96 lines
3.3 KiB
TeX
96 lines
3.3 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=Saturday, January 06, 2007 22:30:18}
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%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
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%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
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%TCIDATA{BibliographyScheme=Manual}
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%TCIDATA{<META NAME="DocumentShell" CONTENT="Standard LaTeX\Blank - Standard LaTeX Article">}
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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 120
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Sei $x=%
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%TCIMACRO{\dsum \limits_{k=0}^{\infty}}%
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%BeginExpansion
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{\displaystyle\sum\limits_{k=0}^{\infty}}
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%EndExpansion
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\left( -1\right) ^{k}\frac{1}{k};$ \ $\widetilde{x}=%
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%TCIMACRO{\dsum \limits_{k=0}^{4}}%
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%BeginExpansion
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{\displaystyle\sum\limits_{k=0}^{4}}
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%EndExpansion
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\left( -1\right) ^{k}\frac{1}{k!}$
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\begin{description}
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\item[a.] Mit Hilfe von welcher speziellen Funktion l\"{a}\ss t sich $x$ genau
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beschreiben? Wie? (Tip: 3.3.5)
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\item[b.] Berechnen Sie $\widetilde{x}$.
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\item[c.] Geben Sie einen absoluten H\"{o}chstfehler von $\widetilde{x}$ an.
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(Tip: 3.2.7)
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\end{description}
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L\"{o}sung:
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\begin{description}
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\item[a.] $\exp(z)=%
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%TCIMACRO{\dsum \limits_{k=0}^{\infty}}%
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%BeginExpansion
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{\displaystyle\sum\limits_{k=0}^{\infty}}
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%EndExpansion
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\frac{1}{k!}z^{k}$ \ $\Longrightarrow$ \ \ $x=%
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%TCIMACRO{\dsum \limits_{k=0}^{\infty}}%
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%BeginExpansion
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{\displaystyle\sum\limits_{k=0}^{\infty}}
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%EndExpansion
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\frac{1}{k!}(-1)^{k}=\exp(-1)=\underline{\underline{\frac{1}{e}}}$
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\item[b.] $\widetilde{x}=%
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%TCIMACRO{\dsum \limits_{k=0}^{4}}%
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%BeginExpansion
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{\displaystyle\sum\limits_{k=0}^{4}}
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%EndExpansion
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\frac{1}{k!}(-1)^{k}=\allowbreak1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}%
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=\frac{12-4+1}{24}=\allowbreak\frac{9}{24}=\frac{3}{8}=\allowbreak
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\underline{\underline{0.375}}\,$
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\item[c.] Da die vorliegende Reihe eine alternierende Reihe ist, gilt $|x-\widetilde{x}|\leq\frac{1}{5!}=\frac{1}{120}=8.\overline{3}\cdot 10^{-3}$.\\
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Damit ist $\underline{\underline{\alpha_x=8.\overline{3}\cdot 10^{-3}}}$ ein absoluter H<>chstfehler von $\widetilde{x}$
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\end{description}
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\end{document} |