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| author | haddoucher <reda.haddouche@ost.ch> | 2022-08-16 14:36:07 +0200 |
|---|---|---|
| committer | haddoucher <reda.haddouche@ost.ch> | 2022-08-16 14:36:07 +0200 |
| commit | f031b148a79d1dafb0e3405643be05e7a7eb1222 (patch) | |
| tree | 3f0c7d27ba90e10874a885cae9026fdbaa0b01e6 | |
| parent | Update tschebyscheff_beispiel.tex (diff) | |
| parent | Merge pull request #5 from haddoucher/sturmliouville/erik-branch (diff) | |
| download | SeminarSpezielleFunktionen-f031b148a79d1dafb0e3405643be05e7a7eb1222.tar.gz SeminarSpezielleFunktionen-f031b148a79d1dafb0e3405643be05e7a7eb1222.zip | |
Merge remote-tracking branch 'origin/master' into sturmliouville/redabranch
165 files changed, 17286 insertions, 1750 deletions
diff --git a/buch/chapters/030-geometrie/hyperbolisch.tex b/buch/chapters/030-geometrie/hyperbolisch.tex index 2938316..d2d0da2 100644 --- a/buch/chapters/030-geometrie/hyperbolisch.tex +++ b/buch/chapters/030-geometrie/hyperbolisch.tex @@ -163,9 +163,9 @@ In der speziellen Relativitätstheorie spielt das Minkowski-Skalarprodukt eine besondere Rolle. Die Koordinaten $x_0$ hat darin die Bedeutung der Zeit, man weiss aus Experimenten wie dem Michelson-Morley-Experiment, -dass die Grösse $\langle x,x\rangle$ ist eine Invariante ist. +dass die Grösse $\langle x,x\rangle$ eine Invariante ist. Die Transformationen mit der Matrix $A$ beschreiben also zulässige -Koordinatentransformationenn, die Invariante erhalten. +Koordinatentransformationen, die Invariante erhalten. Für Transformationen, die zusätzlich die Zeitrichtung erhalten sollen, muss $a_{00}=a_{11}=c>0$ verlangt werden. @@ -174,7 +174,7 @@ muss $a_{00}=a_{11}=c>0$ verlangt werden. Unter der Annahme $c>0$ lässt sich die Matrix vollständig durch den Parameter $t=s/c$ beschreiben. Dividiert man \eqref{buch:geometrie:hyperbolish:eqn:cs} durch $c^2$, -kann $c$ durch $t$ ausdrücken: +kann man $c$ durch $t$ ausdrücken: \[ \frac{1}{c^2} = @@ -199,10 +199,10 @@ H_t t&1 \end{pmatrix}. \] -Diese Formeln erinnern natürlich and die Formeln, mit denen +Diese Formeln erinnern natürlich an die Formeln, mit denen der hyperbolische Sinus und Kosinus aus dem hyperbolischen -Tangens berechnet werden kann. -Dieser Zusammenhang und soll im nächsten Abschnitt hergestellt +Tangens berechnet werden können. +Dieser Zusammenhang soll im nächsten Abschnitt hergestellt werden. % diff --git a/buch/chapters/075-fourier/bessel.tex b/buch/chapters/075-fourier/bessel.tex index 7e978f7..db7880b 100644 --- a/buch/chapters/075-fourier/bessel.tex +++ b/buch/chapters/075-fourier/bessel.tex @@ -454,7 +454,8 @@ Terme mit $\pm n$ können wegen \[ \left. \begin{aligned} -J_{-n}(\xi) &= (-1)^n J_n(\xi) +J_{-n}(\xi) &= (-1)^n J_n(\xi) +\label{buch:fourier:eqn:symetrie} \\ i^{-n}&=(-1)^n i^n \end{aligned} diff --git a/buch/chapters/110-elliptisch/uebungsaufgaben/1.tex b/buch/chapters/110-elliptisch/uebungsaufgaben/1.tex index af094c6..2d08e56 100644 --- a/buch/chapters/110-elliptisch/uebungsaufgaben/1.tex +++ b/buch/chapters/110-elliptisch/uebungsaufgaben/1.tex @@ -25,7 +25,7 @@ Auslenkung. Formulieren Sie den Energieerhaltungssatz für die Gesamtenergie $E$ dieses Oszillators. Leiten Sie daraus eine nichtlineare Differentialgleichung erster Ordnung -for den anharmonischen Oszillator ab, die sie in der Form +für den anharmonischen Oszillator ab, die sie in der Form $\frac12m\dot{x}^2 = f(x)$ schreiben. \item Die Amplitude der Schwingung ist derjenige $x$-Wert, für den die diff --git a/buch/common/macros.tex b/buch/common/macros.tex index bb6e9b0..e37698e 100644 --- a/buch/common/macros.tex +++ b/buch/common/macros.tex @@ -111,6 +111,10 @@ \newtheorem{forderung}{Forderung}[chapter] \newtheorem{konsequenz}[satz]{Konsequenz} \newtheorem{algorithmus}[satz]{Algorithmus} + +% English variants +\newtheorem{theorem}[satz]{Theorem} + \renewcommand{\floatpagefraction}{0.7} \definecolor{darkgreen}{rgb}{0,0.6,0} diff --git a/buch/papers/0f1/images/airy.pdf b/buch/papers/0f1/images/airy.pdf Binary files differdeleted file mode 100644 index 672d789..0000000 --- a/buch/papers/0f1/images/airy.pdf +++ /dev/null diff --git a/buch/papers/0f1/images/konvergenzAiry.pdf b/buch/papers/0f1/images/konvergenzAiry.pdf Binary files differindex 2e635ea..206cd3a 100644 --- a/buch/papers/0f1/images/konvergenzAiry.pdf +++ b/buch/papers/0f1/images/konvergenzAiry.pdf diff --git a/buch/papers/0f1/images/konvergenzNegativ.pdf b/buch/papers/0f1/images/konvergenzNegativ.pdf Binary files differindex 3b58be4..03b2ba1 100644 --- a/buch/papers/0f1/images/konvergenzNegativ.pdf +++ b/buch/papers/0f1/images/konvergenzNegativ.pdf diff --git a/buch/papers/0f1/images/konvergenzPositiv.pdf b/buch/papers/0f1/images/konvergenzPositiv.pdf Binary files differindex 24e3fd5..2e45129 100644 --- a/buch/papers/0f1/images/konvergenzPositiv.pdf +++ b/buch/papers/0f1/images/konvergenzPositiv.pdf diff --git a/buch/papers/0f1/images/stabilitaet.pdf b/buch/papers/0f1/images/stabilitaet.pdf Binary files differindex be4af42..13dea39 100644 --- a/buch/papers/0f1/images/stabilitaet.pdf +++ b/buch/papers/0f1/images/stabilitaet.pdf diff --git a/buch/papers/0f1/listings/kettenbruchIterativ.c b/buch/papers/0f1/listings/kettenbruchIterativ.c index befea8e..d897b8f 100644 --- a/buch/papers/0f1/listings/kettenbruchIterativ.c +++ b/buch/papers/0f1/listings/kettenbruchIterativ.c @@ -1,5 +1,13 @@ -static double fractionRekursion0f1(const double c, const double x, unsigned int n)
+/**
+ * @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 fractionRekursion0f1(const double c, const double z, unsigned int n)
{
+ //declaration
double a = 0.0;
double b = 0.0;
double Ak = 0.0;
@@ -21,15 +29,15 @@ static double fractionRekursion0f1(const double c, const double x, unsigned int else if (k == 1)
{
a = 1.0; //a1
- b = x/c; //b1
+ b = z/c; //b1
//recursion fomula for A1, B1
Ak = a * Ak_1 + b * 1.0;
Bk = a * Bk_1;
}
else
{
- a = 1 + (x / (k * ((k - 1) + c)));//ak
- b = -(x / (k * ((k - 1) + c))); //bk
+ 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;
diff --git a/buch/papers/0f1/listings/kettenbruchRekursion.c b/buch/papers/0f1/listings/kettenbruchRekursion.c index 958d4e1..3caaf43 100644 --- a/buch/papers/0f1/listings/kettenbruchRekursion.c +++ b/buch/papers/0f1/listings/kettenbruchRekursion.c @@ -1,19 +1,27 @@ -static double fractionIter0f1(const double b0, const double z, unsigned int n)
+/**
+ * @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 fractionIter0f1(const double c, const double z, unsigned int k)
{
+ //declaration
double a = 0.0;
double b = 0.0;
- double abn = 0.0;
+ double abk = 0.0;
double temp = 0.0;
- for (; n > 0; --n)
+ for (; k > 0; --k)
{
- abn = z / (n * ((n - 1) + b0)); //abn = ak, bk
+ abk = z / (k * ((k - 1) + c)); //abk = ak, bk
- a = n > 1 ? (1 + abn) : 1; //a0, a1
- b = n > 1 ? -abn : abn; //b1
+ a = k > 1 ? (1 + abk) : 1; //a0, a1
+ b = k > 1 ? -abk : abk; //b1
- temp = b / (a + temp);
+ temp = b / (a + temp); //bk / (ak + last result)
}
- return a + temp; //a0 + temp
+ return a + temp; //a0 + temp
}
\ No newline at end of file diff --git a/buch/papers/0f1/listings/potenzreihe.c b/buch/papers/0f1/listings/potenzreihe.c index bfaa0e3..23fdfea 100644 --- a/buch/papers/0f1/listings/potenzreihe.c +++ b/buch/papers/0f1/listings/potenzreihe.c @@ -1,12 +1,68 @@ #include <math.h>
-static double powerseries(const double b, const double z, unsigned int n)
+/**
+ * @brief Calculates pochhammer
+ * @param (a+n-1)!
+ * @return Result
+ */
+static double pochhammer(const double x, double n)
+{
+ double temp = x;
+
+ if (n > 0)
+ {
+ while (n > 1)
+ {
+ temp *= (x + n - 1);
+ --n;
+ }
+
+ return temp;
+ }
+ else
+ {
+ return 1;
+ }
+}
+
+/**
+ * @brief Calculates the Factorial
+ * @param n!
+ * @return Result
+ */
+static double fac(int n)
+{
+ double temp = n;
+
+ if (n > 0)
+ {
+ while (n > 1)
+ {
+ --n;
+ temp *= n;
+ }
+ return temp;
+ }
+ else
+ {
+ return 1;
+ }
+}
+
+/**
+ * @brief Calculates the Hypergeometric Function 0F1(;b;z)
+ * @param c in 0F1(;c;z)
+ * @param z in 0F1(;c;z)
+ * @param n number of itertions (precision)
+ * @return Result
+ */
+static double powerseries(const double c, const double z, unsigned int n)
{
double temp = 0.0;
for (unsigned int k = 0; k < n; ++k)
{
- temp += pow(z, k) / (factorial(k) * pochhammer(b, k));
+ temp += pow(z, k) / (factorial(k) * pochhammer(c, k));
}
return temp;
diff --git a/buch/papers/0f1/main.tex b/buch/papers/0f1/main.tex index b8cdc21..0b1020f 100644 --- a/buch/papers/0f1/main.tex +++ b/buch/papers/0f1/main.tex @@ -1,24 +1,24 @@ -% -% main.tex -- Paper zum Thema <0f1> -% -% (c) 2020 Hochschule Rapperswil -% -% - - - -\chapter{Algorithmus zur Berechnung von $\mathstrut_0F_1$\label{chapter:0f1}} -\lhead{Algorithmus zur Berechnung von $\mathstrut_0F_1$} -\begin{refsection} -\chapterauthor{Fabian Dünki} - - - - -\input{papers/0f1/teil0.tex} -\input{papers/0f1/teil1.tex} -\input{papers/0f1/teil2.tex} -\input{papers/0f1/teil3.tex} - -\printbibliography[heading=subbibliography] -\end{refsection} +%
+% main.tex -- Paper zum Thema <0f1>
+%
+% (c) 2020 Hochschule Rapperswil
+%
+%
+
+
+
+\chapter{Algorithmus zur Berechnung von $\mathstrut_0F_1$\label{chapter:0f1}}
+\lhead{Algorithmus zur Berechnung von $\mathstrut_0F_1$}
+\begin{refsection}
+\chapterauthor{Fabian Dünki}
+
+
+
+
+\input{papers/0f1/teil0.tex}
+\input{papers/0f1/teil1.tex}
+\input{papers/0f1/teil2.tex}
+\input{papers/0f1/teil3.tex}
+
+\printbibliography[heading=subbibliography]
+\end{refsection}
diff --git a/buch/papers/0f1/references.bib b/buch/papers/0f1/references.bib index fb9cd8b..47555da 100644 --- a/buch/papers/0f1/references.bib +++ b/buch/papers/0f1/references.bib @@ -4,32 +4,82 @@ % (c) 2020 Autor, Hochschule Rapperswil % -@online{0f1:bibtex, - title = {BibTeX}, - url = {https://de.wikipedia.org/wiki/BibTeX}, - date = {2020-02-06}, - year = {2020}, - month = {2}, - day = {6} -} - -@book{0f1: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} -} - -@article{0f1: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{0f1:library-gsl, + title = {GNU Scientific Library}, + url ={https://www.gnu.org/software/gsl/}, + date = {2022-07-07}, + year = {2022}, + month = {7}, + day = {7} } +@online{0f1:wiki-airyFunktion, + title = {Airy-Funktion}, + url ={https://de.wikipedia.org/wiki/Airy-Funktion}, + date = {2022-07-07}, + year = {2022}, + month = {7}, + day = {7} +} + +@online{0f1:wiki-kettenbruch, + title = {Kettenbruch}, + url ={https://de.wikipedia.org/wiki/Kettenbruch}, + date = {2022-07-07}, + year = {2022}, + month = {7}, + day = {25} +} + +@online{0f1:double, + title = {C - Data Types}, + url ={https://www.tutorialspoint.com/cprogramming/c_data_types.htm}, + date = {2022-07-07}, + year = {2022}, + month = {7}, + day = {7} +} + +@online{0f1:wolfram-0f1, + title = {Hypergeometric 0F1}, + url ={https://functions.wolfram.com/webMathematica/FunctionEvaluation.jsp?name=Hypergeometric0F1}, + date = {2022-07-07}, + year = {2022}, + month = {7}, + day = {7} +} + +@online{0f1:wiki-fraction, + title = {Gauss continued fraction}, + url ={https://en.wikipedia.org/wiki/Gauss%27s_continued_fraction}, + date = {2022-07-07}, + year = {2022}, + month = {7}, + day = {7} +} + +@online{0f1:code, + title = {Vollständiger C-Code}, + url ={https://github.com/AndreasFMueller/SeminarSpezielleFunktionen/tree/master/buch/papers/0f1/listings}, + date = {2022-07-07}, + year = {2022}, + month = {7}, + day = {7} +} + +@book{0f1:SeminarNumerik, + title = {Mathematisches Seminar Numerik}, + author = {Andreas Müller et al}, + publisher = {Andreas Müller}, + year = {2022}, +} + +@article{0f1:kettenbrueche, + author = { Benjamin Bouhafs-Keller }, + title = { Kettenbrüche }, + journal = { Mathematisches Seminar Numerik }, + year = 2020, + volume = 13, + pages = {363--376}, + url = {https://github.com/AndreasFMueller/SeminarNumerik} +} diff --git a/buch/papers/0f1/teil0.tex b/buch/papers/0f1/teil0.tex index bfc265f..adccac7 100644 --- a/buch/papers/0f1/teil0.tex +++ b/buch/papers/0f1/teil0.tex @@ -1,15 +1,15 @@ -% -% einleitung.tex -- Einleitung -% -% (c) 2022 Fabian Dünki, Hochschule Rapperswil -% -\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. -In der GNU Scientific Library \cite{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. -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. +%
+% einleitung.tex -- Einleitung
+%
+% (c) 2022 Fabian Dünki, Hochschule Rapperswil
+%
+\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.
+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.
+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 910e8bb..f697f45 100644 --- a/buch/papers/0f1/teil1.tex +++ b/buch/papers/0f1/teil1.tex @@ -1,80 +1,101 @@ -% -% teil1.tex -- Mathematischer Hintergrund -% -% (c) 2022 Fabian Dünki, Hochschule Rapperswil -% -\section{Mathematischer Hintergrund -\label{0f1:section:mathHintergrund}} -\rhead{Mathematischer Hintergrund} - -\subsection{Hypergeometrische Funktion $\mathstrut_0F_1$ -\label{0f1:subsection:0f1}} -Wie in Kapitel \ref{buch:rekursion:section:hypergeometrische-funktion} beschrieben, -wird die Funktion $\mathstrut_0F_1$ folgendermassen definiert. -\begin{definition} - \label{0f1:rekursion:hypergeometrisch:def} - Die hypergeometrische Funktion - $\mathstrut_0F_1$ ist definiert durch die Reihe - \[ - \mathstrut_0F_1 - \biggl( - \begin{matrix} - \\ - b_1 - \end{matrix} - ; - x - \biggr) - = - \mathstrut_0F_1(;b_1;x) - = - \sum_{k=0}^\infty - \frac{1}{(b_1)_k}\frac{x^k}{k!}. - \] -\end{definition} - - -\subsection{Airy Funktion -\label{0f1:subsection:airy}} -Wie in \ref{buch:differentialgleichungen:section:hypergeometrisch} dargestellt, ist die Airy-Differentialgleichung -folgendermassen definiert. -\begin{definition} - y'' - xy = 0 - \label{0f1:airy:eq:differentialgleichung} -\end{definition} - -Daraus ergibt sich wie in Aufgabe~\ref{503} gefundenen Lösungen der -Airy-Differentialgleichung als hypergeometrische Funktionen. - - -\begin{align*} -y_1(x) -= -\sum_{k=0}^\infty -\frac{1}{(\frac23)_k} \frac{1}{k!}\biggl(\frac{x^3}{9}\biggr)^k -= -\mathstrut_0F_1\biggl( -\begin{matrix}\text{---}\\\frac23\end{matrix};\frac{x^3}{9} -\biggr). -\\ -y_2(x) -= -\sum_{k=0}^\infty -\frac{1}{(\frac43)_k} \frac{1}{k!}\biggl(\frac{x^3}{9}\biggr)^k -= -x\cdot\mathstrut_0F_1\biggl( -\begin{matrix}\text{---}\\\frac43\end{matrix}; -\frac{x^3}{9} -\biggr). -\qedhere -\end{align*} - - -\begin{figure} - \centering - \includegraphics{papers/0f1/images/airy.pdf} - \caption{Plot der Lösungen der Airy-Differentialgleichung $y''-xy=0$ - zu den Anfangsbedingungen $y(0)=1$ und $y'(0)=0$ in {\color{red}rot} - und $y(0)=0$ und $y'(0)=1$ in {\color{blue}blau}. - \label{0f1:airy:plot:vorgabe}} -\end{figure}
\ No newline at end of file +%
+% teil1.tex -- Mathematischer Hintergrund
+%
+% (c) 2022 Fabian Dünki, Hochschule Rapperswil
+%
+\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
+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 $\mathstrut_pF_q$.
+
+\begin{definition}
+ \label{0f1:math:qFp:def}
+ Die hypergeometrische Funktion
+ $\mathstrut_pF_q$ ist definiert durch die Reihe
+ \[
+ \mathstrut_pF_q
+ \biggl(
+ \begin{matrix}
+ a_1,\dots,a_p\\
+ b_1,\dots,b_q
+ \end{matrix}
+ ;
+ x
+ \biggr)
+ =
+ \mathstrut_pF_q(a_1,\dots,a_p;b_1,\dots,b_q;x)
+ =
+ \sum_{k=0}^\infty
+ \frac{(a_1)_k\cdots(a_p)_k}{(b_1)_k\cdots(b_q)_k}\frac{x^k}{k!}.
+ \]
+\end{definition}
+
+Angewendet auf die Funktion $\mathstrut_pF_q$ ergibt sich für $\mathstrut_0F_1$:
+
+\begin{equation}
+ \label{0f1:math:0f1:eq}
+ \mathstrut_0F_1
+ \biggl(
+ \begin{matrix}
+ \\-
+ b_1
+ \end{matrix}
+ ;
+ x
+ \biggr)
+ =
+ \mathstrut_0F_1(;b_1;x)
+ =
+ \sum_{k=0}^\infty
+ \frac{x^k}{(b_1)_k \cdot k!}.
+\end{equation}
+
+
+
+
+\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}.
+
+\begin{definition}
+ \label{0f1:airy:differentialgleichung:def}
+ Die Differentialgleichung
+ $y'' - xy = 0$
+ 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 $Ai(0)=1$ 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
+=
+\mathstrut_0F_1\biggl(
+\begin{matrix}\text{---}\\\frac23\end{matrix};\frac{x^3}{9}
+\biggr).
+\\
+Bi(x)
+=&
+\sum_{k=0}^\infty
+\frac{1}{(\frac43)_k} \frac{1}{k!}\biggl(\frac{x^3}{9}\biggr)^k
+=
+x\cdot\mathstrut_0F_1\biggl(
+\begin{matrix}\text{---}\\\frac43\end{matrix};
+\frac{x^3}{9}
+\biggr).
+\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}
+benutzt.
+
+
diff --git a/buch/papers/0f1/teil2.tex b/buch/papers/0f1/teil2.tex index 07e17c0..15a1c44 100644 --- a/buch/papers/0f1/teil2.tex +++ b/buch/papers/0f1/teil2.tex @@ -1,75 +1,171 @@ -% -% teil2.tex -- Umsetzung in C Programmen -% -% (c) 2022 Fabian Dünki, Hochschule Rapperswil -% -\section{Umsetzung -\label{0f1:section:teil2}} -\rhead{Umsetzung} -Zur Umsetzung wurden drei Ansätze gewählt und -Die Unterprogramme wurde jeweils, wie die GNU Scientific Library, in C geschrieben. - -\subsection{Potenzreihe -\label{0f1:subsection:potenzreihe}} -Die naheliegendste Lösung ist die Programmierung der Potenzreihe. - -\begin{equation} - \label{0f1:rekursion:hypergeometrisch:eq} - \mathstrut_0F_1(;b;z) - = - \sum_{k=0}^\infty - \frac{z^k}{(b)_k \cdot k!} -\end{equation} - -\lstinputlisting[style=C,float,caption={Rekursivformel für Kettenbruch.},label={0f1:listing:potenzreihe}]{papers/0f1/listings/potenzreihe.c} - -\subsection{Kettenbruch -\label{0f1:subsection:kettenbruch}} -Ein endlicher Kettenbruch ist ein Bruch der Form -\begin{equation} -a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{\cdots}{\cdots+\cfrac{b_{n-1}}{a_{n-1} + \cfrac{b_n}{a_n}}}}} -\end{equation} -in welchem $a_0, a_1,\dots,a_n$ und $b_1,b_2,\dots,b_n$ ganze Zahlen -darstellen. - -{\color{red}TODO: Bessere Beschreibung mit Verknüpfung zur Potenzreihe} - -%Gauss hat durch - -\lstinputlisting[style=C,float,caption={Rekursivformel für Kettenbruch.},label={0f1:listing:kettenbruchIterativ}]{papers/0f1/listings/kettenbruchIterativ.c} -\subsection{Rekursionsformel -\label{0f1:subsection:rekursionsformel}} -Wesentlich effizienter zur Berechnung eines Kettenbruches ist die Rekursionsformel. - -\begin{align*} -\frac{A_n}{B_n} -= -a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{\cdots}{\cdots+\cfrac{b_{n-1}}{a_{n-1} + \cfrac{b_n}{a_n}}}}} -\end{align*} - -Die Berechnung von $A_n, B_n$ kann man auch ohne die Matrizenschreibweise -aufschreiben: -\begin{itemize} -\item Start: -\begin{align*} -A_{-1} &= 0 & A_0 &= a_0 \\ -B_{-1} &= 1 & B_0 &= 1 -\end{align*} -$\rightarrow$ 0-te Näherung: $\displaystyle\frac{A_0}{B_0} = a_0$ -\item Schritt $k\to k+1$: -\[ -\begin{aligned} -k &\rightarrow k + 1: -& -A_{k+1} &= A_{k-1} \cdot b_k + A_k \cdot a_k \\ -&& -B_{k+1} &= B_{k-1} \cdot b_k + B_k \cdot a_k -\end{aligned} -\] -\item -Näherungsbruch $n$: \qquad$\displaystyle\frac{A_n}{B_n}$ -\end{itemize} -{\color{red}TODO: Verweis Numerik} - - -\lstinputlisting[style=C,float,caption={Rekursivformel für Kettenbruch.},label={0f1:listing:kettenbruchRekursion}]{papers/0f1/listings/kettenbruchRekursion.c}
\ No newline at end of file +%
+% teil2.tex -- Umsetzung in C Programmen
+%
+% (c) 2022 Fabian Dünki, Hochschule Rapperswil
+%
+\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-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}.
+
+\begin{align}
+ \label{0f1:umsetzung:0f1:eq}
+ \mathstrut_0F_1(;c;z)
+ &=
+ \sum_{k=0}^\infty
+ \frac{z^k}{(c)_k \cdot k!}
+ &=
+ \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}}
+\end{align}
+
+\lstinputlisting[style=C,float,caption={Potenzreihe.},label={0f1:listing:potenzreihe}, firstline=59]{papers/0f1/listings/potenzreihe.c}
+
+\subsection{Kettenbruch
+\label{0f1:subsection:kettenbruch}}
+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 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*}
+\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
+\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}
+
+\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}
+
+\subsubsection{Herleitung}
+Ein Näherungsbruch in der Form
+\begin{align*}
+ \cfrac{A_k}{B_k} = a_k + \cfrac{b_{k + 1}}{a_{k + 1} + \cfrac{p}{q}}
+\end{align*}
+lässt sich zu
+\begin{align*}
+ \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:
+\begin{equation*}
+ \begin{pmatrix}
+ A_k\\
+ B_k
+ \end{pmatrix}
+ = \begin{pmatrix}
+ b_{k+1} \cdot q\\
+ a_{k+1} \cdot q + p
+ \end{pmatrix}
+ =\begin{pmatrix}
+ 0& b_{k+1}\\
+ 1& a_{k+1}
+ \end{pmatrix}
+ \begin{pmatrix}
+ p \\
+ q
+ \end{pmatrix}.
+ %\label{0f1:math:rekursionsformel:herleitung}
+\end{equation*}
+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:
+
+\begin{align*}
+ \begin{pmatrix}
+ A_k\\
+ B_k
+ \end{pmatrix}
+ &=
+ \begin{pmatrix}
+ 1& a_0\\
+ 0& 1
+ \end{pmatrix}
+ \begin{pmatrix}
+ 0& b_1\\
+ 1& a_1
+ \end{pmatrix}
+ \cdots
+ \begin{pmatrix}
+ 0& b_{k-1}\\
+ 1& a_{k-1}
+ \end{pmatrix}
+ \begin{pmatrix}
+ b_k\\
+ a_k
+ \end{pmatrix}
+\end{align*}
+Nach vollständiger Induktion ergibt sich für den Schritt $k$, die Matrix
+\begin{equation}
+ \label{0f1:math:matrix:ende:eq}
+ \begin{pmatrix}
+ A_{k}\\
+ B_{k}
+ \end{pmatrix}
+ =
+ \begin{pmatrix}
+ A_{k-2}& A_{k-1}\\
+ B_{k-2}& B_{k-1}
+ \end{pmatrix}
+ \begin{pmatrix}
+ b_k\\
+ a_k
+ \end{pmatrix}.
+\end{equation}
+Und Schlussendlich kann der Näherungsbruch
+\[
+\frac{A_k}{B_k}
+\]
+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}
+\begin{itemize}
+\item Startbedingungen:
+\begin{align*}
+A_{-1} &= 0 & A_0 &= a_0 \\
+B_{-1} &= 1 & B_0 &= 1
+\end{align*}
+\item Schritt $k\to k+1$:
+\[
+\begin{aligned}
+\label{0f1:math:loesung:eq}
+k &\rightarrow k + 1:
+&
+A_{k+1} &= A_{k-1} \cdot b_k + A_k \cdot a_k \\
+&&
+B_{k+1} &= B_{k-1} \cdot b_k + B_k \cdot a_k
+\end{aligned}
+\]
+\item
+Näherungsbruch: \qquad$\displaystyle\frac{A_k}{B_k}$
+\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.
+
+%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 44a4600..72b1b21 100644 --- a/buch/papers/0f1/teil3.tex +++ b/buch/papers/0f1/teil3.tex @@ -1,57 +1,64 @@ -% -% teil3.tex -- Resultate und Ausblick -% -% (c) 2022 Fabian Dünki, Hochschule Rapperswil -% -\section{Resultate -\label{0f1:section:teil3}} -\rhead{Resultate} -Im Verlauf des Seminares 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 x in der Funktion $\mathstrut_0F_1(;b;x)$. -Ebenso wird, je grösser der Wert x wird $\mathstrut_0F_1(;b;x)$, desto mehr weichen die berechneten Resultate -von den erwarteten ab. -{\color{red}TODO cite wolfram alpha rechner} - -\subsection{Auswertung -\label{0f1:subsection:auswertung}} -\begin{figure} - \centering - \includegraphics[width=0.8\textwidth]{papers/0f1/images/konvergenzAiry.pdf} - \caption{Konvergenz nach drei Iterationen, dargestellt anhand der Airy Funktion. - \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. - \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. - \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 GNU Scientific Library. - \label{0f1:ausblick:plot:airy:stabilitaet}} -\end{figure} - -\begin{itemize} - \item Negative Zahlen sind sowohl für die Potenzreihe als auch für den Kettenbruch ein Problem. - \item Die Potenzreihe hat das Problem, je tiefer die Rekursionstiefe, desto mehr machen die Brüche ein Problem. Also der Nenner mit der Fakultät und dem Pochhammer Symbol. - \item Die Rekursionformel liefert für sehr grosse positive Werte die genausten Ergebnisse, verglichen mit der GNU Scientific Library. -\end{itemize} - - -\subsection{Ausblick -\label{0f1:subsection:ausblick}} - - - +%
+% teil3.tex -- Resultate und Ausblick
+%
+% (c) 2022 Fabian Dünki, Hochschule Rapperswil
+%
+\section{Auswertung
+\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.
+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}.
+
+\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 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.
+
+\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.
+
+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 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.
+
+
+
+\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 $Ai'(0)=0$.
+ \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.
+ \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.
+ \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 $Ai(x)$.
+ \label{0f1:ausblick:plot:airy:stabilitaet}}
+\end{figure}
+
diff --git a/buch/papers/fm/00_modulation.tex b/buch/papers/fm/00_modulation.tex new file mode 100644 index 0000000..e2ba39f --- /dev/null +++ b/buch/papers/fm/00_modulation.tex @@ -0,0 +1,32 @@ +% +% teil3.tex -- Beispiel-File für Teil 3 +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\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. +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 +Jedoch ist das für die Vielfalt der Modulationsarten keine Einschrenkung. +Ein Nachrichtensignal kann auch über die Momentanfrequenz (instantenous frequency) \(\omega_i\) eines trägers verändert werden. +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 \(\omega_i\): +\begin{itemize} + \item AM + \item PM + \item FM +\end{itemize} + +To do: Bilder jeder Modulationsart + + + diff --git a/buch/papers/fm/01_AM-FM.tex b/buch/papers/fm/01_AM-FM.tex deleted file mode 100644 index ef55d55..0000000 --- a/buch/papers/fm/01_AM-FM.tex +++ /dev/null @@ -1,36 +0,0 @@ -% -% einleitung.tex -- Beispiel-File für die Einleitung -% -% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{AM - FM\label{fm:section:teil0}} -\rhead{AM- FM} - -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. -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 -Jedoch ist das für die Vilfalt der Modulationsarten keine Einschrenkung. -Ein Nachrichtensignal kann auch über die Momentanfrequenz (instantenous frequency) \(\omega_i\) eines trägers verändert werden. -Mathematisch wird dann daraus -\[ - \omega_i = \omega_c + \frac{d \varphi(t)}{dt} -\] -mit der Ableitung der Phase. -\newline -\newline -TODO: -Hier beschrieib ich was AmplitudenModulation ist und mache dan den link zu Frequenzmodulation inkl Formel \[cos( cos x)\] - - - -%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{fm: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. - - diff --git a/buch/papers/fm/01_AM.tex b/buch/papers/fm/01_AM.tex new file mode 100644 index 0000000..21927f5 --- /dev/null +++ b/buch/papers/fm/01_AM.tex @@ -0,0 +1,29 @@ +% +% einleitung.tex -- Beispiel-File für die Einleitung +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Amplitudenmodulation\label{fm:section:teil0}} +\rhead{AM} + +Das Ziel ist FM zu verstehen doch dazu wird zuerst AM erklärt welches einwenig einfacher zu verstehen ist und erst dann übertragen wir die Ideen in FM. +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 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 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: +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}
\ No newline at end of file diff --git a/buch/papers/fm/02_frequenzyspectrum.tex b/buch/papers/fm/02_FM.tex index 1c6044d..fedfaaa 100644 --- a/buch/papers/fm/02_frequenzyspectrum.tex +++ b/buch/papers/fm/02_FM.tex @@ -3,11 +3,12 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{AM-FM im Frequenzspektrum +\section{FM \label{fm:section:teil1}} -\rhead{Problemstellung} - -Hier Beschreiben ich das Frequenzspektrum und wie AM und FM aussehen und generiert werden. +\rhead{FM} +\subsection{Frequenzspektrum} +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 fdaa0d1..5f85dc6 100644 --- a/buch/papers/fm/03_bessel.tex +++ b/buch/papers/fm/03_bessel.tex @@ -3,26 +3,208 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{FM und Besselfunktion -\label{fm:section:teil2}} -\rhead{Teil 2} +\section{FM und Bessel-Funktion +\label{fm:section:proof}} +\rhead{Herleitung} +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 +\[ +\cos(\omega_c t+\beta\sin(\omega_mt)) +\] +ist. +\subsection{Herleitung} +Das Ziel ist, unser moduliertes Signal mit der Bessel-Funktion so auszudrücken: +\begin{align} + x_c(t) + = + \cos(\omega_ct+\beta\sin(\omega_mt)) + &= + \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} + \cos(A + B) + &= + \cos(A)\cos(B)-\sin(A)\sin(B) + \label{fm:eq:addth1} + \\ + 2\cos (A)\cos (B) + &= + \cos(A-B)+\cos(A+B) + \label{fm:eq:addth2} + \\ + 2\sin(A)\sin(B) + &= + \cos(A-B)-\cos(A+B) + \label{fm:eq:addth3} +\end{align} +und die drei Bessel-Funktionsindentitäten, +\begin{align} + \cos(\beta\sin\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 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} findet. + +\subsubsection{Anwenden des Additionstheorem} +Mit dem \eqref{fm:eq:addth1} wird aus dem modulierten Signal +\[ + x_c(t) + = + \cos(\omega_c t + \beta\sin(\omega_mt)) + = + \cos(\omega_c t)\cos(\beta\sin(\omega_m t))-\sin(\omega_ct)\sin(\beta\sin(\omega_m t)). + \label{fm:eq:start} +\] +%----------------------------------------------------------------------------------------------------------- +\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*} + c(t) + &= + \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*} +%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*} + 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) \} + \\ + &= + \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=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: +\begin{align*} + \sum_{n\, \text{gerade}} J_{n}(\beta) \cos((\omega_c + n \omega_m) t), + \label{fm:eq:gerade} +\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*} + s(t) + &= + -\sin(\omega_c t) \cdot \bigg[ 2 \sum_{k=0}^\infty J_{ 2k + 1}(\beta) \cos(( 2k + 1) \omega_m t) \bigg] + \\ + &= + \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 \(B = n \omega_m t \), +somit wird daraus: +\begin{align*} + s(t) + &= + \sum_{n=0}^\infty J_{-n}(\beta) \{ \underbrace{\cos((\omega_c - n\omega_m) t)} \,-\, \cos((\omega_c + n\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*} +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 +\[ + s(t) + = + \sum_{n\, \text{ungerade}} -1 \cdot J_{n}(\beta) \cos((\omega_c + n\omega_m) t). + \label{fm:eq:ungerade} +\] +schreiben. +%------------------------------------------------------------------------------------------ +\subsubsection{Summe Zusammenführen} +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 +\[ + \sum_{n\, \text{ungerade}} J_{n}(\beta) \cos((\omega_c + n\omega_m) t) +\] +ergeben zusammen +\[ + \cos(\omega_ct+\beta\sin(\omega_mt)) + = + \sum_{k= -\infty}^\infty J_{k}(\beta) \cos((\omega_c+k\omega_m)t). +\] +Somit ist \eqref{fm:eq:proof} bewiesen. +\newpage +%----------------------------------------------------------------------------------------- +\subsection{Bessel und Frequenzspektrum} +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} + \caption{Bessle Funktion \(J_{k}(\beta)\)} + \label{fig:bessel} +\end{figure} +TODO Grafik einfügen, +\newline +Nun einmal das Modulierte FM signal im Frequenzspektrum mit den einzelen Summen dargestellt + +TODO Hier wird beschrieben wie die Bessel Funktion der FM im Frequenzspektrum hilft, wieso diese gebrauch wird und ihre Vorteile. -%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? -% +\begin{itemize} + \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} + + %\subsection{De finibus bonorum et malorum %\label{fm:subsection:bonorum}} diff --git a/buch/papers/fm/04_fazit.tex b/buch/papers/fm/04_fazit.tex index 8c6c002..8d5eca4 100644 --- a/buch/papers/fm/04_fazit.tex +++ b/buch/papers/fm/04_fazit.tex @@ -6,35 +6,7 @@ \section{Fazit \label{fm:section:fazit}} \rhead{Zusamenfassend} -%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{fm: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. + +TODO Anwendungen erklären und Sinn des Ganzen. diff --git a/buch/papers/fm/RS presentation/FM_presentation.pdf b/buch/papers/fm/FM presentation/FM_presentation.pdf Binary files differindex 496e35e..496e35e 100644 --- a/buch/papers/fm/RS presentation/FM_presentation.pdf +++ b/buch/papers/fm/FM presentation/FM_presentation.pdf diff --git a/buch/papers/fm/RS presentation/FM_presentation.tex b/buch/papers/fm/FM presentation/FM_presentation.tex index 92cb501..2801e69 100644 --- a/buch/papers/fm/RS presentation/FM_presentation.tex +++ b/buch/papers/fm/FM presentation/FM_presentation.tex @@ -1,4 +1,4 @@ -%% !TeX root = RS.tex +%% !TeX root = .tex \documentclass[11pt,aspectratio=169]{beamer} \usepackage[utf8]{inputenc} @@ -15,7 +15,7 @@ \logo{} \institute{OST Ostschweizer Fachhochschule} \date{16.5.2022} - \subject{Mathematisches Seminar} + \subject{Mathematisches Seminar - Spezielle Funktionen} %\setbeamercovered{transparent} \setbeamercovered{invisible} \setbeamertemplate{navigation symbols}{} diff --git a/buch/papers/fm/FM presentation/README.txt b/buch/papers/fm/FM presentation/README.txt new file mode 100644 index 0000000..65f390d --- /dev/null +++ b/buch/papers/fm/FM presentation/README.txt @@ -0,0 +1 @@ +Dies ist die Presentation des FM - Bessel
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presentation/images/fm_frequenz.png Binary files differindex 26bfd86..26bfd86 100644 --- a/buch/papers/fm/RS presentation/images/fm_frequenz.png +++ b/buch/papers/fm/FM presentation/images/fm_frequenz.png diff --git a/buch/papers/fm/RS presentation/images/fm_in_time.png b/buch/papers/fm/FM presentation/images/fm_in_time.png Binary files differindex 068eafc..068eafc 100644 --- a/buch/papers/fm/RS presentation/images/fm_in_time.png +++ b/buch/papers/fm/FM presentation/images/fm_in_time.png diff --git a/buch/papers/fm/Makefile b/buch/papers/fm/Makefile index c84963f..f30c4a9 100644 --- a/buch/papers/fm/Makefile +++ b/buch/papers/fm/Makefile @@ -5,8 +5,9 @@ # SOURCES := \ - 01_AM-FM.tex \ - 02_frequenzyspectrum.tex \ + 00_modulation.tex \ + 01_AM.tex \ + 02_FM.tex \ 03_bessel.tex \ 04_fazit.tex \ main.tex @@ -16,15 +17,17 @@ SOURCES := \ #FIGURES := $(patsubst tikz/%.tex, figures/%.pdf, $(TIKZFIGURES)) -#.PHONY: images -#images: $(FIGURES) +all: images standalone + +.PHONY: images +images: $(FIGURES) #figures/%.pdf: tikz/%.tex # mkdir -p figures # pdflatex --output-directory=figures $< .PHONY: standalone -standalone: standalone.tex $(SOURCES) #$(FIGURES) +standalone: standalone.tex $(SOURCES) $(FIGURES) mkdir -p standalone cd ../..; \ pdflatex \ diff --git a/buch/papers/fm/Makefile.inc b/buch/papers/fm/Makefile.inc index e5cd9f6..40f23b1 100644 --- a/buch/papers/fm/Makefile.inc +++ b/buch/papers/fm/Makefile.inc @@ -6,8 +6,9 @@ dependencies-fm = \ papers/fm/packages.tex \ papers/fm/main.tex \ - papers/fm/01_AM-FM.tex \ - papers/fm/02_frequenzyspectrum.tex \ + papers/fm/00_modulation.tex \ + papers/fm/01_AM.tex \ + papers/fm/02_FM.tex \ papers/fm/03_bessel.tex \ papers/fm/04_fazit.tex \ papers/fm/references.bib diff --git a/buch/papers/fm/Python animation/Bessel-FM.ipynb b/buch/papers/fm/Python animation/Bessel-FM.ipynb index bfbb83d..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": 117, + "execution_count": 4, "metadata": {}, "outputs": [], "source": [ @@ -11,6 +11,9 @@ "from scipy.fft import fft, ifft, fftfreq\n", "import scipy.special as sc\n", "import scipy.fftpack\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", "from matplotlib.widgets import Slider\n", "def fm(beta):\n", @@ -67,39 +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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- "text/plain": [ - "<Figure size 432x288 with 1 Axes>" - ] - }, - "metadata": { - "needs_background": "light" - }, - "output_type": "display_data" - } - ], - "source": [ + "plt.show()\n", + "\n", "fm(1)" ] }, { "cell_type": "code", - "execution_count": 122, + "execution_count": 5, "metadata": {}, "outputs": [ { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.7651976865579666\n" + ] + }, + { "data": { - "image/png": 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", + "image/png": 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", "text/plain": [ "<Figure size 432x288 with 1 Axes>" ] @@ -108,29 +98,23 @@ "needs_background": "light" }, "output_type": "display_data" - }, - { - "name": "stdout", - "output_type": "stream", - "text": [ - "0.7651976865579666\n" - ] } ], "source": [ "\n", - "for n in range (-4,4):\n", - " x = np.linspace(0,11,1000)\n", + "for n in range (-2,4):\n", + " x = np.linspace(-11,11,1000)\n", " y = sc.jv(n,x)\n", - " plt.plot(x, y, '-')\n", - "plt.plot([1,1],[sc.jv(0,1),sc.jv(-1,1)],)\n", - "plt.xlim(0,10)\n", + " plt.plot(x, y, '-',label='n='+str(n))\n", + "#plt.plot([1,1],[sc.jv(0,1),sc.jv(-1,1)],)\n", + "plt.xlim(-10,10)\n", "plt.grid(True)\n", - "plt.ylabel('Bessel J_n(b)')\n", - "plt.xlabel('b')\n", + "plt.ylabel('Bessel $J_n(\\\\beta)$')\n", + "plt.xlabel(' $ \\\\beta $ ')\n", "plt.plot(x, y)\n", - "plt.show()\n", - "\n", + "plt.legend()\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{<filename>.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{<path to file>}{<filename>.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}}% 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+\endgroup% diff --git a/buch/papers/fm/Quellen/A2-14.pdf b/buch/papers/fm/Quellen/A2-14.pdf Binary files differnew file mode 100644 index 0000000..7348cca --- /dev/null +++ b/buch/papers/fm/Quellen/A2-14.pdf diff --git a/buch/papers/fm/Quellen/FM_presentation.pdf b/buch/papers/fm/Quellen/FM_presentation.pdf Binary files differnew file mode 100644 index 0000000..496e35e --- /dev/null +++ b/buch/papers/fm/Quellen/FM_presentation.pdf diff --git a/buch/papers/fm/RS presentation/Frequency modulation (FM) and Bessel functions.pdf b/buch/papers/fm/Quellen/Frequency modulation (FM) and Bessel functions.pdf Binary files differindex a6e701c..a6e701c 100644 --- a/buch/papers/fm/RS presentation/Frequency modulation (FM) and Bessel functions.pdf +++ b/buch/papers/fm/Quellen/Frequency modulation (FM) and Bessel functions.pdf diff --git a/buch/papers/fm/Quellen/Seydel2022_Book_HöhereMathematikImAlltag.pdf b/buch/papers/fm/Quellen/Seydel2022_Book_HöhereMathematikImAlltag.pdf Binary files differnew file mode 100644 index 0000000..2a0bddd --- /dev/null +++ b/buch/papers/fm/Quellen/Seydel2022_Book_HöhereMathematikImAlltag.pdf diff --git a/buch/papers/fm/RS presentation/README.txt b/buch/papers/fm/RS presentation/README.txt deleted file mode 100644 index 4d0620f..0000000 --- a/buch/papers/fm/RS presentation/README.txt +++ /dev/null @@ -1 +0,0 @@ -Dies ist die Presentation des Reed-Solomon-Code
\ No newline at end of file diff --git a/buch/papers/fm/RS presentation/RS.tex b/buch/papers/fm/RS presentation/RS.tex deleted file mode 100644 index 8a67619..0000000 --- a/buch/papers/fm/RS presentation/RS.tex +++ /dev/null @@ -1,123 +0,0 @@ -%% !TeX root = RS.tex - -\documentclass[11pt,aspectratio=169]{beamer} -\usepackage[utf8]{inputenc} -\usepackage[T1]{fontenc} -\usepackage{lmodern} -\usepackage[ngerman]{babel} -\usepackage{tikz} -\usetheme{Hannover} - -\begin{document} - \author{Joshua Bär} - \title{FM - Bessel} - \subtitle{} - \logo{} - \institute{OST Ostschweizer Fachhochschule} - \date{16.5.2022} - \subject{Mathematisches Seminar- Spezielle Funktionen} - %\setbeamercovered{transparent} - \setbeamercovered{invisible} - \setbeamertemplate{navigation symbols}{} - \begin{frame}[plain] - \maketitle - \end{frame} -%------------------------------------------------------------------------------- -\section{Einführung} - \begin{frame} - \frametitle{Frequenzmodulation} - - \visible<1->{\begin{equation} \cos(\omega_c t+\beta\sin(\omega_mt))\end{equation}} - - \only<2>{\includegraphics[scale= 0.7]{images/fm_in_time.png}} - \only<3>{\includegraphics[scale= 0.7]{images/fm_frequenz.png}} - \only<4>{\includegraphics[scale= 0.7]{images/bessel_frequenz.png}} - - - \end{frame} -%------------------------------------------------------------------------------- -\section{Proof} -\begin{frame} - \frametitle{Bessel} - - \visible<1->{\begin{align} - \cos(\beta\sin\varphi) - &= - J_0(\beat) + 2\sum_{m=1}^\infty J_{2m}(\beta) \cos(2m\varphi) - \\ - \sin(\beta\sin\varphi) - &= - J_0(\beat) + 2\sum_{m=1}^\infty J_{2m}(\beta) \cos(2m\varphi) - \\ - J_{-n}(\beat) &= (-1)^n J_n(\beta) - \end{align}} - \visible<2->{\begin{align} - \cos(A + B) - &= - \cos(A)\cos(B)-\sin(A)\sin(B) - \\ - 2\cos (A)\cos (B) - &= - \cos(A-B)+\cos(A+B) - \\ - 2\sin(A)\sin(B) - &= - \cos(A-B)-\cos(A+B) - \end{align}} -\end{frame} - -%------------------------------------------------------------------------------- -\begin{frame} - \frametitle{Prof->Done} - \begin{align} - \cos(\omega_ct+\beta\sin(\omega_mt)) - &= - \sum_{k= -\infty}^\infty J_{k}(\beta) \cos((\omega_c+k\omgea_m)t) - \end{align} - \end{frame} -%------------------------------------------------------------------------------- - \begin{frame} - \begin{figure} - \only<1>{\includegraphics[scale = 0.75]{images/fm_frequenz.png}} - \only<2>{\includegraphics[scale = 0.75]{images/bessel_frequenz.png}} - \end{figure} - \end{frame} -%------------------------------------------------------------------------------- -\section{Input Parameter} - \begin{frame} - \frametitle{Träger-Frequenz Parameter} - \onslide<1->{\begin{equation}\cos(\omega_ct+\beta\sin(\omega_mt))\end{equation}} - \only<1>{\includegraphics[scale=0.75]{images/100HZ.png}} - \only<2>{\includegraphics[scale=0.75]{images/200HZ.png}} - \only<3>{\includegraphics[scale=0.75]{images/300HZ.png}} - \only<4>{\includegraphics[scale=0.75]{images/400HZ.png}} - \end{frame} -%------------------------------------------------------------------------------- -\begin{frame} -\frametitle{Modulations-Frequenz Parameter} -\onslide<1->{\begin{equation}\cos(\omega_ct+\beta\sin(\omega_mt))\end{equation}} -\only<1>{\includegraphics[scale=0.75]{images/fm_3Hz.png}} -\only<2>{\includegraphics[scale=0.75]{images/fm_5Hz.png}} -\only<3>{\includegraphics[scale=0.75]{images/fm_7Hz.png}} -\only<4>{\includegraphics[scale=0.75]{images/fm_10Hz.png}} -\only<5>{\includegraphics[scale=0.75]{images/fm_20Hz.png}} -\only<6>{\includegraphics[scale=0.75]{images/fm_30Hz.png}} -\end{frame} -%------------------------------------------------------------------------------- -\begin{frame} -\frametitle{Beta Parameter} - \onslide<1->{\begin{equation}\sum_{k= -\infty}^\infty J_{k}(\beta) \cos((\omega_c+k\omgea_m)t)\end{equation}} - \only<1>{\includegraphics[scale=0.7]{images/beta_0.001.png}} - \only<2>{\includegraphics[scale=0.7]{images/beta_0.1.png}} - \only<3>{\includegraphics[scale=0.7]{images/beta_0.5.png}} - \only<4>{\includegraphics[scale=0.7]{images/beta_1.png}} - \only<5>{\includegraphics[scale=0.7]{images/beta_2.png}} - \only<6>{\includegraphics[scale=0.7]{images/beta_3.png}} - \only<7>{\includegraphics[scale=0.7]{images/bessel.png}} -\end{frame} -%------------------------------------------------------------------------------- -\begin{frame} - \includegraphics[scale=0.5]{images/beta_1.png} - \includegraphics[scale=0.5]{images/bessel.png} -\end{frame} -\end{document} diff --git a/buch/papers/fm/main.tex b/buch/papers/fm/main.tex index fcf4d1a..731f56f 100644 --- a/buch/papers/fm/main.tex +++ b/buch/papers/fm/main.tex @@ -27,10 +27,12 @@ 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. +Nun zur Modulation im nächsten Abschnitt.\cite{fm:NAT} -\input{papers/fm/01_AM-FM.tex} -\input{papers/fm/02_frequenzyspectrum.tex} + +\input{papers/fm/00_modulation.tex} +\input{papers/fm/01_AM.tex} +\input{papers/fm/02_FM.tex} \input{papers/fm/03_bessel.tex} \input{papers/fm/04_fazit.tex} diff --git a/buch/papers/fm/packages.tex b/buch/papers/fm/packages.tex index 4cba2b6..7bbbe35 100644 --- a/buch/papers/fm/packages.tex +++ b/buch/papers/fm/packages.tex @@ -7,4 +7,5 @@ % if your paper needs special packages, add package commands as in the % following example %\usepackage{packagename} - +\usepackage{xcolor} +\usepackage{pgf} diff --git a/buch/papers/fm/references.bib b/buch/papers/fm/references.bib index 76eb265..21b910b 100644 --- a/buch/papers/fm/references.bib +++ b/buch/papers/fm/references.bib @@ -23,6 +23,17 @@ volume = {2} } +@book{fm:NAT, + title = {Nachrichtentechnik 1 + 2}, + author = {Thomas Kneubühler}, + publisher = {None}, + year = {2021}, + isbn = {}, + inseries = {Script for students}, + volume = {} +} + + @article{fm:mendezmueller, author = { Tabea Méndez and Andreas Müller }, title = { Noncommutative harmonic analysis and image registration }, 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..6383984 --- /dev/null +++ b/buch/papers/kra/anwendung.tex @@ -0,0 +1,215 @@ +\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: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 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 +\begin{equation} + x(t) = A \cos(\omega_0 t + \Phi), \quad \omega_0 = \sqrt{\frac{k}{m}} +\end{equation} +\begin{figure} + % 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} +\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 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} + \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 Systems 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:equation:matrixriccati} führt. + +% @TODO Einfluss auf anfangsbedingungen, plots? +% @TODO Fazit ? diff --git a/buch/papers/kra/einleitung.tex b/buch/papers/kra/einleitung.tex new file mode 100644 index 0000000..cde2e66 --- /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 nicht lineare gewöhnliche Differentialgleichung erster Ordnung der Form +\begin{equation} + \label{kra:equation:riccati} + y' = f(x)y + g(x)y^2 + h(x) +\end{equation} +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: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 \cite{kra:ethz} \cite{kra:riccati}. 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/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} 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 diff --git a/buch/papers/kra/images/simple.pdf b/buch/papers/kra/images/simple.pdf Binary files differnew file mode 100644 index 0000000..4351518 --- /dev/null +++ b/buch/papers/kra/images/simple.pdf 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 new file mode 100644 index 0000000..868362d --- /dev/null +++ b/buch/papers/kra/images/simple_mass_spring.tex @@ -0,0 +1,66 @@ +% 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,>=latex] + \newcommand{\ticks}[2] + { + % arguments: x, y coordinates + \draw[thick] (#1, #2 - 0.1 / 2) --++ (0, 0.1); + } + + \tikzmath{ + \hWall = 1.2; + \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) 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=3.5] {$k$}; + + % 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] {$l_0$}; + \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] {$l_{1}$}; + + % 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} diff --git a/buch/papers/kra/loesung.tex b/buch/papers/kra/loesung.tex new file mode 100644 index 0000000..4e0da1c --- /dev/null +++ b/buch/papers/kra/loesung.tex @@ -0,0 +1,86 @@ +\section{Lösungsmethoden} \label{kra:section:loesung} +\rhead{Lösungsmethoden} + +\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:equation: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 sogenannte 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 fcee25b..a84ebaf 100644 --- a/buch/papers/kra/main.tex +++ b/buch/papers/kra/main.tex @@ -3,34 +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} - - 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/einleitung.tex} + \input{papers/kra/loesung.tex} + \input{papers/kra/anwendung.tex} + \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/kra/packages.tex b/buch/papers/kra/packages.tex index df34dcf..56c48d9 100644 --- a/buch/papers/kra/packages.tex +++ b/buch/papers/kra/packages.tex @@ -8,3 +8,11 @@ % following example %\usepackage{packagename} +%\usepackage{physics} +\usepackage[outline]{contour} +\pgfplotsset{compat=1.16} +\usetikzlibrary{patterns} +\usetikzlibrary{snakes} +\usetikzlibrary{arrows.meta} +\usetikzlibrary{decorations} +\usetikzlibrary{decorations.markings} 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<presentation>{ + \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/references.bib b/buch/papers/kra/references.bib index f13c3d8..a9a8ede 100644 --- a/buch/papers/kra/references.bib +++ b/buch/papers/kra/references.bib @@ -4,32 +4,42 @@ % (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} +@misc{kra:riccati, +title = {Riccatische Differentialgleichung}, +url = {https://de.wikipedia.org/wiki/Riccatische_Differentialgleichung}, +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:ethz, +author = {Ch. Roduner}, +title = {Die-Riccati-Gleichung}, +url = {https://www.imrtweb.ethz.ch/users/geering/Riccati.pdf}, +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} +@online{kra:hamilton, + title = {Hamilton-Funktion}, + url = {https://de.wikipedia.org/wiki/Hamilton-Funktion}, + date = {2022-05-26} } +@misc{kra:kanonischegleichungen, + title = {Kanonische Gleichungen}, + url = {https://de.wikipedia.org/wiki/Kanonische_Gleichungen}, + date = {2022-05-26} +} + +@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/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()
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/kreismembran/images/Saite.pdf b/buch/papers/kreismembran/images/Saite.pdf Binary files differnew file mode 100644 index 0000000..0f87c93 --- /dev/null +++ b/buch/papers/kreismembran/images/Saite.pdf diff --git a/buch/papers/kreismembran/images/mask_absorber.png b/buch/papers/kreismembran/images/mask_absorber.png Binary files differnew file mode 100644 index 0000000..5d0cccf --- /dev/null +++ b/buch/papers/kreismembran/images/mask_absorber.png diff --git a/buch/papers/kreismembran/images/mask_disk.png b/buch/papers/kreismembran/images/mask_disk.png Binary files differnew file mode 100644 index 0000000..4b38163 --- /dev/null +++ b/buch/papers/kreismembran/images/mask_disk.png diff --git a/buch/papers/kreismembran/images/sim_1_1.png b/buch/papers/kreismembran/images/sim_1_1.png Binary files differnew file 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Membran\label{chapter:kreismembran}} -\lhead{Schwingungen einer kreisförmligen Membran} +\chapter{Schwingungen einer kreisförmigen Membran\label{chapter:kreismembran}} +\lhead{Schwingungen einer kreisförmigen Membran} \begin{refsection} \chapterauthor{Andrea Mozzini Vellen und Tim Tönz} @@ -12,6 +12,7 @@ \input{papers/kreismembran/teil1.tex} \input{papers/kreismembran/teil2.tex} \input{papers/kreismembran/teil3.tex} +\input{papers/kreismembran/teil4.tex} \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/kreismembran/references.bib b/buch/papers/kreismembran/references.bib index 0b6a683..3d9d0c1 100644 --- a/buch/papers/kreismembran/references.bib +++ b/buch/papers/kreismembran/references.bib @@ -4,6 +4,25 @@ % (c) 2020 Autor, Hochschule Rapperswil % +@online{kreismembran:Duden:Membran, + title = {Duden:Membran}, + url = {https://www.duden.de/rechtschreibung/Membran}, + date = {2022-07-20}, + year = {2022}, + month = {7}, + day = {20} +} + +@online{kreismembran:wellengleichung_herleitung, + title = {Derivation of the 2D Wave Equation}, + author = {Dr. Christopher Lum}, + url = {https://www.youtube.com/watch?v=KAS7JBztw8E&t=0s}, + date = {2022-07-20}, + year = {2022}, + month = {7}, + day = {20} +} + @online{kreismembran:bibtex, title = {BibTeX}, url = {https://de.wikipedia.org/wiki/BibTeX}, @@ -24,7 +43,7 @@ } @article{kreismembran:mendezmueller, - author = { Tabea Méndez and Andreas Müller }, + author = { Tabea Méndez and Andreas Müller }, title = { Noncommutative harmonic analysis and image registration }, journal = { Appl. Comput. Harmon. Anal.}, year = 2019, @@ -33,3 +52,41 @@ 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}, + publisher = {{CRC} Press}, + author = {{Lokenath Debnath} and Dambaru Bhatta}, + date = {2015}, +} + +@thesis{nishanth_p_vibrations_2018, + title = {Vibrations of a Circular Membrane - Some Undergraduadte Exercises}, + type = {phdthesis}, + author = {{Nishanth P.} and {Udayanandan K. M.}}, + date = {2018}, +} + +@thesis{prof_dr_horst_knorrer_kreisformige_2013, + title = {Kreisförmige Membranen}, + institution = {{ETHZ}}, + type = {phdthesis}, + author = {{Prof. Dr. Horst Knörrer}}, + date = {2013}, +} + +@thesis{kreismembran:membrane_vs_thin_plate, + title = {Modeling and Control of SPIDER Satellite Components}, + institution = {{faculty of the Virginia Polytechnic Institute and State University}}, + type = {Dissertation}, + author = {{Eric John Ruggiero Doctor of Philosophy In Mechanical Engineering}}, + date = {2005}, +}
\ No newline at end of file diff --git a/buch/papers/kreismembran/teil0.tex b/buch/papers/kreismembran/teil0.tex index 1552259..c6dac06 100644 --- a/buch/papers/kreismembran/teil0.tex +++ b/buch/papers/kreismembran/teil0.tex @@ -4,7 +4,91 @@ % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % \section{Einleitung\label{kreismembran:section:teil0}} -\rhead{Einleitung} +\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. +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. + +\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 $. + \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 $. + \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. + +\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 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} + + \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} + +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, 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} + 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 +\begin{equation*} + \sum F = m \cdot a +\end{equation*} +befolgen. Die senkrecht wirkenden Kräfte werden mit $ T_1 $ und $ T_2 $ ausgedrückt, die Masse als Funktion der Dichte $ \rho $ und die Beschleunigung in Form der zweiten Ableitung als +\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 +\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*} +vereinfacht als +\begin{equation*} + \tan \beta - \tan \alpha = \frac{\rho dx}{T} \frac{\partial^2 u}{\partial t^2} +\end{equation*} +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} \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} \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. +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. +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. +\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 diff --git a/buch/papers/kreismembran/teil1.tex b/buch/papers/kreismembran/teil1.tex index aef5b79..f6ba7d1 100644 --- a/buch/papers/kreismembran/teil1.tex +++ b/buch/papers/kreismembran/teil1.tex @@ -7,13 +7,14 @@ \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 Separationsmetode gelöst. +An diesem Punkt bleibt also nur noch die Lösung der partiellen Differentialgleichung. In diesem Abschnitt wird sie mit Hilfe der Separationsmethode gelöst. -Wie im vorherigen Kapitel gezeigt, lautet die partielle Differentialgleichung, die die Schwingungen einer Membran beschreibt: +\subsection{Aufgabestellung\label{sub:aufgabestellung}} +Wie im vorherigen Abschnitt gezeigt, lautet die partielle Differentialgleichung, die die Schwingungen einer Membran beschreibt: \begin{equation*} - \frac{1}{c^2}\frac{\partial^2u}{\partial t^2} = \Delta u + \frac{1}{c^2}\frac{\partial^2u}{\partial t^2} = \Delta u. \end{equation*} -Da es sich um eine Kreisscheibe handelt, werden Polarkoordinaten verwendet, so dass sich der Laplaceoperator ergibt: +Da es sich um eine Kreisscheibe handelt, werden Polarkoordinaten verwendet, so dass sich der Laplaceoperator \begin{equation*} \Delta = @@ -22,79 +23,102 @@ Da es sich um eine Kreisscheibe handelt, werden Polarkoordinaten verwendet, so d \frac1r \frac{\partial}{\partial r} + - \frac{1}{r 2} - \frac{\partial^2}{\partial\varphi^2}. + \frac{1}{r^2} + \frac{\partial^2}{\partial\varphi^2} \label{buch:pde:kreis:laplace} \end{equation*} +ergibt. -Es wird eine runde elastische Membran berücksichtigt, die den Gebietbereich $\Omega$ 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 eigespannten homogenen schwingenden Membran. +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}. 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} gilt: -\begin{equation*} - u\big|_{\Gamma} = 0 -\end{equation*} -Um eine eindeutige Lösung bestimmen zu können, werden die folgenden Anfangsbedingungen festgelegt: +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 $t = \text{0}$: \begin{align*} u(r,\varphi, 0) &= f(r,\varphi)\\ - \frac{\partial}{\partial t} u(r,\varphi, 0) &= g(r,\varphi) + u_t(r,\varphi, 0) &= g(r,\varphi). \end{align*} -Daher muss an dieser Stelle von einer Separation der Variablen ausgegangen werden: + +\subsection{Lösung\label{sub:lösung1}} +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}} +Hierfür wird folgenden Ansatz gemacht: \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. Eingesetz in der Differenzialgleichung ergibt: -\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)} -\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 Grunden 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: -\begin{gather*} - 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{gather*} -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: -\begin{gather*} - r^2F''(r) + rF'(r) + (\kappa^2 r^2 - \nu)F(r) = 0 \\ - G''(\varphi) = \nu G(\varphi) -\end{gather*} -$G$ kann in einer Fourierreihe entwickelt werden, so dass man sieht, dass $\nu$ die Form $n^2$ mit einer positiven ganzen Zahl sein muss, also: +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*} - G(\varphi) = C_n \cos(\varphi) + D_n \sin(\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*} -Die Gleichung $F$ hat die Gestalt +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)}. +\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: +\begin{align*} + 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*} - r^2F''(r) + rF'(r) + (\kappa^2 r^2 - n^2)F(r) = 0 \quad (*) + G(\varphi) = C_n \cos(\nu\varphi) + D_n \sin(\nu\varphi) + \label{eq:cos_sin_überlagerung} \end{equation*} -Wir bereits in der Vorlesung von Prof. Müller gezeigt, sind die Besselfunktionen + +\subsubsection{Lösung für $F(r)$\label{subsub:lösung_F}} +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} +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*} -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 $(*)$. 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 ergit sich, dass $\kappa = \frac{\alpha_{mn}}{R}$ für ein $m\geq 1$, und dass +Lösungen der Besselschen Differenzialgleichung \begin{equation*} - F(r) = J_n (\kappa_{mn}r) \quad mit \quad \kappa_{mn}=\frac{\alpha_{mn}}{R} + x^2 y'' + xy' + (\kappa^2 - \nu^2)y = 0 \end{equation*} -Die Differenzialgleichung $T''(t) + c^2\kappa^2T(t) = 0$, wird auf ähnliche Weise gelöst wie $G(\varphi)$. Durch Überlagerung aller Ergebnisse erhält man die Lösung -\begin{equation} - u(r, \varphi, t) = \displaystyle\sum_{m=1}^{\infty}\displaystyle\sum_{n=0}^{\infty} J_n (k_{mn}r)\cos(n\varphi)[a_{mn}\cos(c \kappa_{mn} t)+b_{mn}\sin(c \kappa_{mn} t)] -\end{equation} -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 kmn die Wellenzahl und $r$ der Radius ist. $a_{mn}$ und $b_{mn}$ sind die zu bestimmenden Konstanten. +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 $G(\varphi)$. + +\subsubsection{Zusammenfassung der Lösungen\label{subsub:zusammenfassung_lösungen}} +Durch Überlagerung aller Ergebnisse erhält man die Lösung +\begin{align} + u(r, \varphi, t) = \displaystyle\sum_{m=1}^{\infty}\displaystyle\sum_{n=0}^{\infty} J_n (k_{mn}r)[a_{mn}\cos(n\varphi) + b_{mn}\sin(n\varphi)](n\varphi)[c_{mn}\cos(c \kappa_{mn} t)+d_{mn}\sin(c \kappa_{mn} t)] + \label{eq:lösung_endliche_generelle} +\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}). $J_n(\kappa_{mn}r)$ 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 + \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 und Diskussion mit Prof. Müller 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/teil2.tex b/buch/papers/kreismembran/teil2.tex index 8afe817..ec27bd3 100644 --- a/buch/papers/kreismembran/teil2.tex +++ b/buch/papers/kreismembran/teil2.tex @@ -5,109 +5,103 @@ \section{Die Hankel Transformation \label{kreismembran:section:teil2}} \rhead{Die Hankel Transformation} -Hermann Hankel (1839-1873) war ein deutscher Mathematiker, der für seinen Beitrag zur mathematischen Analyse und insbesondere für seine namensgebende Transformation bekannt ist. -Diese Transformation tritt bei der Untersuchung von funktionen auf, die nur von der Enternung des Ursprungs abhängen. -Er studierte auch funktionen, jetzt Hankel- oder Bessel- Funktionen genannt, der dritten Art. -Die Hankel Transformation mit Bessel Funktionen al Kern taucht natürlich bei achsensymmetrischen Problemen auf, die in Zylindrischen Polarkoordinaten formuliert sind. -In diesem Kapitel werden die Theorie der Transformation und einige Eigenschaften der Grundoperationen erläutert. - - -Wir führen die Definition der Hankel Transformation aus der zweidimensionalen Fourier Transformation und ihrer Umkehrung ein, die durch: +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. +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: \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 $\bm{\kappa}=(k,l)$. Wie bereits erwähnt, sind Polarkoordinaten für diese Art von Problemen am besten geeignet, also 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. + 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, + 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 Integral Darstellung der Besselfunktion vom Ordnung n -\begin{align} - 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 +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{align} -\eqref{equation:F_ohne_bessel} wird sie zu: +\end{equation*} + 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 \label{equation:F_mit_bessel_step_1} \\ + 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), \label{equation:F_mit_bessel_step_2} \end{align} -wo $\tilde{f}_n(\kappa)$ ist die \textit{Hankel Transformation} von $f(r)$ und ist formell definiert durch: +wo $\tilde{f}_n(\kappa)$ ist die \textit{Hankel-Transformation} von $f(r)$ und ist formell definiert durch: \begin{align} - \mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)=\int_{0}^{\infty}rJ_n(\kappa r) f(r) dr. + \mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)=\int_{0}^{\infty}rJ_n(\kappa r) f(r) \; dr. \label{equation:hankel} \end{align} -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}: +\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: \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 \nonumber \\ - &= e^{in\theta}\int_{0}^{\infty}\kappa J_n(\kappa r) \tilde{f}_n(\kappa) d\kappa,\quad \text{von \eqref{equation:bessel_n_ordnung}} -\end{align} - -Also, die inverse \textit{Hankel Transformation} ist 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. + \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. -\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, +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. -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. +\end{align*} +verwendet werden, um die Hankel-Transformation \eqref{equation:hankel} und ihre Umkehrung \eqref{equation:inv_hankel} zu definieren. -\subsection{Operative Eigenschaften 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. +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. -\subsubsection{Theorem 1: Skalierung \label{subsub:skalierung}} -Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann: +\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, diese sind im Buch \textit{Integral Tansforms and Their Applications} \cite{lokenath_debnath_integral_2015} zu finden. -\begin{equation*} - \mathscr{H}_n\{f(ar)\}=\frac{1}{a^{2}}\tilde{f}_n \left(\frac{\kappa}{a}\right), \quad a>0. -\end{equation*} +\begin{satz}{Skalierung:} + 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. + \end{equation*} +\end{satz} -\subsubsection{Theorem 2: Persevalsche Relation \label{subsub:perseval}} -Wenn $\tilde{f}(\kappa)=\mathscr{H}_n\{f(r)\}$ und $\tilde{g}(\kappa)=\mathscr{H}_n\{g(r)\}$, dann: +\begin{satz}{Parsevalsche Relation:} +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) dr = \int_{0}^{\infty}\kappa\tilde{f}(\kappa)\tilde{g}(\kappa) d\kappa. + \int_{0}^{\infty}rf(r)g(r) \; dr = \int_{0}^{\infty}\kappa\tilde{f}(\kappa)\tilde{g}(\kappa) \; d\kappa. \end{equation*} +\end{satz} -\subsubsection{Theorem 3: Hankel Transformationen von Ableitungen \label{subsub:ableitungen}} -Wenn $\tilde{f}_n(\kappa)=\mathscr{H}_n\{f(r)\}$, dann: +\begin{satz}{Hankel-Transformationen von Ableitungen:} +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*} -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} -\subsubsection{Theorem 4 \label{subsub:thorem4}} -Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann: +\begin{satz} +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 als $r\to0$ 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 bef8b5f..a9dcd95 100644 --- a/buch/papers/kreismembran/teil3.tex +++ b/buch/papers/kreismembran/teil3.tex @@ -6,28 +6,27 @@ \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. Wir führen also das Konzept einer unendlichen und achsensymmetrischen Membran ein: -\begin{equation*} +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{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 0<r<\infty, \quad t>0 - \label{eq:PDE_inf_membane} -\end{equation*} - -\begin{align} - u(r,0)=f(r), \quad \frac{\partial}{\partial t} u(r,0) = g(r), \quad \text{für} \quad 0<r<\infty + \frac{\partial u}{\partial r} \right), \quad 0<r<\infty, \quad t>0 \label{eq:PDE_inf_membane} \\ + u(r,0)=f(r), \quad u_t(r,0) = g(r), \quad \text{für} \quad 0<r<\infty. \label{eq:PDE_inf_membane_RB} \end{align} + Mit Anwendung der Hankel-Transformation nullter Ordnung in Abhängigkeit von $r$ auf die Gleichungen \eqref{eq:PDE_inf_membane} und \eqref{eq:PDE_inf_membane_RB}: \begin{align} - \tilde{u}(\kappa,t)=\int_{0}^{\infty}r J_0(\kappa r)u(r,t) dr, + \tilde{u}(\kappa,t)=\int_{0}^{\infty}r J_0(\kappa r)u(r,t) \; dr, \end{align} - bekommt man: \begin{equation*} @@ -36,43 +35,55 @@ bekommt man: \begin{equation*} \tilde{u}(\kappa,0)=\tilde{f}(\kappa), \quad - \frac{\partial}{\partial t}\tilde{u}(\kappa,0)=\tilde{g}(\kappa). + \tilde{u}_t(\kappa,0)=\tilde{g}(\kappa). \end{equation*} - -Die allgemeine Lösung für diese Transformation lautet, wie schon 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). \end{equation*} - -Wendet man an nun die inverse Hankel-Transformation an, so erhält man die formale Lösung +Wendet man nun die inverse Hankel-Transformation an, so erhält man die formale Lösung \begin{align} - u(r,t)=\int_{0}^{\infty}\kappa\tilde{f}(\kappa)\cos(c\kappa t) J_0(\kappa r) d\kappa +\frac{1}{c}\int_{0}^{\infty}\tilde{g}(\kappa)\sin(c\kappa t)J_0(\kappa r) d\kappa. + u(r,t)=\int_{0}^{\infty}\kappa\tilde{f}(\kappa)\cos(c\kappa t) J_0(\kappa r) \; d\kappa +\frac{1}{c}\int_{0}^{\infty}\tilde{g}(\kappa)\sin(c\kappa t)J_0(\kappa r) \; d\kappa. \label{eq:formale_lösung} \end{align} -Es wird daher davon ausgegangen, dass sich die Membran verformt und zum Zeitpunkt $t=0$ freigegeben wird +\subsubsection{Erfüllung der Anfangsbedingungen\label{subsub:erfüllung_AB}} +Es wird im Folgenden davon ausgegangen, dass sich die Membran verformt und zum Zeitpunkt $t=0$ freigegeben wird \begin{equation*} - u(r,0)=f(r)=Aa(r^2 + a^2)^{-\frac{1}{2}}, \quad \frac{d}{dt}(r,0)=g(r)=0 + u(r,0)=f(r)=Aa(r^2 + a^2)^{-\frac{1}{2}}, \quad u_t(r,0)=g(r)=0 \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*} +Aus der Laplace-Transformation und unter Verwendung der Skalierungseigenschaft ergibt sich, dass + +\begin{align*} + \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}} + 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*} +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)] + \label{eq:lösung_unendliche_generelle} +\end{align} +kann man die Lösungsmethoden 1 und 2 vergleichen. -\subsection{Vergleich der Lösungen +\subsection{Vergleich der Analytischen Lösungen \label{kreismembran:vergleich}} -Hier kommt noch der Vergleich der Lösungen ;) +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. diff --git a/buch/papers/kreismembran/teil4.tex b/buch/papers/kreismembran/teil4.tex new file mode 100644 index 0000000..01a6029 --- /dev/null +++ b/buch/papers/kreismembran/teil4.tex @@ -0,0 +1,194 @@ +% +% einleitung.tex -- Beispiel-File für die Einleitung +% +% (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil +% +\section{Lösungsmethode 3: Simulation + \label{kreismembran:section:teil4}} + +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. +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. + +\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. +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*} + 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*} +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. +\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 einem Tiefpassfilter in der Bildverarbeitung entspricht. + +\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, dass 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 $0$ 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. +Der Folgezustand kann also mit den Gleichungen +\begin{align} + \label{kreismembran:eq:folge_U} + 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)\odot M +\end{align} +berechnet werden. +\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. +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} + + \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.} + \label{kreismembran:im:simres_rund} + + \end{center} +\end{figure} +\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. +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. + +\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 $M$ 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 +\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} +\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} + + \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. +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. +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. + + + + + + + diff --git a/buch/papers/kugel/applications.tex b/buch/papers/kugel/applications.tex new file mode 100644 index 0000000..b2f227e --- /dev/null +++ b/buch/papers/kugel/applications.tex @@ -0,0 +1,9 @@ +% vim:ts=2 sw=2 et spell: + +\section{Applications} + +\subsection{Electroencephalography (EEG)} + +\subsection{Measuring Gravitational Fields} + +\subsection{Quantisation of Angular Momentum} diff --git a/buch/papers/kugel/introduction.tex b/buch/papers/kugel/introduction.tex new file mode 100644 index 0000000..5b09e9c --- /dev/null +++ b/buch/papers/kugel/introduction.tex @@ -0,0 +1,35 @@ +% vim:ts=2 sw=2 et spell tw=78: + +\section{Introduction} + +This chapter of the book is devoted to the sef of functions called +\emph{spherical harmonics}. However, before we dive into the topic, we want to +make a few preliminary remarks to avoid ``upsetting'' a certain type of +reader. Specifically, we would like to specify that the authors of this +chapter not mathematicians but engineers, and therefore the text will not be +always complete with sound proofs after every claim. Instead we will go +through the topic in a more intuitive way including rigorous proofs only if +they are enlightening or when they are very short. Where no proofs are given +we will try to give an intuition for why it is true. + +That being said, when talking about spherical harmonics one could start by +describing their name. The latter may be a cause of some confusion because of +the misleading translations in other languages. In German the name for this +set of functions is ``Kugelfunktionen'', which puts the emphasis only on the +spherical context, whereas the English name ``spherical harmonics'' also +contains the \emph{harmonic} part hinting at Fourier theories and harmonic +analysis in general. + +The structure of this chapter is organized in the following way. First, we +will quickly go through some fundamental linear algebra and Fourier theory to +refresh a few important concepts. In principle, we could have written the +whole thing starting from a much more abstract level without much preparation, +but then we would have lost some of the beauty that comes from the +appreciation of the power of some surprisingly simple ideas. Then once the +basics are done, we can explore the main topic of spherical harmonics which as +we will see arises from the eigenfunctions of the Laplacian operator in +spherical coordinates. Finally, after studying what we think are the most +beautiful and interesting properties of the spherical harmonics, to conclude +this journey we will present a few real-world applications, which are of +course most of interest for engineers. + diff --git a/buch/papers/kugel/main.tex b/buch/papers/kugel/main.tex index 06368af..98d9cb2 100644 --- a/buch/papers/kugel/main.tex +++ b/buch/papers/kugel/main.tex @@ -1,39 +1,20 @@ -% +% vim:ts=2 sw=2 et: % main.tex -- Paper zum Thema <kugel> % % (c) 2020 Hochschule Rapperswil % -\chapter{Recurrence Relations for Spherical Harmonics in Quantum Mechanics\label{chapter:kugel}} -\lhead{Recurrence Relations in Quantum Mechanics} +\begin{otherlanguage}{english} +\chapter{Spherical Harmonics\label{chapter:kugel}} +\lhead{Spherical Harmonics} \begin{refsection} \chapterauthor{Manuel Cattaneo, Naoki Pross} -\begin{verbatim} - -Ideas and current research goals --------------------------------- - -- Recurrence relations for spherical harmonics -- Associated Legendre polynomials -- Rodrigues' type formula aka Rodrigues' formula -- Applications: - * Quantization of angular momentum - * Gravitational field measurements (NASA ebb and flow, ESA goce) - * Literally anything that needs basis functions on the surface of a sphere - -Literature ----------- - -- Nichtkommutative Bildverarbeitung, T. Mendez, p57+ -- Linear Algebra Done Right, S. Axler, p212,221,231,237 -- Introduction to Quantum Mechanics, D. J. Griffith, p201+ -- Seminar Quantenmechanik, A. Müller, p101,106,114,121 -- Introduction to Partial Differential Equations, J. Oliver, p510+ -- Partial Differential Equations in Engineering Problems, K. Miller, p175,190 - -\end{verbatim} - +\input{papers/kugel/introduction} +\input{papers/kugel/preliminaries} +\input{papers/kugel/spherical-harmonics} +\input{papers/kugel/applications} \printbibliography[heading=subbibliography] \end{refsection} +\end{otherlanguage} diff --git a/buch/papers/kugel/preliminaries.tex b/buch/papers/kugel/preliminaries.tex new file mode 100644 index 0000000..03cd421 --- /dev/null +++ b/buch/papers/kugel/preliminaries.tex @@ -0,0 +1,346 @@ +% vim:ts=2 sw=2 et spell tw=78: + +\section{Preliminaries} + +The purpose of this section is to dust off some concepts that will become +important later on. This will enable us to be able to get a richer and more +general view of the topic than just liming ourselves to a specific example. + +\subsection{Vectors and inner product spaces} + +We shall start with a few fundamentals of linear algebra. We will mostly work +with complex numbers, but for the sake of generality we will do what most +textbook do, and write \(\mathbb{K}\) instead of \(\mathbb{C}\) since the +theory works the same when we replace \(\mathbb{K}\) with the real +numbers \(\mathbb{R}\). + +\begin{definition}[Vector space] + \label{kugel:def:vector-space} \nocite{axler_linear_2014} + A \emph{vector space} over a field \(\mathbb{K}\) is a set \(V\) with an + addition on \(V\) and a multiplication on \(V\) such that the following + properties hold: + \begin{enumerate}[(a)] + \item (Commutativity) \(u + v = v + u\) for all \(u, v \in V\); + \item (Associativity) \((u + v) + w = u + (v + w)\) and \((ab)v = a(bv)\) + for all \(u, v, w \in V\) and \(a, b \in \mathbb{K}\); + \item (Additive identity) There exists an element \(0 \in V\) such that + \(v + 0 = v\) for all \(v \in V\); + \item (Additive inverse) For every \(v \in V\), there exists a \(w \in V\) + such that \(v + w = 0\); + \item (Multiplicative identity) \(1 v = v\) for all \(v \in V\); + \item (Distributive properties) \(a(u + v) = au + av\) and \((a + b)v = av + + bv\) for all \(a, b \in \mathbb{K}\) and all \(u,v \in V\). + \end{enumerate} +\end{definition} + +\begin{definition}[Dot product] + \label{kugel:def:dot-product} + In the vector field \(\mathbb{K}^n\) the scalar or dot product between two + vectors \(u, v \in \mathbb{K}^n\) is + \( + u \cdot v + = u_1 \overline{v}_1 + u_2 \overline{v}_2 + \cdots + u_n \overline{v}_n + = \sum_{i=1}^n u_i \overline{v}_i. + \) +\end{definition} + +\texttt{TODO: Text here.} + +\begin{definition}[Span] +\end{definition} + +\texttt{TODO: Text here.} + +\begin{definition}[Linear independence] +\end{definition} + + +\texttt{TODO: Text here.} + +\begin{definition}[Basis] +\end{definition} + +\texttt{TODO: Text here.} + +\begin{definition}[Inner product] + \label{kugel:def:inner-product} \nocite{axler_linear_2014} + The \emph{inner product} on \(V\) is a function that takes each ordered pair + \((u, v)\) of elements of \(V\) to a number \(\langle u, v \rangle \in + \mathbb{K}\) and has the following properties: + \begin{enumerate}[(a)] + \item (Positivity) \(\langle v, v \rangle \geq 0\) for all \(v \in V\); + \item (Definiteness) \(\langle v, v \rangle = 0\) iff \(v = 0\); + \item (Additivity) \( + \langle u + v, w \rangle = + \langle u, w \rangle + \langle v, w \rangle + \) for all \(u, v, w \in V\); + \item (Homogeneity) \( + \langle \lambda u, v \rangle = + \lambda \langle u, v \rangle + \) for all \(\lambda \in \mathbb{K}\) and all \(u, v \in V\); + \item (Conjugate symmetry) + \(\langle u, v \rangle = \overline{\langle v, u \rangle}\) for all + \(u, v \in V\). + \end{enumerate} +\end{definition} + +This newly introduced inner product is thus a generalization of the scalar +product that does not explicitly depend on rows or columns of vectors. This +has the interesting consequence that anything that behaves according to the +rules given in definition \ref{kugel:def:inner-product} \emph{is} an inner +product. For example if we say that the vector space \(V = \mathbb{R}^n\), +then the dot product defined in definition \ref{kugel:def:dot-product} +\( + u \cdot v = u_1 \overline{v}_1 + u_2 \overline{v}_2 + \cdots + u_n \overline{v}_n +\) +is an inner product in \(V\), and the two are said to form an \emph{inner +product space}. + +\begin{definition}[Inner product space] + \nocite{axler_linear_2014} + An inner product space is a vector space \(V\) equipped with an inner + product on \(V\). +\end{definition} + +How about a more interesting example: the set of continuous complex valued +functions on the interval \([0; 1]\) can behave like vectors. Functions can +be added, subtracted, multiplied with scalars, are associative and there is +even the identity element (zero function \(f(x) = 0\)), so we can create an +inner product +\[ + \langle f, g \rangle = \int_0^1 f(x) \overline{g(x)} \, dx, +\] +which will indeed satisfy all of the rules for an inner product (in fact this +is called the Hermitian inner product\nocite{allard_mathematics_2009}). If +this last step sounds too good to be true, you are right, because it is not +quite so simple. The problem that we have swept under the rug here is +convergence, which any student who took an analysis class will know is a +rather hairy question. We will not need to go too much into the details since +formally discussing convergence is definitely beyond the scope of this text, +however, for our purposes we will still need to dig a little deeper for a few +more paragraph. + +\subsection{Convergence} + +In the last section we hinted that we can create ``infinite-dimensional'' +vector spaces using functions as vectors, and inner product spaces by +integrating the product of two functions of said vector space. However, there +is a problem with convergence which twofold: the obvious problem is that the +integral of the inner product may not always converge, while the second is a +bit more subtle and will be discussed later. The inner product that does +not converge is a problem because we want a \emph{norm}. + +\begin{definition}[\(L^2\) Norm] + \nocite{axler_linear_2014} + The norm of a vector \(v\) of an inner product space is a number + denoted as \(\| v \|\) that is computed by \(\| v \| = \sqrt{\langle v, v + \rangle}\). +\end{definition} + +In \(\mathbb{R}^n\) with the dot product (Euclidian space) the norm is the +geometric length of a vector, while in a more general inner product space the +norm can be thought of as a more abstract measure of ``length''. In any case +it is rather important that the expression \(\sqrt{\langle v, v \rangle}\), +which when using functions \(f: \mathbb{R} \to \mathbb{C}\) becomes +\[ + \sqrt{\langle f, f \rangle} = + \sqrt{\int_\mathbb{R} f(x) \overline{f(x)} \, dx} = + \sqrt{\int_\mathbb{R} |f(x)|^2 \, dx}, +\] +always exists. So, to fix this problems we do what mathematicians do best: +make up the solution. Since the integrand under the square root is always the +square of the magnitude, we can just specify that the functions must be +\emph{absolutely square integrable}. To be more compact it is common to just +write \(f \in L^2\), where \(L^2\) denotes the set of absolutely square +integrable functions. + +Now we can tackle the second (much more difficult) problem of convergence +mentioned at the beginning. Using the technical jargon, we need that our inner +product space is what is called a \emph{complete metric space}, which just +means that we can measure distances. For the more motivated readers although +not really necessary we can also give a more formal definition, the others can +skip to the next section. + +\begin{definition}[Metric space] + \nocite{tao_analysis_2016} + A metric space \((X, d)\) is a space \(X\) of objects (called points), + together with a distance function or metric \(d: X \times X \to [0, + +\infty)\), which associates to each pair \(x, y\) of points in \(X\) a + non-negative real number \(d(x, y) \geq 0\). Furthermore, the metric must + satisfy the following four axioms: + \begin{enumerate}[(a)] + \item For any \(x\in X\), we have \(d(x, x) = 0\). + \item (Positivity) For any \emph{distinct} \(x, y \in X\), we have + \(d(x,y) > 0\). + \item (Symmetry) For any \(x,y \in X\), we have \(d(x, y) = d(y, x)\). + \item (Triangle inequality) For any \(x, y, z \in X\) we have + \(d(x, z) \leq d(x, y) + d(y, z)\). + \end{enumerate} +\end{definition} + +As is seen in the definition metric spaces are a very abstract concept and +rely on rather weak statements, which makes them very general. Now, the more +intimidating part is the \emph{completeness} which is defined as follows. + +\begin{definition}[Complete metric space] + \label{kugel:def:complete-metric-space} + A metric space \((X, d)\) is said to be \emph{complete} iff every Cauchy + sequence in \((X, d)\) is convergent in \((X, d)\). +\end{definition} + +To fully explain definition \ref{kugel:def:complete-metric-space} it would +take a few more pages, which would get a bit too heavy. So instead we will +give an informal explanation through an counterexample to get a feeling of +what is actually happening. Cauchy sequences is a rather fancy name for a +sequence for example of numbers that keep changing, but in a such a way that +at some point the change keeps getting smaller (the infamous +\(\varepsilon-\delta\) definition). For example consider the sequence of +numbers +\[ + 1, + 1.4, + 1.41, + 1.414, + 1.4142, + 1.41421, + \ldots +\] +in the metric space \((\mathbb{Q}, d)\) with \(d(x, y) = |x - y|\). Each +element of this sequence can be written with some fraction in \(\mathbb{Q}\), +but in \(\mathbb{R}\) the sequence is converging towards the number +\(\sqrt{2}\). However, \(\sqrt{2} \notin \mathbb{Q}\). Since we can find a +sequence of fractions whose distance's limit is not in \(\mathbb{Q}\), the +metric space \((\mathbb{Q}, d)\) is \emph{not} complete. Conversely, +\((\mathbb{R}, d)\) is a complete metric space since \(\sqrt{2} \in +\mathbb{R}\). + +Of course the analogy above also applies to vectors, i.e. if in an inner +product space \(V\) over a field \(\mathbb{K}\) all sequences of vectors have +a distance that is always in \(\mathbb{K}\), then \(V\) is also a complete +metric space. In the jargon, this particular case is what is known as a +Hilbert space, after the incredibly influential German mathematician David +Hilbert. + +\begin{definition}[Hilbert space] + A Hilbert space is a vector space \(H\) with an inner product \(\langle f, g + \rangle\) and a norm \(\sqrt{\langle f, f \rangle}\) defined such that \(H\) + turns into a complete metric space. +\end{definition} + +\subsection{Orthogonal basis and Fourier series} + +Now we finally have almost everything we need to get into the domain of +Fourier theory from the perspective of linear algebra. However, we still need +to briefly discuss the matter of orthogonality\footnote{See chapter +\ref{buch:chapter:orthogonalitaet} for more on orthogonality.} and +periodicity. Both should be very straightforward and already well known. + +\begin{definition}[Orthogonality and orthonormality] + \label{kugel:def:orthogonality} + In an inner product space \(V\) two vectors \(u, v \in V\) are said to be + \emph{orthogonal} if \(\langle u, v \rangle = 0\). Further, if both \(u\) + and \(v\) are of unit length, i.e. \(\| u \| = 1\) and \(\| v \| = 1\), then + they are said to be ortho\emph{normal}. +\end{definition} + +\begin{definition}[1-periodic function and \(C(\mathbb{R}/\mathbb{Z}; \mathbb{C})\)] + A function is said to be 1-periodic if \(f(x + 1) = f(x)\). The set of + 1-periodic function from the real to the complex + numbers is denoted by \(C(\mathbb{R}/\mathbb{Z}; \mathbb{C})\). +\end{definition} + +In the definition above the notation \(\mathbb{R}/\mathbb{Z}\) was borrowed +from group theory, and is what is known as a quotient group; Not really +relevant for our discussion but still a ``good to know''. More importantly, it +is worth noting that we could have also defined more generally \(L\)-periodic +functions with \(L\in\mathbb{R}\), however, this would introduce a few ugly +\(L\)'s everywhere which are not really necessary (it will always be possible +to extend the theorems to \(\mathbb{R} / L\mathbb{Z}\)). Thus, we will +continue without the \(L\)'s, and to simplify the language unless specified +otherwise ``periodic'' will mean 1-periodic. Having said that, we can +officially begin with the Fourier theory. + +\begin{lemma} + The subset of absolutely square integrable functions in + \(C(\mathbb{R}/\mathbb{Z}; \mathbb{C})\) together with the Hermitian inner + product + \[ + \langle f, g \rangle = \int_{[0; 1)} f(x) \overline{g(x)} \, dx + \] + form a Hilbert space. +\end{lemma} +\begin{proof} + It is not too difficult to show that the functions in \(C(\mathbb{R} / + \mathbb{Z}; \mathbb{C})\) are well behaved and form a vector space. Thus, + what remains is that the norm needs to form a complete metric space. + However, this follows from the fact that we defined the functions to be + absolutely square integrable\footnote{For the curious on why, it is because + \(L^2\) is what is known as a \emph{compact metric space}, and compact + metric spaces are always complete (see \cite{eck_metric_2022, + tao_analysis_2016}). To explain compactness and the relationship between + compactness and completeness is definitely beyond the goals of this text.}. +\end{proof} + +This was probably not a very satisfactory proof since we brushed off a lot of +details by referencing other theorems. However, the main takeaway should be +that we have ``constructed'' this new Hilbert space of functions in a such a +way that from now on we will not have to worry about the details of +convergence. + +\begin{lemma} + \label{kugel:lemma:exp-1d} + The set of functions \(E_n(x) = e^{i2\pi nx}\) on the interval + \([0; 1)\) with \(n \in \mathbb{Z} \) are orthonormal. +\end{lemma} +\begin{proof} + We need to show that \(\langle E_m, E_n \rangle\) equals 1 when \(m = n\) + and zero otherwise. This is a straightforward computation: We start by + unpacking the notation to get + \[ + \langle E_m, E_n \rangle + = \int_0^1 e^{i2\pi mx} e^{- i2\pi nx} \, dx + = \int_0^1 e^{i2\pi (m - n)x} \, dx, + \] + then inside the integrand we can see that when \(m = n\) we have \(e^0 = 1\) and + thus \( \int_0^1 dx = 1, \) while when \(m \neq n\) we can just say that we + have a new non-zero integer + \(k := m - n\) and + \[ + \int_0^1 e^{i2\pi kx} \, dx + = \frac{e^{i2\pi k} - e^{0}}{i2\pi k} + = \frac{1 - 1}{i2\pi k} + = 0 + \] + as desired. \qedhere +\end{proof} + +\begin{definition}[Spectrum] +\end{definition} + +\begin{theorem}[Fourier Theorem] + \[ + \lim_{N \to \infty} \left \| + f(x) - \sum_{n = -N}^N \hat{f}(n) E_n(x) + \right \|_2 = 0 + \] +\end{theorem} + +\begin{lemma} + The set of functions \(E_{m, n}(\xi, \eta) = e^{i2\pi m\xi}e^{i2\pi n\eta}\) + on the square \([0; 1)^2\) with \(m, n \in \mathbb{Z} \) are orthonormal. +\end{lemma} +\begin{proof} + The proof is almost identical to lemma \ref{kugel:lemma:exp-1d}, with the + only difference that the inner product is given by + \[ + \langle E_{m,n}, E_{m', n'} \rangle + = \iint_{[0;1)^2} + E_{m, n}(\xi, \eta) \overline{E_{m', n'} (\xi, \eta)} + \, d\xi d\eta + .\qedhere + \] +\end{proof} + +\subsection{Laplacian operator} + +\subsection{Eigenvalue Problem} diff --git a/buch/papers/kugel/references.bib b/buch/papers/kugel/references.bib index 013da60..b74c5cd 100644 --- a/buch/papers/kugel/references.bib +++ b/buch/papers/kugel/references.bib @@ -1,35 +1,195 @@ -% -% references.bib -- Bibliography file for the paper kugel -% -% (c) 2020 Autor, Hochschule Rapperswil -% - -@online{kugel:bibtex, - title = {BibTeX}, - url = {https://de.wikipedia.org/wiki/BibTeX}, - date = {2020-02-06}, - year = {2020}, - month = {2}, - day = {6} -} - -@book{kugel: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} -} - -@article{kugel: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} + +@article{carvalhaes_surface_2015, + title = {The surface Laplacian technique in {EEG}: Theory and methods}, + volume = {97}, + issn = {01678760}, + url = {https://linkinghub.elsevier.com/retrieve/pii/S0167876015001749}, + doi = {10.1016/j.ijpsycho.2015.04.023}, + shorttitle = {The surface Laplacian technique in {EEG}}, + pages = {174--188}, + number = {3}, + journaltitle = {International Journal of Psychophysiology}, + shortjournal = {International Journal of Psychophysiology}, + author = {Carvalhaes, Claudio and de Barros, J. Acacio}, + urldate = {2022-05-16}, + date = {2015-09}, + langid = {english}, + file = {Submitted Version:/Users/npross/Zotero/storage/SN4YUNQC/Carvalhaes and de Barros - 2015 - The surface Laplacian technique in EEG Theory and.pdf:application/pdf}, +} + +@video{minutephysics_better_2021, + title = {A Better Way To Picture Atoms}, + url = {https://www.youtube.com/watch?v=W2Xb2GFK2yc}, + abstract = {Thanks to Google for sponsoring a portion of this video! +Support {MinutePhysics} on Patreon: http://www.patreon.com/minutephysics + +This video is about using Bohmian trajectories to visualize the wavefunctions of hydrogen orbitals, rendered in 3D using custom python code in Blender. + +{REFERENCES} +A Suggested Interpretation of the Quantum Theory in Terms of "Hidden" Variables. I +David Bohm, Physical Review, Vol 85 No. 2, January 15, 1952 + +Speakable and Unspeakable in Quantum Mechanics +J. S. Bell + +Trajectory construction of Dirac evolution +Peter Holland + +The de Broglie-Bohm Causal Interpretation of Quantum Mechanics and its Application to some Simple Systems by Caroline Colijn + +Bohmian Trajectories as the Foundation of Quantum Mechanics +http://arxiv.org/abs/0912.2666v1 + +The Pilot-Wave Perspective on Quantum Scattering and Tunneling +http://arxiv.org/abs/1210.7265v2 + +A Quantum Potential Description of One-Dimensional Time-Dependent Scattering From Square Barriers and Square Wells +Dewdney, Foundations of Physics, {VoL} 12, No. 1, 1982 + +Link to Patreon Supporters: http://www.minutephysics.com/supporters/ + +{MinutePhysics} is on twitter - @minutephysics +And facebook - http://facebook.com/minutephysics + +Minute Physics provides an energetic and entertaining view of old and new problems in physics -- all in a minute! + +Created by Henry Reich}, + author = {{minutephysics}}, + urldate = {2022-05-19}, + date = {2021-05-19}, +} + +@article{ries_role_2013, + title = {Role of the lateral prefrontal cortex in speech monitoring}, + volume = {7}, + issn = {1662-5161}, + url = {http://journal.frontiersin.org/article/10.3389/fnhum.2013.00703/abstract}, + doi = {10.3389/fnhum.2013.00703}, + journaltitle = {Frontiers in Human Neuroscience}, + shortjournal = {Front. Hum. Neurosci.}, + author = {Riès, Stephanie K. and Xie, Kira and Haaland, Kathleen Y. and Dronkers, Nina F. and Knight, Robert T.}, + urldate = {2022-05-16}, + date = {2013}, + file = {Full Text:/Users/npross/Zotero/storage/W7KTJB8E/Riès et al. - 2013 - Role of the lateral prefrontal cortex in speech mo.pdf:application/pdf}, +} + +@online{saylor_academy_atomic_2012, + title = {Atomic Orbitals and Their Energies}, + url = {http://saylordotorg.github.io/text_general-chemistry-principles-patterns-and-applications-v1.0/s10-05-atomic-orbitals-and-their-ener.html}, + author = {{Saylor Academy}}, + urldate = {2022-05-30}, + date = {2012}, + file = {Atomic Orbitals and Their Energies:/Users/npross/Zotero/storage/LJ8DM3YI/s10-05-atomic-orbitals-and-their-ener.html:text/html}, +} + +@inproceedings{schmitz_using_2012, + location = {Santa Clara, {CA}, {USA}}, + title = {Using spherical harmonics for modeling antenna patterns}, + isbn = {978-1-4577-1155-8 978-1-4577-1153-4 978-1-4577-1154-1}, + url = {http://ieeexplore.ieee.org/document/6175298/}, + doi = {10.1109/RWS.2012.6175298}, + eventtitle = {2012 {IEEE} Radio and Wireless Symposium ({RWS})}, + pages = {155--158}, + booktitle = {2012 {IEEE} Radio and Wireless Symposium}, + publisher = {{IEEE}}, + author = {Schmitz, Arne and Karolski, Thomas and Kobbelt, Leif}, + urldate = {2022-05-16}, + date = {2012-01}, +} + +@online{allard_mathematics_2009, + title = {Mathematics 203-204 - Basic Analysis I-{II}}, + url = {https://services.math.duke.edu/~wka/math204/}, + author = {Allard, William K.}, + urldate = {2022-07-25}, + date = {2009}, + file = {Mathematics 203-204 - Basic Analysis I-II:/Users/npross/Zotero/storage/LJISXBCM/math204.html:text/html}, +} + +@book{olver_introduction_2013, + location = {New York, {NY}}, + title = {Introduction to partial differential equations}, + isbn = {978-3-319-02098-3}, + publisher = {Springer Science+Business Media, {LLC}}, + author = {Olver, Peter J.}, + date = {2013}, +} + +@book{miller_partial_2020, + location = {Mineola, New York}, + title = {Partial differential equations in engineering problems}, + isbn = {978-0-486-84329-2}, + abstract = {"Requiring only an elementary knowledge of ordinary differential equations, this concise text begins by deriving common partial differential equations associated with vibration, heat flow, electricity, and elasticity. The treatment discusses and applies the techniques of Fourier analysis to these equations and extends the discussion to the Fourier integral. Final chapters discuss Legendre, Bessel, and Mathieu functions and the general structure of differential operators"--}, + publisher = {Dover Publications, Inc}, + author = {Miller, Kenneth S.}, + date = {2020}, + keywords = {Differential equations, Partial}, +} + +@book{asmar_complex_2018, + location = {Cham}, + title = {Complex analysis with applications}, + isbn = {978-3-319-94062-5}, + series = {Undergraduate texts in mathematics}, + pagetotal = {494}, + publisher = {Springer}, + author = {Asmar, Nakhlé H. and Grafakos, Loukas}, + date = {2018}, + doi = {10.1007/978-3-319-94063-2}, + file = {Table of Contents PDF:/Users/npross/Zotero/storage/G2Q2RDFU/Asmar and Grafakos - 2018 - Complex analysis with applications.pdf:application/pdf}, +} + +@book{adkins_ordinary_2012, + location = {New York}, + title = {Ordinary differential equations}, + isbn = {978-1-4614-3617-1}, + series = {Undergraduate texts in mathematics}, + pagetotal = {799}, + publisher = {Springer}, + author = {Adkins, William A. and Davidson, Mark G.}, + date = {2012}, + keywords = {Differential equations}, +} + +@book{griffiths_introduction_2015, + title = {Introduction to electrodynamics}, + isbn = {978-93-325-5044-5}, + author = {Griffiths, David J}, + date = {2015}, + note = {{OCLC}: 965197645}, +} + +@book{tao_analysis_2016, + title = {Analysis 2}, + isbn = {978-981-10-1804-6}, + url = {https://doi.org/10.1007/978-981-10-1804-6}, + author = {Tao, Terence}, + urldate = {2022-07-25}, + date = {2016}, + note = {{OCLC}: 965325026}, +} + +@book{axler_linear_2015, + location = {Cham}, + title = {Linear Algebra Done Right}, + isbn = {978-3-319-11079-0 978-3-319-11080-6}, + url = {https://link.springer.com/10.1007/978-3-319-11080-6}, + series = {Undergraduate Texts in Mathematics}, + publisher = {Springer International Publishing}, + author = {Axler, Sheldon}, + urldate = {2022-07-25}, + date = {2015}, + langid = {english}, + doi = {10.1007/978-3-319-11080-6}, + file = {Submitted Version:/Users/npross/Zotero/storage/3Y8MX74N/Axler - 2015 - Linear Algebra Done Right.pdf:application/pdf}, } +@online{eck_metric_2022, + title = {Metric Spaces: Completeness}, + url = {https://math.hws.edu/eck/metric-spaces/completeness.html}, + titleaddon = {Math 331: Foundations of Analysis}, + author = {Eck, David J.}, + urldate = {2022-08-01}, + date = {2022}, + file = {Metric Spaces\: Completeness:/Users/npross/Zotero/storage/5JYEE8NF/completeness.html:text/html}, +}
\ No newline at end of file diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex new file mode 100644 index 0000000..6b23ce5 --- /dev/null +++ b/buch/papers/kugel/spherical-harmonics.tex @@ -0,0 +1,13 @@ +% vim:ts=2 sw=2 et spell: + +\section{Spherical Harmonics} + +\subsection{Eigenvalue Problem in Spherical Coordinates} + +\subsection{Properties} + +\subsection{Recurrence Relations} + +\section{Series Expansions in \(C(S^2)\)} + +\nocite{olver_introduction_2013} diff --git a/buch/papers/laguerre/eigenschaften.tex b/buch/papers/laguerre/eigenschaften.tex index 6ba9135..b007c2d 100644 --- a/buch/papers/laguerre/eigenschaften.tex +++ b/buch/papers/laguerre/eigenschaften.tex @@ -97,41 +97,42 @@ Ausserdem ist ersichtlich, dass $p(x)$ die Differentialgleichung \begin{align*} x \frac{dp}{dx} = --(\nu + 1 - x) p +(\nu + 1 - x) p \end{align*} erfüllen muss. Durch Separation erhalten wir dann \begin{align*} \int \frac{dp}{p} & = --\int \frac{\nu + 1 - x}{x} \, dx +\int \frac{\nu + 1 - x}{x} \, dx = --\int \frac{\nu + 1}{x} \, dx - \int 1\, dx +\int \frac{\nu + 1}{x} \, dx - \int 1\, dx \\ \log p & = --(\nu + 1)\log x - x + c +(\nu + 1)\log x - x + c \\ p(x) & = --C x^{\nu + 1} e^{-x} +C x^{\nu + 1} e^{-x} . \end{align*} Eingefügt in Gleichung~\eqref{laguerre:sl-lag} ergibt sich \begin{align*} \frac{C}{w(x)} \left( -x^{\nu+1} e^{-x} \frac{d^2}{dx^2} + +-x^{\nu+1} e^{-x} \frac{d^2}{dx^2} - (\nu + 1 - x) x^{\nu} e^{-x} \frac{d}{dx} \right) = x \frac{d^2}{dx^2} + (\nu + 1 - x) \frac{d}{dx}. \end{align*} Mittels Koeffizientenvergleich kann nun abgelesen werden, -dass $w(x) = x^\nu e^{-x}$ und $C=1$ mit $\nu > -1$. -Die Gewichtsfunktion $w(x)$ wächst für $x\rightarrow-\infty$ sehr schnell an, -deshalb ist die Laguerre-Gewichtsfunktion nur geeignet für den -Definitionsbereich $(0, \infty)$. +dass $w(x) = x^\nu e^{-x}$ und $C=-1$. %mit $\nu \geq 0$. +Die Gewichtsfunktion $w(x)$ wächst für $x\rightarrow-\infty$ sehr schnell an. +Ausserdem hat die Gewichtsfunktion $w(x)$ für negative $\nu$ einen Pol bei $x=0$, +daher ist die Laguerre-Gewichtsfunktion nur für den +Definitionsbereich $(0, \infty)$ geeignet. \subsubsection{Randbedingungen} Bleibt nur noch sicherzustellen, dass die Randbedingungen 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 Binary files differnew file mode 100644 index 0000000..739b02b --- /dev/null +++ b/buch/papers/lambertw/Bilder/Intuition.pdf diff --git a/buch/papers/lambertw/Bilder/Strategie.pdf b/buch/papers/lambertw/Bilder/Strategie.pdf Binary files differnew file mode 100644 index 0000000..b5428f5 --- /dev/null +++ b/buch/papers/lambertw/Bilder/Strategie.pdf diff --git a/buch/papers/lambertw/Bilder/Strategie.py b/buch/papers/lambertw/Bilder/Strategie.py new file mode 100644 index 0000000..975e248 --- /dev/null +++ b/buch/papers/lambertw/Bilder/Strategie.py @@ -0,0 +1,53 @@ +# -*- coding: utf-8 -*- +""" +Created on Fri Jul 29 09:40:11 2022 + +@author: yanik +""" +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]) +VZ = Z-V +vzScale = 0.4 + + +a = [N, N, V] +b = [V, Z, vzScale*VZ] + +X = np.array([i[0] for i in a]) +Y = np.array([i[1] for i in a]) +U = np.array([i[0] for i in b]) +W = np.array([i[1] for i in b]) + +xlim = 6 +ylim = 6 +fig, 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+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/lambertAbstandBauchgefühl.py b/buch/papers/lambertw/Bilder/lambertAbstandBauchgefühl.py new file mode 100644 index 0000000..3a90afa --- /dev/null +++ b/buch/papers/lambertw/Bilder/lambertAbstandBauchgefühl.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-0.08, 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/Bilder/pursuerDGL2.png b/buch/papers/lambertw/Bilder/pursuerDGL2.png Binary files differnew file mode 100644 index 0000000..f41dffe --- /dev/null +++ b/buch/papers/lambertw/Bilder/pursuerDGL2.png diff --git a/buch/papers/lambertw/main.tex b/buch/papers/lambertw/main.tex index 9e6d04f..394963f 100644 --- a/buch/papers/lambertw/main.tex +++ b/buch/papers/lambertw/main.tex @@ -7,7 +7,7 @@ \lhead{Verfolgungskurven} \begin{refsection} \chapterauthor{David Hugentobler und Yanik Kuster} - +% %Ein paar Hinweise für die korrekte Formatierung des Textes %\begin{itemize} %\item @@ -26,12 +26,12 @@ %Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren %Übersicht wegen, aber auch um GIT die Arbeit zu erleichtern. %\end{itemize} - +% \input{papers/lambertw/teil0.tex} %\input{papers/lambertw/teil2.tex} %\input{papers/lambertw/teil3.tex} \input{papers/lambertw/teil4.tex} \input{papers/lambertw/teil1.tex} - +% \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/lambertw/teil0.tex b/buch/papers/lambertw/teil0.tex index 36ef7c3..6632eca 100644 --- a/buch/papers/lambertw/teil0.tex +++ b/buch/papers/lambertw/teil0.tex @@ -6,15 +6,14 @@ \section{Was sind Verfolgungskurven? \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, der der Verfolger während der Verfolgung begeht, wird Verfolgungskurve genannt. +Der Pfad, den der Verfolger während der Verfolgung begeht, wird Verfolgungskurve genannt. Um diese Kurve zu bestimmen, kann das Verfolgungsproblem als Differentialgleichung formuliert werden. Diese Differentialgleichung entspringt der Verfolgungsstrategie des Verfolgers. - - +% \subsection{Verfolger und Verfolgungsstrategie \label{lambertw:subsection:Verfolger}} Wie bereits erwähnt, wird der Verfolger durch seine Verfolgungsstrategie definiert. @@ -25,85 +24,108 @@ Der Verfolger hat nur einen direkten Einfluss auf seinen Geschwindigkeitsvektor. Mit diesem kann er neben Richtung und Betrag auch den Abstand zwischen Verfolger und Ziel kontrollieren. Wenn zwei dieser drei Parameter durch die Strategie definiert werden, ist der dritte nicht mehr frei. Daraus folgt, dass eine Strategie zwei dieser drei Parameter festlegen muss, um den Verfolger komplett zu beschreiben. - +% \begin{table} \centering - \begin{tabular}{|>{$}c<{$}|>{$}c<{$}|>{$}c<{$}|>{$}c<{$}|} + \begin{tabular}{|>{$}l<{$}|>{$}c<{$}|>{$}c<{$}|>{$}c<{$}|} \hline - \text{}&\text{Geschwindigkeit}&\text{Abstand}&\text{Richtung}\\ + \text{Strategie}&\text{Geschwindigkeit}&\text{Abstand}&\text{Richtung}\\ \hline - \text{Strategie 1} - & \text{konstant} & \text{-} & \text{direkt auf Ziel hinzu}\\ + \text{Jagd} + & \text{konstant} & \text{-} & \text{direkt auf Ziel zu}\\ - \text{Strategie 2} - & \text{-} & \text{konstant} & \text{direkt auf Ziel hinzu}\\ + \text{Beschattung} + & \text{-} & \text{konstant} & \text{direkt auf Ziel zu}\\ - \text{Strategie 3} + \text{Vorhalt} & \text{konstant} & \text{-} & \text{etwas voraus Zielen}\\ \hline \end{tabular} \caption{mögliche Verfolgungsstrategien} \label{lambertw:table:Strategien} \end{table} - +% \begin{figure} \centering - \includegraphics[scale=0.1]{./papers/lambertw/Bilder/pursuerDGL2.pdf} - \caption{Vektordarstellung Strategie 1} + \includegraphics[scale=0.6]{./papers/lambertw/Bilder/Strategie.pdf} + \caption{Vektordarstellung Jagdstrategie} \label{lambertw:grafic:pursuerDGL2} \end{figure} - -In der Tabelle \eqref{lambertw:table:Strategien} sind drei mögliche Strategien aufgezählt. -Im Folgenden wird nur noch auf die Strategie 1 eingegangen. +% +In der Tabelle \ref{lambertw:table:Strategien} sind drei mögliche Strategien aufgezählt. +Im Folgenden wird nur noch auf die Jagdstrategie eingegangen. Bei dieser Strategie ist die Geschwindigkeit konstant und der Verfolger bewegt sich immer direkt auf sein Ziel zu. -In der Abbildung \eqref{lambertw:grafic:pursuerDGL2} ist das Problem dargestellt, -wobei $\vec{V}$ der Ortsvektor des Verfolgers, $\vec{Z}$ der Ortsvektor des Ziels und $\dot{\vec{V}}$ der Geschwindigkeitsvektor des Verfolgers ist. -Die konstante Geschwindigkeit kann man mit der Gleichung +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 +% \begin{equation} - |\dot{\vec{V}}| + |\dot{v}| = \operatorname{const} = A - \quad A\in\mathbb{R}>0 + \text{,}\quad A\in\mathbb{R}^+ +\end{equation} +% +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} + \dot{v} + = + |\dot{v}|\cdot e_{z-v} \end{equation} -darstellen. Der Geschwindigkeitsvektor wiederum kann mit der Gleichung +führt. Dies kann noch ausgeschrieben werden zu \begin{equation} - \frac{\vec{Z}-\vec{V}}{|\vec{Z}-\vec{V}|}\cdot|\dot{\vec{V}}| + \dot{v} = - \dot{\vec{V}} + |\dot{v}|\cdot\frac{z-v}{|z-v|} + \text{.} + \label{lambertw:richtungsvektor} \end{equation} -beschrieben werden. -Die Differenz der Ortsvektoren $\vec{V}$ und $\vec{Z}$ ist ein Vektor der vom Punkt $V$ auf $Z$ zeigt. -Da die Länge dieses Vektors beliebig sein kann, wird durch Division durch den Betrag, die Länge auf eins festgelegt. +% 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 $\dot{\vec{V}}$ 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{\vec{Z}-\vec{V}}{|\vec{Z}-\vec{V}|}\cdot|\dot{\vec{V}}|\cdot\dot{\vec{V}} + \frac{z-v}{|z-v|}\cdot|\dot{v}|\cdot\dot{v} &= - |\dot{\vec{V}}|^2 - \\ + |\dot{v}|^2 +\end{align} +was algebraisch zu +\begin{align} \label{lambertw:pursuerDGL} - \frac{\vec{Z}-\vec{V}}{|\vec{Z}-\vec{V}|}\cdot \frac{\dot{\vec{V}}}{|\dot{\vec{V}}|} + \frac{z-v}{|z-v|}\cdot \frac{\dot{v}}{|\dot{v}|} &= - 1 \text{.} + 1 \end{align} -Die Lösungen dieser Differentialgleichung sind die gesuchten Verfolgungskurven, insofern der Verfolger die Strategie 1 verwendet. - +umgeformt werden kann. +Die Lösungen dieser Differentialgleichung sind die gesuchten Verfolgungskurven, sofern 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} - \vec{Z}(t) + z(t) = \left( \begin{array}{c} 0 \\ t \end{array} \right) \end{equation} - +% beschrieben werden könnte. Mit dieser Gleichung ist das Ziel auch schon vollumfänglich definiert. -Die Fluchtkurve kann eine beliebige Form haben, jedoch wird die zu lösende Differentialgleichung für die Verfolgungskurve immer komplexer. +Für die Fluchtkurve kann eine beliebige Form gewählt werden, jedoch wird die zu lösende Differentialgleichung für die Verfolgungskurve komplexer. diff --git a/buch/papers/lambertw/teil1.tex b/buch/papers/lambertw/teil1.tex index fa7deb1..e8eca2c 100644 --- a/buch/papers/lambertw/teil1.tex +++ b/buch/papers/lambertw/teil1.tex @@ -6,134 +6,183 @@ \section{Wird das Ziel erreicht? \label{lambertw:section:Wird_das_Ziel_erreicht}} \rhead{Wird das Ziel erreicht?} - +% Sehr oft kommt es vor, dass bei Verfolgungsproblemen die Frage auftaucht, ob das Ziel überhaupt erreicht wird. 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 an einem Beispiel betrachtet. +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. + +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} + z(t_1)=v(t_1) + \label{bedingung_treffer} +\end{equation} +% +zu lösen. +Die Parametrisierung von $z(t)$ ist im Beispiel definiert als +\begin{equation} + z(t) + = + \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 Bedingung \eqref{bedingung_treffer} jeweils für die unterschiedlichen Startbedingungen separat analysiert. +% +\subsection{Anfangsbedingung im ersten Quadranten} % -%\subsection{Ziel erreichen (überarbeiten) -%\label{lambertw:subsection:ZielErreichen}} -Für diese Betrachtung wird das Beispiel aus \eqref{lambertw:section:teil4} zur Hilfe genommen. -Wir verwenden die hergeleiteten Gleichungen \eqref{lambertw:eqFunkXNachT} für Startbedingung im ersten Quadranten -\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{W\left(\chi\cdot e^{\chi-\frac{4t}{r_0-y_0}}\right)}{\chi}} \\ + 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}\\ + \frac{r_0+y_0}{r_0-y_0}, \quad \eta - &= - \left(\frac{x}{x_0}\right)^2\\ + = + \left(\frac{x}{x_0}\right)^2,\quad r_0 - &= - \sqrt{x_0^2+y_0^2} \text{.}\\ -\end{align*} -% -Das Ziel wird erreicht, wenn die Koordinaten des Verfolgers mit denen des Ziels bei einem diskreten Zeitpunkt $t_1$ übereinstimmen. -Somit gilt es - -\begin{equation*} - \vec{Z}(t_1)=\vec{V}(t_1) -\end{equation*} -% -zu lösen. -Aus dem vorangegangenem Beispiel, ist die Parametrisierung des Verfolgers und des Ziels bekannt. -Das Ziel wird parametrisiert durch - -\begin{equation} - \vec{Z}(t) = - \left( \begin{array}{c} 0 \\ t \end{array} \right) -\end{equation} + \sqrt{x_0^2+y_0^2} + \text{.} +\end{align} % -und der Verfolger durch - +Der Verfolger ist durch \begin{equation} - \vec{V}(t) + v(t) = \left( \begin{array}{c} x(t) \\ y(t) \end{array} \right) \text{.} \end{equation} % - Da $y(t)$ viel komplexer ist als $x(t)$ wird das Problem in zwei einzelne Teilprobleme zerlegt. Wobei die Bedingung der x- und y-Koordinaten einzeln überprüft werden. Es entstehen daher folgende Bedingungen - -\begin{align*} +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 die Bedingungen +% +\begin{align} 0 &= x(t) = - x_0\sqrt{\frac{W\left(\chi\cdot e^{\chi-\frac{4t}{r_0-y_0}}\right)}{\chi}} + x_0\sqrt{\frac{1}{\chi}W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right)\right)} \\ t &= 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) - \\ -\end{align*} + \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. -Zuerst wird die Bedingung der x-Koordinate betrachtet. -Diese kann durch dividieren durch $x_0$, anschliessendes quadrieren und multiplizieren von $\chi$ vereinfacht werden. Daraus folgt +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 $\chi \neq 0$ kann \begin{equation} - 0 - = - W\left(\chi\cdot e^{\chi-\frac{4t}{r_0-y_0}}\right) - \text{.} + 0 + = + x_0\sqrt{\frac{1}{\chi}W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right)\right)} \end{equation} -% +algebraisch zu +\begin{equation} + 0 + = + W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right)\right) +\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 - +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*} -% -besitzt, kann die Bedingung weiter vereinfacht werden zu - +kann die Bedingung weiter vereinfacht werden zu \begin{equation} 0 = - \chi\cdot e^{\chi-\frac{4t}{r_0-y_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. -Somit kann nach den Gestellten Bedingungen das Ziel nie erreicht werden. -Aus der Symmetrie des Problems an der y-Achse können auch alle Anfangspunkte im zweiten Quadranten die Bedingungen nicht erfüllen. -Bei allen Anfangspunkten mit $y_0<0$ ist ein Einholen unmöglich, da die Geschwindigkeit des Verfolgers und Ziels übereinstimmen und der Verfolger dem Ziel bereits am Anfang nachgeht. -Wenn die Wertemenge der Anfangsbedingung um die positive y-Achse erweitert wird, kann das Ziel wiederum erreicht werden. -Sobald der Verfolger auf der positiven y-Achse startet, bewegen sich Verfolger und Ziel aufeinander zu, da der Geschwindigkeitsvektor des Verfolgers auf das Ziel zeigt und der Verfolger sich auf der Fluchtgeraden befindet. -Dies führt zwingend dazu, dass der Verfolger das Ziel erreichen wird. -Die Verfolgungskurve kann in diesem Fall mit - +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. +% +% +% +%Diese kann durch dividieren durch $x_0$, anschliessendes quadrieren und multiplizieren von $\chi$ vereinfacht werden. Daraus folgt +%\begin{equation} +% 0 +% = +% W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right)\right) +% \text{.} +%5\end{equation} +% +%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*} +% W(0)=0 +%\end{equation*} +% +%besitzt, 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. +%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 Verfolger niemals das Ziel einholen. +Dies kann veranschaulicht werden anhand +% +\begin{equation} + v(t)\cdot \left( \begin{array}{c} 0 \\ 1 \end{array}\right) + \leq + z(t)\cdot \left( \begin{array}{c} 0 \\ 1 \end{array}\right) + = + 1\text{.} +\end{equation} +% +Da der $y$-Anteil der Geschwindigkeit des Ziels grösser-gleich der 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. +Dies führt dazu, dass der Verfolger und das Ziel sich direkt aufeinander zu bewegen, da der Geschwindigkeitsvektor des Verfolgers auf das Ziel zeigt. +Die Folge ist, dass das Ziel zwingend erreicht wird. +Um $t_1$ zu bestimmen, kann die Verfolgungskurve in diesem Fall mit +% \begin{equation} - \vec{V}(t) + v(t) = \left( \begin{array}{c} 0 \\ y_0-t \end{array} \right) \end{equation} % parametrisiert werden. Nun kann der Abstand zwischen Verfolger und Ziel leicht bestimmt und nach 0 aufgelöst werden. -Daraus folgt - +Woraus folgt +% \begin{equation} 0 = - |\vec{V}(t_1)-\vec{Z}(t_1)| + |v(t_1)-z(t_1)| = - y_0-2t_1 + y_0-2t_1\text{,} \end{equation} % -, was aufgelöst zu - +was aufgelöst zu +% \begin{equation} t_1 = @@ -141,29 +190,160 @@ Daraus folgt \end{equation} % führt. -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. +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 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. 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 - +% \begin{equation} - |\vec{V}-\vec{Z}|<a_{min} \quad a_{min}\in\mathbb{R}>0 + |v-z|<a_{\text{min}} \text{,}\quad a_{\text{min}}\in\mathbb{R}^+ \end{equation} % -beschrieben werden, wobei $a_{min}$ dem Trefferradius entspricht. +beschrieben werden, wobei $a_{\text{min}}$ dem Trefferradius entspricht. Durch quadrieren verschwindet die Wurzel des Betrages, womit - +% \begin{equation} - |\vec{V}-\vec{Z}|^2<a_{min}^2 \quad a_{min}\in \mathbb{R} > 0 + |v-z|^2<a_{\text{min}}^2 \text{,}\quad a_{\text{min}}\in \mathbb{R}^+ + \label{lambertw:minimumAbstand} \end{equation} % die neue Bedingung ist. -Da sowohl der Betrag als auch $a_{min}$ grösser null sind, bleibt die Aussage unverändert. - - - +Da sowohl der Betrag als auch $a_{\text{min}}$ grösser null sind, bleibt die Aussage unverändert. +% +\subsection{trügerische Intuition}%verleitende/trügerische/verführerisch +In der Grafik \ref{lambertw:grafic:intuition} ist eine Mögliche Verfolgungskurve dargestellt, wobei für die Startbedingung der erste-Quadrant verwendet wurde. +Als erste Intuition für den Punkt bei dem $|v-z|$ minimal ist bietet sich der tiefste Punkt der Verfolgungskurve an, bei dem der y-Anteil des Richtungsvektors null entspricht. +Es kann argumentiert werden, dass weil die Geschwindigkeiten gleich gross sind und $\dot{v}$ sich aus einem $y$- als auch einem $x$-Anteil zusammensetzt und $\dot{z}$ nur ein $y$-Anteil besitzt, der Abstand nur grösser werden kann, wenn $e_y\cdot z>e_y\cdot v$. +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} + \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 $\alpha \in[ 0, 2\pi)$ +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. +% +\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|)<a_\text{min} + \label{lambertw:Bedingung:abstandMinimal} +\end{equation} +erfüllt ist. +Für folgende Betrachtung wurde für den Verfolger die Jagdstrategie mit $|\dot{v}|=|\dot{z}|$ gewählt. +Das Minimum des Abstandes kann mit +\begin{equation} + 0=\frac{d|z-v|}{dt} +\end{equation} +gefunden werden. +Mithilfe $(z-v)(z-v)=|z-v|^2$ kann die Gleichung umgeformt werden zu +\begin{equation} + 0=\frac{d(\sqrt{(z-v)(z-v)})}{dt} + \text{.} +\end{equation} +Jetzt kann die Ableitung leicht ausgeführt werden, womit +\begin{equation} + 0=(\dot{z}-\dot{v})\frac{z-v}{\sqrt{(z-v)(z-v)}} +\end{equation} +entsteht. +In dieser Gleichung kann $(z-v)(z-v)=|z-v|^2$ nochmals angewendet werden, wodurch die Gleichung zu +\begin{equation} + 0=(\dot{z}-\dot{v})\frac{z-v}{|z-v|} +\end{equation} +umgeformt werden kann. +Nun ist die Struktur der Gleichung \eqref{lambertw:richtungsvektor} erkennbar. +Wird dies ausgenutzt folgt +\begin{equation} + 0=(\dot{z}-\dot{v})\frac{\dot{v}}{|\dot{v}|} + \text{.} +\end{equation} +Durch algebraische Umwandlung kann die Gleichung in die Form +\begin{equation} + \dot{z}\dot{v}=|\dot{v}|^2 +\end{equation} +gebracht werden. +Da $|\dot{v}|=|\dot{z}|$ folgt +\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{,} +\end{equation} +woraus folgt, 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. +Daraus folgt, dass die Bedingung \eqref{lambertw:Bedingung:abstandMinimal} lediglich für den Abstand bei $t=\{0, \infty\}$ überprüft werden muss.
\ No newline at end of file diff --git a/buch/papers/lambertw/teil4.tex b/buch/papers/lambertw/teil4.tex index 84a0ec7..1053dd1 100644 --- a/buch/papers/lambertw/teil4.tex +++ b/buch/papers/lambertw/teil4.tex @@ -6,29 +6,29 @@ \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 werden 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 \(\vec{Z}\) bewegt sich entlang der \(y\)-Achse mit konstanter Geschwindigkeit \(v = 1\), beginnend beim Ursprung des Kartesischen Koordinatensystems. Der Verfolger \(\vec{V}\) 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} - \vec{Z} + 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) ,\: - \vec{V} + V = \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 @@ -38,57 +38,76 @@ Diese DGL haben wir bereits in Kapitel \ref{lambertw:subsection:Verfolger} defin \label{lambertw:eqMitAnfangsbed} \end{equation} -\subsection{DGL vereinfachen +\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. -Auf dem Weg dahin muss die definierte DGL zuerst wesentlich vereinfacht werden, sei es mittels algebraische Umformungen oder mit den Tools aus der Analysis. Also legen wir los! +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. -Zuerst müssen wir den Bruch in \eqref{lambertw:eqMitAnfangsbed} los werden, der sieht so nicht handlich aus. Dafür multiplizieren wir beidseitig mit dem Nenner: -\begin{equation} - \left( \begin{array}{c} 0-x \\ t-y \end{array} \right) - \cdot - \left(\begin{array}{c} \dot{x} \\ \dot{y} \end{array}\right) - = \sqrt{x^2 + (t-y)^2}. - \label{lambertw:eqOhneBruch} -\end{equation} -In einem weiteren Schritt, lösen wir das Skalarprodukt auf und erhalten folgende Gleichung \eqref{lambertw:eqOhneSkalarprod} ohne vektorielle Grössen: +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 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. Meiner Meinung ziemlich elegant und nicht selbstverständlich in der Lage zu sein, das Problem auf eine einzige Gleichung reduzieren zu können. +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. -Die nächsten Schritte sind sehr algebralastig und würden das lesen dieses Papers einfach nur mühsam machen, also werde ich diese auslassen. Hingegen werden ich die algebraische Hauptschritte erwähnen, die notwendig wären falls man es trotzdem selber ausprobieren möchte: -\begin{itemize} - \item - Quadrieren und erweitern. - \item - Gruppieren. - \item - Substitution von einzelnen Thermen mittels der Beziehung \(\dot{x}^2 + \dot{y}^2 = 1\). - \item - Und das erkennen des Musters einer Binomischen Formel. -\end{itemize} -Das Resultat aller dieser Vereinfachungen führen zu folgender Gleichung \eqref{lambertw:eqAlgVerinfacht}, die viel handhabbarer ist als zuvor: +\subsubsection{Quadrieren und Gruppieren + \label{lambertw:subsubsection:QuadUndGrup}} +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 + \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 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. 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. + \label{lambertw:eqGeschwSubstituiert} +\end{equation} +Diese unscheinbare Substitution führt dazu, dass weitere Vereinfachungen durchgeführt werden können. + +\subsubsection{Binom erkennen und vereinfachen + \label{lambertw:subsubsection:BinomVereinfach}} +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 einfachere 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 schient 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. -Der erste Schritt auf dem Weg dahin, ist es die zeitlichen Ableitung los zu werden, dafür wird \eqref{lambertw:eqGanzVerinfacht} beidseitig mit \(\dot{x}\) dividiert, was erlaubt ist, weil diese Änderung ungleich Null ist: +\subsubsection{Zeitliche Ableitungen loswerden + \label{lambertw:subsubsection:ZeitAbleit}} +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. @@ -103,32 +122,40 @@ 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 diese Eigenschaft \eqref{lambertw:eqQuotZeitAbleit} in \eqref{lambertw:eqVorKeineZeitAbleit} eingesetzt wird und vereinfacht wurde, entsteht folgende neue Gleichung: +Nach dem die Eigenschaft \eqref{lambertw:eqQuotZeitAbleit} in \eqref{lambertw:eqVorKeineZeitAbleit} eingesetzt wird und vereinfacht wurde, entsteht die neue Gleichung \begin{equation} x y^{\prime} + t - y = 0. \label{lambertw:DGLmitT} \end{equation} -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 aus der Analysis zurückgreifen, wobei die Strecke \(s\) folgendem entspricht: + +\subsubsection{Variable \(t\) eliminieren + \label{lambertw:subsubsection:ZeitAbleit}} +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} -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: +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 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 \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, \\ @@ -136,18 +163,25 @@ 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{DGL lösen +\subsection{Differentialgleichung lösen \label{lambertw:subsection:DGLloes}} -Die Gleichung \eqref{lambertw:DGLohneInt} ist eine DGL zweiter Ordnung und 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} -Diese \eqref{lambertw:DGLmitU} zu lösen ist ziemlich einfach da sie separierbar ist, aus diesem Grund werde ich direkt zur Lösung \eqref{lambertw:loesDGLmitU} übergehen: +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} + = + \int{\frac{1}{x}\:dx}, +\end{equation} +lässt sich die Gleichung mittels einer Integrationstabelle sehr rasch lösen. +Das Ergebnis ist \begin{align} \operatorname{arsinh}(u) &= @@ -157,20 +191,23 @@ Diese \eqref{lambertw:DGLmitU} zu lösen ist ziemlich einfach da sie separierbar \operatorname{sinh}(\operatorname{ln}(x) + C). \label{lambertw:loesDGLmitU} \end{align} -Indem man die Substitution rückgängig macht, erhält man eine weitere DGL erster Ordnung die bereits separiert ist und erhält folgende Gleichung: +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} -Diese \eqref{lambertw:loesDGLmitY} kann mit den selben Methoden gelöst werden wie \eqref{lambertw:DGLmitU}, diesmal aber in Kombination mit der exponentiellen Definition der \(\operatorname{sinh}\)-Funktion: +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 in nächsten Abschnitt \ref{lambertw:subsection:LoesAnalys} 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}} @@ -178,36 +215,39 @@ Nun haben wir eine Lösung, aber wie es immer mit Lösungen ist, stellt sich die \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{ln(x)}-Teil entspricht. + \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. \label{lambertw:BildFunkLoes} } \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. - \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 im nachgeht. + 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 - 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. + 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. \end{itemize} -Alle diese Eigenschafte stimmen mit dem überein, was man von einer Kurve dieser Art erwarten würde, welche durch die Grafik \ref{lambertw:BildFunkLoes} repräsentiert wurde. Nun stellt sich die Frage wie die Kurve wirklich aussieht. Dies wird im folgenden Abschnitt \ref{lambertw:subsection:AllgLoes} behandelt. +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}} -Wie üblich bei der Suche nach einer exakten Lösung, kommt ein Anfangswertproblem vor. Um dieses zu lösen, müssen wir zuerst die Anfangswerte definieren. Da wir das Problem allgemein lösen wollen, ergeben sich folgende zwei Anfangswerte: +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 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 \begin{equation} y(x)\big \vert_{t=0} = @@ -222,141 +262,170 @@ 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 \(\Delta y = y_0\). -Das Lösen des Anfangswertproblems ist ein Problem aus der Algebra, auf welches ich nicht unbedingt eingehen möchte. Zur Vollständigkeit und Nachvollziehbarkeit, werde ich aber das Gleichungssystem \eqref{lambertw:eqGleichungssystem} präsentieren, welches notwendig ist um das Anfangswertproblem zu lösen, sowie auch die allgemeine Lösung \eqref{lambertw:eqAllgLoes} die sich nach dem einsetzen der Koeffizienten \(C_1\) und \(C_2\) in die Funktion \eqref{lambertw:funkLoes} ergibt. - -\begin{itemize} - \item - Gleichungssystem: - \begin{subequations} - \begin{align} - y_0 - &= - C_1 + C_2 x^2_0 - \frac{\operatorname{ln}(x_0)}{8 \cdot C_2}, \\ - \frac{y_0}{x_0} - &= - 2 \cdot C_2 x_0 - \frac{1}{8 \cdot C_2 \cdot x_0}. - \end{align} - \label{lambertw:eqGleichungssystem} - \end{subequations} - \item - Die allgemeine Funktion: - \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) - \label{lambertw:eqAllgLoes} - \end{equation} - Damit die Funkion \eqref{lambertw:eqAllgLoes} trotzdem noch übersichtlich bleibt, wurden \(\eta\) und \(r_0\) wie folgt definiert: - \begin{equation} - \eta - = - \left(\frac{x}{x_0}\right)^2 - \:\:\text{und}\:\: - r_0 - = - \sqrt{x_0^2+y_0^2}. - \end{equation} -\end{itemize} -Diese neue allgemein Funktion \eqref{lambertw:eqAllgLoes} weist immer noch die selbe Struktur wie die vorherig hergeleitete Funktion \eqref{lambertw:funkLoes} auf, einerseits einen quadratischen Teil der in \(\eta\) enthalten ist, anderseits den \(\operatorname{ln}\)-Teil. Aus dieser Ähnlichkeit kann geschlossen werden, dass sich \eqref{lambertw:eqAllgLoes} auf eine ähnliche Art verhalten wird. +\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 das Gleichungssystem +\begin{subequations} + \label{lambertw:eqGleichungssystem} + \begin{align} + y_0 + &= + C_1 + C_2 x^2_0 - \frac{\operatorname{ln}(x_0)}{8 \cdot C_2}, \\ + \frac{y_0}{x_0} + &= + 2 \cdot C_2 x_0 - \frac{1}{8 \cdot C_2 \cdot x_0}. + \end{align} +\end{subequations} +Damit die gesuchte Funktion im ersten Quadranten bleibt, werden nur die positiven Lösungen +\begin{subequations} + \begin{align} + \label{lambertw:eqKoeff1} + C_1 + &= + \frac{2\cdot\operatorname{ln}(x_0)\left(\sqrt{x_0^2 + y_0^2} - y_0 \right) - \sqrt{x_0^2 + y_0^2} + 3 y_0}{4}, \\ + \label{lambertw:eqKoeff2} + C_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 beim Vereinfachen die gesuchte Parameterfunktion +\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). + \label{lambertw:eqAllgLoes} +\end{equation} +Damit die Funktion \eqref{lambertw:eqAllgLoes} trotzdem übersichtlich bleibt, wurden Anfangssteigung \(\eta\) und Anfangsentfernung \(r_0\) wie folgt definiert: +\begin{equation} + \eta + = + \left(\frac{x}{x_0}\right)^2 + \:\:\text{und}\:\: + r_0 + = + \sqrt{x_0^2+y_0^2}. +\end{equation} +Diese neue allgemeine Funktion \eqref{lambertw:eqAllgLoes} weist immer noch die selbe Struktur wie die vorher hergeleitete Funktion \eqref{lambertw:funkLoes} auf. Sie enthält einerseits einen quadratischen Teil, der in \(\eta\) enthalten ist, anderseits den \(\operatorname{ln}\)-Teil. Aus dieser Ähnlichkeit kann geschlossen werden, dass sich \eqref{lambertw:eqAllgLoes} auf eine ähnliche Art verhalten wird. -Nun sind wir soweit, dass wir eine \(y(x)\)-Beziehung für beliebige Anfangswerte darstellen können, unser erstes Ziel wurde erreicht. Ist das alles? Nein, wir können einen Schritt weiter gehen und uns Fragen: Ist es analytisch möglich herauszufinden, wo sich Verfolger und Ziel zu jedem Zeitpunkt befinden? Dieser Frage werden wir im nächsten Abschnitt nachgehen. +Nun sind wir soweit, dass wir eine \(y(x)\)-Beziehung für beliebige Anfangswerte darstellen können, unser erstes Ziel wurde erreicht. Wir können aber einen Schritt weiter gehen und uns Fragen: Ist es analytisch möglich herauszufinden, wo sich Verfolger und Ziel zu jedem Zeitpunkt befinden? Dieser Frage werden wir im nächsten Abschnitt nachgehen. \subsection{Funktion nach der Zeit \label{lambertw:subsection:FunkNachT}} -Lieber Leser sei mir nicht böse, aber in diesem Abschnitt werde ich ein wenig mehr bei den algebraischen Umformungen ins Detail gehen. Dies hat auch einen bestimmten Grund, ich möchte 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 zu kurz zu beantworten, es ist "YouTube's favorite special function" laut dem Mathematiker Michael Penn, die Lambert-W-Funktion \(W(x)\) welche übrigens 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? 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. -Also fangen wir an. Der erste Schritt ist es herauszufinden, wie die Zeitabhängigkeit wieder hinein gebracht 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: +\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 \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 = \left(y_0+r_0\right)\left(\eta-1\right)+\left(r_0-y_0\right)\operatorname{ln}\left(\eta\right). \label{lambertw:eqFunkUndAbleitEingefuegt} \end{equation} -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} + +\subsubsection{Umformungen die zur Funktion nach der Zeit führen + \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 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} -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. +\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: \begin{equation} - e^{\displaystyle \frac{-4t}{r_0-y_0}+\frac{y_0+r_0}{r_0-y_0}} + \operatorname{exp}\left(\displaystyle \frac{-4t}{r_0-y_0}+\frac{y_0+r_0}{r_0-y_0}\right) = - \eta\cdot e^{\displaystyle \frac{y_0+r_0}{r_0-y_0}\eta} . + \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 das selbe 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 e^{\displaystyle \chi-\frac{4t}{r_0-y_0}} + \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} -Schön oder? 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 e^{\displaystyle \chi-\frac{4t}{r_0-y_0}}\right) + 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) &= - x_0\cdot\sqrt{\frac{W\left(\chi\cdot e^{\displaystyle \chi-\frac{4t}{r_0-y_0}}\right)}{\chi}}, \\ + x_0\cdot\sqrt{\frac{W\left(\chi\cdot \operatorname{exp}\left(\displaystyle \chi-\frac{4t}{r_0-y_0}\right)\right)}{\chi}}, \\ \label{lambertw:eqFunkYNachT} y(x(t)) = 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. - -Wir sind aber noch nicht ganz fertig, ich muss gestehen, dass ich in diesem Abschnitt einen wichtigen Teil verschwiegen habe. Und zwar wieso, dass ich schon bei der Gleichung \eqref{lambertw:eqFunkUndAbleitEingefuegt} wusste, dass man nach einigen Umformungen die Lambert-W-Funktion eingesetzt werden kann. -Der Grund dafür ist die Struktur + +\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. +Nun, der Grund dafür ist die Struktur \begin{equation} y = @@ -365,4 +434,4 @@ 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 diff --git a/buch/papers/parzyl/img/koordinaten.png b/buch/papers/parzyl/img/koordinaten.png Binary files differnew file mode 100644 index 0000000..3ee582d --- /dev/null +++ b/buch/papers/parzyl/img/koordinaten.png diff --git a/buch/papers/parzyl/main.tex b/buch/papers/parzyl/main.tex index ff21c9f..528a2e2 100644 --- a/buch/papers/parzyl/main.tex +++ b/buch/papers/parzyl/main.tex @@ -8,29 +8,11 @@ \begin{refsection} \chapterauthor{Thierry Schwaller, Alain Keller} -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/parzyl/teil0.tex} \input{papers/parzyl/teil1.tex} \input{papers/parzyl/teil2.tex} -\input{papers/parzyl/teil3.tex} \printbibliography[heading=subbibliography] \end{refsection} diff --git a/buch/papers/parzyl/teil0.tex b/buch/papers/parzyl/teil0.tex index 09b4024..4b251db 100644 --- a/buch/papers/parzyl/teil0.tex +++ b/buch/papers/parzyl/teil0.tex @@ -3,20 +3,239 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Teil 0\label{parzyl:section:teil0}} +\section{Einleitung\label{parzyl: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{parzyl: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. +Die Laplace-Gleichung ist eine wichtige Gleichung in der Physik. +Mit ihr lässt sich zum Beispiel das elektrische Feld in einem ladungsfreien Raum bestimmen. +In diesem Kapitel wird die Lösung der Laplace-Gleichung im +parabolischen Zylinderkoordinatensystem genauer untersucht. +\subsection{Laplace Gleichung} +Die partielle Differentialgleichung +\begin{equation} + \Delta f = 0 +\end{equation} +ist als Laplace-Gleichung bekannt. +Sie ist eine spezielle Form der Poisson-Gleichung +\begin{equation} + \Delta f = g +\end{equation} +mit g als beliebige Funktion. +In der Physik hat die Laplace-Gleichung in verschieden Gebieten +verwendet, zum Beispiel im Elektromagnetismus. +Das Gaussche Gesetz in den Maxwellgleichungen +\begin{equation} + \nabla \cdot E = \frac{\varrho}{\epsilon_0} +\label{parzyl:eq:max1} +\end{equation} +besagt das die Divergenz eines Elektrischen Feldes an einem +Punkt gleich der Ladung an diesem Punkt ist. +Das elektrische Feld ist hierbei der Gradient des elektrischen +Potentials +\begin{equation} + \nabla \phi = E. +\end{equation} +Eingesetzt in \eqref{parzyl:eq:max1} resultiert +\begin{equation} + \nabla \cdot \nabla \phi = \Delta \phi = \frac{\varrho}{\epsilon_0}, +\end{equation} +was eine Possion-Gleichung ist. +An Ladungsfreien Stellen, ist der rechte Teil der Gleichung $0$. +\subsection{Parabolische Zylinderkoordinaten +\label{parzyl:subsection:finibus}} +Im parabolischen Zylinderkoordinatensystem bilden parabolische Zylinder die Koordinatenflächen. +Die Koordinate $(\sigma, \tau, z)$ sind in kartesischen Koordinaten ausgedrückt mit +\begin{align} + x & = \sigma \tau \\ + \label{parzyl:coordRelationsa} + y & = \frac{1}{2}\left(\tau^2 - \sigma^2\right) \\ + z & = z. + \label{parzyl:coordRelationse} +\end{align} +Wird $\tau$ oder $\sigma$ konstant gesetzt resultieren die Parabeln +\begin{equation} + y = \frac{1}{2} \left( \frac{x^2}{\sigma^2} - \sigma^2 \right) +\end{equation} +und +\begin{equation} + y = \frac{1}{2} \left( -\frac{x^2}{\tau^2} + \tau^2 \right). +\end{equation} + +\begin{figure} + \centering + \includegraphics[scale=0.4]{papers/parzyl/img/koordinaten.png} + \caption{Das parabolische Koordinatensystem. Die roten Parabeln haben ein + konstantes $\sigma$ und die grünen ein konstantes $\tau$.} + \label{parzyl:fig:cordinates} +\end{figure} + +Abbildung \ref{parzyl:fig:cordinates} zeigt das Parabolische Koordinatensystem. +Das parabolische Zylinderkoordinatensystem entsteht wenn die Parabeln aus der +Ebene gezogen werden. + +Um in diesem Koordinatensystem integrieren und differenzieren zu +können braucht es die Skalierungsfaktoren $h_{\tau}$, $h_{\sigma}$ und $h_{z}$. + +\dots + +Wird eine infinitessimal kleine Distanz $ds$ zwischen zwei Punkten betrachtet +kann dies im kartesischen Koordinatensystem mit +\begin{equation} + \left(ds\right)^2 = \left(dx\right)^2 + \left(dy\right)^2 + + \left(dz\right)^2 + \label{parzyl:eq:ds} +\end{equation} +ausgedrückt werden. +Das Skalierungsfaktoren werden so bestimmt, dass +\begin{equation} + \left(ds\right)^2 = \left(h_{\sigma}d\sigma\right)^2 + + \left(h_{\tau}d\tau\right)^2 + \left(h_z dz\right)^2 +\label{parzyl:eq:dspara} +\end{equation} +gilt. +Dafür werden $dx$, $dy$, und $dz$ in \eqref{parzyl:eq:ds} mit den Beziehungen +von \eqref{parzyl:coordRelationsa} - \eqref{parzyl:coordRelationse} als +\begin{align} + dx &= \frac{\partial x }{\partial \sigma} d\sigma + + \frac{\partial x }{\partial \tau} d\tau + + \frac{\partial x }{\partial \tilde{z}} d \tilde{z} + = \tau d\sigma + \sigma d \tau \\ + dy &= \frac{\partial y }{\partial \sigma} d\sigma + + \frac{\partial y }{\partial \tau} d\tau + + \frac{\partial y }{\partial \tilde{z}} d \tilde{z} + = \tau d\tau - \sigma d \sigma \\ + dz &= \frac{\partial \tilde{z} }{\partial \sigma} d\sigma + + \frac{\partial \tilde{z} }{\partial \tau} d\tau + + \frac{\partial \tilde{z} }{\partial \tilde{z}} d \tilde{z} + = d \tilde{z} \\ +\end{align} +substituiert. +Wird diese Gleichung in der Form von \eqref{parzyl:eq:dspara} +geschrieben, resultiert +\begin{equation} + \left(d s\right)^2 = + \left(\sigma^2 + \tau^2\right)\left(d\sigma\right)^2 + + \left(\sigma^2 + \tau^2\right)\left(d\tau\right)^2 + + \left(d \tilde{z}\right)^2. +\end{equation} +Daraus ergeben sich die Skalierungsfaktoren +\begin{align} + h_{\sigma} &= \sqrt{\sigma^2 + \tau^2}\\ + h_{\sigma} &= \sqrt{\sigma^2 + \tau^2}\\ + h_{z} &= 1. +\end{align} +\subsection{Differentialgleichung} +Möchte man eine Differentialgleichung im parabolischen +Zylinderkoordinatensystem aufstellen müssen die Skalierungsfaktoren +mitgerechnet werden. +Der Laplace Operator ist dadurch gegeben als +\begin{equation} + \Delta f = \frac{1}{\sigma^2 + \tau^2} + \left( + \frac{\partial^2 f}{\partial \sigma ^2} + + \frac{\partial^2 f}{\partial \tau ^2} + \right) + + \frac{\partial^2 f}{\partial z}. + \label{parzyl:eq:laplaceInParZylCor} +\end{equation} +\subsubsection{Lösung der Helmholtz-Gleichung im parabolischen Zylinderfunktion} +Die Differentialgleichungen, welche zu den parabolischen Zylinderfunktionen führen, tauchen +%, wie bereits erwähnt, +dann auf, wenn die Helmholtz-Gleichung +\begin{equation} + \Delta f(x,y,z) = \lambda f(x,y,z) +\end{equation} +im parabolischen Zylinderkoordinatensystem +\begin{equation} + \Delta f(\sigma,\tau,z) = \lambda f(\sigma,\tau,z) +\end{equation} +gelöst wird. +%Wobei der Laplace Operator $\Delta$ im parabolischen Zylinderkoordinatensystem gegeben ist als +%\begin{equation} +% \Delta +% = +% \frac{1}{\sigma^2 + \tau^2} +% \left ( +% \frac{\partial^2}{\partial \sigma^2} +% + +% \frac{\partial^2}{\partial \tau^2} +% \right ) +% + +% \frac{\partial^2}{\partial z^2}. +%\end{equation} +Mit dem Laplace Operator aus \eqref{parzyl:eq:laplaceInParZylCor} lautet die Helmholtz Gleichung +\begin{equation} + \Delta f(\sigma, \tau, z) + = + \frac{1}{\sigma^2 + \tau^2} + \left ( + \frac{\partial^2 f(\sigma,\tau,z)}{\partial \sigma^2} + + + \frac{\partial^2 f(\sigma,\tau,z)}{\partial \tau^2} + \right ) + + + \frac{\partial^2 f(\sigma,\tau,z)}{\partial z^2} + = + \lambda f(\sigma,\tau,z). +\end{equation} +Diese partielle Differentialgleichung kann mit Hilfe von Separation gelöst werden, dazu wird +\begin{equation} + f(\sigma,\tau,z) = g(\sigma)h(\tau)i(z) +\end{equation} +gesetzt. +Was dann schlussendlich zu den Differentialgleichungen +\begin{equation}\label{parzyl:sep_dgl_1} + g''(\sigma) + - + \left ( + \lambda\sigma^2 + + + \mu + \right ) + g(\sigma) + = + 0, +\end{equation} +\begin{equation}\label{parzyl:sep_dgl_2} + h''(\tau) + - + \left ( + \lambda\tau^2 + - + \mu + \right ) + h(\tau) + = + 0 +\end{equation} +und +\begin{equation}\label{parzyl:sep_dgl_3} + i''(z) + + + \left ( + \lambda + + + \mu + \right ) + i(\tau) + = + 0 +\end{equation} +führt. +Wobei die Lösung von \eqref{parzyl:sep_dgl_3} +\begin{equation} + i(z) + = + A\cos{ + \left ( + \sqrt{\lambda + \mu}z + \right )} + + + B\sin{ + \left ( + \sqrt{\lambda + \mu}z + \right )} +\end{equation} +ist und \eqref{parzyl:sep_dgl_1} und \eqref{parzyl:sep_dgl_2} die sogenannten Weberschen Differentialgleichungen sind, welche die parabolischen Zylinder Funktionen als Lösung haben. + diff --git a/buch/papers/parzyl/teil1.tex b/buch/papers/parzyl/teil1.tex index 9ea60e2..f297189 100644 --- a/buch/papers/parzyl/teil1.tex +++ b/buch/papers/parzyl/teil1.tex @@ -3,53 +3,26 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Teil 1 +\section{Lösung \label{parzyl: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{parzyl: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. +Die Differentialgleichungen \eqref{parzyl:sep_dgl_1} und \eqref{parzyl:sep_dgl_2} können mit einer Substitution +in die Whittaker Gleichung gelöst werden. +\begin{definition} + Die Funktion + \begin{equation*} + W_{k,m}(z) = + e^{-z/2} z^{m+1/2} \, + {}_{1} F_{1}(\frac{1}{2} + m - k, 1 + 2m; z) + \end{equation*} + heisst Whittaker Funktion und ist eine Lösung + von + \begin{equation} + \frac{d^2W}{d z^2} + + \left(-\frac{1}{4} + \frac{k}{z} + \frac{\frac{1}{4} - m^2}{z^2} \right) W = 0. + \end{equation} +\end{definition} -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{parzyl: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{parzyl: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{parzyl: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. +Lösung Folgt\dots diff --git a/buch/papers/parzyl/teil2.tex b/buch/papers/parzyl/teil2.tex index 75ba259..3f890d0 100644 --- a/buch/papers/parzyl/teil2.tex +++ b/buch/papers/parzyl/teil2.tex @@ -3,38 +3,89 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Teil 2 +\section{Anwendung in der Physik \label{parzyl: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 + +\subsection{Elektrisches Feld einer semi-infiniten Platte \label{parzyl: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. +Die parabolischen Zylinderkoordinaten tauchen auf, wenn man das elektrische Feld einer semi-infiniten Platte finden will. +Das dies so ist kann im zwei Dimensionalen mit Hilfe von komplexen Funktionen gezeigt werden. Wobei die Platte dann nur eine Linie ist. +Jede komplexe Funktion $F(z)$ kann geschrieben werden als +\begin{equation} + F(z) = U(x,y) + iV(x,y) \qquad z \in \mathbb{C}; x,y \in \mathbb{R}. +\end{equation} +Dabei muss gelten, falls die Funktion differenzierbar ist, dass +\begin{equation} + \frac{\partial U(x,y)}{\partial x} + = + \frac{\partial V(x,y)}{\partial y} + \qquad + \frac{\partial V(x,y)}{\partial x} + = + -\frac{\partial U(x,y)}{\partial y}. +\end{equation} +Aus dieser Bedingung folgt +\begin{equation} + \label{parzyl_e_feld_zweite_ab} + \underbrace{ + \frac{\partial^2 U(x,y)}{\partial x^2} + + + \frac{\partial^2 U(x,y)}{\partial y^2} + = + 0 + }_{\nabla^2U(x,y)=0} + \qquad + \underbrace{ + \frac{\partial^2 V(x,y)}{\partial x^2} + + + \frac{\partial^2 V(x,y)}{\partial y^2} + = + 0 + }_{\nabla^2V(x,y) = 0}. +\end{equation} +Zusätzlich zeigen diese Bedingungen auch, dass die zwei Funktionen $U(x,y)$ und $V(x,y)$ orthogonal zueinander sind. +Der Zusammenhang zum elektrischen Feld ist jetzt, dass das Potential an einem quellenfreien Punkt gegeben ist als +\begin{equation} + \nabla^2\phi(x,y) = 0. +\end{equation} +Da dies bei komplexen differenzierbaren Funktionen gilt, wie Gleichung \ref{parzyl_e_feld_zweite_ab} zeigt, kann entweder $U(x,y)$ oder $V(x,y)$ von einer solchen Funktion als das Potential angesehen werden. Im weiteren wird für das Potential $U(x,y)$ verwendet. +Da die Funktion, welche nicht das Potential beschreibt, in weiteren angenommen als $V(x,y)$, orthogonal zum Potential ist, zeigt dies das Verhalten des elektrischen Feldes. +Um nun zu den parabolische Zylinderkoordinaten zu gelangen muss nur noch eine geeignete komplexe Funktion $F(z)$ gefunden werden, welche eine semi-infinite Platte beschreiben kann. Man könnte natürlich auch nach anderen Funktionen suchen, welche andere Bedingungen erfüllen und würde dann auf andere Koordinatensysteme stossen. Die gesuchte Funktion in diesem Fall ist +\begin{equation} + F(z) + = + \sqrt{z} + = + \sqrt{x + iy}. +\end{equation} +Dies kann umgeformt werden zu +\begin{equation} + F(z) + = + \underbrace{\sqrt{\frac{\sqrt{x^2+y^2} + x}{2}}}_{U(x,y)} + + + i\underbrace{\sqrt{\frac{\sqrt{x^2+y^2} - x}{2}}}_{V(x,y)} + . +\end{equation} +Die Äquipotentialflächen können nun betrachtet werden, indem man die Funktion welche das Potential beschreibt gleich eine Konstante setzt, +\begin{equation} + \sigma = U(x,y) = \sqrt{\frac{\sqrt{x^2+y^2} + x}{2}}, +\end{equation} +und die Flächen mit der gleichen elektrischen Feldstärke können als +\begin{equation} + \tau = V(x,y) = \sqrt{\frac{\sqrt{x^2+y^2} - x}{2}} +\end{equation} +beschrieben werden. Diese zwei Gleichungen zeigen nun wie man vom kartesischen Koordinatensystem ins parabolische Zylinderkoordinatensystem kommt. Werden diese Formeln nun nach x und y aufgelöst so beschreibe sie, wie man aus dem parabolischen Zylinderkoordinatensystem zurück ins kartesische rechnen kann +\begin{equation} + x = \sigma \tau, +\end{equation} +\begin{equation} + y = \frac{1}{2}\left ( \tau^2 - \sigma^2 \right ) +\end{equation} + + + diff --git a/buch/papers/parzyl/teil3.tex b/buch/papers/parzyl/teil3.tex index 72c23ca..4e44bd6 100644 --- a/buch/papers/parzyl/teil3.tex +++ b/buch/papers/parzyl/teil3.tex @@ -6,35 +6,3 @@ \section{Teil 3 \label{parzyl: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{parzyl: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/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 diff --git a/buch/papers/sturmliouville/eigenschaften.tex b/buch/papers/sturmliouville/eigenschaften.tex index 9f20070..fda8be6 100644 --- a/buch/papers/sturmliouville/eigenschaften.tex +++ b/buch/papers/sturmliouville/eigenschaften.tex @@ -1,9 +1,78 @@ % -% 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 im Folgenden +\[ + L_0 + = + -\frac{d}{dx}p(x)\frac{d}{dx} +\] +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 +\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 +\cite{sturmliouville:spektralsatz-wiki}. +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 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}, diff --git a/buch/papers/sturmliouville/waermeleitung_beispiel.tex b/buch/papers/sturmliouville/waermeleitung_beispiel.tex index 14fca40..b22d5f5 100644 --- a/buch/papers/sturmliouville/waermeleitung_beispiel.tex +++ b/buch/papers/sturmliouville/waermeleitung_beispiel.tex @@ -6,16 +6,16 @@ \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 -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} + \label{sturmliouville:eq:example-fourier-heat-equation} \frac{\partial u}{\partial t} = \kappa \frac{\partial^{2}u}{{\partial x}^{2}} \end{equation} @@ -26,16 +26,17 @@ 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} + \label{sturmliouville:eq:example-fourier-boundary-condition-ends-constant} u(t,0) = u(t,l) @@ -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 @@ -59,7 +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) @@ -68,13 +71,12 @@ Somit folgen \end{equation} als Randbedingungen. -%%%%%%%%%%% Lösung der Differenzialgleichung %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +% +% Lösung der Differenzialgleichung mittels Separation +% \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 +85,9 @@ Dazu wird = T(t)X(x) \] -in die ursprüngliche Differenzialgleichung eingesetzt. +in die partielle +Differenzialgleichung~\eqref{sturmliouville:eq:example-fourier-heat-equation} +eingesetzt. Daraus ergibt sich \[ T^{\prime}(t)X(x) @@ -95,61 +99,70 @@ 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} - \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 \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 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 + \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} \end{equation} 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} \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 +Um zu verifizieren, ob die Randbedingungen erfüllt sind, wird zunächst +$p(x)$ +benötigt. +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 \[ \begin{aligned} k_a y(0) + h_a y'(0) &= h_a y'(0) = 0 \\ @@ -157,10 +170,10 @@ Werden nun $p(x)$ und die Randbedingungen \end{aligned} \] 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 @@ -169,7 +182,12 @@ 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. +% +% 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 \[ X^{\prime \prime}(x) - \mu X(x) @@ -181,186 +199,429 @@ 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 -\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 \[ X^{\prime}(x) = - \alpha A \cos \left( \alpha x \right) - - \beta B \sin \left( \beta x \right) + - \alpha A \sin \left( \alpha x \right) + + \beta B \cos \left( \beta x \right) \] 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 +Eingesetzt in Gleichung~\eqref{sturmliouville:eq: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 $ 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. -Zu bemerken ist, dass die Randbedingungen nur Anforderungen in $x$ stellen und -somit direkt für $X(x)$ übernomen werden können. +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 \[ X(0) = - A \sin(0 \alpha) + B \cos(0 \beta) + A \cos(0 \alpha) + B \sin(0 \beta) = 0. \] -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 -\begin{equation} - \mu_1 = -\alpha^{2} = -\frac{n^{2}\pi^{2}}{l^{2}} -\end{equation} +Es folgt nun wegen $\mu = -\beta^{2}$, dass +\[ + \mu_1 = -\beta^{2} = -\frac{n^{2}\pi^{2}}{l^{2}} +\] 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 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 +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 \[ X^{\prime}(0) = - \alpha A \cos(0 \alpha) - \beta B \sin(0 \beta) + -\alpha A \sin(0 \alpha) + \beta B \cos(0 \beta) = 0. \] -In diesem Fall muss $A = 0$ gelten. +In diesem Fall muss $B = 0$ gelten. Zusammen mit der Bedignung für $x = l$ folgt nun \[ X^{\prime}(l) = - 0 \alpha \cos(\alpha l) - \beta B \sin(\beta l) + - \alpha A \sin(\alpha l) + 0 \beta \cos(\beta l) = - -\beta B \sin(\beta l) + - \alpha A \sin(\alpha l) = 0. \] -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 \[ \begin{aligned} - \sin(\beta l) &= 0 \\ - \beta l &= n \pi \qquad n \in \mathbb{N} \\ - \beta &= \frac{n \pi}{l} \qquad n \in \mathbb{N} + \sin(\alpha l) &= 0 \\ + \alpha l &= n \pi \qquad n \in \mathbb{N} \\ + \alpha &= \frac{n \pi}{l} \qquad n \in \mathbb{N} \end{aligned} \] 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 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}}. \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 $A$ 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 \[ X(x) = - a_n\sin\left(\frac{n\pi}{l}x\right) + a_0 + - b_n\cos\left(\frac{n\pi}{l}x\right) + \sum_{n = 1}^{\infty} a_n\cos\left(\frac{n\pi}{l}x\right) + + + \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right). \] -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}. -Diese Lösen wir über das charakteristische Polynom +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{sturmliouville:eq:example-fourier-initial-conditions} + u(0, x) + = + 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) +\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{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. + +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{sturmliouville:eq:dot-product-cosine} + \langle u(0, x), \cos\left(\frac{m \pi}{l}x\right) \rangle + = + \langle 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), + \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 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: +\[ +\begin{aligned} + \hat{u}_c(0, x) + &= + \begin{cases} + u(0, -x) & -l \leq x < 0 + \\ + u(0, x) & 0 \leq x \leq l + \end{cases} + \\ + \hat{u}_s(0, x) + &= + \begin{cases} + -u(0, -x) & -l \leq x < 0 + \\ + u(0, x) & 0 \leq x \leq l + \end{cases}. +\end{aligned} +\] + +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{sturmliouville:eq:dot-product-cosine} +berechnet: +\[ +\begin{aligned} + \int_{-l}^{l}\hat{u}_c(0, x)\cos\left(\frac{m \pi}{l}x\right)dx + =& + \int_{-l}^{l} \left[a_0 + + + \sum_{n = 1}^{\infty} a_n\cos\left(\frac{n\pi}{l}x\right) + + + \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right)\right] + \cos\left(\frac{m \pi}{l}x\right) dx + \\ + 2\int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx + =& + a_0 \int_{-l}^{l}cos\left(\frac{m \pi}{l}x\right) dx + + + \sum_{n = 1}^{\infty}\left[a_n\int_{-l}^{l}\cos\left(\frac{n\pi}{l}x\right) + \cos\left(\frac{m \pi}{l}x\right)dx\right] + \\ + &+ + \sum_{n = 1}^{\infty}\left[b_n\int_{-l}^{l}\sin\left(\frac{n\pi}{l}x\right) + \cos\left(\frac{m \pi}{l}x\right)dx\right]. +\end{aligned} +\] + +Betrachtet man nun die Summanden auf der rechten Seite stellt man fest, dass +nahezu alle Terme verschwinden, denn +\[ + \int_{-l}^{l}cos\left(\frac{m \pi}{l}x\right) dx + = + 0 +\] +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 sind. + +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: +\[ + \begin{aligned} + 2\int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx + &= + a_m\frac{l}{m\pi}\int_{-m\pi}^{m\pi}\cos^2\left(u\right)du + \\ + &= + a_m\frac{l}{m\pi}\left[\frac{u}{2} + + \frac{\sin\left(2u\right)}{4}\right]_{u=-m\pi}^{m\pi} + \\ + &= + a_m\frac{l}{m\pi}\left(\frac{m\pi}{2} + + \underbrace{\frac{\sin\left(2m\pi\right)}{4}}_{\displaystyle = 0} - + \frac{-m\pi}{2} - + \underbrace{\frac{\sin\left(-2m\pi\right)}{4}}_{\displaystyle = 0}\right) + \\ + &= + a_m l + \\ + a_m + &= + \frac{2}{l} \int_{0}^{l}u(0, x)\cos\left(\frac{m \pi}{l}x\right)dx + \end{aligned} +\] + +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{sturmliouville:eq:example-fourier-initial-conditions} das +Skalarprodukt mit der konstanten Basisfunktion $1$ gebildet: +\[ +\begin{aligned} + \int_{-l}^{l}\hat{u}_c(0, x)dx + &= + \int_{-l}^{l} a_0 + + + \sum_{n = 1}^{\infty} a_n\cos\left(\frac{n\pi}{l}x\right) + + + \sum_{n = 1}^{\infty} b_n\sin\left(\frac{n\pi}{l}x\right)dx + \\ + 2\int_{0}^{l}u(0, x)dx + &= + a_0 \int_{-l}^{l}dx + + + \sum_{n = 1}^{\infty}\left[a_n\int_{-l}^{l}\cos\left(\frac{n\pi}{l}x\right) + dx\right] + + \sum_{n = 1}^{\infty}\left[b_n\int_{-l}^{l}\sin\left(\frac{n\pi}{l}x\right) + dx\right]. +\end{aligned} +\] + +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: +\[ +\begin{aligned} + 2\int_{0}^{l}u(0, x)dx + &= + a_0 \int_{-l}^{l}dx + \\ + &= + a_0 \left[x\right]_{x=-l}^{l} + \\ + &= + a_0(l - (-l)) + \\ + &= + a_0 \cdot 2l + \\ + a_0 + &= + \frac{1}{l} \int_{0}^{l}u(0, x)dx +\end{aligned} +\] + +% +% Lösung von T(t) +% + +\subsubsection{Lösund 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 \[ \lambda - \kappa \mu = - 0. + 0 \] +gelöst. + Es ist direkt ersichtlich, dass $\lambda = \kappa \mu$ gelten muss, was zur Lösung \[ @@ -368,8 +629,7 @@ Lösung = e^{-\kappa \mu t} \] -führt. -Und mit dem Resultat \eqref{eq:slp-example-fourier-mu-solution} +führt und mit dem Resultat~\eqref{sturmliouville:eq:example-fourier-mu-solution} \[ T(t) = @@ -377,22 +637,24 @@ Und mit dem Resultat \eqref{eq:slp-example-fourier-mu-solution} \] ergibt. -% TODO: Rechenweg -TODO: Rechenweg... Enden auf konstanter Temperatur: +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} \[ \begin{aligned} u(t,x) &= - \sum_{n=1}^{\infty}a_{n}e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t} + \sum_{n=1}^{\infty}b_{n}e^{-\frac{n^{2}\pi^{2}\kappa}{l^{2}}t} \sin\left(\frac{n\pi}{l}x\right) \\ - a_{n} + b_{n} &= \frac{2}{l}\int_{0}^{l}u(0,x)sin\left(\frac{n\pi}{l}x\right) dx \end{aligned} \] -TODO: Rechenweg... Enden isoliert: +\subsubsection{Lösung für einen Stab mit isolierten Enden} \[ \begin{aligned} u(t,x) diff --git a/buch/papers/zeta/analytic_continuation.tex b/buch/papers/zeta/analytic_continuation.tex index 0ccc116..d45a6ae 100644 --- a/buch/papers/zeta/analytic_continuation.tex +++ b/buch/papers/zeta/analytic_continuation.tex @@ -3,16 +3,16 @@ 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 $\Re(s) = 0.5$. -Diese sind relevant für die Primzahlverteilung und sind Gegenstand der Riemannschen Vermutung. +So liegen zum Beispiel unendlich viele Nullstellen der Zetafunktion bei $\Re(s) = \frac{1}{2}$. +Wie bereits erwähnt sind diese 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 - \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. @@ -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}} + }_{\displaystyle{-\frac{1}{2^s}}} + + + {\color{red}\frac{1}{3^s}} + \underbrace{- + {\color{blue}\frac{2}{4^s}} + + + {\color{red}\frac{1}{4^s}} + }_{\displaystyle{-\frac{1}{4^s}}} \ldots \\ &= \eta(s). @@ -75,7 +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) = @@ -87,86 +101,33 @@ 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 $\sum_{n=1}^{\infty}$ -\begin{equation} +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) - = + &= \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}} - \left( - 2 - \sum_{n=1}^{\infty} - e^{\frac{-n^2 \pi}{x}} - + - 1 - \right) + \,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} @@ -178,109 +139,20 @@ Zunächst teilen wir nun das Integral aus \eqref{zeta:equation:integral1} auf al 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} - = - I_1 + I_2, -\end{equation} -wobei wir uns nun 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} - \left( - - \frac{1}{2} - + \frac{\psi\left(\frac{1}{x} \right)}{\sqrt{x}} - + \frac{1}{2 \sqrt{x}} - \right) - \,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} -Dabei 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 + }_{\displaystyle{I_2}} = - \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 von \eqref{zeta:equation:integral2} sind. -Wir setzen beide Lösungen ein in Gleichung \eqref{zeta:equation:integral3} 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} @@ -294,8 +166,8 @@ Wir setzen beide Lösungen ein in Gleichung \eqref{zeta:equation:integral3} 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) @@ -356,17 +228,21 @@ 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} -%TODO Definitionen und Gleichungen klarer unterscheiden +was einer Spiegelung an der $\Re(s) = \frac{1}{2}$ Geraden entspricht. +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} -\subsection{Poissonsche Summenformel} \label{zeta:subsec:poisson_summation} +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. -Der Beweis für Gleichung \ref{zeta:equation:psi} folgt direkt durch die poissonsche Summenformel. -Um diese zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta Funktion. +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 + 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) @@ -437,8 +313,8 @@ Um diese zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta F \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} @@ -454,7 +330,7 @@ Um diese zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta F \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 in Gleichung \eqref{zeta:equation:1934} erhalten wir \begin{equation} \sum_{k=-\infty}^{\infty} F(k) @@ -472,6 +348,190 @@ Um diese zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta F \, 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} +\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 + }_{\displaystyle{I_3}} + + + \underbrace{ + \frac{1}{2} + \int_0^1 + x^{\frac{s}{2}-\frac{3}{2}} + - + x^{\frac{s}{2}-1} + \,dx + }_{\displaystyle{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}. diff --git a/buch/papers/zeta/einleitung.tex b/buch/papers/zeta/einleitung.tex index 3b70531..828678d 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 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/primzahlfunktion2.tex} + \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 a6ed512..9c08dd2 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 @@ -28,9 +28,9 @@ Die Verteilung der Primzahlen ist Gegenstand der Riemannschen Vermutung, welche = \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} @@ -53,33 +53,34 @@ Die Verteilung der Primzahlen ist Gegenstand der Riemannschen Vermutung, welche \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 \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 $n$. - Da die Summen alle möglichen Kombinationen von Exponenten und Primzahlen in \eqref{zeta:equation:eulerprodukt2} enthält haben wir + 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} \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) + \zeta(s), \end{equation} + wodurch das Eulerprodukt bewiesen ist. \end{proof} diff --git a/buch/papers/zeta/fazit.tex b/buch/papers/zeta/fazit.tex new file mode 100644 index 0000000..027f324 --- /dev/null +++ b/buch/papers/zeta/fazit.tex @@ -0,0 +1,94 @@ +\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 $s=-1$. +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) + &= + \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} + \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 $\Gamma(-\frac{1}{2})$. + +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} +\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}}_{\displaystyle{\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 $\Gamma(-\frac{1}{2})$ 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 $\Gamma(\frac{3}{2})$ verwendet. +Es ergeben sich die Werte +\begin{align*} + \Gamma(1) + &= 1\\ + \Gamma\biggl(-\frac{1}{2}\biggr) + &= \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{\pi^{-\frac{1}{2}}}{ + -\frac{\sqrt{\pi}}{2}} + \\ + &= + -\frac{1}{12}. +\end{align*} + +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. +\begin{figure} + \centering + \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} +\end{figure} diff --git a/buch/papers/zeta/images/Makefile b/buch/papers/zeta/images/Makefile new file mode 100644 index 0000000..611662d --- /dev/null +++ b/buch/papers/zeta/images/Makefile @@ -0,0 +1,10 @@ +# +# Makefile to build images +# +all: primzahlfunktion2.pdf zetaplot.pdf + +primzahlfunktion2.pdf: primzahlfunktion2.tex + pdflatex primzahlfunktion2.tex + +zetaplot.pdf: zetaplot.tex zetapath.tex + pdflatex zetaplot.tex diff --git a/buch/papers/zeta/continuation_overview.tikz.tex b/buch/papers/zeta/images/continuation_overview.tikz.tex index 836ab1d..836ab1d 100644 --- a/buch/papers/zeta/continuation_overview.tikz.tex +++ b/buch/papers/zeta/images/continuation_overview.tikz.tex 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{<filename>.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{<path to file>}{<filename>.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}% 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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} + diff --git a/buch/papers/zeta/images/primzahlfunktion_paper.pgf b/buch/papers/zeta/images/primzahlfunktion_paper.pgf new file mode 100644 index 0000000..b9d67d3 --- /dev/null +++ b/buch/papers/zeta/images/primzahlfunktion_paper.pgf @@ -0,0 +1,505 @@ +%% Creator: Matplotlib, PGF backend +%% +%% To include the figure in your LaTeX document, write +%% \input{<filename>.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{<path to file>}{<filename>.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}% 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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{<filename>.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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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{<filename>.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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For loading figures +%% from other directories you can use the `import` package +%% \usepackage{import} +%% +%% and then include the figures with +%% \import{<path to file>}{<filename>.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}% 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+\endgroup% diff --git a/buch/papers/zeta/images/zeta_re_0_plot.pgf b/buch/papers/zeta/images/zeta_re_0_plot.pgf new file mode 100644 index 0000000..29a844e --- /dev/null +++ b/buch/papers/zeta/images/zeta_re_0_plot.pgf @@ -0,0 +1,1242 @@ +%% Creator: Matplotlib, PGF backend +%% +%% To include the figure in your LaTeX document, write +%% \input{<filename>.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/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.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 Binary files differnew file mode 100644 index 0000000..c6d3693 --- /dev/null +++ b/buch/papers/zeta/images/zetaplot.pdf diff --git a/buch/papers/zeta/images/zetaplot.tex b/buch/papers/zeta/images/zetaplot.tex new file mode 100644 index 0000000..521bb1a --- /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{papers/zeta/images/zetapath.tex} + +\draw[color=blue,line width=1pt] \zetapath; + +\end{tikzpicture} +\end{document} + 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 new file mode 100644 index 0000000..53fd305 --- /dev/null +++ b/buch/papers/zeta/presentation/presentation.tex @@ -0,0 +1,368 @@ +\documentclass[ngerman, aspectratio=169]{beamer} + +%style +\mode<presentation>{ + \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}} +\newcommand*{\YE}{\textcolor{violet}} + + + + +\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]{../images/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{../images/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} + \begin{frame} + \frametitle{Fortsetzung auf $\Re(s) > 0$} + \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}: + \begin{align*} + \left(1 - \frac{1}{2^{s-1}} \right) + \zeta(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) + \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{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) + \ldots + \\ + = + 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{../images/primzahlfunktion.pgf}} + \end{center} + \end{frame} + + + \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{../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{../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{../images/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 Binary files differnew file mode 100644 index 0000000..b8c7298 --- /dev/null +++ b/buch/papers/zeta/presentation/zeta_color_plot-img0.png 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{<filename>.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{<path to file>}{<filename>.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}% 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+\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/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() + diff --git a/buch/papers/zeta/references.bib b/buch/papers/zeta/references.bib index a4f2521..f2a2f31 100644 --- a/buch/papers/zeta/references.bib +++ b/buch/papers/zeta/references.bib @@ -4,32 +4,58 @@ % (c) 2020 Autor, Hochschule Rapperswil % -@online{zeta:bibtex, - title = {BibTeX}, - 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} +} + +@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} } -@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:mryoumath, + title = {Riemann Zeta Function Playlist}, + url = {https://www.youtube.com/playlist?list=PL32446FDD4DA932C9}, + author = {MrYouMath}, + year = {2022}, + 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} +} diff --git a/buch/papers/zeta/zeta_color_plot-img0.png b/buch/papers/zeta/zeta_color_plot-img0.png Binary files differnew file mode 100644 index 0000000..b8c7298 --- /dev/null +++ b/buch/papers/zeta/zeta_color_plot-img0.png 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{<filename>.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{<path to file>}{<filename>.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}% 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+\endgroup% diff --git a/buch/papers/zeta/zeta_gamma.tex b/buch/papers/zeta/zeta_gamma.tex index db41676..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) &= @@ -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} = @@ -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} |
