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-rw-r--r--vorlesungen/slides/10/matrix-vektor-dgl.tex44
-rw-r--r--vorlesungen/slides/10/so2.tex138
-rw-r--r--vorlesungen/slides/10/taylor.tex2
-rw-r--r--vorlesungen/slides/10/vektorfelder.mp266
-rw-r--r--vorlesungen/slides/5/potenzreihenmethode.tex2
5 files changed, 406 insertions, 46 deletions
diff --git a/vorlesungen/slides/10/matrix-vektor-dgl.tex b/vorlesungen/slides/10/matrix-vektor-dgl.tex
index d9bd97c..f7bd995 100644
--- a/vorlesungen/slides/10/matrix-vektor-dgl.tex
+++ b/vorlesungen/slides/10/matrix-vektor-dgl.tex
@@ -6,50 +6,6 @@
%
% !TeX spellcheck = de_CH
\bgroup
-%\begin{frame}[t]
-%\setlength{\abovedisplayskip}{5pt}
-%\setlength{\belowdisplayskip}{5pt}
-%\frametitle{Matrix-Vektor-DGL}
-%\vspace{-20pt}
-%\begin{columns}[t,onlytextwidth]
-%\begin{column}{0.48\textwidth}
-% \begin{block}{Bekannt}
-% Vorgehen für DGL 1.~Ordnung mit Skalaren.
-% Aufgabe: Sei $a, x, x_0 \in \mathbb R$,
-% \[
-% \dot x = ax,
-% \quad
-% x(0) = x_0
-% \]
-% Lösung: $x(t) = \exp(at) x_0$, wobei
-% \begin{align*}
-% \exp(at)
-% &= 1 + at + \frac{a^2t^2}{2!} + \ldots\\
-% &= e^{at}
-% \end{align*}
-% \end{block}
-%\end{column}
-%\begin{column}{0.48\textwidth}
-% \begin{block}{Mit Matrizen}
-% Wir können:
-% \begin{itemize}
-% \item Matrizen potenzieren: $A$, $A^2$, $A^3$
-% \item Matrizen skalieren: $At$
-% \item Matrizen addieren: $A_1 + A_2$
-% \end{itemize}
-% Also ist auch
-% \[
-% \exp(At) = 1 + At + \frac{A^2t^2}{2!} + \ldots
-% \]
-% wohldefiniert.
-% \end{block}
-%\end{column}
-%\end{columns}
-%Folglich, sei $A \in M_n$ und $x \in \mathbb R^n$,
-%\[ \dot x = Ax, \quad x(0) = x_0, \]
-%dann ist
-%\[ x = \exp(At)x_0. \]
-%\end{frame}
\begin{frame}[t]
\setlength{\abovedisplayskip}{5pt}
diff --git a/vorlesungen/slides/10/so2.tex b/vorlesungen/slides/10/so2.tex
new file mode 100644
index 0000000..e3f74ae
--- /dev/null
+++ b/vorlesungen/slides/10/so2.tex
@@ -0,0 +1,138 @@
+%
+% so2.tex -- Illustration of so(2) -> SO(2)
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+% Erstellt durch Roy Seitz
+%
+% !TeX spellcheck = de_CH
+\bgroup
+
+\newcommand{\gSL}[2]{\ensuremath{\text{SL}(#1, \mathbb{#2})}}
+\newcommand{\gSO}[1]{\ensuremath{\text{SO}(#1)}}
+\newcommand{\gGL}[2]{\ensuremath{\text{GL}(#1, \mathbb #2)}}
+
+\newcommand{\asl}[2]{\ensuremath{\mathfrak{sl}(#1, \mathbb{#2})}}
+\newcommand{\aso}[1]{\ensuremath{\mathfrak{so}(#1)}}
+\newcommand{\agl}[2]{\ensuremath{\mathfrak{gl}(#1, \mathbb #2)}}
+
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Von der Lie-Gruppe zur -Algebra}
+\vspace{-20pt}
+\begin{columns}[t,onlytextwidth]
+\begin{column}{0.48\textwidth}
+ \begin{block}{Lie-Gruppe}
+ Darstellung von \gSO2:
+ \begin{align*}
+ \mathbb R
+ &\to
+ \gSO2
+ \\
+ t
+ &\mapsto
+ \begin{pmatrix}
+ \cos t & -\sin t \\
+ \sin t & \phantom-\cos t
+ \end{pmatrix}
+ \end{align*}
+ \end{block}
+ \begin{block}{Ableitung am neutralen Element}
+ \begin{align*}
+ \frac{d}{d t}
+ &
+ \left.
+ \begin{pmatrix}
+ \cos t & -\sin t \\
+ \sin t & \phantom-\cos t
+ \end{pmatrix}
+ \right|_{ t = 0}
+ \\
+ =
+ &
+ \begin{pmatrix} -\sin0 & -\cos0 \\ \phantom-\cos0 & -\sin0 \end{pmatrix}
+ =
+ \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix}
+ \end{align*}
+ \end{block}
+\end{column}
+\begin{column}{0.48\textwidth}
+ \begin{block}{Lie-Algebra}
+ Darstellung von \aso2:
+ \begin{align*}
+ \mathbb R
+ &\to
+ \aso2
+ \\
+ t
+ &\mapsto
+ \begin{pmatrix}
+ 0 & -t \\
+ t & \phantom-0
+ \end{pmatrix}
+ \end{align*}
+ \end{block}
+\end{column}
+\end{columns}
+\end{frame}
+
+
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Von der Lie-Algebra zur -Gruppe}
+\vspace{-20pt}
+\begin{columns}[t,onlytextwidth]
+\begin{column}{0.48\textwidth}
+ \begin{block}{Differentialgleichung}
+ Gegeben:
+ \[
+ A
+ =
+ \dot\gamma(0) = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix}
+ \]
+ Gesucht:
+ \[ \dot \gamma (t) = \gamma(t) A \qquad \gamma \in \gSO2 \]
+ \[ \Rightarrow \gamma(t) = \exp(At) \gamma(0) = \exp(At) \]
+ \end{block}
+\end{column}
+\begin{column}{0.48\textwidth}
+ \begin{block}{Lie-Algebra}
+ Potenzen von A:
+ \begin{align*}
+ A^2 &= -I &
+ A^3 &= -A &
+ A^4 &= I &
+ \ldots
+ \end{align*}
+ \end{block}
+\end{column}
+\end{columns}
+Folglich:
+\begin{align*}
+ \exp(At)
+ &= I + At
+ + A^2\frac{t^2}{2!}
+ + A^3\frac{t^3}{3!}
+ + A^4\frac{t^4}{4!}
+ + A^5\frac{t^5}{5!}
+ + \ldots \\
+ &= \begin{pmatrix}
+ \vspace*{3pt}
+ 1 - \frac{t^2}{2} + \frac{t^4}{4!} - \ldots
+ &
+ -t + \frac{t^3}{3!} - \frac{t^5}{5!} + \ldots
+ \\
+ t - \frac{t^3}{3!} + \frac{t^5}{5!} - \ldots
+ &
+ 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \ldots
+ \end{pmatrix}
+ =
+ \begin{pmatrix}
+ \cos t & -\sin t \\
+ \sin t & \phantom-\cos t
+ \end{pmatrix}
+\end{align*}
+
+\end{frame}
+\egroup
diff --git a/vorlesungen/slides/10/taylor.tex b/vorlesungen/slides/10/taylor.tex
index bbd1126..920470f 100644
--- a/vorlesungen/slides/10/taylor.tex
+++ b/vorlesungen/slides/10/taylor.tex
@@ -1,5 +1,5 @@
%
-% eindiomensional.tex -- Lösung der eindimensionalen DGL
+% taylor.tex -- Repetition Taylot-Reihen
%
% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
% Erstellt durch Roy Seitz
diff --git a/vorlesungen/slides/10/vektorfelder.mp b/vorlesungen/slides/10/vektorfelder.mp
new file mode 100644
index 0000000..f488327
--- /dev/null
+++ b/vorlesungen/slides/10/vektorfelder.mp
@@ -0,0 +1,266 @@
+%
+% Stroemungsfelder linearer Differentialgleichungen
+%
+% (c) 2015 Prof Dr Andreas Mueller, Hochschule Rapperswil
+% 2021-04-14, Roy Seitz, Copied for SeminarMatrizen
+%
+verbatimtex
+\documentclass{book}
+\usepackage{times}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{amsfonts}
+\usepackage{txfonts}
+\begin{document}
+etex;
+
+input TEX;
+
+TEXPRE("%&latex" & char(10) &
+"\documentclass{book}" &
+"\usepackage{times}" &
+"\usepackage{amsmath}" &
+"\usepackage{amssymb}" &
+"\usepackage{amsfonts}" &
+"\usepackage{txfonts}" &
+"\begin{document}");
+TEXPOST("\end{document}");
+
+%
+% Vektorfeld in der Ebene mit Lösungskurve
+% so(2)
+%
+beginfig(1)
+
+% Scaling parameter
+numeric unit;
+unit := 150;
+
+% Some points
+z1 = (-1.5, 0) * unit;
+z2 = ( 1.5, 0) * unit;
+z3 = ( 0, -1.5) * unit;
+z4 = ( 0, 1.5) * unit;
+
+pickup pencircle scaled 1pt;
+drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0));
+drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10));
+label.top(btex $x_1$ etex, z2 shifted (10,0));
+label.rt(btex $x_2$ etex, z4 shifted (0,10));
+
+% Draw circles
+for x = 0.2 step 0.2 until 1.4:
+ path p;
+ p = (x,0);
+ for a = 5 step 5 until 355:
+ p := p--(x*cosd(a), x*sind(a));
+ endfor;
+ p := p--cycle;
+ pickup pencircle scaled 1pt;
+ draw p scaled unit withcolor red;
+endfor;
+
+% Define DGL
+def dglField(expr x, y) =
+ %(-0.5 * (x + y), -0.5 * (y - x))
+ (-y, x)
+enddef;
+
+% Draw arrows for each grid point
+pickup pencircle scaled 0.5pt;
+for x = -1.5 step 0.1 until 1.55:
+ for y = -1.5 step 0.1 until 1.55:
+ drawarrow ((x, y) * unit)
+ --(((x,y) * unit) shifted (8 * dglField(x,y)))
+ withcolor blue;
+ endfor;
+endfor;
+
+endfig;
+
+
+
+%
+% Vektorfeld in der Ebene mit Lösungskurve
+% X \in sl(2, R)
+%
+beginfig(2)
+
+numeric unit;
+unit := 150;
+
+z0 = ( 0, 0);
+z1 = (-1.5, 0) * unit;
+z2 = ( 1.5, 0) * unit;
+z3 = ( 0, -1.5) * unit;
+z4 = ( 0, 1.5) * unit;
+
+pickup pencircle scaled 1pt;
+drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0));
+drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10));
+label.top(btex $x_1$ etex, z2 shifted (10,0));
+label.rt(btex $x_2$ etex, z4 shifted (0,10));
+
+% Draw flow lines
+for y = -1.4 step 0.2 until 1.4:
+ path p;
+ p = (-1.5,y) -- (1.5, y);
+ pickup pencircle scaled 1pt;
+ draw p scaled unit withcolor red;
+endfor;
+
+def dglField(expr x, y) =
+ (y, 0)
+enddef;
+
+% Draw arrows for each grid point
+pickup pencircle scaled 0.5pt;
+for x = -1.5 step 0.1 until 1.55:
+ for y = -1.5 step 0.1 until 1.55:
+ drawarrow ((x, y) * unit)
+ --(((x,y) * unit) shifted (8 * dglField(x,y)))
+ withcolor blue;
+ endfor;
+endfor;
+
+endfig;
+
+
+
+
+%
+% Vektorfeld in der Ebene mit Lösungskurve
+% Y \in sl(2, R)
+%
+beginfig(3)
+
+numeric unit;
+unit := 150;
+
+z0 = ( 0, 0);
+z1 = (-1.5, 0) * unit;
+z2 = ( 1.5, 0) * unit;
+z3 = ( 0, -1.5) * unit;
+z4 = ( 0, 1.5) * unit;
+
+pickup pencircle scaled 1pt;
+drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0));
+drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10));
+label.top(btex $x_1$ etex, z2 shifted (10,0));
+label.rt(btex $x_2$ etex, z4 shifted (0,10));
+
+% Draw flow lines
+for x = -1.4 step 0.2 until 1.4:
+ path p;
+ p = (x, -1.5) -- (x, 1.5);
+ pickup pencircle scaled 1pt;
+ draw p scaled unit withcolor red;
+endfor;
+
+def dglField(expr x, y) =
+ (0, x)
+enddef;
+
+% def dglFieldp(expr z) =
+% dglField(xpart z, ypart z)
+% enddef;
+%
+% def curve(expr z, l) =
+% path p;
+% p := z;
+% for t = 0 step 1 until l:
+% p := p--((point (length p) of p) shifted (0.01 * dglFieldp(point (length p) of p)));
+% endfor;
+% draw p scaled unit withcolor red;
+% enddef;
+%
+% numeric outerlength;
+% outerlength = 200;
+% curve(( 0.1, 0), outerlength);
+% curve(( 0.2, 0), outerlength);
+%
+% numeric innerlength;
+% innerlength = 500;
+%
+% for a = 0 step 30 until 330:
+% curve(0.05 * (cosd(a), sind(a)), innerlength);
+% endfor;
+
+% Draw arrows for each grid point
+pickup pencircle scaled 0.5pt;
+for x = -1.5 step 0.1 until 1.55:
+ for y = -1.5 step 0.1 until 1.55:
+ drawarrow ((x, y) * unit)
+ --(((x,y) * unit) shifted (8 * dglField(x,y)))
+ withcolor blue;
+ endfor;
+endfor;
+
+endfig;
+
+
+%
+% Vektorfeld in der Ebene mit Lösungskurve
+% H \in sl(2, R)
+%
+beginfig(4)
+
+numeric unit;
+unit := 150;
+
+z0 = ( 0, 0);
+z1 = (-1.5, 0) * unit;
+z2 = ( 1.5, 0) * unit;
+z3 = ( 0, -1.5) * unit;
+z4 = ( 0, 1.5) * unit;
+
+pickup pencircle scaled 1pt;
+drawarrow (z1 shifted (-25,0))--(z2 shifted (25,0));
+drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10));
+label.top(btex $x_1$ etex, z2 shifted (25,0));
+label.rt(btex $x_2$ etex, z4 shifted (0,10));
+
+def dglField(expr x, y) =
+ (x, -y)
+enddef;
+
+def dglFieldp(expr z) =
+ dglField(xpart z, ypart z)
+enddef;
+
+def curve(expr z, l) =
+ path p;
+ p := z;
+ for t = 0 step 1 until l:
+ p := p--((point (length p) of p) shifted (0.01 * dglFieldp(point (length p) of p)));
+ endfor;
+ draw p scaled unit withcolor red;
+enddef;
+
+for i = -1 step 2 until 1:
+ for k = -1 step 2 until 1:
+ curve((1.3 * i, 1.5 * k), 18);
+ curve((1.1 * i, 1.5 * k), 35);
+ curve((0.9 * i, 1.5 * k), 55);
+ curve((0.7 * i, 1.5 * k), 80);
+ curve((0.5 * i, 1.5 * k), 114);
+ curve((0.3 * i, 1.5 * k), 165);
+ curve((0.1 * i, 1.5 * k), 275);
+ endfor;
+endfor;
+
+% Draw arrows for each grid point
+pickup pencircle scaled 0.5pt;
+for x = -1.5 step 0.1 until 1.55:
+ for y = -1.5 step 0.1 until 1.55:
+ drawarrow ((x, y) * unit)
+ --(((x,y) * unit) shifted (8 * dglField(x,y)))
+ withcolor blue;
+ endfor;
+endfor;
+
+endfig;
+
+
+
+end;
diff --git a/vorlesungen/slides/5/potenzreihenmethode.tex b/vorlesungen/slides/5/potenzreihenmethode.tex
index 0c3503d..12d3fa5 100644
--- a/vorlesungen/slides/5/potenzreihenmethode.tex
+++ b/vorlesungen/slides/5/potenzreihenmethode.tex
@@ -79,7 +79,7 @@ a_k=\frac1{k!}a^kC}
\\
\uncover<4->{
\Rightarrow y(x) &= C}\uncover<8->{+Cax}\uncover<9->{ + C\frac12(ax)^2}
-\uncover<10->{ + C \frac16(ac)^3}
+\uncover<10->{ + C \frac16(ax)^3}
\uncover<11->{ + \dots+C\frac{1}{k!}(ax)^k+\dots}
\ifthenelse{\boolean{presentation}}{
\only<12>{
generated by cgit v1.2.1 (git 2.18.0) at 2024-11-23 03:34:21 +0100