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%TCIDATA{Created=Saturday, December 30, 2006 12:55:45}
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\begin{document}
Aufgabe 125

Gegeben sei das eindeutig l\"{o}sbare lineare Gleichungssystem $\ A\cdot
\overrightarrow{x}=\overrightarrow{b}$ mit

$A=\left(
\begin{array}
[c]{cccccc}%
4 & -1 & 0 & -1 & 0 & 0\\
-1 & 4 & -1 & 0 & -1 & 0\\
0 & -1 & 4 & 0 & 0 & -1\\
-1 & 0 & 0 & 4 & -1 & 0\\
0 & -1 & 0 & -1 & 4 & -1\\
0 & 0 & -1 & 0 & -1 & 4
\end{array}
\right)  $, $\overrightarrow{b}=\left(
\begin{array}
[c]{c}%
2\\
1\\
2\\
2\\
1\\
2
\end{array}
\right)  $

\begin{itemize}
\item[a.] Sei $\overrightarrow{x}^{\left(  0\right)  }=\overrightarrow{0}$.
Berechnen Sie die N\"{a}herungsl\"{o}sung $\overrightarrow{x}^{\left(
3\right)  }$\ des Systems, die man nach 3 Schritten des
Gesamtschrittverfahrens erh\"{a}lt.

\item[b.] Zeigen Sie, da\ss \ das Gesamtschrittverfahren konvergiert.

\item[c.] F\"{u}hren Sie eine Apeoteriori-Fehlerabsch\"{a}tzung f\"{u}r
$\overrightarrow{x}^{\left(  3\right)  }$\ durch.

\item[d.] F\"{u}hren Sie eine Apriori-Fehlerabsch\"{a}tzung f\"{u}r
$\overrightarrow{x}^{\left(  10\right)  }$\ durch.
\end{itemize}

L\"{o}sung:

\begin{itemize}
\item[a.] Rechenvorschriften:\newline$x_{1}^{\left(  Z\right)  }=\frac{1}%
{4}\left(  2+x_{2}^{\left(  Z-1\right)  }+x_{4}^{\left(  Z-1\right)  }\right)
=\frac{1}{2}+\frac{1}{4}x_{2}^{\left(  Z-1\right)  }+\frac{1}{4}x_{4}^{\left(
Z-1\right)  }$\newline$x_{2}^{\left(  Z\right)  }=\frac{1}{4}\left(
1+x_{1}^{\left(  Z-1\right)  }+x_{3}^{\left(  Z-1\right)  }+x_{5}^{\left(
Z-1\right)  }\right)  =\frac{1}{4}+\frac{1}{4}x_{1}^{\left(  Z-1\right)
}+\frac{1}{4}x_{3}^{\left(  Z-1\right)  }+\frac{1}{4}x_{5}^{\left(
Z-1\right)  }$\newline$x_{3}^{\left(  Z\right)  }=\frac{1}{4}\left(
2+x_{2}^{\left(  Z-1\right)  }+x_{6}^{\left(  Z-1\right)  }\right)  =\frac
{1}{2}+\frac{1}{4}x_{2}^{\left(  Z-1\right)  }+\frac{1}{4}x_{6}^{\left(
Z-1\right)  }$\newline$x_{4}^{\left(  Z\right)  }=\frac{1}{4}\left(
2+x_{1}^{\left(  Z-1\right)  }+x_{5}^{\left(  Z-1\right)  }\right)  =\frac
{1}{2}+\frac{1}{4}x_{1}^{\left(  Z-1\right)  }+\frac{1}{4}x_{5}^{\left(
Z-1\right)  }$\newline$x_{5}^{\left(  Z\right)  }=\frac{1}{4}\left(
1+x_{2}^{\left(  Z-1\right)  }+x_{4}^{\left(  Z-1\right)  }+x_{6}^{\left(
Z-1\right)  }\right)  =\frac{1}{4}+\frac{1}{4}x_{2}^{\left(  Z-1\right)
}+\frac{1}{4}x_{4}^{\left(  Z-1\right)  }+\frac{1}{4}x_{6}^{\left(
Z-1\right)  }$\newline$x_{6}^{\left(  Z\right)  }=\frac{1}{4}\left(
2+x_{3}^{\left(  Z-1\right)  }+x_{5}^{\left(  Z-1\right)  }\right)  =\frac
{1}{2}+\frac{1}{4}x_{3}^{\left(  Z-1\right)  }+\frac{1}{4}x_{5}^{\left(
Z-1\right)  }$\newline%
\begin{tabular}
[c]{l||llllll}%
z & $x_{1}^{\left(  Z\right)  }$ & $x_{2}^{\left(  Z\right)  }$ &
$x_{3}^{\left(  Z\right)  }$ & $x_{4}^{\left(  Z\right)  }$ & $x_{5}^{\left(
Z\right)  }$ & $x_{6}^{\left(  Z\right)  }$\\\hline\hline
1 & $0.5$ & $0.25$ & $0.5$ & $0.5$ & $0.25$ & $0.5$\\
2 & $0.6875$ & $0.5625$ & $0.6875$ & $0.6875$ & $0.5625$ & $0.6875$\\
3 & $0.8125$ & $0.734325$ & $0.8125$ & $0.8125$ & $0.734375$ & $0.8125$%
\end{tabular}
\newline

$\overrightarrow{x}=\overrightarrow{x}^{\left(  3\right)  }=\left(
0.8125;\text{ }0.734325;\text{ }0.8125;\text{ }0.8125;\text{ }0.734375;\text{
}0.8125\right)  $

\item[b.] Berechnung der Kontraktionszahl $\lambda$\newline\newline$%
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq1}}^{6}}%
%BeginExpansion
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq1}}^{6}}
%EndExpansion
\left\vert \frac{a_{1j}}{a_{11}}\right\vert =0.5;$ \ \ $%
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq2}}^{6}}%
%BeginExpansion
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq2}}^{6}}
%EndExpansion
\left\vert \frac{a_{2j}}{a_{22}}\right\vert =0.75;$ \ \ $%
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq3}}^{6}}%
%BeginExpansion
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq3}}^{6}}
%EndExpansion
\left\vert \frac{a_{3j}}{a_{33}}\right\vert =0.5;$\newline\newline$%
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq4}}^{6}}%
%BeginExpansion
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq4}}^{6}}
%EndExpansion
\left\vert \frac{a_{4j}}{a_{44}}\right\vert =0.5;$ \ \ $%
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq5}}^{6}}%
%BeginExpansion
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq5}}^{6}}
%EndExpansion
\left\vert \frac{a_{5j}}{a_{55}}\right\vert =0.75;$ \ \ $%
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq6}}^{6}}%
%BeginExpansion
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq6}}^{6}}
%EndExpansion
\left\vert \frac{a_{6j}}{a_{66}}\right\vert =0.5;$\newline\newline%
$\lambda=\max\{0.5;0.75\}=\underline{\underline{0.75}}$\newline$\lambda
<1\Longrightarrow$ \underline{das Gesamtschrittverfahren konvergiert} f\"{u}r
jeden Startvektor

\item[c.] $\underset{i=1,...,6}{\max}\left\vert x_{i}^{\left(  3\right)
}-x_{i}\right\vert \leq\frac{\lambda}{1-\lambda}$ \ \ \ $\underset
{i=1,...,6}{\max}\left\vert x_{i}^{\left(  3\right)  }-x_{i}^{\left(
2\right)  }\right\vert =\frac{0.75}{0.25}\cdot0.171875=\allowbreak
\underline{\underline{0.515\,625}}\,$

\item[d.] $\underset{i=1,...,6}{\max}\left\vert x_{i}^{\left(  10\right)
}-x_{i}\right\vert \leq\frac{\lambda^{10}}{1-\lambda}$ \ \ $\underset
{i=1,...,6}{\max}\left\vert x_{i}^{\left(  1\right)  }-x_{i}^{\left(
0\right)  }\right\vert =\frac{0.75^{10}}{0.25}\cdot0.5=\underline{\underline{\allowbreak
0.112\,627\,029\,\allowbreak4}}$
\end{itemize}


\end{document}