128 lines
5.6 KiB
TeX
128 lines
5.6 KiB
TeX
\section{Aufgabe 125}
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Gegeben sei das eindeutig lösbare lineare Gleichungssystem $\displaystyle A\cdot
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\overrightarrow{x}=\overrightarrow{b}$ mit
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$\displaystyle A=\left(
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\begin{array}
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[c]{cccccc}%
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4 & -1 & 0 & -1 & 0 & 0\\
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-1 & 4 & -1 & 0 & -1 & 0\\
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0 & -1 & 4 & 0 & 0 & -1\\
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-1 & 0 & 0 & 4 & -1 & 0\\
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0 & -1 & 0 & -1 & 4 & -1\\
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0 & 0 & -1 & 0 & -1 & 4
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\end{array}
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\right) $, $\overrightarrow{b}=\left(
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\begin{array}
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[c]{c}%
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2\\
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1\\
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2\\
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2\\
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1\\
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2
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\end{array}
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\right) $
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\begin{itemize}
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\item[a.] Sei $\displaystyle \overrightarrow{x}^{\left( 0\right) }=\overrightarrow{0}$.
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Berechnen Sie die Näherungslösung $\displaystyle \overrightarrow{x}^{\left(
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3\right) }$ des Systems, die man nach 3 Schritten des
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Gesamtschrittverfahrens erhält.
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\item[b.] Zeigen Sie, daß das Gesamtschrittverfahren konvergiert.
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\item[c.] Führen Sie eine Apeoteriori-Fehlerabschätzung für
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$\displaystyle \overrightarrow{x}^{\left( 3\right) }$\ durch.
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\item[d.] Führen Sie eine Apriori-Fehlerabschätzung für
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$\displaystyle \overrightarrow{x}^{\left( 10\right) }$ durch.
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\end{itemize}
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\subsection{Lösung}
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\begin{itemize}
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\item[a.] Rechenvorschriften:\newline$\displaystyle x_{1}^{\left( Z\right) }=\frac{1}%
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{4}\left( 2+x_{2}^{\left( Z-1\right) }+x_{4}^{\left( Z-1\right) }\right)
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=\frac{1}{2}+\frac{1}{4}x_{2}^{\left( Z-1\right) }+\frac{1}{4}x_{4}^{\left(
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Z-1\right) }$\newline$\displaystyle x_{2}^{\left( Z\right) }=\frac{1}{4}\left(
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1+x_{1}^{\left( Z-1\right) }+x_{3}^{\left( Z-1\right) }+x_{5}^{\left(
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Z-1\right) }\right) =\frac{1}{4}+\frac{1}{4}x_{1}^{\left( Z-1\right)
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}+\frac{1}{4}x_{3}^{\left( Z-1\right) }+\frac{1}{4}x_{5}^{\left(
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Z-1\right) }$\newline$\displaystyle x_{3}^{\left( Z\right) }=\frac{1}{4}\left(
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2+x_{2}^{\left( Z-1\right) }+x_{6}^{\left( Z-1\right) }\right) =\frac
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{1}{2}+\frac{1}{4}x_{2}^{\left( Z-1\right) }+\frac{1}{4}x_{6}^{\left(
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Z-1\right) }$\newline$\displaystyle x_{4}^{\left( Z\right) }=\frac{1}{4}\left(
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2+x_{1}^{\left( Z-1\right) }+x_{5}^{\left( Z-1\right) }\right) =\frac
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{1}{2}+\frac{1}{4}x_{1}^{\left( Z-1\right) }+\frac{1}{4}x_{5}^{\left(
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Z-1\right) }$\newline$\displaystyle x_{5}^{\left( Z\right) }=\frac{1}{4}\left(
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1+x_{2}^{\left( Z-1\right) }+x_{4}^{\left( Z-1\right) }+x_{6}^{\left(
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Z-1\right) }\right) =\frac{1}{4}+\frac{1}{4}x_{2}^{\left( Z-1\right)
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}+\frac{1}{4}x_{4}^{\left( Z-1\right) }+\frac{1}{4}x_{6}^{\left(
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Z-1\right) }$\newline$\displaystyle x_{6}^{\left( Z\right) }=\frac{1}{4}\left(
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2+x_{3}^{\left( Z-1\right) }+x_{5}^{\left( Z-1\right) }\right) =\frac
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{1}{2}+\frac{1}{4}x_{3}^{\left( Z-1\right) }+\frac{1}{4}x_{5}^{\left(
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Z-1\right) }$\newline\newline%
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\begin{tabular}
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[c]{l||llllll}%
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z & $\displaystyle x_{1}^{\left( Z\right) }$ & $\displaystyle x_{2}^{\left( Z\right) }$ &
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$x_{3}^{\left( Z\right) }$ & $\displaystyle x_{4}^{\left( Z\right) }$ & $\displaystyle x_{5}^{\left(
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Z\right) }$ & $x_{6}^{\left( Z\right) }$\\\hline\hline
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1 & $0.5$ & $0.25$ & $0.5$ & $0.5$ & $0.25$ & $0.5$\\
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2 & $0.6875$ & $0.5625$ & $0.6875$ & $0.6875$ & $0.5625$ & $0.6875$\\
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3 & $0.8125$ & $0.734325$ & $0.8125$ & $0.8125$ & $0.734375$ & $0.8125$%
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\end{tabular}
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\newline
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$\overrightarrow{x}=\overrightarrow{x}^{\left( 3\right) }=\left(
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0.8125;\text{ }0.734325;\text{ }0.8125;\text{ }0.8125;\text{ }0.734375;\text{
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}0.8125\right) $
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\item[b.] Berechnung der Kontraktionszahl $\lambda$\newline\newline$%
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%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq1}}^{6}}%
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%BeginExpansion
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{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq1}}^{6}}
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%EndExpansion
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\left\vert \frac{a_{1j}}{a_{11}}\right\vert =0.5;$ \ \ $%
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%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq2}}^{6}}%
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%BeginExpansion
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{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq2}}^{6}}
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%EndExpansion
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\left\vert \frac{a_{2j}}{a_{22}}\right\vert =0.75;$ \ \ $%
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%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq3}}^{6}}%
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%BeginExpansion
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{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq3}}^{6}}
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%EndExpansion
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\left\vert \frac{a_{3j}}{a_{33}}\right\vert =0.5;$\newline\newline$%
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%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq4}}^{6}}%
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%BeginExpansion
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{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq4}}^{6}}
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%EndExpansion
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\left\vert \frac{a_{4j}}{a_{44}}\right\vert =0.5;$ \ \ $%
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%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq5}}^{6}}%
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%BeginExpansion
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{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq5}}^{6}}
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%EndExpansion
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\left\vert \frac{a_{5j}}{a_{55}}\right\vert =0.75;$ \ \ $%
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%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq6}}^{6}}%
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%BeginExpansion
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{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq6}}^{6}}
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%EndExpansion
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\left\vert \frac{a_{6j}}{a_{66}}\right\vert =0.5;$\newline\newline%
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$\lambda=\max\{0.5;0.75\}=\underline{\underline{0.75}}$\newline$\lambda
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<1\Longrightarrow$ \underline{das Gesamtschrittverfahren konvergiert} für
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jeden Startvektor
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\item[c.] $\underset{i=1,...,6}{\max}\left\vert x_{i}^{\left( 3\right)
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}-x_{i}\right\vert \leq\frac{\lambda}{1-\lambda}$ \ \ \ $\underset
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{i=1,...,6}{\max}\left\vert x_{i}^{\left( 3\right) }-x_{i}^{\left(
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2\right) }\right\vert =\frac{0.75}{0.25}\cdot0.171875=\allowbreak
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\underline{\underline{0.515\,625}}\,$
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\item[d.] $\underset{i=1,...,6}{\max}\left\vert x_{i}^{\left( 10\right)
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}-x_{i}\right\vert \leq\frac{\lambda^{10}}{1-\lambda}$ \ \ $\underset
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{i=1,...,6}{\max}\left\vert x_{i}^{\left( 1\right) }-x_{i}^{\left(
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0\right) }\right\vert =\frac{0.75^{10}}{0.25}\cdot0.5=\underline{\underline{\allowbreak
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0.112\,627\,029\,\allowbreak4}}$
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\end{itemize} |