update to current state of book

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