From cb562f607b18540df333d6d9a911cf51c91884d0 Mon Sep 17 00:00:00 2001 From: Roy Seitz Date: Thu, 15 Apr 2021 23:49:49 +0200 Subject: Mehr Folien und einige Typos. --- vorlesungen/slides/10/matrix-vektor-dgl.tex | 44 ----- vorlesungen/slides/10/so2.tex | 138 ++++++++++++++ vorlesungen/slides/10/taylor.tex | 2 +- vorlesungen/slides/10/vektorfelder.mp | 266 +++++++++++++++++++++++++++ vorlesungen/slides/5/potenzreihenmethode.tex | 2 +- 5 files changed, 406 insertions(+), 46 deletions(-) create mode 100644 vorlesungen/slides/10/so2.tex create mode 100644 vorlesungen/slides/10/vektorfelder.mp (limited to 'vorlesungen/slides') diff --git a/vorlesungen/slides/10/matrix-vektor-dgl.tex b/vorlesungen/slides/10/matrix-vektor-dgl.tex index d9bd97c..f7bd995 100644 --- a/vorlesungen/slides/10/matrix-vektor-dgl.tex +++ b/vorlesungen/slides/10/matrix-vektor-dgl.tex @@ -6,50 +6,6 @@ % % !TeX spellcheck = de_CH \bgroup -%\begin{frame}[t] -%\setlength{\abovedisplayskip}{5pt} -%\setlength{\belowdisplayskip}{5pt} -%\frametitle{Matrix-Vektor-DGL} -%\vspace{-20pt} -%\begin{columns}[t,onlytextwidth] -%\begin{column}{0.48\textwidth} -% \begin{block}{Bekannt} -% Vorgehen für DGL 1.~Ordnung mit Skalaren. -% Aufgabe: Sei $a, x, x_0 \in \mathbb R$, -% \[ -% \dot x = ax, -% \quad -% x(0) = x_0 -% \] -% Lösung: $x(t) = \exp(at) x_0$, wobei -% \begin{align*} -% \exp(at) -% &= 1 + at + \frac{a^2t^2}{2!} + \ldots\\ -% &= e^{at} -% \end{align*} -% \end{block} -%\end{column} -%\begin{column}{0.48\textwidth} -% \begin{block}{Mit Matrizen} -% Wir können: -% \begin{itemize} -% \item Matrizen potenzieren: $A$, $A^2$, $A^3$ -% \item Matrizen skalieren: $At$ -% \item Matrizen addieren: $A_1 + A_2$ -% \end{itemize} -% Also ist auch -% \[ -% \exp(At) = 1 + At + \frac{A^2t^2}{2!} + \ldots -% \] -% wohldefiniert. -% \end{block} -%\end{column} -%\end{columns} -%Folglich, sei $A \in M_n$ und $x \in \mathbb R^n$, -%\[ \dot x = Ax, \quad x(0) = x_0, \] -%dann ist -%\[ x = \exp(At)x_0. \] -%\end{frame} \begin{frame}[t] \setlength{\abovedisplayskip}{5pt} diff --git a/vorlesungen/slides/10/so2.tex b/vorlesungen/slides/10/so2.tex new file mode 100644 index 0000000..e3f74ae --- /dev/null +++ b/vorlesungen/slides/10/so2.tex @@ -0,0 +1,138 @@ +% +% so2.tex -- Illustration of so(2) -> SO(2) +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% Erstellt durch Roy Seitz +% +% !TeX spellcheck = de_CH +\bgroup + +\newcommand{\gSL}[2]{\ensuremath{\text{SL}(#1, \mathbb{#2})}} +\newcommand{\gSO}[1]{\ensuremath{\text{SO}(#1)}} +\newcommand{\gGL}[2]{\ensuremath{\text{GL}(#1, \mathbb #2)}} + +\newcommand{\asl}[2]{\ensuremath{\mathfrak{sl}(#1, \mathbb{#2})}} +\newcommand{\aso}[1]{\ensuremath{\mathfrak{so}(#1)}} +\newcommand{\agl}[2]{\ensuremath{\mathfrak{gl}(#1, \mathbb #2)}} + +\begin{frame}[t] +\setlength{\abovedisplayskip}{5pt} +\setlength{\belowdisplayskip}{5pt} +\frametitle{Von der Lie-Gruppe zur -Algebra} +\vspace{-20pt} +\begin{columns}[t,onlytextwidth] +\begin{column}{0.48\textwidth} + \begin{block}{Lie-Gruppe} + Darstellung von \gSO2: + \begin{align*} + \mathbb R + &\to + \gSO2 + \\ + t + &\mapsto + \begin{pmatrix} + \cos t & -\sin t \\ + \sin t & \phantom-\cos t + \end{pmatrix} + \end{align*} + \end{block} + \begin{block}{Ableitung am neutralen Element} + \begin{align*} + \frac{d}{d t} + & + \left. + \begin{pmatrix} + \cos t & -\sin t \\ + \sin t & \phantom-\cos t + \end{pmatrix} + \right|_{ t = 0} + \\ + = + & + \begin{pmatrix} -\sin0 & -\cos0 \\ \phantom-\cos0 & -\sin0 \end{pmatrix} + = + \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix} + \end{align*} + \end{block} +\end{column} +\begin{column}{0.48\textwidth} + \begin{block}{Lie-Algebra} + Darstellung von \aso2: + \begin{align*} + \mathbb R + &\to + \aso2 + \\ + t + &\mapsto + \begin{pmatrix} + 0 & -t \\ + t & \phantom-0 + \end{pmatrix} + \end{align*} + \end{block} +\end{column} +\end{columns} +\end{frame} + + +\begin{frame}[t] +\setlength{\abovedisplayskip}{5pt} +\setlength{\belowdisplayskip}{5pt} +\frametitle{Von der Lie-Algebra zur -Gruppe} +\vspace{-20pt} +\begin{columns}[t,onlytextwidth] +\begin{column}{0.48\textwidth} + \begin{block}{Differentialgleichung} + Gegeben: + \[ + A + = + \dot\gamma(0) = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix} + \] + Gesucht: + \[ \dot \gamma (t) = \gamma(t) A \qquad \gamma \in \gSO2 \] + \[ \Rightarrow \gamma(t) = \exp(At) \gamma(0) = \exp(At) \] + \end{block} +\end{column} +\begin{column}{0.48\textwidth} + \begin{block}{Lie-Algebra} + Potenzen von A: + \begin{align*} + A^2 &= -I & + A^3 &= -A & + A^4 &= I & + \ldots + \end{align*} + \end{block} +\end{column} +\end{columns} +Folglich: +\begin{align*} + \exp(At) + &= I + At + + A^2\frac{t^2}{2!} + + A^3\frac{t^3}{3!} + + A^4\frac{t^4}{4!} + + A^5\frac{t^5}{5!} + + \ldots \\ + &= \begin{pmatrix} + \vspace*{3pt} + 1 - \frac{t^2}{2} + \frac{t^4}{4!} - \ldots + & + -t + \frac{t^3}{3!} - \frac{t^5}{5!} + \ldots + \\ + t - \frac{t^3}{3!} + \frac{t^5}{5!} - \ldots + & + 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \ldots + \end{pmatrix} + = + \begin{pmatrix} + \cos t & -\sin t \\ + \sin t & \phantom-\cos t + \end{pmatrix} +\end{align*} + +\end{frame} +\egroup diff --git a/vorlesungen/slides/10/taylor.tex b/vorlesungen/slides/10/taylor.tex index bbd1126..920470f 100644 --- a/vorlesungen/slides/10/taylor.tex +++ b/vorlesungen/slides/10/taylor.tex @@ -1,5 +1,5 @@ % -% eindiomensional.tex -- Lösung der eindimensionalen DGL +% taylor.tex -- Repetition Taylot-Reihen % % (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule % Erstellt durch Roy Seitz diff --git a/vorlesungen/slides/10/vektorfelder.mp b/vorlesungen/slides/10/vektorfelder.mp new file mode 100644 index 0000000..f488327 --- /dev/null +++ b/vorlesungen/slides/10/vektorfelder.mp @@ -0,0 +1,266 @@ +% +% Stroemungsfelder linearer Differentialgleichungen +% +% (c) 2015 Prof Dr Andreas Mueller, Hochschule Rapperswil +% 2021-04-14, Roy Seitz, Copied for SeminarMatrizen +% +verbatimtex +\documentclass{book} +\usepackage{times} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{amsfonts} +\usepackage{txfonts} +\begin{document} +etex; + +input TEX; + +TEXPRE("%&latex" & char(10) & +"\documentclass{book}" & +"\usepackage{times}" & +"\usepackage{amsmath}" & +"\usepackage{amssymb}" & +"\usepackage{amsfonts}" & +"\usepackage{txfonts}" & +"\begin{document}"); +TEXPOST("\end{document}"); + +% +% Vektorfeld in der Ebene mit Lösungskurve +% so(2) +% +beginfig(1) + +% Scaling parameter +numeric unit; +unit := 150; + +% Some points +z1 = (-1.5, 0) * unit; +z2 = ( 1.5, 0) * unit; +z3 = ( 0, -1.5) * unit; +z4 = ( 0, 1.5) * unit; + +pickup pencircle scaled 1pt; +drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0)); +drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); +label.top(btex $x_1$ etex, z2 shifted (10,0)); +label.rt(btex $x_2$ etex, z4 shifted (0,10)); + +% Draw circles +for x = 0.2 step 0.2 until 1.4: + path p; + p = (x,0); + for a = 5 step 5 until 355: + p := p--(x*cosd(a), x*sind(a)); + endfor; + p := p--cycle; + pickup pencircle scaled 1pt; + draw p scaled unit withcolor red; +endfor; + +% Define DGL +def dglField(expr x, y) = + %(-0.5 * (x + y), -0.5 * (y - x)) + (-y, x) +enddef; + +% Draw arrows for each grid point +pickup pencircle scaled 0.5pt; +for x = -1.5 step 0.1 until 1.55: + for y = -1.5 step 0.1 until 1.55: + drawarrow ((x, y) * unit) + --(((x,y) * unit) shifted (8 * dglField(x,y))) + withcolor blue; + endfor; +endfor; + +endfig; + + + +% +% Vektorfeld in der Ebene mit Lösungskurve +% X \in sl(2, R) +% +beginfig(2) + +numeric unit; +unit := 150; + +z0 = ( 0, 0); +z1 = (-1.5, 0) * unit; +z2 = ( 1.5, 0) * unit; +z3 = ( 0, -1.5) * unit; +z4 = ( 0, 1.5) * unit; + +pickup pencircle scaled 1pt; +drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0)); +drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); +label.top(btex $x_1$ etex, z2 shifted (10,0)); +label.rt(btex $x_2$ etex, z4 shifted (0,10)); + +% Draw flow lines +for y = -1.4 step 0.2 until 1.4: + path p; + p = (-1.5,y) -- (1.5, y); + pickup pencircle scaled 1pt; + draw p scaled unit withcolor red; +endfor; + +def dglField(expr x, y) = + (y, 0) +enddef; + +% Draw arrows for each grid point +pickup pencircle scaled 0.5pt; +for x = -1.5 step 0.1 until 1.55: + for y = -1.5 step 0.1 until 1.55: + drawarrow ((x, y) * unit) + --(((x,y) * unit) shifted (8 * dglField(x,y))) + withcolor blue; + endfor; +endfor; + +endfig; + + + + +% +% Vektorfeld in der Ebene mit Lösungskurve +% Y \in sl(2, R) +% +beginfig(3) + +numeric unit; +unit := 150; + +z0 = ( 0, 0); +z1 = (-1.5, 0) * unit; +z2 = ( 1.5, 0) * unit; +z3 = ( 0, -1.5) * unit; +z4 = ( 0, 1.5) * unit; + +pickup pencircle scaled 1pt; +drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0)); +drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); +label.top(btex $x_1$ etex, z2 shifted (10,0)); +label.rt(btex $x_2$ etex, z4 shifted (0,10)); + +% Draw flow lines +for x = -1.4 step 0.2 until 1.4: + path p; + p = (x, -1.5) -- (x, 1.5); + pickup pencircle scaled 1pt; + draw p scaled unit withcolor red; +endfor; + +def dglField(expr x, y) = + (0, x) +enddef; + +% def dglFieldp(expr z) = +% dglField(xpart z, ypart z) +% enddef; +% +% def curve(expr z, l) = +% path p; +% p := z; +% for t = 0 step 1 until l: +% p := p--((point (length p) of p) shifted (0.01 * dglFieldp(point (length p) of p))); +% endfor; +% draw p scaled unit withcolor red; +% enddef; +% +% numeric outerlength; +% outerlength = 200; +% curve(( 0.1, 0), outerlength); +% curve(( 0.2, 0), outerlength); +% +% numeric innerlength; +% innerlength = 500; +% +% for a = 0 step 30 until 330: +% curve(0.05 * (cosd(a), sind(a)), innerlength); +% endfor; + +% Draw arrows for each grid point +pickup pencircle scaled 0.5pt; +for x = -1.5 step 0.1 until 1.55: + for y = -1.5 step 0.1 until 1.55: + drawarrow ((x, y) * unit) + --(((x,y) * unit) shifted (8 * dglField(x,y))) + withcolor blue; + endfor; +endfor; + +endfig; + + +% +% Vektorfeld in der Ebene mit Lösungskurve +% H \in sl(2, R) +% +beginfig(4) + +numeric unit; +unit := 150; + +z0 = ( 0, 0); +z1 = (-1.5, 0) * unit; +z2 = ( 1.5, 0) * unit; +z3 = ( 0, -1.5) * unit; +z4 = ( 0, 1.5) * unit; + +pickup pencircle scaled 1pt; +drawarrow (z1 shifted (-25,0))--(z2 shifted (25,0)); +drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10)); +label.top(btex $x_1$ etex, z2 shifted (25,0)); +label.rt(btex $x_2$ etex, z4 shifted (0,10)); + +def dglField(expr x, y) = + (x, -y) +enddef; + +def dglFieldp(expr z) = + dglField(xpart z, ypart z) +enddef; + +def curve(expr z, l) = + path p; + p := z; + for t = 0 step 1 until l: + p := p--((point (length p) of p) shifted (0.01 * dglFieldp(point (length p) of p))); + endfor; + draw p scaled unit withcolor red; +enddef; + +for i = -1 step 2 until 1: + for k = -1 step 2 until 1: + curve((1.3 * i, 1.5 * k), 18); + curve((1.1 * i, 1.5 * k), 35); + curve((0.9 * i, 1.5 * k), 55); + curve((0.7 * i, 1.5 * k), 80); + curve((0.5 * i, 1.5 * k), 114); + curve((0.3 * i, 1.5 * k), 165); + curve((0.1 * i, 1.5 * k), 275); + endfor; +endfor; + +% Draw arrows for each grid point +pickup pencircle scaled 0.5pt; +for x = -1.5 step 0.1 until 1.55: + for y = -1.5 step 0.1 until 1.55: + drawarrow ((x, y) * unit) + --(((x,y) * unit) shifted (8 * dglField(x,y))) + withcolor blue; + endfor; +endfor; + +endfig; + + + +end; diff --git a/vorlesungen/slides/5/potenzreihenmethode.tex b/vorlesungen/slides/5/potenzreihenmethode.tex index 0c3503d..12d3fa5 100644 --- a/vorlesungen/slides/5/potenzreihenmethode.tex +++ b/vorlesungen/slides/5/potenzreihenmethode.tex @@ -79,7 +79,7 @@ a_k=\frac1{k!}a^kC} \\ \uncover<4->{ \Rightarrow y(x) &= C}\uncover<8->{+Cax}\uncover<9->{ + C\frac12(ax)^2} -\uncover<10->{ + C \frac16(ac)^3} +\uncover<10->{ + C \frac16(ax)^3} \uncover<11->{ + \dots+C\frac{1}{k!}(ax)^k+\dots} \ifthenelse{\boolean{presentation}}{ \only<12>{ -- cgit v1.2.1