From 588ba7c095b15178deedd7aed4713d66a8cf4da7 Mon Sep 17 00:00:00 2001
From: Roy Seitz <roy.seitz@ost.ch>
Date: Wed, 14 Apr 2021 22:29:43 +0200
Subject: =?UTF-8?q?Slides=20f=C3=BCr=20Vorlesung=20DGL=20begonnen.?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
vorlesungen/slides/10/n-zu-1.tex | 54 +++++++++++
vorlesungen/slides/10/taylor.tex | 195 +++++++++++++++++++++++++++++++++++++++
2 files changed, 249 insertions(+)
create mode 100644 vorlesungen/slides/10/n-zu-1.tex
create mode 100644 vorlesungen/slides/10/taylor.tex
(limited to 'vorlesungen/slides/10')
diff --git a/vorlesungen/slides/10/n-zu-1.tex b/vorlesungen/slides/10/n-zu-1.tex
new file mode 100644
index 0000000..e3fffe9
--- /dev/null
+++ b/vorlesungen/slides/10/n-zu-1.tex
@@ -0,0 +1,54 @@
+%
+% n-zu-1.tex -- Umwandlend einer DGL n-ter Ordnung in ein System 1. Ordnung
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+% Erstellt: 2021-04-14, Roy Seitz
+%
+% !TeX spellcheck = de_CH
+\bgroup
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Reicht $1.$ Ordnung?}
+\vspace{-20pt}
+\begin{columns}[t,onlytextwidth]
+\begin{column}{0.48\textwidth}
+\begin{block}{Beispiel: DGL 3.~Ordnung} \vspace*{-1ex}
+ \begin{align*}
+ x^{(3)} + a_2 \ddot x + a_1 \dot x + a_0 x = 0 \\
+ \Rightarrow
+ x^{(3)} = -a_2 \ddot x - a_1 \dot x - a_0 x
+ \end{align*}
+\end{block}
+\begin{block}{Ziel: Nur noch 1.~Ableitungen}
+ Einführen neuer Variablen:
+ \begin{align*}
+ x_0 &\coloneqq x &
+ x_1 &\coloneqq \dot x &
+ x_2 &\coloneqq \ddot x
+ \end{align*}
+System von Gleichungen 1.~Ordnung
+ \begin{align*}
+ \dot x_0 &= x_1 \\
+ \dot x_1 &= x_2 \\
+ \dot x_2 &= -a_2 x_2 - a_1 x_1 - a_0 x_0
+\end{align*}
+\end{block}
+\end{column}
+\begin{column}{0.48\textwidth}
+\begin{block}{Als Vektor-Gleichung} \vspace*{-1ex}
+ \begin{align*}
+ \frac{d}{dt}
+ \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix}
+ = \begin{pmatrix}
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ -a_0 & -a_1 & -a_2
+ \end{pmatrix}
+ \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix}
+ \end{align*}
+\end{block}
+\end{column}
+\end{columns}
+\end{frame}
+\egroup
diff --git a/vorlesungen/slides/10/taylor.tex b/vorlesungen/slides/10/taylor.tex
new file mode 100644
index 0000000..8912cb7
--- /dev/null
+++ b/vorlesungen/slides/10/taylor.tex
@@ -0,0 +1,195 @@
+%
+% eindiomensional.tex -- Lösung der eindimensionalen DGL
+%
+% (c) 2021 Roy Seitz, Hochschule Rapperswil
+%
+% !TeX spellcheck = de_CH
+\bgroup
+
+\begin{frame}[t]
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ \frametitle{Beispiel $\sin x$}
+ \vspace{-20pt}
+ %\onslide<+->
+ \begin{block}{Taylor-Approximationen von $\sin x$}
+ \begin{align*}
+ p_n(x)
+ &=
+ \uncover<1->{0}
+ \uncover<2->{+ x}
+ \uncover<3->{+ 0 \frac{x^2}{2!}}
+ \uncover<4->{- 1 \frac{x^3}{3!}}
+ \uncover<5->{+ 0 \frac{x^4}{4!}}
+ \uncover<6->{+ 1 \frac{x^5}{5!}}
+ \uncover<7->{+ \ldots}
+ \uncover<8->{
+ = \sum_{k=0}^{n/2} (-1)^{2k + 1}\frac{x^{2k+1}}{(2k+1)!}
+ }
+ \end{align*}
+ \end{block}
+ \begin{center}
+ \begin{tikzpicture}[>=latex,thick,scale=1.3]
+ \draw[->] (-5.0, 0.0) -- (5.0,0.0) coordinate[label=$x$];
+ \draw[->] ( 0.0,-1.5) -- (0.0,1.5);
+ \clip (-5,-1.5) rectangle (5,1.5);
+ \draw[domain=-4:4, samples=50, smooth, blue]
+ plot ({\x}, {sin(180/3.1415968*\x)})
+ node[above right] {$\sin(x)$};
+ \uncover<1>{
+ \draw[domain=-4:4, samples=2, smooth, red]
+ plot ({\x}, {0})
+ node[above right] {$p_0(x)$};}
+ \uncover<2>{
+ \draw[domain=-1.5:1.5, samples=2, smooth, red]
+ plot ({\x}, {\x})
+ node[below right] {$p_1(x)$};}
+ \uncover<3>{
+ \draw[domain=-1.5:1.5, samples=2, smooth, red]
+ plot ({\x}, {\x})
+ node[below right] {$p_2(x)$};}
+ \uncover<4>{
+ \draw[domain=-3:3, samples=50, smooth, red]
+ plot ({\x}, {\x - \x*\x*\x/6})
+ node[above right] {$p_3(x)$};}
+ \uncover<5>{
+ \draw[domain=-3:3, samples=50, smooth, red]
+ plot ({\x}, {\x - \x*\x*\x/6})
+ node[above right] {$p_4(x)$};}
+ \uncover<6>{
+ \draw[domain=-3.9:3.9, samples=50, smooth, red]
+ plot ({\x}, {\x - \x*\x*\x/6 + \x*\x*\x*\x*\x/120})
+ node[below right] {$p_5(x)$};}
+ \uncover<7>{
+ \draw[domain=-3.9:3.9, samples=50, smooth, red]
+ plot ({\x}, {\x - \x*\x*\x/6 + \x*\x*\x*\x*\x/120})
+ node[below right] {$p_6(x)$};}
+ \uncover<8->{
+ \draw[domain=-4:4, samples=50, smooth, red]
+ plot ({\x}, {\x - \x*\x*\x/6 + \x*\x*\x*\x*\x/120 -
+ \x*\x*\x*\x*\x*\x*\x/5040})
+ node[above right] {$p_7(x)$};}
+ \end{tikzpicture}
+ \end{center}
+\end{frame}
+
+
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Taylor-Reihen}
+\vspace{-20pt}
+\onslide<+->
+ \begin{block}{Polynom-Approximationen von $f(t)$}
+ \vspace{-15pt}
+ \begin{align*}
+ p_n(t)
+ &=
+ f(0)
+ + f'(0) t
+ + f''(0)\frac{t^2}{2}
+ + f^{(3)}(0)\frac{t^3}{3!}
+ + \ldots
+ + f^{(n)}(0) \frac{t^n}{n!}
+ =
+ \sum_{k=0}^{n} f^{(k)} \frac{t^k}{k!}
+ \end{align*}
+ \end{block}
+ \begin{block}{Die ersten $n$ Ableitungen von $f(0)$ und $p_n(0)$ sind gleich!}
+ \vspace{-15pt}
+ \begin{align*}
+ p'_n(t)
+ &=
+ f'(0)
+ + f''(0)t
+ + f^{(3)}(0) \frac{t^2}{2!}
+ + \mathcal O(t^3)
+ &\Rightarrow&&
+ p'_n(0) = f'(0)
+ \\
+ p''_n(0)
+ &=
+ f''(0) + f^{(3)}(0)t + \ldots + f^{(n)}(0) \frac{t^{n-2}}{(n-2)!}
+ &\Rightarrow&&
+ p''_n(0) = f''(0)
+ \end{align*}
+ \end{block}
+ \begin{block}{Für unendlich lange Polynome stimmen alle Ableitungen überein!}
+ \vspace{-15pt}
+ \begin{align*}
+ \lim_{n\to \infty} p_n(t)
+ =
+ f(t)
+ \end{align*}
+ \end{block}
+\end{frame}
+
+
+\begin{frame}[t]
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ \frametitle{Beispiel $\exp x$}
+ \vspace{-20pt}
+ %\onslide<+->
+ \begin{block}{Taylor-Approximationen von $\exp x$}
+ \begin{align*}
+ p_n(x)
+ =
+ 1
+ \uncover<1->{+ x}
+ \uncover<2->{+ \frac{x^2}{2}}
+ \uncover<3->{+ \frac{x^3}{3!}}
+ \uncover<4->{+ \frac{x^4}{4!}}
+ \uncover<5->{+ \frac{x^5}{5!}}
+ \uncover<6->{+ \frac{x^6}{6!}}
+ \uncover<7->{+ \ldots
+ = \sum_{k=0}^{n} \frac{x^k}{k!}}
+ \end{align*}
+ \end{block}
+ \begin{center}
+ \begin{tikzpicture}[>=latex,thick,scale=1.3]
+ \draw[->] (-4.0, 0.0) -- (4.0,0.0) coordinate[label=$x$];
+ \draw[->] ( 0.0,-0.5) -- (0.0,2.5);
+ \clip (-3,-0.5) rectangle (3,2.5);
+ \draw[domain=-4:1, samples=50, smooth, blue]
+ plot ({\x}, {exp(\x)})
+ node[above right] {$\exp(x)$};
+ \uncover<1>{
+ \draw[domain=-4:1.5, samples=10, smooth, red]
+ plot ({\x}, {1 + \x})
+ node[below right] {$p_1(x)$};}
+ \uncover<2>{
+ \draw[domain=-4:1, samples=50, smooth, red]
+ plot ({\x}, {1 + \x + \x*\x/2})
+ node[below right] {$p_2(x)$};}
+ \uncover<3>{
+ \draw[domain=-4:1, samples=50, smooth, red]
+ plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6})
+ node[below right] {$p_3(x)$};}
+ \uncover<4>{
+ \draw[domain=-4:0.9, samples=50, smooth, red]
+ plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24})
+ node[below left] {$p_4(x)$};}
+ \uncover<5>{
+ \draw[domain=-4:0.9, samples=50, smooth, red]
+ plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24
+ + \x*\x*\x*\x*\x/120})
+ node[below left] {$p_5(x)$};}
+ \uncover<6>{
+ \draw[domain=-4:0.9, samples=50, smooth, red]
+ plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24
+ + \x*\x*\x*\x*\x/120
+ + \x*\x*\x*\x*\x*\x/720})
+ node[below left] {$p_6(x)$};}
+ \uncover<7>{
+ \draw[domain=-4:0.9, samples=50, smooth, red]
+ plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24
+ + \x*\x*\x*\x*\x/120
+ + \x*\x*\x*\x*\x*\x/720
+ + \x*\x*\x*\x*\x*\x*\x/5040})
+ node[below left] {$p_7(x)$};}
+ \end{tikzpicture}
+ \end{center}
+\end{frame}
+
+\egroup
--
cgit v1.2.1
From bd913de11450cc6294e874f2c38a838b0c1e6cd5 Mon Sep 17 00:00:00 2001
From: Roy Seitz <roy.seitz@ost.ch>
Date: Thu, 15 Apr 2021 17:46:09 +0200
Subject: Skalar-zu-Matrix DGL Folie.
---
vorlesungen/slides/10/matrix-vektor-dgl.tex | 127 ++++++++++++++++++++++++++++
vorlesungen/slides/10/n-zu-1.tex | 5 +-
vorlesungen/slides/10/taylor.tex | 3 +-
vorlesungen/slides/10/template.tex | 21 +++++
4 files changed, 154 insertions(+), 2 deletions(-)
create mode 100644 vorlesungen/slides/10/matrix-vektor-dgl.tex
create mode 100644 vorlesungen/slides/10/template.tex
(limited to 'vorlesungen/slides/10')
diff --git a/vorlesungen/slides/10/matrix-vektor-dgl.tex b/vorlesungen/slides/10/matrix-vektor-dgl.tex
new file mode 100644
index 0000000..d9bd97c
--- /dev/null
+++ b/vorlesungen/slides/10/matrix-vektor-dgl.tex
@@ -0,0 +1,127 @@
+%
+% matrix-vektor-dgl.tex -- DGL mit Matrix-Koeffizienten und Vektor-Variablen
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+% Erstellt durch Roy Seitz
+%
+% !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}
+ \setlength{\belowdisplayskip}{5pt}
+ \frametitle{1.~Ordnung mit Skalaren}
+ \vspace{-20pt}
+ \begin{columns}[t,onlytextwidth]
+ \begin{column}{0.48\textwidth}
+ \begin{block}{Aufgabe}
+ Sei $a, x(t), x_0 \in \mathbb R$,
+ \[
+ \dot x(t) = ax(t),
+ \quad
+ x(0) = x_0
+ \]
+ \end{block}
+ \begin{block}{Potenzreihen-Ansatz}
+ Sei $a_k \in \mathbb R$,
+ \[
+ x(t) = a_0 + a_1t + a_2t^2 + a_3t^3 \ldots
+ \]
+ \end{block}
+ \end{column}
+ \begin{column}{0.48\textwidth}
+ \begin{block}{Lösung}
+ Einsetzen in DGL, Koeffizientenvergleich liefert
+ \[ x(t) = \exp(at) \, x_0, \]
+ wobei
+ \begin{align*}
+ \exp(at)
+ &= 1 + at + \frac{a^2t^2}{2} + \frac{a^3t^3}{3!} + \ldots \\
+ &{\color{gray}(= e^{at}.)}
+ \end{align*}
+ \end{block}
+ \end{column}
+ \end{columns}
+\end{frame}
+
+\begin{frame}[t]
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ \frametitle{1.~Ordnung mit Matrizen}
+ \vspace{-20pt}
+ \begin{columns}[t,onlytextwidth]
+ \begin{column}{0.48\textwidth}
+ \begin{block}{Aufgabe}
+ Sei $A \in M_n$, $x(t), x_0 \in \mathbb R^n$,
+ \[
+ \dot x(t) = Ax(t),
+ \quad
+ x(0) = x_0
+ \]
+ \end{block}
+ \begin{block}{Potenzreihen-Ansatz}
+ Sei $A_k \in \mathbb M_n$,
+ \[
+ x(t) = A_0 + A_1t + A_2t^2 + A_3t^3 \ldots
+ \]
+ \end{block}
+ \end{column}
+ \begin{column}{0.48\textwidth}
+ \begin{block}{Lösung}
+ Einsetzen in DGL, Koeffizientenvergleich liefert
+ \[ x(t) = \exp(At) \, x_0, \]
+ wobei
+ \[
+ \exp(At)
+ = 1 + At + \frac{A^2t^2}{2} + \frac{A^3t^3}{3!} + \ldots
+ \]
+ \end{block}
+ \end{column}
+ \end{columns}
+\end{frame}
+
+\egroup
diff --git a/vorlesungen/slides/10/n-zu-1.tex b/vorlesungen/slides/10/n-zu-1.tex
index e3fffe9..737df03 100644
--- a/vorlesungen/slides/10/n-zu-1.tex
+++ b/vorlesungen/slides/10/n-zu-1.tex
@@ -2,7 +2,7 @@
% n-zu-1.tex -- Umwandlend einer DGL n-ter Ordnung in ein System 1. Ordnung
%
% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
-% Erstellt: 2021-04-14, Roy Seitz
+% Erstellt durch Roy Seitz
%
% !TeX spellcheck = de_CH
\bgroup
@@ -47,6 +47,9 @@ System von Gleichungen 1.~Ordnung
\end{pmatrix}
\begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix}
\end{align*}
+
+ Geht für jede lineare Differentialgleichung!
+
\end{block}
\end{column}
\end{columns}
diff --git a/vorlesungen/slides/10/taylor.tex b/vorlesungen/slides/10/taylor.tex
index 8912cb7..bbd1126 100644
--- a/vorlesungen/slides/10/taylor.tex
+++ b/vorlesungen/slides/10/taylor.tex
@@ -1,7 +1,8 @@
%
% eindiomensional.tex -- Lösung der eindimensionalen DGL
%
-% (c) 2021 Roy Seitz, Hochschule Rapperswil
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+% Erstellt durch Roy Seitz
%
% !TeX spellcheck = de_CH
\bgroup
diff --git a/vorlesungen/slides/10/template.tex b/vorlesungen/slides/10/template.tex
new file mode 100644
index 0000000..50f0a3b
--- /dev/null
+++ b/vorlesungen/slides/10/template.tex
@@ -0,0 +1,21 @@
+%
+% template.tex -- slide template
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+% Erstellt durch Roy Seitz
+%
+% !TeX spellcheck = de_CH
+\bgroup
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Template}
+\vspace{-20pt}
+\begin{columns}[t,onlytextwidth]
+\begin{column}{0.48\textwidth}
+\end{column}
+\begin{column}{0.48\textwidth}
+\end{column}
+\end{columns}
+\end{frame}
+\egroup
--
cgit v1.2.1
From cb562f607b18540df333d6d9a911cf51c91884d0 Mon Sep 17 00:00:00 2001
From: Roy Seitz <roy.seitz@ost.ch>
Date: Thu, 15 Apr 2021 23:49:49 +0200
Subject: Mehr Folien und einige Typos.
---
vorlesungen/slides/10/matrix-vektor-dgl.tex | 44 -----
vorlesungen/slides/10/so2.tex | 138 +++++++++++++++
vorlesungen/slides/10/taylor.tex | 2 +-
vorlesungen/slides/10/vektorfelder.mp | 266 ++++++++++++++++++++++++++++
4 files changed, 405 insertions(+), 45 deletions(-)
create mode 100644 vorlesungen/slides/10/so2.tex
create mode 100644 vorlesungen/slides/10/vektorfelder.mp
(limited to 'vorlesungen/slides/10')
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;
--
cgit v1.2.1
From 8358d9bc031913305a52c6c2ab05184b89f7678f Mon Sep 17 00:00:00 2001
From: Roy Seitz <roy.seitz@ost.ch>
Date: Sat, 17 Apr 2021 22:00:36 +0200
Subject: Slides erweitert.
---
vorlesungen/slides/10/ableitung-exp.tex | 60 +++++++++
vorlesungen/slides/10/matrix-dgl.tex | 83 ++++++++++++
vorlesungen/slides/10/matrix-vektor-dgl.tex | 83 ------------
vorlesungen/slides/10/potenzreihenmethode.tex | 91 +++++++++++++
vorlesungen/slides/10/repetition.tex | 151 ++++++++++++++++++++++
vorlesungen/slides/10/so2.tex | 8 --
vorlesungen/slides/10/taylor.tex | 176 ++++++++++++++------------
7 files changed, 483 insertions(+), 169 deletions(-)
create mode 100644 vorlesungen/slides/10/ableitung-exp.tex
create mode 100644 vorlesungen/slides/10/matrix-dgl.tex
delete mode 100644 vorlesungen/slides/10/matrix-vektor-dgl.tex
create mode 100644 vorlesungen/slides/10/potenzreihenmethode.tex
create mode 100644 vorlesungen/slides/10/repetition.tex
(limited to 'vorlesungen/slides/10')
diff --git a/vorlesungen/slides/10/ableitung-exp.tex b/vorlesungen/slides/10/ableitung-exp.tex
new file mode 100644
index 0000000..10ce191
--- /dev/null
+++ b/vorlesungen/slides/10/ableitung-exp.tex
@@ -0,0 +1,60 @@
+%
+% ableitung-exp.tex -- Ableitung von exp(x)
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+% Erstellt durch Roy Seitz
+%
+% !TeX spellcheck = de_CH
+\bgroup
+\begin{frame}[t]
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ %\frametitle{Ableitung von $\exp(x)$}
+ %\vspace{-20pt}
+ \begin{columns}[t,onlytextwidth]
+ \begin{column}{0.48\textwidth}
+ \begin{block}{Ableitung von $\exp(at)$}
+ \begin{align*}
+ \frac{d}{dt} \exp(at)
+ &=
+ \frac{d}{dt} \sum_{k=0}^{\infty} a^k \frac{t^k}{k!}
+ \\
+ &\uncover<2->{
+ = \sum_{k=0}^{\infty} a^k\frac{kt^{k-1}}{k(k-1)!}
+ }
+ \\
+ &\uncover<3->{
+ = a \sum_{k=1}^{\infty}
+ a^{k-1}\frac{t^{k-1}}{(k-1)!}
+ }
+ \\
+ &\uncover<4->{
+ = a \exp(at)
+ }
+ \end{align*}
+ \end{block}
+ \end{column}
+ \begin{column}{0.48\textwidth}
+ \uncover<5->{
+ \begin{block}{Ableitung von $\exp(At)$}
+ \begin{align*}
+ \frac{d}{dt} \exp(At)
+ &=
+ \frac{d}{dt} \sum_{k=0}^{\infty} A^k \frac{t^k}{k!}
+ \\
+ &=
+ \sum_{k=0}^{\infty} A^k\frac{kt^{k-1}}{k(k-1)!}
+ \\
+ &=
+ A \sum_{k=1}^{\infty} A^{k-1}\frac{t^{k-1}}{(k-1)!}
+ \\
+ &=
+ A \exp(At)
+ \end{align*}
+ \end{block}
+ }
+ \end{column}
+ \end{columns}
+\end{frame}
+
+\egroup
diff --git a/vorlesungen/slides/10/matrix-dgl.tex b/vorlesungen/slides/10/matrix-dgl.tex
new file mode 100644
index 0000000..ae68fb1
--- /dev/null
+++ b/vorlesungen/slides/10/matrix-dgl.tex
@@ -0,0 +1,83 @@
+%
+% matrix-dgl.tex -- Matrix-Differentialgleichungen
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+% Erstellt durch Roy Seitz
+%
+% !TeX spellcheck = de_CH
+\bgroup
+
+\begin{frame}[t]
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ \frametitle{1.~Ordnung mit Skalaren}
+ \vspace{-20pt}
+ \begin{columns}[t,onlytextwidth]
+ \begin{column}{0.48\textwidth}
+ \begin{block}{Aufgabe}
+ Sei $a, x(t), x_0 \in \mathbb R$,
+ \[
+ \dot x(t) = ax(t),
+ \quad
+ x(0) = x_0
+ \]
+ \end{block}
+ \begin{block}{Potenzreihen-Ansatz}
+ Sei $a_k \in \mathbb R$,
+ \[
+ x(t) = a_0 + a_1t + a_2t^2 + a_3t^3 \ldots
+ \]
+ \end{block}
+ \end{column}
+ \begin{column}{0.48\textwidth}
+ \begin{block}{Lösung}
+ Einsetzen in DGL, Koeffizientenvergleich liefert
+ \[ x(t) = \exp(at) \, x_0, \]
+ wobei
+ \begin{align*}
+ \exp(at)
+ &= 1 + at + \frac{a^2t^2}{2} + \frac{a^3t^3}{3!} + \ldots \\
+ &{\color{gray}(= e^{at}.)}
+ \end{align*}
+ \end{block}
+ \end{column}
+ \end{columns}
+\end{frame}
+
+\begin{frame}[t]
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ \frametitle{1.~Ordnung mit Matrizen}
+ \vspace{-20pt}
+ \begin{columns}[t,onlytextwidth]
+ \begin{column}{0.48\textwidth}
+ \begin{block}{Aufgabe}
+ Sei $A \in M_n$, $x(t), x_0 \in \mathbb R^n$,
+ \[
+ \dot x(t) = Ax(t),
+ \quad
+ x(0) = x_0
+ \]
+ \end{block}
+ \begin{block}{Potenzreihen-Ansatz}
+ Sei $A_k \in \mathbb M_n$,
+ \[
+ x(t) = A_0 + A_1t + A_2t^2 + A_3t^3 \ldots
+ \]
+ \end{block}
+ \end{column}
+ \begin{column}{0.48\textwidth}
+ \begin{block}{Lösung}
+ Einsetzen in DGL, Koeffizientenvergleich liefert
+ \[ x(t) = \exp(At) \, x_0, \]
+ wobei
+ \[
+ \exp(At)
+ = 1 + At + \frac{A^2t^2}{2} + \frac{A^3t^3}{3!} + \ldots
+ \]
+ \end{block}
+ \end{column}
+ \end{columns}
+\end{frame}
+
+\egroup
diff --git a/vorlesungen/slides/10/matrix-vektor-dgl.tex b/vorlesungen/slides/10/matrix-vektor-dgl.tex
deleted file mode 100644
index f7bd995..0000000
--- a/vorlesungen/slides/10/matrix-vektor-dgl.tex
+++ /dev/null
@@ -1,83 +0,0 @@
-%
-% matrix-vektor-dgl.tex -- DGL mit Matrix-Koeffizienten und Vektor-Variablen
-%
-% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
-% Erstellt durch Roy Seitz
-%
-% !TeX spellcheck = de_CH
-\bgroup
-
-\begin{frame}[t]
- \setlength{\abovedisplayskip}{5pt}
- \setlength{\belowdisplayskip}{5pt}
- \frametitle{1.~Ordnung mit Skalaren}
- \vspace{-20pt}
- \begin{columns}[t,onlytextwidth]
- \begin{column}{0.48\textwidth}
- \begin{block}{Aufgabe}
- Sei $a, x(t), x_0 \in \mathbb R$,
- \[
- \dot x(t) = ax(t),
- \quad
- x(0) = x_0
- \]
- \end{block}
- \begin{block}{Potenzreihen-Ansatz}
- Sei $a_k \in \mathbb R$,
- \[
- x(t) = a_0 + a_1t + a_2t^2 + a_3t^3 \ldots
- \]
- \end{block}
- \end{column}
- \begin{column}{0.48\textwidth}
- \begin{block}{Lösung}
- Einsetzen in DGL, Koeffizientenvergleich liefert
- \[ x(t) = \exp(at) \, x_0, \]
- wobei
- \begin{align*}
- \exp(at)
- &= 1 + at + \frac{a^2t^2}{2} + \frac{a^3t^3}{3!} + \ldots \\
- &{\color{gray}(= e^{at}.)}
- \end{align*}
- \end{block}
- \end{column}
- \end{columns}
-\end{frame}
-
-\begin{frame}[t]
- \setlength{\abovedisplayskip}{5pt}
- \setlength{\belowdisplayskip}{5pt}
- \frametitle{1.~Ordnung mit Matrizen}
- \vspace{-20pt}
- \begin{columns}[t,onlytextwidth]
- \begin{column}{0.48\textwidth}
- \begin{block}{Aufgabe}
- Sei $A \in M_n$, $x(t), x_0 \in \mathbb R^n$,
- \[
- \dot x(t) = Ax(t),
- \quad
- x(0) = x_0
- \]
- \end{block}
- \begin{block}{Potenzreihen-Ansatz}
- Sei $A_k \in \mathbb M_n$,
- \[
- x(t) = A_0 + A_1t + A_2t^2 + A_3t^3 \ldots
- \]
- \end{block}
- \end{column}
- \begin{column}{0.48\textwidth}
- \begin{block}{Lösung}
- Einsetzen in DGL, Koeffizientenvergleich liefert
- \[ x(t) = \exp(At) \, x_0, \]
- wobei
- \[
- \exp(At)
- = 1 + At + \frac{A^2t^2}{2} + \frac{A^3t^3}{3!} + \ldots
- \]
- \end{block}
- \end{column}
- \end{columns}
-\end{frame}
-
-\egroup
diff --git a/vorlesungen/slides/10/potenzreihenmethode.tex b/vorlesungen/slides/10/potenzreihenmethode.tex
new file mode 100644
index 0000000..1715134
--- /dev/null
+++ b/vorlesungen/slides/10/potenzreihenmethode.tex
@@ -0,0 +1,91 @@
+%
+% potenzreihenmethode.tex
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+% Bearbeitet durch Roy Seitz
+%
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Potenzreihenmethode}
+\vspace{-15pt}
+\begin{columns}[t,onlytextwidth]
+\begin{column}{0.48\textwidth}
+\begin{block}{Lineare Differentialgleichung}
+\begin{align*}
+x'&=ax&&\Rightarrow&x'-ax&=0
+\\
+x(0)&=C
+\end{align*}
+\end{block}
+\end{column}
+\begin{column}{0.48\textwidth}
+\uncover<2->{%
+\begin{block}{Potenzreihenansatz}
+\begin{align*}
+x(t)
+&=
+a_0+ a_1t + a_2t^2 + \dots
+\\
+x(0)&=a_0=C
+\end{align*}
+\end{block}}
+\end{column}
+\end{columns}
+\uncover<3->{%
+\begin{block}{Lösung}
+\[
+\arraycolsep=1.4pt
+\begin{array}{rcrcrcrcrcr}
+\uncover<3->{ x'(t)}
+ \uncover<5->{
+ &=&\phantom{(} a_1\phantom{\mathstrut-aa_0)}
+ &+& 2a_2\phantom{\mathstrut-aa_1)}t
+ &+& 3a_3\phantom{\mathstrut-aa_2)}t^2
+ &+& 4a_4\phantom{\mathstrut-aa_3)}t^3
+ &+& \dots}\\
+\uncover<3->{-ax(t)}
+ \uncover<6->{
+ &=&\mathstrut-aa_0 \phantom{)}
+ &-& aa_1\phantom{)}t
+ &-& aa_2\phantom{)}t^2
+ &-& aa_3\phantom{)}t^3
+ &-& \dots}\\[2pt]
+\hline
+\\[-10pt]
+\uncover<3->{0}
+ \uncover<7->{
+ &=&(a_1-aa_0)
+ &+& (2a_2-aa_1)t
+ &+& (3a_3-aa_2)t^2
+ &+& (4a_4-aa_3)t^3
+ &+& \dots}\\
+\end{array}
+\]
+\begin{align*}
+\uncover<4->{
+a_0&=C}\uncover<8->{,
+\quad
+a_1=aa_0=aC}\uncover<9->{,
+\quad
+a_2=\frac12a^2C}\uncover<10->{,
+\quad
+a_3=\frac16a^3C}\uncover<11->{,
+\ldots,
+a_k=\frac1{k!}a^kC}
+\hspace{3cm}
+\\
+\uncover<4->{
+\Rightarrow x(t) &= C}\uncover<8->{+Cat}\uncover<9->{ + C\frac12(at)^2}
+\uncover<10->{ + C \frac16(at)^3}
+\uncover<11->{ + \dots+C\frac{1}{k!}(at)^k+\dots}
+\ifthenelse{\boolean{presentation}}{
+\only<12>{
+=
+C\sum_{k=0}^\infty \frac{(at)^k}{k!}}
+}{}
+\uncover<13->{=
+C\exp(at)}
+\end{align*}
+\end{block}}
+\end{frame}
diff --git a/vorlesungen/slides/10/repetition.tex b/vorlesungen/slides/10/repetition.tex
new file mode 100644
index 0000000..c45d47b
--- /dev/null
+++ b/vorlesungen/slides/10/repetition.tex
@@ -0,0 +1,151 @@
+%
+% intro.tex -- Repetition Lie-Gruppen und -Algebren
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+% Erstellt durch Roy Seitz
+%
+% !TeX spellcheck = de_CH
+\bgroup
+
+\begin{frame}[t]
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ \frametitle{Repetition}
+ \vspace{-20pt}
+ \begin{columns}[t,onlytextwidth]
+ \begin{column}{0.48\textwidth}
+ \uncover<1->{
+ \begin{block}{Lie-Gruppe}
+ Kontinuierliche Matrix-Gruppe $G$ mit bestimmter Eigenschaft
+ \end{block}
+ }
+ \uncover<3->{
+ \begin{block}{Ein-Parameter-Untergruppe}
+ Darstellung der Lie-Gruppe $G$:
+ \[
+ \gamma \colon \mathbb R \to G
+ : \quad
+ t \mapsto \gamma(t),
+ \]
+ so dass
+ \[ \gamma(s + t) = \gamma(t) \gamma(s). \]
+ \end{block}
+ }
+ \end{column}
+ \begin{column}{0.48\textwidth}
+ \uncover<2->{
+ \begin{block}{Beispiel}
+ Volumen-erhaltende Abbildungen:
+ \[ \gSL2R= \{A \in M_2 \,|\, \det(A) = 1\} .\]
+ \begin{align*}
+ \uncover<4->{ \gamma_x(t) }
+ &
+ \uncover<4->{= \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix} }
+ \\
+ \uncover<5->{ \gamma_y(t) }
+ &
+ \uncover<5->{= \begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix} }
+ \\
+ \uncover<6->{ \gamma_h(t)}
+ &
+ \uncover<6->{= \begin{pmatrix} e^t & 0 \\ 0 & e^{-t} \end{pmatrix} }
+ \end{align*}
+ \end{block}
+ }
+ \end{column}
+ \end{columns}
+\end{frame}
+
+
+\begin{frame}[t]
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ \frametitle{Repetition}
+ \vspace{-20pt}
+ \begin{columns}[t,onlytextwidth]
+ \begin{column}{0.48\textwidth}
+ \uncover<1->{
+ \begin{block}{Lie-Algebra aus Lie-Gruppe}
+ Ableitungen der Ein-Parameter-Untergruppen:
+ \begin{align*}
+ G &\to \mathcal A \\
+ \gamma &\mapsto \dot\gamma(0)
+ \end{align*}
+ \uncover<3->{
+ Lie-Klammer als Produkt:
+ \[ [A, B] = AB - BA \in \mathcal A \]
+ }
+ \end{block}
+ }
+ \uncover<7->{\vspace*{-4ex}
+ \begin{block}{Lie-Gruppe aus Lie-Algebra}
+ Lösung der Differentialgleichung:
+ \[
+ \dot\gamma(t) = A\gamma(t)
+ \quad \text{mit} \quad
+ A = \dot\gamma(0)
+ \]
+ \[
+ \Rightarrow \gamma(t) = \exp(At)
+ \]
+ \end{block}
+ }
+ \end{column}
+ \begin{column}{0.48\textwidth}
+ \uncover<2->{
+ \begin{block}{Beispiel}
+ Lie-Algebra von \gSL2R:
+ \[ \asl2R = \{ A \in M_2 \,|\, \Spur(A) = 0 \} \]
+ \end{block}
+ }
+ \begin{align*}
+ \uncover<4->{ X(t) }
+ &
+ \uncover<4->{= \begin{pmatrix} 0 & t \\ 0 & 0 \end{pmatrix} }
+ \\
+ \uncover<5->{ Y(t) }
+ &
+ \uncover<5->{= \begin{pmatrix} 0 & 0 \\ t & 0 \end{pmatrix} }
+ \\
+ \uncover<6->{ H(t) }
+ &
+ \uncover<6->{= \begin{pmatrix} t & 0 \\ 0 & -t \end{pmatrix} }
+ \end{align*}
+
+ \end{column}
+ \end{columns}
+\end{frame}
+
+\begin{frame}[t]
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ \frametitle{Repetition}
+ \vspace{-20pt}
+ \begin{block}{Offene Fragen}
+ \begin{itemize}[<+->]
+ \item Woher kommt die Exponentialfunktion?
+ \begin{fleqn}
+ \[
+ \exp(At)
+ =
+ 1
+ + At
+ + A^2\frac{t^2}{2}
+ + A^3\frac{t^3}{3!}
+ + \ldots
+ \]
+ \end{fleqn}
+ \item Wie löst man eine Matrix-DGL?
+ \begin{fleqn}
+ \[
+ \dot\gamma(t) = A\gamma(t),
+ \qquad
+ \gamma(t) \in G \subset M_n
+ \]
+ \end{fleqn}
+ \item Was bedeutet $\exp(At)$?
+ \end{itemize}
+ \end{block}
+\end{frame}
+
+\egroup
diff --git a/vorlesungen/slides/10/so2.tex b/vorlesungen/slides/10/so2.tex
index e3f74ae..b63a67e 100644
--- a/vorlesungen/slides/10/so2.tex
+++ b/vorlesungen/slides/10/so2.tex
@@ -7,14 +7,6 @@
% !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}
diff --git a/vorlesungen/slides/10/taylor.tex b/vorlesungen/slides/10/taylor.tex
index 920470f..25745f5 100644
--- a/vorlesungen/slides/10/taylor.tex
+++ b/vorlesungen/slides/10/taylor.tex
@@ -10,12 +10,19 @@
\begin{frame}[t]
\setlength{\abovedisplayskip}{5pt}
\setlength{\belowdisplayskip}{5pt}
- \frametitle{Beispiel $\sin x$}
+ \frametitle{Beispiel $\sin(x)$}
\vspace{-20pt}
- %\onslide<+->
- \begin{block}{Taylor-Approximationen von $\sin x$}
+ \begin{block}{Taylor-Approximationen von $\sin(x)$}
\begin{align*}
- p_n(x)
+ p_{
+ \only<1>{0}
+ \only<2>{1}
+ \only<3>{2}
+ \only<4>{3}
+ \only<5>{4}
+ \only<6>{5}
+ \only<7->{n}
+ }(x)
&=
\uncover<1->{0}
\uncover<2->{+ x}
@@ -74,121 +81,134 @@
\end{center}
\end{frame}
-
\begin{frame}[t]
-\setlength{\abovedisplayskip}{5pt}
-\setlength{\belowdisplayskip}{5pt}
-\frametitle{Taylor-Reihen}
-\vspace{-20pt}
-\onslide<+->
- \begin{block}{Polynom-Approximationen von $f(t)$}
- \vspace{-15pt}
- \begin{align*}
- p_n(t)
- &=
- f(0)
- + f'(0) t
- + f''(0)\frac{t^2}{2}
- + f^{(3)}(0)\frac{t^3}{3!}
- + \ldots
- + f^{(n)}(0) \frac{t^n}{n!}
- =
- \sum_{k=0}^{n} f^{(k)} \frac{t^k}{k!}
- \end{align*}
- \end{block}
- \begin{block}{Die ersten $n$ Ableitungen von $f(0)$ und $p_n(0)$ sind gleich!}
- \vspace{-15pt}
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ \frametitle{Taylor-Reihen}
+ \vspace{-20pt}
+ \begin{block}{Polynom-Approximationen von $f(t)$}
+ \begin{align*}
+ p_n(t)
+ &=
+ f(0)
+ \uncover<2->{ + f' (0) t }
+ \uncover<3->{ + f''(0)\frac{t^2}{2} }
+ \uncover<4->{ + \ldots + f^{(n)}(0) \frac{t^n}{n!} }
+ \uncover<5->{ = \sum_{k=0}^{n} f^{(k)} \frac{t^k}{k!} }
+ \end{align*}
+ \end{block}
+ \uncover<6->{
+ \begin{block}{Erste $n$ Ableitungen von $f(0)$ und $p_n(0)$ sind gleich!}}
\begin{align*}
- p'_n(t)
- &=
- f'(0)
- + f''(0)t
- + f^{(3)}(0) \frac{t^2}{2!}
- + \mathcal O(t^3)
- &\Rightarrow&&
- p'_n(0) = f'(0)
+ \uncover<6->{ p'_n(t) }
+ &
+ \uncover<7->{
+ = f'(0)
+ + f''(0)t
+ + \mathcal O(t^2)
+ }
+ &\uncover<8->{\Rightarrow}&&
+ \uncover<8->{p'_n(0) = f'(0)}
\\
- p''_n(0)
- &=
- f''(0) + f^{(3)}(0)t + \ldots + f^{(n)}(0) \frac{t^{n-2}}{(n-2)!}
- &\Rightarrow&&
- p''_n(0) = f''(0)
- \end{align*}
- \end{block}
- \begin{block}{Für unendlich lange Polynome stimmen alle Ableitungen überein!}
- \vspace{-15pt}
- \begin{align*}
- \lim_{n\to \infty} p_n(t)
- =
- f(t)
+ \uncover<9->{ p''_n(t) }
+ &
+ \uncover<10->{
+ = f''(0)
+ + \mathcal O(t)
+ }
+ &\uncover<11->{\Rightarrow}&&
+ \uncover<11->{ p''_n(0) = f''(0) }
\end{align*}
\end{block}
+ \uncover<12->{
+ \begin{block}{Für alle praktisch relevanten Funktionen $f(t)$ gilt:}
+ \begin{align*}
+ \lim_{n\to \infty} p_n(t)
+ =
+ f(t)
+ \end{align*}
+ \end{block}
+ }
\end{frame}
\begin{frame}[t]
\setlength{\abovedisplayskip}{5pt}
\setlength{\belowdisplayskip}{5pt}
- \frametitle{Beispiel $\exp x$}
- \vspace{-20pt}
- %\onslide<+->
- \begin{block}{Taylor-Approximationen von $\exp x$}
+% \frametitle{Beispiel $e^t$}
+% \vspace{-20pt}
+ \begin{block}{Taylor-Approximationen von $e^{at}$}
\begin{align*}
- p_n(x)
- =
+ p_{
+ \only<1>{0}
+ \only<2>{1}
+ \only<3>{2}
+ \only<4>{3}
+ \only<5>{4}
+ \only<6>{5}
+ \only<7->{n}
+ }(t)
+ &=
1
- \uncover<1->{+ x}
- \uncover<2->{+ \frac{x^2}{2}}
- \uncover<3->{+ \frac{x^3}{3!}}
- \uncover<4->{+ \frac{x^4}{4!}}
- \uncover<5->{+ \frac{x^5}{5!}}
- \uncover<6->{+ \frac{x^6}{6!}}
- \uncover<7->{+ \ldots
- = \sum_{k=0}^{n} \frac{x^k}{k!}}
+ \uncover<2->{+ a t}
+ \uncover<3->{+ a^2 \frac{t^2}{2}}
+ \uncover<4->{+ a^3 \frac{t^3}{3!}}
+ \uncover<5->{+ a^4 \frac{t^4}{4!}}
+ \uncover<6->{+ a^5 \frac{t^5}{5!}}
+ \uncover<7->{+ a^6 \frac{t^6}{6!}}
+ \uncover<8->{+ \ldots
+ = \sum_{k=0}^{n} a^k \frac{t^k}{k!}}
+ \\
+ &
+ \uncover<9->{= \exp(at)}
\end{align*}
\end{block}
\begin{center}
\begin{tikzpicture}[>=latex,thick,scale=1.3]
- \draw[->] (-4.0, 0.0) -- (4.0,0.0) coordinate[label=$x$];
+ \draw[->] (-4.0, 0.0) -- (4.0,0.0) coordinate[label=$t$];
\draw[->] ( 0.0,-0.5) -- (0.0,2.5);
\clip (-3,-0.5) rectangle (3,2.5);
\draw[domain=-4:1, samples=50, smooth, blue]
plot ({\x}, {exp(\x)})
- node[above right] {$\exp(x)$};
+ node[above right] {$\exp(t)$};
\uncover<1>{
- \draw[domain=-4:1.5, samples=10, smooth, red]
- plot ({\x}, {1 + \x})
- node[below right] {$p_1(x)$};}
+ \draw[domain=-4:4, samples=12, smooth, red]
+ plot ({\x}, {1})
+ node[below right] {$p_0(t)$};}
\uncover<2>{
+ \draw[domain=-4:1.5, samples=10, smooth, red]
+ plot ({\x}, {1 + \x})
+ node[below right] {$p_1(t)$};}
+ \uncover<3>{
\draw[domain=-4:1, samples=50, smooth, red]
plot ({\x}, {1 + \x + \x*\x/2})
- node[below right] {$p_2(x)$};}
- \uncover<3>{
+ node[below right] {$p_2(t)$};}
+ \uncover<4>{
\draw[domain=-4:1, samples=50, smooth, red]
plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6})
- node[below right] {$p_3(x)$};}
- \uncover<4>{
+ node[below right] {$p_3(t)$};}
+ \uncover<5>{
\draw[domain=-4:0.9, samples=50, smooth, red]
plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24})
- node[below left] {$p_4(x)$};}
- \uncover<5>{
+ node[below left] {$p_4(t)$};}
+ \uncover<6>{
\draw[domain=-4:0.9, samples=50, smooth, red]
plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24
+ \x*\x*\x*\x*\x/120})
- node[below left] {$p_5(x)$};}
- \uncover<6>{
+ node[below left] {$p_5(t)$};}
+ \uncover<7>{
\draw[domain=-4:0.9, samples=50, smooth, red]
plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24
+ \x*\x*\x*\x*\x/120
+ \x*\x*\x*\x*\x*\x/720})
- node[below left] {$p_6(x)$};}
- \uncover<7>{
+ node[below left] {$p_6(t)$};}
+ \uncover<8->{
\draw[domain=-4:0.9, samples=50, smooth, red]
plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24
+ \x*\x*\x*\x*\x/120
+ \x*\x*\x*\x*\x*\x/720
+ \x*\x*\x*\x*\x*\x*\x/5040})
- node[below left] {$p_7(x)$};}
+ node[below left] {$p_7(t)$};}
\end{tikzpicture}
\end{center}
\end{frame}
--
cgit v1.2.1
From 4313f2c207d5d60171898ccfd4c3b3d0d2fb4a75 Mon Sep 17 00:00:00 2001
From: Roy Seitz <roy.seitz@ost.ch>
Date: Sun, 18 Apr 2021 17:49:56 +0200
Subject: =?UTF-8?q?Pr=C3=A4sentation=20feritg.?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
vorlesungen/slides/10/intro.tex | 45 ++++++
vorlesungen/slides/10/n-zu-1.tex | 98 +++++++-------
vorlesungen/slides/10/repetition.tex | 40 +-----
vorlesungen/slides/10/so2.tex | 237 ++++++++++++++++----------------
vorlesungen/slides/10/vektorfelder.mp | 241 +++++++++++++++++++++++----------
vorlesungen/slides/10/vektorfelder.tex | 82 +++++++++++
6 files changed, 475 insertions(+), 268 deletions(-)
create mode 100644 vorlesungen/slides/10/intro.tex
create mode 100644 vorlesungen/slides/10/vektorfelder.tex
(limited to 'vorlesungen/slides/10')
diff --git a/vorlesungen/slides/10/intro.tex b/vorlesungen/slides/10/intro.tex
new file mode 100644
index 0000000..276bf49
--- /dev/null
+++ b/vorlesungen/slides/10/intro.tex
@@ -0,0 +1,45 @@
+%
+% intro.tex -- Repetition Lie-Gruppen und -Algebren
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+% Erstellt durch Roy Seitz
+%
+% !TeX spellcheck = de_CH
+\bgroup
+
+
+
+\begin{frame}[t]
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+% \frametitle{Repetition}
+% \vspace{-20pt}
+ \begin{block}{Offene Fragen}
+ \begin{itemize}[<+->]
+ \item Woher kommt die Exponentialfunktion?
+ \begin{fleqn}
+ \[
+ \exp(At)
+ =
+ 1
+ + At
+ + A^2\frac{t^2}{2}
+ + A^3\frac{t^3}{3!}
+ + \ldots
+ \]
+ \end{fleqn}
+ \item Wie löst man eine Matrix-DGL?
+ \begin{fleqn}
+ \[
+ \dot\gamma(t) = A\gamma(t),
+ \qquad
+ \gamma(t) \in G \subset M_n
+ \]
+ \end{fleqn}
+ \item Lie-Gruppen und Lie-Algebren
+ \item Was bedeutet $\exp(At)$?
+ \end{itemize}
+ \end{block}
+\end{frame}
+
+\egroup
diff --git a/vorlesungen/slides/10/n-zu-1.tex b/vorlesungen/slides/10/n-zu-1.tex
index 737df03..09475ad 100644
--- a/vorlesungen/slides/10/n-zu-1.tex
+++ b/vorlesungen/slides/10/n-zu-1.tex
@@ -7,51 +7,57 @@
% !TeX spellcheck = de_CH
\bgroup
\begin{frame}[t]
-\setlength{\abovedisplayskip}{5pt}
-\setlength{\belowdisplayskip}{5pt}
-\frametitle{Reicht $1.$ Ordnung?}
-\vspace{-20pt}
-\begin{columns}[t,onlytextwidth]
-\begin{column}{0.48\textwidth}
-\begin{block}{Beispiel: DGL 3.~Ordnung} \vspace*{-1ex}
- \begin{align*}
- x^{(3)} + a_2 \ddot x + a_1 \dot x + a_0 x = 0 \\
- \Rightarrow
- x^{(3)} = -a_2 \ddot x - a_1 \dot x - a_0 x
- \end{align*}
-\end{block}
-\begin{block}{Ziel: Nur noch 1.~Ableitungen}
- Einführen neuer Variablen:
- \begin{align*}
- x_0 &\coloneqq x &
- x_1 &\coloneqq \dot x &
- x_2 &\coloneqq \ddot x
- \end{align*}
-System von Gleichungen 1.~Ordnung
- \begin{align*}
- \dot x_0 &= x_1 \\
- \dot x_1 &= x_2 \\
- \dot x_2 &= -a_2 x_2 - a_1 x_1 - a_0 x_0
-\end{align*}
-\end{block}
-\end{column}
-\begin{column}{0.48\textwidth}
-\begin{block}{Als Vektor-Gleichung} \vspace*{-1ex}
- \begin{align*}
- \frac{d}{dt}
- \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix}
- = \begin{pmatrix}
- 0 & 1 & 0 \\
- 0 & 0 & 1 \\
- -a_0 & -a_1 & -a_2
- \end{pmatrix}
- \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix}
- \end{align*}
-
- Geht für jede lineare Differentialgleichung!
-
-\end{block}
-\end{column}
-\end{columns}
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ %\frametitle{Reicht $1.$ Ordnung?}
+ %\vspace{-20pt}
+ \begin{columns}[t,onlytextwidth]
+ \begin{column}{0.48\textwidth}
+ \uncover<1->{
+ \begin{block}{Beispiel: DGL 3.~Ordnung} \vspace*{-1ex}
+ \begin{align*}
+ x^{(3)} + a_2 \ddot x + a_1 \dot x + a_0 x = 0 \\
+ \Rightarrow
+ x^{(3)} = -a_2 \ddot x - a_1 \dot x - a_0 x
+ \end{align*}
+ \end{block}
+ }
+ \uncover<2->{
+ \begin{block}{Ziel: Nur noch 1.~Ableitungen}
+ Einführen neuer Variablen:
+ \begin{align*}
+ x_0 &\coloneqq x &
+ x_1 &\coloneqq \dot x &
+ x_2 &\coloneqq \ddot x
+ \end{align*}
+ System von Gleichungen 1.~Ordnung
+ \begin{align*}
+ \dot x_0 &= x_1 \\
+ \dot x_1 &= x_2 \\
+ \dot x_2 &= -a_2 x_2 - a_1 x_1 - a_0 x_0
+ \end{align*}
+ \end{block}
+ }
+ \end{column}
+ \uncover<3->{
+ \begin{column}{0.48\textwidth}
+ \begin{block}{Als Vektor-Gleichung} \vspace*{-1ex}
+ \begin{align*}
+ \frac{d}{dt}
+ \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix}
+ = \begin{pmatrix}
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ -a_0 & -a_1 & -a_2
+ \end{pmatrix}
+ \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix}
+ \end{align*}
+
+ \uncover<4->{Geht für jede lineare Differentialgleichung!}
+
+ \end{block}
+ \end{column}
+ }
+ \end{columns}
\end{frame}
\egroup
diff --git a/vorlesungen/slides/10/repetition.tex b/vorlesungen/slides/10/repetition.tex
index c45d47b..7c007ca 100644
--- a/vorlesungen/slides/10/repetition.tex
+++ b/vorlesungen/slides/10/repetition.tex
@@ -1,5 +1,5 @@
%
-% intro.tex -- Repetition Lie-Gruppen und -Algebren
+% repetition.tex -- Repetition Lie-Gruppen und -Algebren
%
% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
% Erstellt durch Roy Seitz
@@ -101,51 +101,19 @@
\begin{align*}
\uncover<4->{ X(t) }
&
- \uncover<4->{= \begin{pmatrix} 0 & t \\ 0 & 0 \end{pmatrix} }
+ \uncover<4->{= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} }
\\
\uncover<5->{ Y(t) }
&
- \uncover<5->{= \begin{pmatrix} 0 & 0 \\ t & 0 \end{pmatrix} }
+ \uncover<5->{= \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} }
\\
\uncover<6->{ H(t) }
&
- \uncover<6->{= \begin{pmatrix} t & 0 \\ 0 & -t \end{pmatrix} }
+ \uncover<6->{= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} }
\end{align*}
\end{column}
\end{columns}
\end{frame}
-\begin{frame}[t]
- \setlength{\abovedisplayskip}{5pt}
- \setlength{\belowdisplayskip}{5pt}
- \frametitle{Repetition}
- \vspace{-20pt}
- \begin{block}{Offene Fragen}
- \begin{itemize}[<+->]
- \item Woher kommt die Exponentialfunktion?
- \begin{fleqn}
- \[
- \exp(At)
- =
- 1
- + At
- + A^2\frac{t^2}{2}
- + A^3\frac{t^3}{3!}
- + \ldots
- \]
- \end{fleqn}
- \item Wie löst man eine Matrix-DGL?
- \begin{fleqn}
- \[
- \dot\gamma(t) = A\gamma(t),
- \qquad
- \gamma(t) \in G \subset M_n
- \]
- \end{fleqn}
- \item Was bedeutet $\exp(At)$?
- \end{itemize}
- \end{block}
-\end{frame}
-
\egroup
diff --git a/vorlesungen/slides/10/so2.tex b/vorlesungen/slides/10/so2.tex
index b63a67e..dcbcdc8 100644
--- a/vorlesungen/slides/10/so2.tex
+++ b/vorlesungen/slides/10/so2.tex
@@ -8,123 +8,134 @@
\bgroup
\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}
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ \frametitle{Von der Lie-Gruppe zur -Algebra}
+ \vspace{-20pt}
+ \begin{columns}[t,onlytextwidth]
+ \begin{column}{0.48\textwidth}
+ \uncover<1->{
+ \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}
+ }
+ \uncover<2->{
+ \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}
+ \uncover<3->{
+ \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
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ \frametitle{Von der Lie-Algebra zur -Gruppe}
+ \vspace{-20pt}
+ \begin{columns}[t,onlytextwidth]
+ \begin{column}{0.48\textwidth}
+ \uncover<1->{
+ \begin{block}{Differentialgleichung}
+ Gegeben:
+ \[
+ J
+ =
+ \dot\gamma(0) = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix}
+ \]
+ Gesucht:
+ \[ \dot \gamma (t) = J \gamma(t) \qquad \gamma \in \gSO2 \]
+ \[ \Rightarrow \gamma(t) = \exp(Jt) \gamma(0) = \exp(Jt) \]
+ \end{block}
+ }
+ \end{column}
+ \begin{column}{0.48\textwidth}
+ \uncover<2->{
+ \begin{block}{Lie-Algebra}
+ Potenzen von $J$:
+ \begin{align*}
+ J^2 &= -I &
+ J^3 &= -J &
+ J^4 &= I &
+ \ldots
+ \end{align*}
+ \end{block}
+ }
+ \end{column}
+ \end{columns}
+\uncover<3->{
+ Folglich:
+ \begin{align*}
+ \exp(Jt)
+ &= I + Jt
+ + J^2\frac{t^2}{2!}
+ + J^3\frac{t^3}{3!}
+ + J^4\frac{t^4}{4!}
+ + J^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}
=
- \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*}
-
+ \begin{pmatrix}
+ \cos t & -\sin t \\
+ \sin t & \phantom-\cos t
+ \end{pmatrix}
+ \end{align*}
+ }
\end{frame}
\egroup
diff --git a/vorlesungen/slides/10/vektorfelder.mp b/vorlesungen/slides/10/vektorfelder.mp
index f488327..e63b2d5 100644
--- a/vorlesungen/slides/10/vektorfelder.mp
+++ b/vorlesungen/slides/10/vektorfelder.mp
@@ -48,17 +48,17 @@ 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;
+% % 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) =
@@ -66,6 +66,10 @@ def dglField(expr x, y) =
(-y, x)
enddef;
+pair A;
+A := (1, 0);
+draw A scaled unit withpen pencircle scaled 8bp withcolor red;
+
% Draw arrows for each grid point
pickup pencircle scaled 0.5pt;
for x = -1.5 step 0.1 until 1.55:
@@ -78,11 +82,9 @@ endfor;
endfig;
-
-
%
% Vektorfeld in der Ebene mit Lösungskurve
-% X \in sl(2, R)
+% Euler(1)
%
beginfig(2)
@@ -101,18 +103,28 @@ 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:
+def dglField(expr x, y) =
+ (-y, x)
+enddef;
+
+def dglFieldp(expr z) =
+ dglField(xpart z, ypart z)
+enddef;
+
+def curve(expr z, l, s) =
path p;
- p = (-1.5,y) -- (1.5, y);
- pickup pencircle scaled 1pt;
+ p := z;
+ for t = 0 step 1 until l:
+ p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p)));
+ endfor;
draw p scaled unit withcolor red;
-endfor;
-
-def dglField(expr x, y) =
- (y, 0)
enddef;
+pair A;
+A := (1, 0);
+draw A scaled unit withpen pencircle scaled 8bp withcolor red;
+curve(A, 0, 1);
+
% Draw arrows for each grid point
pickup pencircle scaled 0.5pt;
for x = -1.5 step 0.1 until 1.55:
@@ -125,12 +137,9 @@ endfor;
endfig;
-
-
-
%
% Vektorfeld in der Ebene mit Lösungskurve
-% Y \in sl(2, R)
+% Euler(2)
%
beginfig(3)
@@ -149,42 +158,82 @@ 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:
+def dglField(expr x, y) =
+ (-y, x)
+enddef;
+
+def dglFieldp(expr z) =
+ dglField(xpart z, ypart z)
+enddef;
+
+def curve(expr z, l, s) =
path p;
- p = (x, -1.5) -- (x, 1.5);
- pickup pencircle scaled 1pt;
+ p := z;
+ for t = 0 step 1 until l:
+ p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p)));
+ endfor;
draw p scaled unit withcolor red;
+enddef;
+
+pair A;
+A := (1, 0);
+draw A scaled unit withpen pencircle scaled 8bp withcolor red;
+curve(A, 1, 0.5);
+
+% 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;
-def dglField(expr x, y) =
- (0, x)
-enddef;
+endfig;
-% 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);
+% Vektorfeld in der Ebene mit Lösungskurve
+% Euler(3)
%
-% numeric innerlength;
-% innerlength = 500;
-%
-% for a = 0 step 30 until 330:
-% curve(0.05 * (cosd(a), sind(a)), innerlength);
-% endfor;
+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 (-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));
+
+def dglField(expr x, y) =
+ (-y, x)
+enddef;
+
+def dglFieldp(expr z) =
+ dglField(xpart z, ypart z)
+enddef;
+
+def curve(expr z, l, s) =
+ path p;
+ p := z;
+ for t = 0 step 1 until l:
+ p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p)));
+ endfor;
+ draw p scaled unit withcolor red;
+enddef;
+
+pair A;
+A := (1, 0);
+draw A scaled unit withpen pencircle scaled 8bp withcolor red;
+curve(A, 3, 0.25);
% Draw arrows for each grid point
pickup pencircle scaled 0.5pt;
@@ -198,12 +247,11 @@ endfor;
endfig;
-
%
% Vektorfeld in der Ebene mit Lösungskurve
-% H \in sl(2, R)
+% Euler(4)
%
-beginfig(4)
+beginfig(5)
numeric unit;
unit := 150;
@@ -215,40 +263,88 @@ z3 = ( 0, -1.5) * unit;
z4 = ( 0, 1.5) * unit;
pickup pencircle scaled 1pt;
-drawarrow (z1 shifted (-25,0))--(z2 shifted (25,0));
+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 (25,0));
+label.top(btex $x_1$ etex, z2 shifted (10,0));
label.rt(btex $x_2$ etex, z4 shifted (0,10));
def dglField(expr x, y) =
- (x, -y)
+ (-y, x)
enddef;
def dglFieldp(expr z) =
dglField(xpart z, ypart z)
enddef;
-def curve(expr z, l) =
+def curve(expr z, l, s) =
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)));
+ p := p--((point (length p) of p) shifted (s * 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);
+pair A;
+A := (1, 0);
+draw A scaled unit withpen pencircle scaled 8bp withcolor red;
+curve(A, 7, 0.125);
+
+% 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
+% Euler(5)
+%
+beginfig(6)
+
+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));
+
+def dglField(expr x, y) =
+ (-y, x)
+enddef;
+
+def dglFieldp(expr z) =
+ dglField(xpart z, ypart z)
+enddef;
+
+def curve(expr z, l, s) =
+ path p;
+ p := z;
+ for t = 0 step 1 until l:
+ p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p)));
+ endfor;
+ draw p scaled unit withcolor red;
+enddef;
+
+pair A;
+A := (1, 0);
+draw A scaled unit withpen pencircle scaled 8bp withcolor red;
+curve(A, 99, 0.01);
+
% Draw arrows for each grid point
pickup pencircle scaled 0.5pt;
for x = -1.5 step 0.1 until 1.55:
@@ -262,5 +358,4 @@ endfor;
endfig;
-
end;
diff --git a/vorlesungen/slides/10/vektorfelder.tex b/vorlesungen/slides/10/vektorfelder.tex
new file mode 100644
index 0000000..a4612aa
--- /dev/null
+++ b/vorlesungen/slides/10/vektorfelder.tex
@@ -0,0 +1,82 @@
+%
+% iterativ.tex -- Iterative Approximation in \dot x = J x
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+% Erstellt durch Roy Seitz
+%
+% !TeX spellcheck = de_CH
+\bgroup
+\begin{frame}[t]
+ \setlength{\abovedisplayskip}{5pt}
+ \setlength{\belowdisplayskip}{5pt}
+ \frametitle{Als Strömungsfeld}
+ \vspace{-20pt}
+ \begin{columns}[t,onlytextwidth]
+ \begin{column}{0.48\textwidth}
+ \vfil
+ \only<1>{
+ \includegraphics[width=\linewidth,keepaspectratio]
+ {../slides/10/vektorfelder-1.pdf}
+ }
+ \only<2>{
+ \includegraphics[width=\linewidth,keepaspectratio]
+ {../slides/10/vektorfelder-2.pdf}
+ }
+ \only<3>{
+ \includegraphics[width=\linewidth,keepaspectratio]
+ {../slides/10/vektorfelder-3.pdf}
+ }
+ \only<4>{
+ \includegraphics[width=\linewidth,keepaspectratio]
+ {../slides/10/vektorfelder-4.pdf}
+ }
+ \only<5>{
+ \includegraphics[width=\linewidth,keepaspectratio]
+ {../slides/10/vektorfelder-5.pdf}
+ }
+ \only<6->{
+ \includegraphics[width=\linewidth,keepaspectratio]
+ {../slides/10/vektorfelder-6.pdf}
+ }
+ \vfil
+ \end{column}
+ \begin{column}{0.48\textwidth}
+ \begin{block}{Differentialgleichung}
+ \[
+ \dot x(t) = J x(t)
+ \quad
+ J = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix}
+ \quad
+ x_0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}
+ \]
+ \end{block}
+
+ \only<2>{
+ Nach einem Schritt der Länge $t$:
+ \[
+ x(t) = x_0 + \dot x t = x_0 + Jx_0t = (1 + Jt)x_0
+ \]
+ }
+
+ \only<3>{
+ Nach zwei Schritten der Länge $t/2$:
+ \[
+ x(t) = \left(1 + \frac{Jt}{2}\right)^2x_0
+ \]
+ }
+
+ \only<4->{
+ Nach n Schritten der Länge $t/n$:
+ \[
+ x(t) = \left(1 + \frac{Jt}{n}\right)^nx_0
+ \]
+ }
+ \only<6->{
+ \[
+ \lim_{n\to\infty}\left(1 + \frac{At}{n}\right)^n = \exp(At)
+ \]
+ }
+ \end{column}
+ \end{columns}
+\end{frame}
+\egroup
--
cgit v1.2.1
From 8a32e72ace1b6442b2601b821e4d7ed24047939e Mon Sep 17 00:00:00 2001
From: Roy Seitz <roy.seitz@ost.ch>
Date: Mon, 19 Apr 2021 11:10:24 +0200
Subject: Titleseite und Handout.
---
vorlesungen/slides/10/taylor.tex | 36 +++++++++++++++++-----------------
vorlesungen/slides/10/vektorfelder.tex | 14 ++++++-------
2 files changed, 25 insertions(+), 25 deletions(-)
(limited to 'vorlesungen/slides/10')
diff --git a/vorlesungen/slides/10/taylor.tex b/vorlesungen/slides/10/taylor.tex
index 25745f5..8c71965 100644
--- a/vorlesungen/slides/10/taylor.tex
+++ b/vorlesungen/slides/10/taylor.tex
@@ -11,7 +11,7 @@
\setlength{\abovedisplayskip}{5pt}
\setlength{\belowdisplayskip}{5pt}
\frametitle{Beispiel $\sin(x)$}
- \vspace{-20pt}
+ \ifthenelse{\boolean{presentation}}{\vspace{-20pt}}{\vspace{-8pt}}
\begin{block}{Taylor-Approximationen von $\sin(x)$}
\begin{align*}
p_{
@@ -44,15 +44,15 @@
\draw[domain=-4:4, samples=50, smooth, blue]
plot ({\x}, {sin(180/3.1415968*\x)})
node[above right] {$\sin(x)$};
- \uncover<1>{
+ \uncover<1|handout:0>{
\draw[domain=-4:4, samples=2, smooth, red]
plot ({\x}, {0})
node[above right] {$p_0(x)$};}
- \uncover<2>{
+ \uncover<2|handout:0>{
\draw[domain=-1.5:1.5, samples=2, smooth, red]
plot ({\x}, {\x})
node[below right] {$p_1(x)$};}
- \uncover<3>{
+ \uncover<3|handout:0>{
\draw[domain=-1.5:1.5, samples=2, smooth, red]
plot ({\x}, {\x})
node[below right] {$p_2(x)$};}
@@ -60,19 +60,19 @@
\draw[domain=-3:3, samples=50, smooth, red]
plot ({\x}, {\x - \x*\x*\x/6})
node[above right] {$p_3(x)$};}
- \uncover<5>{
+ \uncover<5|handout:0>{
\draw[domain=-3:3, samples=50, smooth, red]
plot ({\x}, {\x - \x*\x*\x/6})
node[above right] {$p_4(x)$};}
- \uncover<6>{
+ \uncover<6|handout:0>{
\draw[domain=-3.9:3.9, samples=50, smooth, red]
plot ({\x}, {\x - \x*\x*\x/6 + \x*\x*\x*\x*\x/120})
node[below right] {$p_5(x)$};}
- \uncover<7>{
+ \uncover<7|handout:0>{
\draw[domain=-3.9:3.9, samples=50, smooth, red]
plot ({\x}, {\x - \x*\x*\x/6 + \x*\x*\x*\x*\x/120})
node[below right] {$p_6(x)$};}
- \uncover<8->{
+ \uncover<8-|handout:0>{
\draw[domain=-4:4, samples=50, smooth, red]
plot ({\x}, {\x - \x*\x*\x/6 + \x*\x*\x*\x*\x/120 -
\x*\x*\x*\x*\x*\x*\x/5040})
@@ -85,7 +85,7 @@
\setlength{\abovedisplayskip}{5pt}
\setlength{\belowdisplayskip}{5pt}
\frametitle{Taylor-Reihen}
- \vspace{-20pt}
+ \ifthenelse{\boolean{presentation}}{\vspace{-20pt}}{\vspace{-8pt}}
\begin{block}{Polynom-Approximationen von $f(t)$}
\begin{align*}
p_n(t)
@@ -135,8 +135,8 @@
\begin{frame}[t]
\setlength{\abovedisplayskip}{5pt}
\setlength{\belowdisplayskip}{5pt}
-% \frametitle{Beispiel $e^t$}
-% \vspace{-20pt}
+ \frametitle{Beispiel $e^t$}
+ \ifthenelse{\boolean{presentation}}{\vspace{-20pt}}{\vspace{-8pt}}
\begin{block}{Taylor-Approximationen von $e^{at}$}
\begin{align*}
p_{
@@ -171,15 +171,15 @@
\draw[domain=-4:1, samples=50, smooth, blue]
plot ({\x}, {exp(\x)})
node[above right] {$\exp(t)$};
- \uncover<1>{
+ \uncover<1|handout:0>{
\draw[domain=-4:4, samples=12, smooth, red]
plot ({\x}, {1})
node[below right] {$p_0(t)$};}
- \uncover<2>{
+ \uncover<2|handout:0>{
\draw[domain=-4:1.5, samples=10, smooth, red]
plot ({\x}, {1 + \x})
node[below right] {$p_1(t)$};}
- \uncover<3>{
+ \uncover<3|handout:0>{
\draw[domain=-4:1, samples=50, smooth, red]
plot ({\x}, {1 + \x + \x*\x/2})
node[below right] {$p_2(t)$};}
@@ -187,22 +187,22 @@
\draw[domain=-4:1, samples=50, smooth, red]
plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6})
node[below right] {$p_3(t)$};}
- \uncover<5>{
+ \uncover<5|handout:0>{
\draw[domain=-4:0.9, samples=50, smooth, red]
plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24})
node[below left] {$p_4(t)$};}
- \uncover<6>{
+ \uncover<6|handout:0>{
\draw[domain=-4:0.9, samples=50, smooth, red]
plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24
+ \x*\x*\x*\x*\x/120})
node[below left] {$p_5(t)$};}
- \uncover<7>{
+ \uncover<7|handout:0>{
\draw[domain=-4:0.9, samples=50, smooth, red]
plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24
+ \x*\x*\x*\x*\x/120
+ \x*\x*\x*\x*\x*\x/720})
node[below left] {$p_6(t)$};}
- \uncover<8->{
+ \uncover<8-|handout:0>{
\draw[domain=-4:0.9, samples=50, smooth, red]
plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6 + \x*\x*\x*\x/24
+ \x*\x*\x*\x*\x/120
diff --git a/vorlesungen/slides/10/vektorfelder.tex b/vorlesungen/slides/10/vektorfelder.tex
index a4612aa..3ba7cda 100644
--- a/vorlesungen/slides/10/vektorfelder.tex
+++ b/vorlesungen/slides/10/vektorfelder.tex
@@ -14,11 +14,11 @@
\begin{columns}[t,onlytextwidth]
\begin{column}{0.48\textwidth}
\vfil
- \only<1>{
+ \only<1|handout:0>{
\includegraphics[width=\linewidth,keepaspectratio]
{../slides/10/vektorfelder-1.pdf}
}
- \only<2>{
+ \only<2|handout:0>{
\includegraphics[width=\linewidth,keepaspectratio]
{../slides/10/vektorfelder-2.pdf}
}
@@ -26,15 +26,15 @@
\includegraphics[width=\linewidth,keepaspectratio]
{../slides/10/vektorfelder-3.pdf}
}
- \only<4>{
+ \only<4|handout:0>{
\includegraphics[width=\linewidth,keepaspectratio]
{../slides/10/vektorfelder-4.pdf}
}
- \only<5>{
+ \only<5|handout:0>{
\includegraphics[width=\linewidth,keepaspectratio]
{../slides/10/vektorfelder-5.pdf}
}
- \only<6->{
+ \only<6-|handout:0>{
\includegraphics[width=\linewidth,keepaspectratio]
{../slides/10/vektorfelder-6.pdf}
}
@@ -51,14 +51,14 @@
\]
\end{block}
- \only<2>{
+ \only<2|handout:0>{
Nach einem Schritt der Länge $t$:
\[
x(t) = x_0 + \dot x t = x_0 + Jx_0t = (1 + Jt)x_0
\]
}
- \only<3>{
+ \only<3|handout:0>{
Nach zwei Schritten der Länge $t/2$:
\[
x(t) = \left(1 + \frac{Jt}{2}\right)^2x_0
--
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