Merge branch 'master' of github.com:AndreasFMueller/SeminarMatrizen

10 files changed, 848 insertions, 327 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/intro.tex b/vorlesungen/slides/10/intro.tex
new file mode 100644
index 0000000..276bf49
--- /dev/null
+++ b/vorlesungen/slides/10/intro.tex
@@ -0,0 +1,45 @@
+%
+% 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{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 Lie-Gruppen und Lie-Algebren
+      \item Was bedeutet $\exp(At)$?
+    \end{itemize}
+  \end{block}
+\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/n-zu-1.tex b/vorlesungen/slides/10/n-zu-1.tex
index 737df03..09475ad 100644
--- a/vorlesungen/slides/10/n-zu-1.tex
+++ b/vorlesungen/slides/10/n-zu-1.tex
@@ -7,51 +7,57 @@
 % !TeX spellcheck = de_CH
 \bgroup
 \begin{frame}[t]
-\setlength{\abovedisplayskip}{5pt}
-\setlength{\belowdisplayskip}{5pt}
-\frametitle{Reicht $1.$ Ordnung?}
-\vspace{-20pt}
-\begin{columns}[t,onlytextwidth]
-\begin{column}{0.48\textwidth}
-\begin{block}{Beispiel: DGL 3.~Ordnung} \vspace*{-1ex}
-  \begin{align*}
-    x^{(3)} + a_2 \ddot x + a_1 \dot x + a_0 x = 0 \\
-    \Rightarrow
-     x^{(3)} = -a_2 \ddot x - a_1 \dot x - a_0 x
-  \end{align*}
-\end{block}
-\begin{block}{Ziel: Nur noch 1.~Ableitungen}
-  Einführen neuer Variablen:
-    \begin{align*}
-    x_0 &\coloneqq x &
-    x_1 &\coloneqq \dot  x &
-    x_2 &\coloneqq \ddot x
-  \end{align*}
-System von Gleichungen 1.~Ordnung
-  \begin{align*}
-  \dot x_0 &= x_1 \\
-  \dot x_1 &= x_2 \\
-  \dot x_2 &= -a_2 x_2 - a_1 x_1 - a_0 x_0
-\end{align*}
-\end{block}
-\end{column}
-\begin{column}{0.48\textwidth}
-\begin{block}{Als Vektor-Gleichung} \vspace*{-1ex}
-  \begin{align*}
-    \frac{d}{dt}
-    \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix}
-    = \begin{pmatrix}
-      0     & 1     & 0   \\
-      0     & 0     & 1   \\
-      -a_0  & -a_1  & -a_2 
-    \end{pmatrix}
-    \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix}    
-  \end{align*}
-
-  Geht für jede lineare Differentialgleichung!
-  
-\end{block}
-\end{column}
-\end{columns}
+  \setlength{\abovedisplayskip}{5pt}
+  \setlength{\belowdisplayskip}{5pt}
+  %\frametitle{Reicht $1.$ Ordnung?}
+  %\vspace{-20pt}
+  \begin{columns}[t,onlytextwidth]
+    \begin{column}{0.48\textwidth}
+      \uncover<1->{
+        \begin{block}{Beispiel: DGL 3.~Ordnung} \vspace*{-1ex}
+          \begin{align*}
+            x^{(3)} + a_2 \ddot x + a_1 \dot x + a_0 x = 0 \\
+            \Rightarrow
+            x^{(3)} = -a_2 \ddot x - a_1 \dot x - a_0 x
+          \end{align*}
+        \end{block}
+      }
+      \uncover<2->{
+        \begin{block}{Ziel: Nur noch 1.~Ableitungen}
+          Einführen neuer Variablen:
+          \begin{align*}
+            x_0 &\coloneqq x &
+            x_1 &\coloneqq \dot  x &
+            x_2 &\coloneqq \ddot x
+          \end{align*}
+          System von Gleichungen 1.~Ordnung
+          \begin{align*}
+            \dot x_0 &= x_1 \\
+            \dot x_1 &= x_2 \\
+            \dot x_2 &= -a_2 x_2 - a_1 x_1 - a_0 x_0
+          \end{align*}
+        \end{block}
+      }
+    \end{column}
+    \uncover<3->{
+      \begin{column}{0.48\textwidth}
+        \begin{block}{Als Vektor-Gleichung} \vspace*{-1ex}
+          \begin{align*}
+            \frac{d}{dt}
+            \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix}
+            = \begin{pmatrix}
+              0     & 1     & 0   \\
+              0     & 0     & 1   \\
+              -a_0  & -a_1  & -a_2 
+            \end{pmatrix}
+            \begin{pmatrix} x_0 \\ x_1 \\ x_2 \end{pmatrix}    
+          \end{align*}
+          
+          \uncover<4->{Geht für jede lineare Differentialgleichung!}
+          
+        \end{block}
+      \end{column}
+    }
+  \end{columns}
 \end{frame}
 \egroup
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..7c007ca
--- /dev/null
+++ b/vorlesungen/slides/10/repetition.tex
@@ -0,0 +1,119 @@
+%
+% repetition.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 & 1 \\ 0 & 0 \end{pmatrix} }
+          \\
+          \uncover<5->{ Y(t) }
+          &
+          \uncover<5->{= \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} }
+          \\
+          \uncover<6->{ H(t) }
+          &
+          \uncover<6->{= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} }
+        \end{align*}
+
+    \end{column}
+  \end{columns}
+\end{frame}
+
+\egroup
diff --git a/vorlesungen/slides/10/so2.tex b/vorlesungen/slides/10/so2.tex
index e3f74ae..dcbcdc8 100644
--- a/vorlesungen/slides/10/so2.tex
+++ b/vorlesungen/slides/10/so2.tex
@@ -7,132 +7,135 @@
 % !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)}}
+\begin{frame}[t]
+  \setlength{\abovedisplayskip}{5pt}
+  \setlength{\belowdisplayskip}{5pt}
+  \frametitle{Von der Lie-Gruppe zur -Algebra}
+  \vspace{-20pt}
+  \begin{columns}[t,onlytextwidth]
+    \begin{column}{0.48\textwidth}
+      \uncover<1->{
+        \begin{block}{Lie-Gruppe}
+          Darstellung von \gSO2:
+          \begin{align*}
+            \mathbb R 
+            &\to 
+            \gSO2
+            \\
+            t
+            &\mapsto
+            \begin{pmatrix}
+              \cos t &         -\sin t \\ 
+              \sin t & \phantom-\cos t
+            \end{pmatrix}
+          \end{align*}
+        \end{block}
+      }
+      \uncover<2->{
+        \begin{block}{Ableitung am neutralen Element}
+          \begin{align*}
+            \frac{d}{d t}
+            &
+            \left.
+            \begin{pmatrix}
+              \cos t &         -\sin t \\ 
+              \sin t & \phantom-\cos t
+            \end{pmatrix}
+            \right|_{ t = 0}
+            \\
+            =
+            & 
+            \begin{pmatrix} -\sin0 & -\cos0 \\ \phantom-\cos0 & -\sin0 \end{pmatrix}
+            = 
+            \begin{pmatrix} 0 & -1 \\ 1 &  \phantom-0 \end{pmatrix}
+          \end{align*}
+        \end{block}
+      }
+    \end{column}
+    \begin{column}{0.48\textwidth}
+      \uncover<3->{
+        \begin{block}{Lie-Algebra}
+          Darstellung von \aso2:
+          \begin{align*}
+            \mathbb R 
+            &\to 
+            \aso2
+            \\
+            t
+            &\mapsto
+            \begin{pmatrix}
+              0 &         -t \\ 
+              t & \phantom-0
+            \end{pmatrix}
+          \end{align*}
+        \end{block}
+      }
+    \end{column}
+  \end{columns}
+\end{frame}
 
-\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}
-\frametitle{Von der Lie-Gruppe zur -Algebra}
-\vspace{-20pt}
-\begin{columns}[t,onlytextwidth]
-\begin{column}{0.48\textwidth}
-  \begin{block}{Lie-Gruppe}
-    Darstellung von \gSO2:
-    \begin{align*}
-      \mathbb R 
-      &\to 
-      \gSO2
-      \\
-      t
-      &\mapsto
-      \begin{pmatrix}
-        \cos t &         -\sin t \\ 
-        \sin t & \phantom-\cos t
-      \end{pmatrix}
-    \end{align*}
-  \end{block}
-  \begin{block}{Ableitung am neutralen Element}
-    \begin{align*}
-    \frac{d}{d t}
-    &
-    \left.
+  \setlength{\abovedisplayskip}{5pt}
+  \setlength{\belowdisplayskip}{5pt}
+  \frametitle{Von der Lie-Algebra zur -Gruppe}
+  \vspace{-20pt}
+  \begin{columns}[t,onlytextwidth]
+    \begin{column}{0.48\textwidth}
+      \uncover<1->{
+      \begin{block}{Differentialgleichung}
+        Gegeben:
+        \[
+        J
+        =
+        \dot\gamma(0) = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix}
+        \]
+        Gesucht:
+        \[ \dot \gamma (t) = J \gamma(t) \qquad \gamma \in \gSO2 \]
+        \[ \Rightarrow \gamma(t) = \exp(Jt) \gamma(0) = \exp(Jt) \]
+      \end{block}
+    }
+    \end{column}
+    \begin{column}{0.48\textwidth}
+      \uncover<2->{
+      \begin{block}{Lie-Algebra}
+        Potenzen von $J$:
+        \begin{align*}
+          J^2 &= -I &
+          J^3 &= -J &
+          J^4 &=  I &
+          \ldots
+        \end{align*}
+      \end{block}
+    }
+    \end{column}
+  \end{columns}
+\uncover<3->{
+  Folglich:
+  \begin{align*}
+    \exp(Jt)
+    &= I + Jt 
+    + J^2\frac{t^2}{2!} 
+    + J^3\frac{t^3}{3!}
+    + J^4\frac{t^4}{4!}
+    + J^5\frac{t^5}{5!}
+    + \ldots \\
+    &= \begin{pmatrix}
+      \vspace*{3pt}
+      1 - \frac{t^2}{2} + \frac{t^4}{4!} - \ldots
+      &
+      -t + \frac{t^3}{3!} - \frac{t^5}{5!} + \ldots
+      \\ 
+      t - \frac{t^3}{3!} + \frac{t^5}{5!} - \ldots
+      &
+      1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \ldots
+    \end{pmatrix}
+    =
     \begin{pmatrix}
       \cos t &         -\sin t \\ 
       \sin t & \phantom-\cos t
     \end{pmatrix}
-    \right|_{ t = 0}
-    \\
-    =
-    & 
-    \begin{pmatrix} -\sin0 & -\cos0 \\ \phantom-\cos0 & -\sin0 \end{pmatrix}
-    = 
-    \begin{pmatrix} 0 & -1 \\ 1 &  \phantom-0 \end{pmatrix}
-    \end{align*}
-  \end{block}
-\end{column}
-\begin{column}{0.48\textwidth}
-  \begin{block}{Lie-Algebra}
-    Darstellung von \aso2:
-    \begin{align*}
-      \mathbb R 
-      &\to 
-      \aso2
-      \\
-      t
-      &\mapsto
-      \begin{pmatrix}
-        0 &         -t \\ 
-        t & \phantom-0
-      \end{pmatrix}
-    \end{align*}
-  \end{block}
-\end{column}
-\end{columns}
-\end{frame}
-
-
-\begin{frame}[t]
-\setlength{\abovedisplayskip}{5pt}
-\setlength{\belowdisplayskip}{5pt}
-\frametitle{Von der Lie-Algebra zur -Gruppe}
-\vspace{-20pt}
-\begin{columns}[t,onlytextwidth]
-\begin{column}{0.48\textwidth}
-  \begin{block}{Differentialgleichung}
-    Gegeben:
-    \[
-    A
-    =
-    \dot\gamma(0) = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix}
-    \]
-    Gesucht:
-    \[ \dot \gamma (t) = \gamma(t) A \qquad \gamma \in \gSO2 \]
-    \[ \Rightarrow \gamma(t) = \exp(At) \gamma(0) = \exp(At) \]
-  \end{block}
-\end{column}
-\begin{column}{0.48\textwidth}
-  \begin{block}{Lie-Algebra}
-    Potenzen von A:
-    \begin{align*}
-      A^2 &= -I &
-      A^3 &= -A &
-      A^4 &=  I &
-      \ldots
-    \end{align*}
-  \end{block}
-\end{column}
-\end{columns}
-Folglich:
-\begin{align*}
-  \exp(At)
-  &= I + At 
-  + A^2\frac{t^2}{2!} 
-  + A^3\frac{t^3}{3!}
-  + A^4\frac{t^4}{4!}
-  + A^5\frac{t^5}{5!}
-  + \ldots \\
-  &= \begin{pmatrix}
-    \vspace*{3pt}
-    1 - \frac{t^2}{2} + \frac{t^4}{4!} - \ldots
-    &
-    -t + \frac{t^3}{3!} - \frac{t^5}{5!} + \ldots
-    \\ 
-    t - \frac{t^3}{3!} + \frac{t^5}{5!} - \ldots
-    &
-    1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \ldots
-  \end{pmatrix}
-  =
-  \begin{pmatrix}
-    \cos t &         -\sin t \\ 
-    \sin t & \phantom-\cos t
-  \end{pmatrix}
-\end{align*}
-
+  \end{align*}
+  }
 \end{frame}
 \egroup
diff --git a/vorlesungen/slides/10/taylor.tex b/vorlesungen/slides/10/taylor.tex
index 920470f..8c71965 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 $\sin x$}
-  \vspace{-20pt}
-  %\onslide<+->
-  \begin{block}{Taylor-Approximationen von $\sin x$}
+  \frametitle{Beispiel $\sin(x)$}
+  \ifthenelse{\boolean{presentation}}{\vspace{-20pt}}{\vspace{-8pt}}
+  \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}
@@ -37,15 +44,15 @@
       \draw[domain=-4:4, samples=50, smooth, blue]
       plot ({\x}, {sin(180/3.1415968*\x)})
       node[above right] {$\sin(x)$};
-      \uncover<1>{
+      \uncover<1|handout:0>{
         \draw[domain=-4:4, samples=2, smooth, red]
         plot ({\x}, {0})
         node[above right] {$p_0(x)$};}
-      \uncover<2>{
+      \uncover<2|handout:0>{
         \draw[domain=-1.5:1.5, samples=2, smooth, red]
         plot ({\x}, {\x})
         node[below right] {$p_1(x)$};}
-      \uncover<3>{
+      \uncover<3|handout:0>{
         \draw[domain=-1.5:1.5, samples=2, smooth, red]
         plot ({\x}, {\x})
         node[below right] {$p_2(x)$};}
@@ -53,19 +60,19 @@
         \draw[domain=-3:3, samples=50, smooth, red]
         plot ({\x}, {\x - \x*\x*\x/6})
         node[above right] {$p_3(x)$};}
-      \uncover<5>{
+      \uncover<5|handout:0>{
         \draw[domain=-3:3, samples=50, smooth, red]
         plot ({\x}, {\x - \x*\x*\x/6})
         node[above right] {$p_4(x)$};}
-      \uncover<6>{
+      \uncover<6|handout:0>{
         \draw[domain=-3.9:3.9, samples=50, smooth, red]
         plot ({\x}, {\x - \x*\x*\x/6 + \x*\x*\x*\x*\x/120})
         node[below right] {$p_5(x)$};}
-      \uncover<7>{
+      \uncover<7|handout:0>{
         \draw[domain=-3.9:3.9, samples=50, smooth, red]
         plot ({\x}, {\x - \x*\x*\x/6 + \x*\x*\x*\x*\x/120})
         node[below right] {$p_6(x)$};}
-      \uncover<8->{
+      \uncover<8-|handout:0>{
         \draw[domain=-4:4, samples=50, smooth, red]
         plot ({\x}, {\x - \x*\x*\x/6 + \x*\x*\x*\x*\x/120 -
           \x*\x*\x*\x*\x*\x*\x/5040})
@@ -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}
+  \ifthenelse{\boolean{presentation}}{\vspace{-20pt}}{\vspace{-8pt}}
+    \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 $e^t$}
+  \ifthenelse{\boolean{presentation}}{\vspace{-20pt}}{\vspace{-8pt}}
+  \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] {$\exp(x)$};
-      \uncover<1>{
+        node[above right] {$\exp(t)$};
+      \uncover<1|handout:0>{
+      \draw[domain=-4:4, samples=12, smooth, red]
+        plot ({\x}, {1})
+        node[below right] {$p_0(t)$};}
+      \uncover<2|handout:0>{
       \draw[domain=-4:1.5, samples=10, smooth, red]
-        plot ({\x}, {1 + \x})
-        node[below right] {$p_1(x)$};}
-      \uncover<2>{
+      plot ({\x}, {1 + \x})
+      node[below right] {$p_1(t)$};}
+      \uncover<3|handout:0>{
       \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|handout:0>{
       \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|handout:0>{
       \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|handout:0>{
       \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-|handout:0>{
       \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}
diff --git a/vorlesungen/slides/10/vektorfelder.mp b/vorlesungen/slides/10/vektorfelder.mp
index f488327..e63b2d5 100644
--- a/vorlesungen/slides/10/vektorfelder.mp
+++ b/vorlesungen/slides/10/vektorfelder.mp
@@ -48,17 +48,17 @@ drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10));
 label.top(btex $x_1$ etex, z2 shifted (10,0));
 label.rt(btex $x_2$ etex, z4 shifted (0,10));
 
-% Draw circles
-for x = 0.2 step 0.2 until 1.4:
-  path p;
-  p = (x,0);
-  for a = 5 step 5 until 355:
-    p := p--(x*cosd(a), x*sind(a));
-  endfor;
-  p := p--cycle;
-  pickup pencircle scaled 1pt;
-  draw p scaled unit withcolor red;
-endfor;
+% % Draw circles
+% for x = 0.2 step 0.2 until 1.4:
+%   path p;
+%   p = (x,0);
+%   for a = 5 step 5 until 355:
+%     p := p--(x*cosd(a), x*sind(a));
+%   endfor;
+%   p := p--cycle;
+%   pickup pencircle scaled 1pt;
+%   draw p scaled unit withcolor red;
+% endfor;
 
 % Define DGL
 def dglField(expr x, y) =
@@ -66,6 +66,10 @@ def dglField(expr x, y) =
   (-y, x)
 enddef;
 
+pair A;
+A := (1, 0);
+draw A scaled unit withpen pencircle scaled 8bp withcolor red;
+
 % Draw arrows for each grid point
 pickup pencircle scaled 0.5pt;
 for x = -1.5 step 0.1 until 1.55:
@@ -78,11 +82,9 @@ endfor;
 
 endfig;
 
-
-
 %
 % Vektorfeld in der Ebene mit Lösungskurve
-% X \in sl(2, R)
+% Euler(1)
 %
 beginfig(2)
 
@@ -101,18 +103,28 @@ drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10));
 label.top(btex $x_1$ etex, z2 shifted (10,0));
 label.rt(btex $x_2$ etex, z4 shifted (0,10));
 
-% Draw flow lines
-for y = -1.4 step 0.2 until 1.4:
+def dglField(expr x, y) =
+  (-y, x)
+enddef;
+
+def dglFieldp(expr z) =
+  dglField(xpart z, ypart z)
+enddef;
+
+def curve(expr z, l, s) =
   path p;
-  p = (-1.5,y) -- (1.5, y);
-  pickup pencircle scaled 1pt;
+  p := z;
+  for t = 0 step 1 until l:
+    p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p)));
+  endfor;
   draw p scaled unit withcolor red;
-endfor;
-
-def dglField(expr x, y) =
-  (y, 0)
 enddef;
 
+pair A;
+A := (1, 0);
+draw A scaled unit withpen pencircle scaled 8bp withcolor red;
+curve(A, 0, 1);
+
 % Draw arrows for each grid point
 pickup pencircle scaled 0.5pt;
 for x = -1.5 step 0.1 until 1.55:
@@ -125,12 +137,9 @@ endfor;
 
 endfig;
 
-
-
-
 %
 % Vektorfeld in der Ebene mit Lösungskurve
-% Y \in sl(2, R)
+% Euler(2)
 %
 beginfig(3)
 
@@ -149,42 +158,82 @@ drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10));
 label.top(btex $x_1$ etex, z2 shifted (10,0));
 label.rt(btex $x_2$ etex, z4 shifted (0,10));
 
-% Draw flow lines
-for x = -1.4 step 0.2 until 1.4:
+def dglField(expr x, y) =
+  (-y, x)
+enddef;
+
+def dglFieldp(expr z) =
+  dglField(xpart z, ypart z)
+enddef;
+
+def curve(expr z, l, s) =
   path p;
-  p = (x, -1.5) -- (x, 1.5);
-  pickup pencircle scaled 1pt;
+  p := z;
+  for t = 0 step 1 until l:
+    p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p)));
+  endfor;
   draw p scaled unit withcolor red;
+enddef;
+
+pair A;
+A := (1, 0);
+draw A scaled unit withpen pencircle scaled 8bp withcolor red;
+curve(A, 1, 0.5);
+
+% Draw arrows for each grid point
+pickup pencircle scaled 0.5pt;
+for x = -1.5 step 0.1 until 1.55:
+  for y = -1.5 step 0.1 until 1.55:
+    drawarrow ((x, y) * unit)
+      --(((x,y) * unit) shifted (8 * dglField(x,y)))
+        withcolor blue;
+  endfor;
 endfor;
 
-def dglField(expr x, y) =
-  (0, x)
-enddef;
+endfig;
 
-% def dglFieldp(expr z) =
-%   dglField(xpart z, ypart z)
-% enddef;
-% 
-% def curve(expr z, l) =
-%   path p;
-%   p := z;
-%   for t = 0 step 1 until l:
-%     p := p--((point (length p) of p) shifted (0.01 * dglFieldp(point (length p) of p)));
-%   endfor;
-%   draw p scaled unit withcolor red;
-% enddef;
 %
-% numeric outerlength;
-% outerlength = 200;
-% curve(( 0.1, 0), outerlength);
-% curve(( 0.2, 0), outerlength);
+% Vektorfeld in der Ebene mit Lösungskurve
+% Euler(3)
 %
-% numeric innerlength;
-% innerlength = 500;
-% 
-% for a = 0 step 30 until 330:
-%   curve(0.05 * (cosd(a), sind(a)), innerlength);
-% endfor;
+beginfig(4)
+
+numeric unit;
+unit := 150;
+
+z0 = (   0,    0);
+z1 = (-1.5,    0) * unit;
+z2 = ( 1.5,    0) * unit;
+z3 = (   0, -1.5) * unit;
+z4 = (   0,  1.5) * unit;
+
+pickup pencircle scaled 1pt;
+drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0));
+drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10));
+label.top(btex $x_1$ etex, z2 shifted (10,0));
+label.rt(btex $x_2$ etex, z4 shifted (0,10));
+
+def dglField(expr x, y) =
+  (-y, x)
+enddef;
+
+def dglFieldp(expr z) =
+  dglField(xpart z, ypart z)
+enddef;
+
+def curve(expr z, l, s) =
+  path p;
+  p := z;
+  for t = 0 step 1 until l:
+    p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p)));
+  endfor;
+  draw p scaled unit withcolor red;
+enddef;
+
+pair A;
+A := (1, 0);
+draw A scaled unit withpen pencircle scaled 8bp withcolor red;
+curve(A, 3, 0.25);
 
 % Draw arrows for each grid point
 pickup pencircle scaled 0.5pt;
@@ -198,12 +247,11 @@ endfor;
 
 endfig;
 
-
 %
 % Vektorfeld in der Ebene mit Lösungskurve
-% H \in sl(2, R)
+% Euler(4)
 %
-beginfig(4)
+beginfig(5)
 
 numeric unit;
 unit := 150;
@@ -215,40 +263,88 @@ z3 = (   0, -1.5) * unit;
 z4 = (   0,  1.5) * unit;
 
 pickup pencircle scaled 1pt;
-drawarrow (z1 shifted (-25,0))--(z2 shifted (25,0));
+drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0));
 drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10));
-label.top(btex $x_1$ etex, z2 shifted (25,0));
+label.top(btex $x_1$ etex, z2 shifted (10,0));
 label.rt(btex $x_2$ etex, z4 shifted (0,10));
 
 def dglField(expr x, y) =
-  (x, -y)
+  (-y, x)
 enddef;
 
 def dglFieldp(expr z) =
   dglField(xpart z, ypart z)
 enddef;
 
-def curve(expr z, l) =
+def curve(expr z, l, s) =
   path p;
   p := z;
   for t = 0 step 1 until l:
-    p := p--((point (length p) of p) shifted (0.01 * dglFieldp(point (length p) of p)));
+    p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p)));
   endfor;
   draw p scaled unit withcolor red;
 enddef;
 
-for i = -1 step 2 until 1:
-  for k = -1 step 2 until 1:
-    curve((1.3 * i,  1.5 * k), 18);
-    curve((1.1 * i,  1.5 * k), 35);
-    curve((0.9 * i,  1.5 * k), 55);
-    curve((0.7 * i,  1.5 * k), 80);
-    curve((0.5 * i,  1.5 * k), 114);
-    curve((0.3 * i,  1.5 * k), 165);
-    curve((0.1 * i,  1.5 * k), 275);
+pair A;
+A := (1, 0);
+draw A scaled unit withpen pencircle scaled 8bp withcolor red;
+curve(A, 7, 0.125);
+
+% Draw arrows for each grid point
+pickup pencircle scaled 0.5pt;
+for x = -1.5 step 0.1 until 1.55:
+  for y = -1.5 step 0.1 until 1.55:
+    drawarrow ((x, y) * unit)
+      --(((x,y) * unit) shifted (8 * dglField(x,y)))
+        withcolor blue;
   endfor;
 endfor;
 
+endfig;
+
+%
+% Vektorfeld in der Ebene mit Lösungskurve
+% Euler(5)
+%
+beginfig(6)
+
+numeric unit;
+unit := 150;
+
+z0 = (   0,    0);
+z1 = (-1.5,    0) * unit;
+z2 = ( 1.5,    0) * unit;
+z3 = (   0, -1.5) * unit;
+z4 = (   0,  1.5) * unit;
+
+pickup pencircle scaled 1pt;
+drawarrow (z1 shifted (-10,0))--(z2 shifted (10,0));
+drawarrow (z3 shifted (0,-10))--(z4 shifted (0,10));
+label.top(btex $x_1$ etex, z2 shifted (10,0));
+label.rt(btex $x_2$ etex, z4 shifted (0,10));
+
+def dglField(expr x, y) =
+  (-y, x)
+enddef;
+
+def dglFieldp(expr z) =
+  dglField(xpart z, ypart z)
+enddef;
+
+def curve(expr z, l, s) =
+  path p;
+  p := z;
+  for t = 0 step 1 until l:
+    p := p--((point (length p) of p) shifted (s * dglFieldp(point (length p) of p)));
+  endfor;
+  draw p scaled unit withcolor red;
+enddef;
+
+pair A;
+A := (1, 0);
+draw A scaled unit withpen pencircle scaled 8bp withcolor red;
+curve(A, 99, 0.01);
+
 % Draw arrows for each grid point
 pickup pencircle scaled 0.5pt;
 for x = -1.5 step 0.1 until 1.55:
@@ -262,5 +358,4 @@ endfor;
 endfig;
 
 
-
 end;
diff --git a/vorlesungen/slides/10/vektorfelder.tex b/vorlesungen/slides/10/vektorfelder.tex
new file mode 100644
index 0000000..3ba7cda
--- /dev/null
+++ b/vorlesungen/slides/10/vektorfelder.tex
@@ -0,0 +1,82 @@
+%
+% iterativ.tex -- Iterative Approximation in \dot x = J 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{Als Strömungsfeld}
+  \vspace{-20pt}
+  \begin{columns}[t,onlytextwidth]
+    \begin{column}{0.48\textwidth}
+      \vfil
+      \only<1|handout:0>{
+        \includegraphics[width=\linewidth,keepaspectratio]
+        {../slides/10/vektorfelder-1.pdf}
+      }
+      \only<2|handout:0>{
+        \includegraphics[width=\linewidth,keepaspectratio]
+        {../slides/10/vektorfelder-2.pdf}
+      }
+      \only<3>{
+        \includegraphics[width=\linewidth,keepaspectratio]
+        {../slides/10/vektorfelder-3.pdf}
+      }
+      \only<4|handout:0>{
+        \includegraphics[width=\linewidth,keepaspectratio]
+        {../slides/10/vektorfelder-4.pdf}
+      }
+      \only<5|handout:0>{
+        \includegraphics[width=\linewidth,keepaspectratio]
+        {../slides/10/vektorfelder-5.pdf}
+      }
+      \only<6-|handout:0>{
+        \includegraphics[width=\linewidth,keepaspectratio]
+        {../slides/10/vektorfelder-6.pdf}
+      }
+      \vfil
+    \end{column}
+    \begin{column}{0.48\textwidth}
+      \begin{block}{Differentialgleichung}
+        \[ 
+        \dot x(t) = J x(t)
+        \quad
+        J = \begin{pmatrix} 0 & -1 \\ 1 & \phantom-0 \end{pmatrix}
+        \quad
+        x_0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}
+        \]
+      \end{block}
+    
+      \only<2|handout:0>{
+        Nach einem Schritt der Länge $t$:
+        \[
+        x(t) = x_0 + \dot x t = x_0 + Jx_0t = (1 + Jt)x_0
+        \]
+      }
+
+      \only<3|handout:0>{
+        Nach zwei Schritten der Länge $t/2$:
+        \[
+        x(t) = \left(1 + \frac{Jt}{2}\right)^2x_0
+        \]
+      }
+
+      \only<4->{
+        Nach n Schritten der Länge $t/n$:
+        \[
+        x(t) = \left(1 + \frac{Jt}{n}\right)^nx_0
+        \]
+      }
+      \only<6->{
+      \[
+      \lim_{n\to\infty}\left(1 + \frac{At}{n}\right)^n = \exp(At)
+      \]
+      }
+    \end{column}
+  \end{columns}
+\end{frame}
+\egroup