74 lines
2.4 KiB
TeX
74 lines
2.4 KiB
TeX
|
|
\documentclass{article}
|
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
%TCIDATA{OutputFilter=LATEX.DLL}
|
|
%TCIDATA{Version=5.50.0.2890}
|
|
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
|
|
%TCIDATA{BibliographyScheme=Manual}
|
|
%TCIDATA{Created=Saturday, December 30, 2006 12:55:45}
|
|
%TCIDATA{LastRevised=Saturday, December 30, 2006 13:08:01}
|
|
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
|
|
%TCIDATA{<META NAME="DocumentShell" CONTENT="Standard LaTeX\Blank - Standard LaTeX Article">}
|
|
%TCIDATA{CSTFile=40 LaTeX article.cst}
|
|
|
|
\newtheorem{theorem}{Theorem}
|
|
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
|
|
\newtheorem{algorithm}[theorem]{Algorithm}
|
|
\newtheorem{axiom}[theorem]{Axiom}
|
|
\newtheorem{case}[theorem]{Case}
|
|
\newtheorem{claim}[theorem]{Claim}
|
|
\newtheorem{conclusion}[theorem]{Conclusion}
|
|
\newtheorem{condition}[theorem]{Condition}
|
|
\newtheorem{conjecture}[theorem]{Conjecture}
|
|
\newtheorem{corollary}[theorem]{Corollary}
|
|
\newtheorem{criterion}[theorem]{Criterion}
|
|
\newtheorem{definition}[theorem]{Definition}
|
|
\newtheorem{example}[theorem]{Example}
|
|
\newtheorem{exercise}[theorem]{Exercise}
|
|
\newtheorem{lemma}[theorem]{Lemma}
|
|
\newtheorem{notation}[theorem]{Notation}
|
|
\newtheorem{problem}[theorem]{Problem}
|
|
\newtheorem{proposition}[theorem]{Proposition}
|
|
\newtheorem{remark}[theorem]{Remark}
|
|
\newtheorem{solution}[theorem]{Solution}
|
|
\newtheorem{summary}[theorem]{Summary}
|
|
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
|
|
\input{tcilatex}
|
|
|
|
\begin{document}
|
|
|
|
|
|
Aufgabe 125
|
|
|
|
Gegeben sei das eindeutig l\"{o}sbare lineare Gleichungssystem $\ A\cdot
|
|
\overrightarrow{x}=\overrightarrow{b}$ mit
|
|
|
|
$A=\left(
|
|
\begin{array}{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}
|
|
2 \\
|
|
1 \\
|
|
2 \\
|
|
2 \\
|
|
1 \\
|
|
2%
|
|
\end{array}%
|
|
\right) $
|
|
|
|
a.\qquad 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.
|
|
|
|
b.
|
|
|
|
\end{document}
|