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Diffstat (limited to '')
-rw-r--r-- | vorlesungen/slides/7/dg.tex | 184 |
1 files changed, 92 insertions, 92 deletions
diff --git a/vorlesungen/slides/7/dg.tex b/vorlesungen/slides/7/dg.tex index 4447bac..446b2ab 100644 --- a/vorlesungen/slides/7/dg.tex +++ b/vorlesungen/slides/7/dg.tex @@ -1,92 +1,92 @@ -% -% dg.tex -- Differentialgleichung für die Exponentialabbildung -% -% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule -% -\bgroup -\begin{frame}[t] -\setlength{\abovedisplayskip}{5pt} -\setlength{\belowdisplayskip}{5pt} -\frametitle{Zurück zur Lie-Gruppe} -\vspace{-20pt} -\begin{columns}[t,onlytextwidth] -\begin{column}{0.48\textwidth} -\begin{block}{Tangentialvektor im Punkt $\gamma(t)$} -Ableitung von $\gamma(t)$ an der Stelle $t$: -\begin{align*} -\dot{\gamma}(t) -&\uncover<2->{= -\frac{d}{d\tau}\gamma(\tau)\bigg|_{\tau=t} -} -\\ -&\uncover<3->{= -\frac{d}{ds} -\gamma(t+s) -\bigg|_{s=0} -} -\\ -&\uncover<4->{= -\frac{d}{ds} -\gamma(t)\gamma(s) -\bigg|_{s=0} -} -\\ -&\uncover<5->{= -\gamma(t) -\frac{d}{ds} -\gamma(s) -\bigg|_{s=0} -} -\uncover<6->{= -\gamma(t) \dot{\gamma}(0) -} -\end{align*} -\end{block} -\vspace{-10pt} -\uncover<7->{% -\begin{block}{Differentialgleichung} -\vspace{-10pt} -\[ -\dot{\gamma}(t) = \gamma(t) A -\quad -\text{mit} -\quad -A=\dot{\gamma}(0)\in LG -\] -\end{block}} -\end{column} -\begin{column}{0.50\textwidth} -\uncover<8->{% -\begin{block}{Lösung} -Exponentialfunktion -\[ -\exp\colon LG\to G : A \mapsto \exp(At) = \sum_{k=0}^\infty \frac{t^k}{k!}A^k -\] -\end{block}} -\vspace{-5pt} -\uncover<9->{% -\begin{block}{Kontrolle: Tangentialvektor berechnen} -\vspace{-10pt} -\begin{align*} -\frac{d}{dt}e^{At} -&\uncover<10->{= -\sum_{k=1}^\infty A^k \frac{d}{dt} \frac{t^k}{k!} -} -\\ -&\uncover<11->{= -\sum_{k=1}^\infty A^{k-1}\frac{t^{k-1}}{(k-1)!} A -} -\\ -&\uncover<12->{= -\sum_{k=0} A^k\frac{t^k}{k!} -A -} -\uncover<13->{= -e^{At} A -} -\end{align*} -\end{block}} -\end{column} -\end{columns} -\end{frame} -\egroup +%
+% dg.tex -- Differentialgleichung für die Exponentialabbildung
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+%
+\bgroup
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Zurück zur Lie-Gruppe}
+\vspace{-20pt}
+\begin{columns}[t,onlytextwidth]
+\begin{column}{0.48\textwidth}
+\begin{block}{Tangentialvektor im Punkt $\gamma(t)$}
+Ableitung von $\gamma(t)$ an der Stelle $t$:
+\begin{align*}
+\dot{\gamma}(t)
+&\uncover<2->{=
+\frac{d}{d\tau}\gamma(\tau)\bigg|_{\tau=t}
+}
+\\
+&\uncover<3->{=
+\frac{d}{ds}
+\gamma(t+s)
+\bigg|_{s=0}
+}
+\\
+&\uncover<4->{=
+\frac{d}{ds}
+\gamma(t)\gamma(s)
+\bigg|_{s=0}
+}
+\\
+&\uncover<5->{=
+\gamma(t)
+\frac{d}{ds}
+\gamma(s)
+\bigg|_{s=0}
+}
+\uncover<6->{=
+\gamma(t) \dot{\gamma}(0)
+}
+\end{align*}
+\end{block}
+\vspace{-10pt}
+\uncover<7->{%
+\begin{block}{Differentialgleichung}
+\vspace{-10pt}
+\[
+\dot{\gamma}(t) = \gamma(t) A
+\quad
+\text{mit}
+\quad
+A=\dot{\gamma}(0)\in LG
+\]
+\end{block}}
+\end{column}
+\begin{column}{0.50\textwidth}
+\uncover<8->{%
+\begin{block}{Lösung}
+Exponentialfunktion
+\[
+\exp\colon LG\to G : A \mapsto \exp(At) = \sum_{k=0}^\infty \frac{t^k}{k!}A^k
+\]
+\end{block}}
+\vspace{-5pt}
+\uncover<9->{%
+\begin{block}{Kontrolle: Tangentialvektor berechnen}
+\vspace{-10pt}
+\begin{align*}
+\frac{d}{dt}e^{At}
+&\uncover<10->{=
+\sum_{k=1}^\infty A^k \frac{d}{dt} \frac{t^k}{k!}
+}
+\\
+&\uncover<11->{=
+\sum_{k=1}^\infty A^{k-1}\frac{t^{k-1}}{(k-1)!} A
+}
+\\
+&\uncover<12->{=
+\sum_{k=0} A^k\frac{t^k}{k!}
+A
+}
+\uncover<13->{=
+e^{At} A
+}
+\end{align*}
+\end{block}}
+\end{column}
+\end{columns}
+\end{frame}
+\egroup
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