171 lines
7.1 KiB
TeX
171 lines
7.1 KiB
TeX
\documentclass{article}%
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{graphicx}%
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\setcounter{MaxMatrixCols}{30}
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%TCIDATA{OutputFilter=latex2.dll}
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%TCIDATA{Version=5.50.0.2953}
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%TCIDATA{CSTFile=40 LaTeX article.cst}
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%TCIDATA{Created=Saturday, December 30, 2006 12:55:45}
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%TCIDATA{LastRevised=Friday, January 05, 2007 12:09:26}
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%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
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%TCIDATA{BibliographyScheme=Manual}
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%TCIDATA{ComputeGeneralSettings=0,10,6,0,0,0,0}
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%BeginMSIPreambleData
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\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
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%EndMSIPreambleData
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\newtheorem{theorem}{Theorem}
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\newtheorem{acknowledgement}[theorem]{Acknowledgement}
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\newtheorem{algorithm}[theorem]{Algorithm}
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\newtheorem{axiom}[theorem]{Axiom}
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\newtheorem{case}[theorem]{Case}
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\newtheorem{claim}[theorem]{Claim}
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\newtheorem{conclusion}[theorem]{Conclusion}
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\newtheorem{condition}[theorem]{Condition}
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\newtheorem{conjecture}[theorem]{Conjecture}
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\newtheorem{corollary}[theorem]{Corollary}
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\newtheorem{criterion}[theorem]{Criterion}
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\newtheorem{definition}[theorem]{Definition}
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\newtheorem{example}[theorem]{Example}
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\newtheorem{exercise}[theorem]{Exercise}
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\newtheorem{lemma}[theorem]{Lemma}
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\newtheorem{notation}[theorem]{Notation}
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\newtheorem{problem}[theorem]{Problem}
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\newtheorem{proposition}[theorem]{Proposition}
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\newtheorem{remark}[theorem]{Remark}
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\newtheorem{solution}[theorem]{Solution}
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\newtheorem{summary}[theorem]{Summary}
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\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
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\begin{document}
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Aufgabe 125
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Gegeben sei das eindeutig lösbare lineare Gleichungssystem $\ A\cdot
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\overrightarrow{x}=\overrightarrow{b}$ mit
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$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 $\overrightarrow{x}^{\left( 0\right) }=\overrightarrow{0}$. Berechnen Sie die Näherungslösung $\overrightarrow{x}^{\left(3\right) }$\ 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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$\overrightarrow{x}^{\left( 3\right) }$\ durch.
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\item[d.] F\"{u}hren Sie eine Apriori-Fehlerabsch\"{a}tzung f\"{u}r
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$\overrightarrow{x}^{\left( 10\right) }$\ durch.
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\end{itemize}
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Lösung:
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\begin{itemize}
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\item[a.] Rechenvorschriften:\newline$x_{1}^{\left( Z\right) }=\frac{1}%
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{4}\left( 2+x_{2}^{\left( Z-1\right) }+x_{4}^{\left( Z-1\right) }\right)
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=\frac{1}{2}+\frac{1}{4}x_{2}^{\left( Z-1\right) }+\frac{1}{4}x_{4}^{\left(
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Z-1\right) }$\newline$x_{2}^{\left( Z\right) }=\frac{1}{4}\left(
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1+x_{1}^{\left( Z-1\right) }+x_{3}^{\left( Z-1\right) }+x_{5}^{\left(
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Z-1\right) }\right) =\frac{1}{4}+\frac{1}{4}x_{1}^{\left( Z-1\right)
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}+\frac{1}{4}x_{3}^{\left( Z-1\right) }+\frac{1}{4}x_{5}^{\left(
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Z-1\right) }$\newline$x_{3}^{\left( Z\right) }=\frac{1}{4}\left(
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2+x_{2}^{\left( Z-1\right) }+x_{6}^{\left( Z-1\right) }\right) =\frac
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{1}{2}+\frac{1}{4}x_{2}^{\left( Z-1\right) }+\frac{1}{4}x_{6}^{\left(
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Z-1\right) }$\newline$x_{4}^{\left( Z\right) }=\frac{1}{4}\left(
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2+x_{1}^{\left( Z-1\right) }+x_{5}^{\left( Z-1\right) }\right) =\frac
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{1}{2}+\frac{1}{4}x_{1}^{\left( Z-1\right) }+\frac{1}{4}x_{5}^{\left(
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Z-1\right) }$\newline$x_{5}^{\left( Z\right) }=\frac{1}{4}\left(
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1+x_{2}^{\left( Z-1\right) }+x_{4}^{\left( Z-1\right) }+x_{6}^{\left(
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Z-1\right) }\right) =\frac{1}{4}+\frac{1}{4}x_{2}^{\left( Z-1\right)
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}+\frac{1}{4}x_{4}^{\left( Z-1\right) }+\frac{1}{4}x_{6}^{\left(
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Z-1\right) }$\newline$x_{6}^{\left( Z\right) }=\frac{1}{4}\left(
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2+x_{3}^{\left( Z-1\right) }+x_{5}^{\left( Z-1\right) }\right) =\frac
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{1}{2}+\frac{1}{4}x_{3}^{\left( Z-1\right) }+\frac{1}{4}x_{5}^{\left(
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Z-1\right) }$\newline%
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\begin{tabular}
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[c]{l||llllll}%
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z & $x_{1}^{\left( Z\right) }$ & $x_{2}^{\left( Z\right) }$ &
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$x_{3}^{\left( Z\right) }$ & $x_{4}^{\left( Z\right) }$ & $x_{5}^{\left(
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Z\right) }$ & $x_{6}^{\left( Z\right) }$\\\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} f\"{u}r
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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}
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\end{document} |