\documentclass[11pt,a4paper]{article}
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\date{\today{}}
\author{Thomas Kuster}
\title{UWIS, Mathematik 1, Lösung Serie 10}

\begin{document}
\markright{UWIS, Mathematik 1, Lösung Serie 10}
\pagestyle{myheadings}

\maketitle

\section{Differentialgleichungen}
\subsection{}
\begin{eqnarray*}
x\frac{d}{dx}(1+t) & = & x^2\\
(1+t) dx & = & x^2 dt\\
\int 1+t\,dx & = & \int x^2\,dt\\
-\frac{1}{x}+c_1 & = & \ln{(1+t)} + c_2 \quad c = c_2 - c_1\\
x & = & \frac{-1}{\ln{(1+t)+c}}
\end{eqnarray*}

\subsection{}
\begin{eqnarray*}
\dot{x} & = & -2x + e^{3t}
\end{eqnarray*}
homogene Lösung:
\begin{eqnarray*}
\dot{x}_h & = & -2x_h\\
x_h & = & c_1 e^{-2t}
\end{eqnarray*}
Ansatz $C(t) c_1 e^{-2t}$ (auch $x= c_1 e^{-2t} + Ae^{3t}$ möglich):
\begin{eqnarray*}
\dot{C}(t) c_1 e^{-2t} + C(t) c_1 (-2) e^{-2t} & = & -2 C(t) c_1 e^{-2t} + e^{3t}\\
\dot{C}(t) c_1 e^{-2t} & = & e^{3t}\\
\dot{C}(t) & = & \frac{e^{3t}}{c_1 e^{-2t}} \quad c_2 = \frac{1}{c_1}\\
\dot{C}(t) & = & c_2 e^{3t}e^{2t}\\
\dot{C}(t) & = & c_2 e^{5t}\\
C(t) & = & \frac{c_2}{5} e^{5t}\\
\end{eqnarray*}
Lösung:
\begin{eqnarray*}
x & = & \frac{c_2}{5} e^{5t} c_1 e^{-2t} \quad c = c_1 c_2\\
x & = & \frac{c}{5}e^{3t}\\
\end{eqnarray*}

\subsection{}
\begin{eqnarray*}
\dot{x} + x \sin{(t)} & = & \sin{(t)}\\
\dot{x} & = & - x \sin{(t)} + \sin{(t)}\\
\dot{x} & = & \sin{(t)}(1-x)\\
\frac{\dot{x}}{(1-x)} & = & \sin{(t)}\\
\frac{dx}{(1-x)dt} & = & \sin{(t)}\\
\frac{1}{(1-x)}dx & = & \sin{(t)}dt\\
\int \frac{1}{(1-x)}\,dx & = & \int \sin{(t)}\,dt\\
\ln{(1-t)} & = & -\cos{(t)} + c_1\\
1-x & = & e^{-\cos{(t)} + c_1}\\
x & = & 1 - e^{-\cos{(t)} + c_1}\\
x & = & 1 - ce^{-\cos{(t)}}
\end{eqnarray*}

\section{Newtonverfahren}
Die Tangente der Funktion $f$ an der Stelle $x_0$ (nullte Näherung der Nullstelle der Funktion $f$) wird mit der $x$-Achse geschnitten, dieser Schnittpunkt ergibt den neuen Punkt $x_1$ (erste Näherung der Nullstelle der Funktion $f$).
\begin{eqnarray*}
f(x_n) + f'(x_n)(x_{n+1}-x_n) & = & 0\\
f(x_n) + f'(x_n)x_{n+1}-f'(x_n)x_n & = & 0\\
f(x_n) -f'(x_n)x_n & = & -f'(x_n)x_{n+1}\\
\frac{f'(x_n)x_n-f(x_n)}{f'(x_n)} & = & x_{n+1}\\
\end{eqnarray*}
\begin{eqnarray*}
f(x) & = & \tan{x} - x\\
f'(x) & = & 1 + tan^2{x} - 1 = tan^2{x}\\
\end{eqnarray*}
\begin{eqnarray*}
\frac{\left(tan^2{x_n}\right)x_n - \left(\tan{x_n} - x_n\right)}{tan^2{x}} & = & x_{n+1}\\
\frac{\left(tan^2{4.5}\right)4.5 - \left(\tan{4.5} - 4.5\right)}{tan^2{4.5}} & = & 4.4936\\
\frac{\left(tan^2{4.49}\right)4.49 - \left(\tan{4.49} - 4.49\right)}{tan^2{4.49}} & = & 4.4934\\
\end{eqnarray*}


\section{Mac Laurinsche Reihen}
\subsection{}
\subsubsection{}\label{i}
\begin{eqnarray*}
f(t) & = & (1+t)^{\frac{1}{2}}\\
f'(t) & = & \frac{1}{2}(1+t)^{-\frac{1}{2}}\\
f''(t) & = & -\frac{1}{4}(1+t)^{-\frac{3}{2}}\\
f'''(t) & = & \frac{3}{8}(1+t)^{-\frac{5}{2}}\\
f^{(4)}(t) & = & -\frac{15}{16}(1+t)^{-\frac{7}{2}}
\end{eqnarray*}
\begin{equation*}\begin{split}
T(t) = & (1+0)^{\frac{1}{2}} + \frac{1}{2}(1+0)^{-\frac{1}{2}}(t-0) -\frac{1}{4}(1+0)^{-\frac{3}{2}}(t-0)^2\frac{1}{2}\\
&+ \frac{3}{8}(1+0)^{-\frac{5}{2}}(t-0)^3\frac{1}{6} -\frac{15}{16}(1+0)^{-\frac{7}{2}}(t-0)^4\frac{1}{24}\\
T(t) = & 1 + \frac{1}{2}t -\frac{1}{8}t^2 + \frac{1}{16}t^3 -\frac{5}{128}t^4
\end{split}\end{equation*}

\subsubsection{}
eine Ableitung mehr als in \ref{i} und $g^{(n)}(t) = f^{(n+1)}(t)$
\begin{eqnarray*}
g(t) & = & (1+t)^{-\frac{1}{2}}\\
g'(t) & = & -\frac{1}{2}(1+t)^{-\frac{3}{2}}\\
g''(t) & = & \frac{3}{4}(1+t)^{-\frac{5}{2}}\\
g'''(t) & = & -\frac{15}{8}(1+t)^{-\frac{7}{2}}\\
g^{4}(t) & = & \frac{105}{16}(1+t)^{-\frac{9}{2}}
\end{eqnarray*}
\begin{eqnarray*}
T(t) & = & 1 - \frac{1}{2}t + \frac{3}{8}t^2 - \frac{5}{16}t^3 + \frac{35}{128}t^4
\end{eqnarray*}

\subsubsection{}
\begin{eqnarray*}
f(t) & = & \ln{\left(\frac{1+t}{1-t}\right)} = \ln{(1+t)}-\ln{(1-t)}\\
f'(t) & = & \frac{1}{1+t}+\frac{1}{1-t}\\
f''(t) & = & -\frac{1}{(1+t)^2}+\frac{1}{(1-t)^2}\\
f'''(t) & = & \frac{2}{(1+t)^3}+\frac{2}{(1-t)^3}\\
f^{(4)}(t) & = &-\frac{6}{(1+t)^4}+\frac{6}{(1-t)^4}\\
f^{(5)}(t) & = & \frac{24}{(1+t)^5}+\frac{24}{(1-t)^5}\\
\end{eqnarray*}
\begin{eqnarray*}
T(t) & = & 2t + \frac{2}{3}t^3 + \frac{2}{5}t^5
\end{eqnarray*}

\subsubsection{}
\begin{eqnarray*}
f(t) & = & \tan{(t)}\\
f'(t) & = & 1 + \tan^2{(t)}\\
f''(t) & = & 2\tan{(t)}\left(1 + \tan^2{(t)}\right) = 2\tan{(t)} + 2\tan^3{(t)}\\
f'''(t) & = & 2 + 2\tan^2{(t)} + 6\tan^2{(t)}\left(1+\tan^2{(t)}\right)\\
 & = & 2 + 8\tan^2{(t)} + 6\tan^4{(t)}\\
f^{(4)}(t) & = & 16\tan{(t)}\left(1+\tan^2{(t)}\right) + 24\tan^3{(t)}\left(1+\tan^2{(t)}\right)\\
 & = & 16\tan{(t)} + 40\tan^3{(t)} + 24\tan^5{(t)}\\
f^{(5)}(t) & = & 16\left(1+\tan^2{(t)}\right) + 120\tan^2{(t)}\left(1+\tan^2{(t)}\right) +\\
 & & 120\tan^4{(t)}\left(1+\tan^2{(t)}\right)\\
 & = & 16+136\tan^2{(t)}+240\tan^4{(t)}+120\tan^6{(t)}
\end{eqnarray*}

\subsection{}
\begin{eqnarray*}
\frac{1+t}{1-t} & = & 2\\
1+t & = & 2 - 2t\\
3t & = & 1\\
t & = & \frac{1}{3}
\end{eqnarray*}
\begin{eqnarray*}
T\left(\frac{1}{3}\right) & = & 2\frac{1}{3} + \frac{2}{3}\left(\frac{1}{3}\right)^3 + \frac{2}{5}\left(\frac{1}{3}\right)^5 = \frac{2}{3} + \frac{2}{81} + {2}{1215}\\
T\left(\frac{1}{3}\right) & = & \frac{842}{1215} = 0.69300
\end{eqnarray*}
Zum Vergleich $\ln(2) = 0.69315$ der Fehler ist sicher kleiner $t^7 < 0.00046$.

\section{Taylorpolynome verschiedenen Grades}
\begin{verbatim}
> series(sin(t),t=0,1);
                         O(t)

> series(sin(t),t=0,3);
                              3
                       t + O(t )

> series(sin(t),t=0,5);
                           3      5
                  t - 1/6 t  + O(t )

> series(sin(t),t=0,7);
                      3          5      7
             t - 1/6 t  + 1/120 t  + O(t )
\end{verbatim}
allgemein:
\begin{displaymath}
\sin{t} = \sum_{k=0}^{\infty}{(-1)^k\frac{t^{2k+1}}{(2k+1)!}}
\end{displaymath}
%Graph

\section{komplexe Zahlen}
\subsection{}
\includegraphics[width=\linewidth]{serie10_loesung_komplexe_zahlen.eps}

\subsection{}
\subsubsection{}
\begin{eqnarray*}
z & = & \frac{5-5i}{-1+2i} = \frac{(5-5i)(-1-2i)}{(-1+2i)(-1-2i)} = \frac{-5-10i+5i+10i^2}{1-4i^2} = \frac{-15-5i}{5}\\
z & = & -3-i
\end{eqnarray*}

\subsubsection{}
\begin{eqnarray*}
z & = & \frac{1}{(1+2i)^3} = \frac{(1-2i)^3}{\left((1+2i)(1-2i)\right)^3} = \frac{(-3-4i)(1-2i)}{(1-4i^2)^3} = \frac{-11+2i}{5^3}\\
z & = & \frac{-11}{125}+\frac{2}{125}i
\end{eqnarray*}

\end{document}
