From 31cddc1e5890d5acf33e12078c7cdc311e92031c Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Andreas=20M=C3=BCller?= <andreas.mueller@ost.ch>
Date: Thu, 11 Mar 2021 17:11:17 +0100
Subject: add new slides

---
 vorlesungen/slides/5/hyperbolisch.tex | 105 ++++++++++++++++++++++++++++++++++
 1 file changed, 105 insertions(+)
 create mode 100644 vorlesungen/slides/5/hyperbolisch.tex

(limited to 'vorlesungen/slides/5/hyperbolisch.tex')

diff --git a/vorlesungen/slides/5/hyperbolisch.tex b/vorlesungen/slides/5/hyperbolisch.tex
new file mode 100644
index 0000000..905082a
--- /dev/null
+++ b/vorlesungen/slides/5/hyperbolisch.tex
@@ -0,0 +1,105 @@
+%
+% hyperbolisch.tex
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+%
+\bgroup
+\definecolor{darkgreen}{rgb}{0,0.6,0}
+\begin{frame}[t]
+\setlength{\abovedisplayskip}{5pt}
+\setlength{\belowdisplayskip}{5pt}
+\frametitle{Hyperbolische Funktionen}
+\vspace{-15pt}
+\begin{columns}[t,onlytextwidth]
+\begin{column}{0.48\textwidth}
+\begin{block}{Differentialgleichung}
+\vspace{-10pt}
+\begin{align*}
+\ddot{y} &= y
+\;\Rightarrow\;
+\frac{d}{dt}
+\begin{pmatrix}y\\y_1\end{pmatrix}
+=
+\begin{pmatrix}0&1\\1&0\end{pmatrix}
+\begin{pmatrix}y\\y_1\end{pmatrix}
+\\
+y(0)&=a,\qquad y'(0)=b
+\end{align*}
+\end{block}
+\vspace{-10pt}
+\uncover<2->{%
+\begin{block}{Lösung}
+\vspace{-13pt}
+\begin{align*}
+\lambda^2-1&=0
+\uncover<3->{
+\qquad\Rightarrow\qquad \lambda=\pm 1
+}
+\\
+\uncover<4->{
+y(t)&=Ae^t+Be^{-t}}
+\uncover<5->{
+\Rightarrow
+\left\{
+\arraycolsep=1.4pt
+\begin{array}{rcrcr}
+A&+&B&=&a\\
+A&-&B&=&b
+\end{array}
+\right.}
+\\
+&\uncover<6->{
+=\frac{a+b}2e^t + \frac{a-b}2e^{-t}}
+\\
+&\uncover<7->{=
+a{\color{darkgreen}\frac{e^t+e^{-t}}2} + b{\color{red}\frac{e^t-e^{-t}}2}}
+\end{align*}
+\end{block}}
+\end{column}
+\begin{column}{0.49\textwidth}
+\uncover<8->{%
+\begin{block}{Potenzreihe}
+\vspace{-12pt}
+\begin{align*}
+K&=\begin{pmatrix}0&1\\1&0\end{pmatrix}
+\uncover<10->{\quad\Rightarrow\quad K^2=I}
+\\
+\uncover<9->{
+e^{Kt}
+&=
+I+K+\frac1{2!}K^2 + \frac{1}{3!}K^3 + \frac{1}{4!}K^4+\dots
+}
+\\
+\uncover<11->{
+&=
+\biggl( 1+\frac{t^2}{2!} + \frac{t^4}{4!}+\dots \biggr)I
+}
+\\
+\uncover<11->{
+&\qquad
++\biggl(t+\frac{t^3}{3!}+\frac{t^5}{5!}+\dots\biggr)K
+}
+\\
+\uncover<12->{
+&=
+I{\,\color{darkgreen}\cosh t}  + K{\,\color{red}\sinh t}
+}
+\\
+\uncover<13->{
+\begin{pmatrix}y(t)\\y_1(t)\end{pmatrix}
+&=
+e^{Kt}\begin{pmatrix}a\\b\end{pmatrix}
+}
+\uncover<14->{
+=
+\begin{pmatrix}
+a{\,\color{darkgreen}\cosh t} + b{\,\color{red}\sinh t}\\
+a{\,\color{red}\sinh t} + b{\,\color{darkgreen}\cosh t}
+\end{pmatrix}
+}
+\end{align*}
+\end{block}}
+\end{column}
+\end{columns}
+\end{frame}
+\egroup
-- 
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