Merge remote-tracking branch 'upstream/master' into mceliece

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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-dgl.tex b/vorlesungen/slides/10/matrix-dgl.tex
new file mode 100644
index 0000000..ae68fb1
--- /dev/null
+++ b/vorlesungen/slides/10/matrix-dgl.tex
@@ -0,0 +1,83 @@
+%
+% matrix-dgl.tex -- Matrix-Differentialgleichungen
+%
+% (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{1.~Ordnung mit Skalaren}
+  \vspace{-20pt}
+  \begin{columns}[t,onlytextwidth]
+  \begin{column}{0.48\textwidth}
+    \begin{block}{Aufgabe}
+      Sei $a, x(t), x_0 \in \mathbb R$,
+      \[
+      \dot x(t) = ax(t),
+      \quad
+      x(0) = x_0
+      \]
+    \end{block}
+    \begin{block}{Potenzreihen-Ansatz}
+      Sei $a_k \in \mathbb R$,
+      \[
+      x(t) = a_0 + a_1t + a_2t^2 + a_3t^3 \ldots
+      \]
+    \end{block}
+  \end{column}
+  \begin{column}{0.48\textwidth}
+    \begin{block}{Lösung}
+      Einsetzen in DGL, Koeffizientenvergleich liefert
+      \[ x(t) = \exp(at) \, x_0, \]
+      wobei
+      \begin{align*}
+      \exp(at)
+      &= 1 + at + \frac{a^2t^2}{2} + \frac{a^3t^3}{3!} + \ldots \\
+      &{\color{gray}(= e^{at}.)}
+      \end{align*}
+    \end{block}
+  \end{column}
+  \end{columns}
+\end{frame}
+
+\begin{frame}[t]
+  \setlength{\abovedisplayskip}{5pt}
+  \setlength{\belowdisplayskip}{5pt}
+  \frametitle{1.~Ordnung mit Matrizen}
+  \vspace{-20pt}
+  \begin{columns}[t,onlytextwidth]
+    \begin{column}{0.48\textwidth}
+      \begin{block}{Aufgabe}
+        Sei $A \in M_n$, $x(t), x_0 \in \mathbb R^n$,
+        \[
+        \dot x(t) = Ax(t),
+        \quad
+        x(0) = x_0
+        \]
+      \end{block}
+      \begin{block}{Potenzreihen-Ansatz}
+        Sei $A_k \in \mathbb M_n$,
+        \[
+        x(t) = A_0 + A_1t + A_2t^2 + A_3t^3 \ldots
+        \]
+      \end{block}
+    \end{column}
+    \begin{column}{0.48\textwidth}
+      \begin{block}{Lösung}
+        Einsetzen in DGL, Koeffizientenvergleich liefert
+        \[ x(t) = \exp(At) \, x_0, \]
+        wobei
+        \[
+        \exp(At)
+        = 1 + At + \frac{A^2t^2}{2} + \frac{A^3t^3}{3!} + \ldots
+        \]
+      \end{block}
+    \end{column}
+  \end{columns}
+\end{frame}
+
+\egroup
diff --git a/vorlesungen/slides/10/n-zu-1.tex b/vorlesungen/slides/10/n-zu-1.tex
new file mode 100644
index 0000000..09475ad
--- /dev/null
+++ b/vorlesungen/slides/10/n-zu-1.tex
@@ -0,0 +1,63 @@
+%
+% n-zu-1.tex -- Umwandlend einer DGL n-ter Ordnung in ein System 1. Ordnung
+%
+% (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{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
new file mode 100644
index 0000000..dcbcdc8
--- /dev/null
+++ b/vorlesungen/slides/10/so2.tex
@@ -0,0 +1,141 @@
+%
+% so2.tex -- Illustration of so(2) -> SO(2)
+%
+% (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{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}
+
+
+\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}
+      \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}
+  \end{align*}
+  }
+\end{frame}
+\egroup
diff --git a/vorlesungen/slides/10/taylor.tex b/vorlesungen/slides/10/taylor.tex
new file mode 100644
index 0000000..8c71965
--- /dev/null
+++ b/vorlesungen/slides/10/taylor.tex
@@ -0,0 +1,216 @@
+%
+% taylor.tex -- Repetition Taylot-Reihen
+%
+% (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{Beispiel $\sin(x)$}
+  \ifthenelse{\boolean{presentation}}{\vspace{-20pt}}{\vspace{-8pt}}
+  \begin{block}{Taylor-Approximationen von $\sin(x)$}
+    \begin{align*}
+      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}
+      \uncover<3->{+ 0 \frac{x^2}{2!}}
+      \uncover<4->{- 1 \frac{x^3}{3!}}
+      \uncover<5->{+ 0 \frac{x^4}{4!}}
+      \uncover<6->{+ 1 \frac{x^5}{5!}}
+      \uncover<7->{+ \ldots}
+      \uncover<8->{
+        = \sum_{k=0}^{n/2} (-1)^{2k + 1}\frac{x^{2k+1}}{(2k+1)!}
+      }
+    \end{align*}
+  \end{block}
+  \begin{center}
+    \begin{tikzpicture}[>=latex,thick,scale=1.3]
+      \draw[->] (-5.0, 0.0) -- (5.0,0.0) coordinate[label=$x$];
+      \draw[->] ( 0.0,-1.5) -- (0.0,1.5);
+      \clip (-5,-1.5) rectangle (5,1.5);
+      \draw[domain=-4:4, samples=50, smooth, blue]
+      plot ({\x}, {sin(180/3.1415968*\x)})
+      node[above right] {$\sin(x)$};
+      \uncover<1|handout:0>{
+        \draw[domain=-4:4, samples=2, smooth, red]
+        plot ({\x}, {0})
+        node[above right] {$p_0(x)$};}
+      \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|handout:0>{
+        \draw[domain=-1.5:1.5, samples=2, smooth, red]
+        plot ({\x}, {\x})
+        node[below right] {$p_2(x)$};}
+      \uncover<4>{
+        \draw[domain=-3:3, samples=50, smooth, red]
+        plot ({\x}, {\x - \x*\x*\x/6})
+        node[above right] {$p_3(x)$};}
+      \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|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|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-|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})
+        node[above right] {$p_7(x)$};}
+    \end{tikzpicture}
+  \end{center}
+\end{frame}
+
+\begin{frame}[t]
+  \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*}
+      \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)}
+      \\
+      \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 $e^t$}
+  \ifthenelse{\boolean{presentation}}{\vspace{-20pt}}{\vspace{-8pt}}
+  \begin{block}{Taylor-Approximationen von $e^{at}$}
+    \begin{align*}
+      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<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=$t$];
+      \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] {$\exp(t)$};
+      \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] {$p_1(t)$};}
+      \uncover<3|handout:0>{
+      \draw[domain=-4:1, samples=50, smooth, red]
+        plot ({\x}, {1 + \x + \x*\x/2})
+        node[below right] {$p_2(t)$};}
+      \uncover<4>{
+      \draw[domain=-4:1, samples=50, smooth, red]
+        plot ({\x}, {1 + \x + \x*\x/2 + \x*\x*\x/6})
+        node[below right] {$p_3(t)$};}
+      \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] {$p_4(t)$};}
+      \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] {$p_5(t)$};}
+      \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] {$p_6(t)$};}
+      \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] {$p_7(t)$};}
+    \end{tikzpicture}
+  \end{center}
+\end{frame}
+
+\egroup
diff --git a/vorlesungen/slides/10/template.tex b/vorlesungen/slides/10/template.tex
new file mode 100644
index 0000000..50f0a3b
--- /dev/null
+++ b/vorlesungen/slides/10/template.tex
@@ -0,0 +1,21 @@
+%
+% template.tex -- slide template
+%
+% (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{Template}
+\vspace{-20pt}
+\begin{columns}[t,onlytextwidth]
+\begin{column}{0.48\textwidth}
+\end{column}
+\begin{column}{0.48\textwidth}
+\end{column}
+\end{columns}
+\end{frame}
+\egroup
diff --git a/vorlesungen/slides/10/vektorfelder.mp b/vorlesungen/slides/10/vektorfelder.mp
new file mode 100644
index 0000000..e63b2d5
--- /dev/null
+++ b/vorlesungen/slides/10/vektorfelder.mp
@@ -0,0 +1,361 @@
+%
+% Stroemungsfelder linearer Differentialgleichungen
+%
+% (c) 2015 Prof Dr Andreas Mueller, Hochschule Rapperswil
+% 2021-04-14, Roy Seitz, Copied for SeminarMatrizen
+%
+verbatimtex
+\documentclass{book}
+\usepackage{times}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{amsfonts}
+\usepackage{txfonts}
+\begin{document}
+etex;
+
+input TEX;
+
+TEXPRE("%&latex" & char(10) &
+"\documentclass{book}" &
+"\usepackage{times}" &
+"\usepackage{amsmath}" &
+"\usepackage{amssymb}" &
+"\usepackage{amsfonts}" &
+"\usepackage{txfonts}" &
+"\begin{document}");
+TEXPOST("\end{document}");
+
+%
+% Vektorfeld in der Ebene mit Lösungskurve
+% so(2)
+%
+beginfig(1)
+
+% Scaling parameter
+numeric unit;
+unit := 150;
+
+% Some points
+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));
+
+% % 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) =
+  %(-0.5 * (x + y), -0.5 * (y - x))
+  (-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:
+  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(1)
+%
+beginfig(2)
+
+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, 0, 1);
+
+% 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(2)
+%
+beginfig(3)
+
+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, 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;
+
+endfig;
+
+%
+% Vektorfeld in der Ebene mit Lösungskurve
+% Euler(3)
+%
+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;
+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(4)
+%
+beginfig(5)
+
+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, 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:
+  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;
+
+
+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