 0 + y(a) = y(b) = 0, \end{equation} -, so spricht man von einer Dirichlet-Randbedingung\footnote{Die Dirichlet-Randbedingung oder auch Randbedingung des ersten Typs genannt ist nach dem deutschen Mathematiker Peter Gstav Lejeune Dirichlet benannt. Sie findet Anwendung auf gewöhnliche oder patielle Differentialgleichungen und gibt mit der Bedingung die Werte an, die für die abgeleitete Lösung innerhalb der Domänengrenze gelten.}, und von einer Neumann-Randbedingung\footnote{Die Neumann-Randbedingung oder auch Randbedingung des zweiten Typs genannt, ist nach dem deutschen Mathematiker Carl Neumann benannt. Sie legt die Werte fest, die eine Lösung entlang der Domänengrenze annehmen muss, wenn eine gewöhnliche oder partielle Differentialgleichung gestellt wird.} spricht man, wenn +so spricht man von einer Dirichlet-Randbedingung\footnote{Die Dirichlet-Randbedingung oder auch Randbedingung des ersten Typs genannt ist nach dem deutschen Mathematiker Peter Gstav Lejeune Dirichlet benannt. Sie findet Anwendung auf gewöhnliche oder patielle Differentialgleichungen und gibt mit der Bedingung die Werte an, die für die abgeleitete Lösung innerhalb der Domänengrenze gelten.}, und von einer Neumann-Randbedingung\footnote{Die Neumann-Randbedingung oder auch Randbedingung des zweiten Typs genannt, ist nach dem deutschen Mathematiker Carl Neumann benannt. Sie legt die Werte fest, die eine Lösung entlang der Domänengrenze annehmen muss, wenn eine gewöhnliche oder partielle Differentialgleichung gestellt wird.} spricht man, wenn \begin{equation} y'(a) = y'(b) = 0 \end{equation} @@ -53,15 +53,16 @@ Es ist genau das gleiche Prinzip wie bei den Matrizen, andere Eigenwerte ergeben Es besteht eine Korrespondenz zwischen den Eigenwerten und den Eigenvektoren. Das gleiche gilt auch beim Sturm-Liouville-Problem, und zwar \begin{equation} - \lambda \overset{Korrespondenz}\leftrightarrow y -\end{equation}. + \lambda \overset{Korrespondenz}\leftrightarrow y. +\end{equation} Die Theorie besagt, wenn $y_m$, $y_n$ Eigenfuktionen des Sturm-Liouville-Problems sind, die verschiedene Eigenwerte $\lambda_m$, $\lambda_n$ ($\lambda_m \neq \lambda_n$) entsprechen, so sind $y_m$, $y_n$ orthogonal zu y - dies gilt für das Intervall (a,b). Somit ergibt die Gleichung \begin{equation} - \int_{a}^{b} w(x)y_m y_n = 0 -\end{equation}. + \label{eq:skalar-sturm-liouville} + \int_{a}^{b} w(x)y_m y_n = 0. +\end{equation} \subsection{Koeffizientenfunktionen} Die Funktionen $p(x)$, $q(x)$ und $w(x)$ werden als Koeffizientenfunktionen mit ihren freien Variablen $x$ bezeichnet. diff --git a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex index a18684f..3817dc0 100644 --- a/buch/papers/sturmliouville/tschebyscheff_beispiel.tex +++ b/buch/papers/sturmliouville/tschebyscheff_beispiel.tex @@ -4,8 +4,9 @@ % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\subsection{Tschebyscheff-Polynome\label{sub:tschebyscheff-polynome}} -Im Kapitel \ref{sub:beispiele_sturm_liouville_problem} sind die Koeffizientenfunktionen die man braucht schon aufgeliste, und zwar mit +\subsection{Sind Tschebyscheff-Polynome orthogonal zueinander?\label{sub:tschebyscheff-polynome}} +\subsubsection*{Definition der Koeffizientenfunktion} +Im Kapitel \ref{sub:beispiele_sturm_liouville_problem} sind die Koeffizientenfunktionen, die man braucht, schon aufgeliste, und zwar mit \begin{align*} w(x) &= \frac{1}{\sqrt{1-x^2}} \\ p(x) &= \sqrt{1-x^2} \\ @@ -14,10 +15,12 @@ Im Kapitel \ref{sub:beispiele_sturm_liouville_problem} sind die Koeffizientenfun Da die Sturm-Liouville-Gleichung \begin{equation} \label{eq:sturm-liouville-equation-tscheby} - \frac{d}{dx}\lbrack \sqrt{1-x^2} \frac{dy}{dx} \rbrack + \lbrack 0 + \lambda \frac{1}{\sqrt{1-x^2}} \rbrack y = 0 + \frac{d}{dx} (\sqrt{1-x^2} \frac{dy}{dx}) + (0 + \lambda \frac{1}{\sqrt{1-x^2}}) y = 0 \end{equation} nun mit den Koeffizientenfunktionen aufgestellt werden kann, bleibt die Frage, ob es sich um ein reguläres oder singuläres Sturm-Liouville-Problem handelt. -Für das reguläre Problem laut der Definition \ref{def:reguläres_sturm-liouville-problem} muss die funktion $p(x) = \sqrt{1-x^2}$, $p'(x) = -2x$, $q(x) = 0$ und $w(x) = \frac{1}{\sqrt{1-x^2}}$ stetig und reell sein - und sie sind es auch. + +\subsubsection*{regulär oder singulär?} +Für das reguläre Problem laut der Definition \ref{def:reguläres_sturm-liouville-problem} muss die funktion $p(x) = \sqrt{1-x^2}$, $p'(x) = -2x$, $q(x) = 0$ und $w(x) = \frac{1}{\sqrt{1-x^2}}$ stetig und reell sein --- und sie sind es auch. Auf dem Intervall $(-1,1)$ sind die Tschebyscheff-Polynome erster Art mit Hilfe von Hyperbelfunktionen \begin{equation} T_n(x) = \cos n (\arccos x) @@ -28,22 +31,23 @@ Für $x>1$ und $x<-1$ sehen die Polynome wie folgt aus: (-1)^n \cosh (n \arccos (-x)), & x<-1 \end{array}\right. \end{equation}, jedoch ist die Orthogonalität nur auf dem Intervall $[ -1, 1]$ sichergestellt. -Die nächste Bedingung beinhaltet, dass die Funktion $p(x)^{-1}$ und $w(x)>0$ sein müssen. +Die nächste Bedingung beinhaltet, dass die Funktion $p(x)$ und $w(x)>0$ sein müssen. Die Funktion \begin{equation*} p(x)^{-1} = \frac{1}{\sqrt{1-x^2}} \end{equation*} -ist die gleiche wie $w(x)$. +ist die gleiche wie $w(x)$ und erfüllt die Bedingung. +\subsubsection*{Randwertproblem} Für die Verifizierung der Randbedingungen benötigt man erneut $p(x)$. Da sich die Polynome nur auf dem Intervall $[ -1,1 ]$ orthogonal verhalten, sind $a = -1$ und $b = 1$ gesetzt. Beim einsetzen in die Randbedingung \ref{eq:randbedingungen}, erhält man \begin{equation} \begin{aligned} k_a y(-1) + h_a y'(-1) &= 0 - k_b y(-1) + h_b y'(-1) &= 0 + k_b y(-1) + h_b y'(-1) &= 0. \end{aligned} -\end{equation}. +\end{equation} Die Funktion $y(x)$ und $y'(x)$ sind in diesem Fall die Tschebyscheff Polynome (siehe \label{sub:definiton_der_tschebyscheff-Polynome}). Es gibt zwei Arten von Tschebyscheff Polynome: die erste Art $T_n(x)$ und die zweite Art $U_n(x)$. Jedoch beachtet man in diesem Kapitel nur die Tschebyscheff Polynome erster Art (\ref{eq:tschebyscheff-polynome}). @@ -52,12 +56,19 @@ Somit erhält man \begin{equation} \begin{aligned} k_a T_2(-1) + h_a T_{2}'(-1) &= k_a = 0\\ - k_b T_2(1) + h_b T_{2}'(1) &= k_b = 0 + k_b T_2(1) + h_b T_{2}'(1) &= k_b = 0. \end{aligned} -\end{equation}. +\end{equation} Ähnlich wie beim Beispiel der Wärmeleitung in einem homogenen Stab kann man, damit die Bedingung $|k_i|^2 + |h_i|^2\ne 0$ erfüllt ist, können beliebige $h_a \ne 0$ und $h_b \ne 0$ gewählt werden. Somit ist erneut gezeigt, dass die Randbedingungen der Tschebyscheff-Polynome auf die Sturm-Liouville-Randbedingungen erfüllt und alle daraus resultierenden Lösungen orthogonal sind. +\begin{beispiel} + Die Gleichung \ref{eq:skalar-sturm-liouville} mit $y_m = T_1(x)$ und $y_n(x) = T_2(x)$ eingesetzt sowie $a=-1$ und $b = 1$ ergibt + \[ + \int_{-1}^{1} w(x) x (2x^2-1) dx = 0. + \] +\end{beispiel} + -- cgit v1.2.1 From 2180deab391444f58ce2a2f20d13f01c0cb69be7 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Andreas=20M=C3=BCller?= Date: Thu, 18 Aug 2022 20:09:42 +0200 Subject: new images --- buch/papers/parzyl/images/Makefile | 16 +++ buch/papers/parzyl/images/common.inc | 64 ++++++++++ buch/papers/parzyl/images/halfplane.jpg | Bin 0 -> 200681 bytes buch/papers/parzyl/images/halfplane.pdf | Bin 0 -> 208606 bytes buch/papers/parzyl/images/halfplane.png | Bin 0 -> 473623 bytes buch/papers/parzyl/images/halfplane.pov | 201 ++++++++++++++++++++++++++++++++ buch/papers/parzyl/images/halfplane.tex | 41 +++++++ 7 files changed, 322 insertions(+) create mode 100644 buch/papers/parzyl/images/Makefile create mode 100644 buch/papers/parzyl/images/common.inc create mode 100644 buch/papers/parzyl/images/halfplane.jpg create mode 100644 buch/papers/parzyl/images/halfplane.pdf create mode 100644 buch/papers/parzyl/images/halfplane.png create mode 100644 buch/papers/parzyl/images/halfplane.pov create mode 100644 buch/papers/parzyl/images/halfplane.tex (limited to 'buch/papers') diff --git a/buch/papers/parzyl/images/Makefile b/buch/papers/parzyl/images/Makefile new file mode 100644 index 0000000..4bd13ec --- /dev/null +++ b/buch/papers/parzyl/images/Makefile @@ -0,0 +1,16 @@ +# +# Makefile to build 3d images +# +# (c) 2022 Prof Dr Andreas Müller +# + +all: halfplane.pdf + +halfplane.pdf: halfplane.tex halfplane.jpg + pdflatex halfplane.tex +halfplane.png: halfplane.pov + povray +A0.1 -W1920 -H1080 -Ohalfplane.png halfplane.pov +halfplane.jpg: halfplane.png Makefile + convert -extract 1280x1080+340+0 halfplane.png \ + -density 300 -units PixelsPerInch halfplane.jpg + diff --git a/buch/papers/parzyl/images/common.inc b/buch/papers/parzyl/images/common.inc new file mode 100644 index 0000000..28aed2b --- /dev/null +++ b/buch/papers/parzyl/images/common.inc @@ -0,0 +1,64 @@ +// +// common.inc -- some common useful tools for drawing 3d images +// +// (c) 2018 Prof Dr Andreas Müller, Hochschule Rapperswil +// + +// +// draw a right angle quarter circle at point with legs and and +// color +// +#declare rechterwinkelradius = 0.5; +#declare rechterwinkelthickness = 0.01; +#macro rechterwinkel(o, v1, v2, c) +intersection { + sphere { o, rechterwinkelradius } + #declare rnormale = vnormalize(vcross(v1, v2)); + plane { rnormale, vdot(o, rnormale) + rechterwinkelthickness * rechterwinkelradius / 0.5 } + plane { -rnormale, -vdot(o, rnormale) + rechterwinkelthickness * rechterwinkelradius / 0.5 } + plane { -v1, -vdot(o, v1) } + plane { -v2, -vdot(o, v2) } + pigment { + color c + } +} +sphere { o + 0.45 * (vnormalize(v1) +vnormalize(v2)) * rechterwinkelradius, + 0.05 * rechterwinkelradius / 0.5 + pigment { + color c + } +} +#end + +// +// draw an arrow from to with thickness with +// color +// +#macro arrow(from, to, arrowthickness, c) + #declare arrowdirection = vnormalize(to - from); + #declare arrowlength = vlength(to - from); + union { + sphere { + from, 1.1 * arrowthickness + } + cylinder { + from, + from + (arrowlength - 5 * arrowthickness) * arrowdirection, + arrowthickness + } + cone { + from + (arrowlength - 5 * arrowthickness) * arrowdirection, + 2 * arrowthickness, + to, + 0 + } + pigment { + color c + } + finish { + specular 0.9 + metallic + } + } +#end + diff --git a/buch/papers/parzyl/images/halfplane.jpg b/buch/papers/parzyl/images/halfplane.jpg new file mode 100644 index 0000000..8cb5ae3 Binary files /dev/null and b/buch/papers/parzyl/images/halfplane.jpg differ diff --git a/buch/papers/parzyl/images/halfplane.pdf b/buch/papers/parzyl/images/halfplane.pdf new file mode 100644 index 0000000..7275810 Binary files /dev/null and b/buch/papers/parzyl/images/halfplane.pdf differ diff --git a/buch/papers/parzyl/images/halfplane.png b/buch/papers/parzyl/images/halfplane.png new file mode 100644 index 0000000..5beefa0 Binary files /dev/null and b/buch/papers/parzyl/images/halfplane.png differ diff --git a/buch/papers/parzyl/images/halfplane.pov b/buch/papers/parzyl/images/halfplane.pov new file mode 100644 index 0000000..419bb67 --- /dev/null +++ b/buch/papers/parzyl/images/halfplane.pov @@ -0,0 +1,201 @@ +// +// 3dimage.pov +// +// (c) 2022 Prof Dr Andreas Müller +// +#version 3.7; +#include "colors.inc" +#include "skies.inc" +#include "common.inc" + +global_settings { + assumed_gamma 1 +} + +#declare imagescale = 0.63; +#declare ar = 0.02; + +#declare Cameracenter = <5,3,-4>; +#declare Worldpoint = <0,-0.80, 0>; +#declare Lightsource = < 7,10,-3>; +#declare Lightdirection = vnormalize(Lightsource - Worldpoint); +#declare Lightaxis1 = vnormalize(vcross(Lightdirection, <0,1,0>)); +#declare Lightaxis2 = vnormalize(vcross(Lightaxis1, Lightdirection)); + +camera { + location Cameracenter + look_at Worldpoint + right 16/9 * x * imagescale + up y * imagescale +} + +light_source { + Lightsource color White + area_light Lightaxis1 Lightaxis2, 10, 10 + adaptive 1 + jitter +} + +sky_sphere { + pigment { + color White + } +} + +arrow( <-2.1, 0, 0 >, < 2.2, 0, 0 >, ar, White) +arrow( < 0, -1.1, 0 >, < 0, 1.3, 0 >, ar, White) +arrow( < 0, 0, -2 >, < 0, 0, 2.2 >, ar, White) + +#declare planecolor = rgb<0.2,0.6,1.0>; +#declare r = 0.01; + +#macro planebox() + box { <-2.1,-1.1,-2.1>, <0,1.1,2.1> } +#end + +intersection { + plane { <0, 0, 1>, 0.001 } + plane { <0, 0, -1>, 0.001 } + planebox() + pigment { + color planecolor transmit 0.3 + } + finish { + metallic + specular 0.95 + } +} + +#declare Xstep = 0.2; + +intersection { + union { + #declare X = 0; + #while (X > -2.5) + cylinder { , , r } + #declare X = X - Xstep; + #end + + #declare Y = Xstep; + #while (Y < 2.5) + cylinder { <-3, Y, 0>, <0, Y, 0>, r } + cylinder { <-3, -Y, 0>, <0, -Y, 0>, r } + #declare Y = Y + Xstep; + #end + } + planebox() + pigment { + color planecolor + } + finish { + metallic + specular 0.95 + } +} + +#declare parammin = -4; +#declare parammax = 4; +#declare paramsteps = 100; +#declare paramstep = (parammax - parammin) / paramsteps; + +#macro punkt(sigma, tau, Z) + < + 0.5 * (tau*tau - sigma*sigma) + Z, + sigma * tau, + > +#end + +#macro sigmasurface(sigma, farbe) + #declare taumin1 = 2/sigma; + #declare taumin2 = sqrt(4+sigma*sigma); + #if (taumin1 > taumin2) + #declare taumin = -taumin2; + #else + #declare taumin = -taumin1; + #end + + mesh { + #declare tau = taumin; + #declare taumax = -taumin; + #declare taustep = (taumax - taumin) / paramsteps; + #while (tau < taumax - taustep/2) + triangle { + punkt(sigma, tau, -1), + punkt(sigma, tau, 0), + punkt(sigma, tau + taustep, -1) + } + triangle { + punkt(sigma, tau + taustep, -1), + punkt(sigma, tau + taustep, 0), + punkt(sigma, tau, 0) + } + #declare tau = tau + taustep; + #end + pigment { + color farbe + } + finish { + specular 0.9 + metallic + } + } + + union { + #declare tau = taumin; + #declare taumax = -taumin; + #declare taustep = (taumax - taumin) / paramsteps; + #while (tau < taumax - taustep/2) + sphere { punkt(sigma, tau, 0), r } + cylinder { + punkt(sigma, tau, 0), + punkt(sigma, tau + taustep, 0), + r + } + #declare tau = tau + taustep; + #end + sphere { punkt(sigma, tau, 0), r } + pigment { + color farbe + } + finish { + specular 0.9 + metallic + } + + } +#end + +#declare greensurfacecolor = rgb<0.6,1.0,0.6>; +#declare redsurfacecolor = rgb<1.0,0.6,0.6>; + +sigmasurface(0.25, greensurfacecolor) +sigmasurface(0.5, greensurfacecolor) +sigmasurface(0.75, greensurfacecolor) +sigmasurface(1, greensurfacecolor) +sigmasurface(1.25, greensurfacecolor) +sigmasurface(1.5, greensurfacecolor) +sigmasurface(1.75, greensurfacecolor) +sigmasurface(2, greensurfacecolor) + +union { + sigmasurface(0.25, redsurfacecolor) + sigmasurface(0.5, redsurfacecolor) + sigmasurface(0.75, redsurfacecolor) + sigmasurface(1.00, redsurfacecolor) + sigmasurface(1.25, redsurfacecolor) + sigmasurface(1.5, redsurfacecolor) + sigmasurface(1.75, redsurfacecolor) + sigmasurface(2, redsurfacecolor) + rotate <0, 180, 0> +} + +box { <-2,-1,-2>, <2,-0.99,2> + pigment { + color rgb<1.0,0.8,0.6> transmit 0.8 + } + finish { + specular 0.9 + metallic + } +} diff --git a/buch/papers/parzyl/images/halfplane.tex b/buch/papers/parzyl/images/halfplane.tex new file mode 100644 index 0000000..e470057 --- /dev/null +++ b/buch/papers/parzyl/images/halfplane.tex @@ -0,0 +1,41 @@ +% +% halfplane.tex +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{times} +\usepackage{amsmath} +\usepackage{txfonts} +\usepackage[utf8]{inputenc} +\usepackage{graphics} +\usetikzlibrary{arrows,intersections,math} +\usepackage{ifthen} +\begin{document} + +\newboolean{showgrid} +\setboolean{showgrid}{false} +\def\breite{5} +\def\hoehe{4} + +\begin{tikzpicture}[>=latex,thick] + +% Povray Bild +\node at (0,0) {\includegraphics[width=10cm]{halfplane.jpg}}; + +% Gitter +\ifthenelse{\boolean{showgrid}}{ +\draw[step=0.1,line width=0.1pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw[step=0.5,line width=0.4pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw (-\breite,-\hoehe) grid (\breite, \hoehe); +\fill (0,0) circle[radius=0.05]; +}{} + +\node at (0,3.7) {$z$}; +\node at (3.3,-0.3) {$x$}; +\node at (2.7,2.5) {$y$}; + +\end{tikzpicture} + +\end{document} + -- cgit v1.2.1