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\begin{document}
Aufgabe 120

Sei $x=%
%TCIMACRO{\dsum \limits_{k=0}^{\infty}}%
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{\displaystyle\sum\limits_{k=0}^{\infty}}
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\left(  -1\right)  ^{k}\frac{1}{k};$ \ $\widetilde{x}=%
%TCIMACRO{\dsum \limits_{k=0}^{4}}%
%BeginExpansion
{\displaystyle\sum\limits_{k=0}^{4}}
%EndExpansion
\left(  -1\right)  ^{k}\frac{1}{k!}$

\begin{description}
\item[a.] Mit Hilfe von welcher speziellen Funktion l\"{a}\ss t sich $x$ genau
beschreiben? Wie? (Tip: 3.3.5)

\item[b.] Berechnen Sie $\widetilde{x}$.

\item[c.] Geben Sie einen absoluten H\"{o}chstfehler von $\widetilde{x}$ an.
(Tip: 3.2.7)
\end{description}

L\"{o}sung:

\begin{description}
\item[a.] $\exp(z)=%
%TCIMACRO{\dsum \limits_{k=0}^{\infty}}%
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{\displaystyle\sum\limits_{k=0}^{\infty}}
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\frac{1}{k!}z^{k}$ \ $\Longrightarrow$ \ \ $x=%
%TCIMACRO{\dsum \limits_{k=0}^{\infty}}%
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{\displaystyle\sum\limits_{k=0}^{\infty}}
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\frac{1}{k!}(-1)^{k}=\exp(-1)=\underline{\underline{\frac{1}{e}}}$

\item[b.] $\widetilde{x}=%
%TCIMACRO{\dsum \limits_{k=0}^{4}}%
%BeginExpansion
{\displaystyle\sum\limits_{k=0}^{4}}
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\frac{1}{k!}(-1)^{k}=\allowbreak1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}%
=\frac{12-4+1}{24}=\allowbreak\frac{9}{24}=\frac{3}{8}=\allowbreak
\underline{\underline{0.375}}\,$

\item[c.] Da die vorliegende Reihe eine alternierende Reihe ist, gilt $|x-\widetilde{x}|\leq\frac{1}{5!}=\frac{1}{120}=8.\overline{3}\cdot 10^{-3}$.\\
    Damit ist $\underline{\underline{\alpha_x=8.\overline{3}\cdot 10^{-3}}}$ ein absoluter Höchstfehler von $\widetilde{x}$
\end{description}


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