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-rw-r--r--vorlesungen/slides/10/ableitung-exp.tex60
-rw-r--r--vorlesungen/slides/10/matrix-dgl.tex (renamed from vorlesungen/slides/10/matrix-vektor-dgl.tex)2
-rw-r--r--vorlesungen/slides/10/potenzreihenmethode.tex91
-rw-r--r--vorlesungen/slides/10/repetition.tex151
-rw-r--r--vorlesungen/slides/10/so2.tex8
-rw-r--r--vorlesungen/slides/10/taylor.tex176
6 files changed, 401 insertions, 87 deletions
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-vektor-dgl.tex b/vorlesungen/slides/10/matrix-dgl.tex
index f7bd995..ae68fb1 100644
--- a/vorlesungen/slides/10/matrix-vektor-dgl.tex
+++ b/vorlesungen/slides/10/matrix-dgl.tex
@@ -1,5 +1,5 @@
%
-% matrix-vektor-dgl.tex -- DGL mit Matrix-Koeffizienten und Vektor-Variablen
+% matrix-dgl.tex -- Matrix-Differentialgleichungen
%
% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
% Erstellt durch Roy Seitz
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}