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+%
+% 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)
+&=
+\frac{d}{d\tau}\gamma(\tau)\bigg|_{\tau=t}
+\\
+&=
+\frac{d}{ds}
+\gamma(t+s)
+\bigg|_{s=0}
+\\
+&=
+\frac{d}{ds}
+\gamma(t)\gamma(s)
+\bigg|_{s=0}
+\\
+&=
+\gamma(t)
+\frac{d}{ds}
+\gamma(s)
+\bigg|_{s=0}
+=
+\gamma(t) \dot{\gamma}(0)
+\end{align*}
+\end{block}
+\vspace{-10pt}
+\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}
+\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}
+\begin{block}{Kontrolle: Tangentialvektor berechnen}
+\vspace{-10pt}
+\begin{align*}
+\frac{d}{dt}e^{At}
+&=
+\sum_{k=1}^\infty A^k \frac{d}{dt} t^{k}{k!}
+\\
+&=
+\sum_{k=1}^\infty A^{k-1}\frac{t^{k-1}}{(k-1)!} A
+\\
+&=
+\sum_{k=0} A^k\frac{t^k}{k!}
+A
+=
+e^{At} A
+\end{align*}
+\end{block}
+\end{column}
+\end{columns}
+\end{frame}
+\egroup