From 2db90bfe4b174570424c408f04000902411d8755 Mon Sep 17 00:00:00 2001
From: Joshua Baer <the.baer.joshua@gmail.ch>
Date: Mon, 12 Apr 2021 21:51:55 +0200
Subject: update to current state of book

---
 vorlesungen/slides/7/dg.tex | 184 ++++++++++++++++++++++----------------------
 1 file changed, 92 insertions(+), 92 deletions(-)

(limited to 'vorlesungen/slides/7/dg.tex')

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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