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\documentclass{article}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%TCIDATA{OutputFilter=LATEX.DLL}
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%TCIDATA{Version=5.50.0.2890}
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%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
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%TCIDATA{BibliographyScheme=Manual}
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%TCIDATA{Created=Saturday, April 30, 2011 16:26:47}
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%TCIDATA{LastRevised=Saturday, April 30, 2011 16:44:31}
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%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
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%TCIDATA{<META NAME="DocumentShell" CONTENT="Standard LaTeX\Blank - Standard LaTeX Article">}
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%TCIDATA{CSTFile=40 LaTeX article.cst}
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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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\input{tcilatex}
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\begin{document}
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\section{Aufgabe 3}
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Gegeben sind: $\vec{a}=\left(3,-1,2\right)\text{, }\vec{b}=\left(1,2,4\right)%
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\text{, }\vec{c}=\left(1,1,1\right)$.
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Berechnen Sie:
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\begin{description}
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\item[a.] ds\"{o}lfkj\"{o}dsak
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\item[b.] fsdfsd
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\item[c.]
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\end{description}
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\subsection{L\"{o}sung}
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\begin{list}{\ding{42}}
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{\setlength{\topsep}{0.3cm}
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\setlength{\itemsep}{0.3cm}
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\setlength{\leftmargin}{6mm}
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\setlength{\labelwidth}{4mm}
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\setlength{\parsep}{1mm}
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\setlength{\labelsep}{2mm}
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\renewcommand{\makelabel}[1]{\textbf{#1}}}
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\item[a.]$\underline{\underline{\vec{a}+\vec{b}}}=\left(3,-1,2\right)+\left(1,2,4\right)=\left(3+1,-1+2,2+4\right)=\underline{\underline{\left(4,1,6\right)}}$ \\
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$\underline{\underline{\vec{a}-\vec{b}}}=\left(3,-1,2\right)-\left(1,2,4\right)=\left(3-1,-1-2,2-4\right)=\underline{\underline{\left(2,-3,-2\right)}}$ \\
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$\underline{\underline {4 \cdot \vec a}} = 4 \cdot \left( {3. - 1,2} \right) = \left( {4 \cdot 3,4 \cdot \left( { - 1} \right),4 \cdot 2} \right) = \underline{\underline {\left( {12, - 4,8} \right)}} $\\
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$ \underline{\underline { - \frac{1}{4} \cdot \vec b}} = - \frac{1}{4} \cdot \left( {1,2,4} \right) = \left( {\left( { - \frac{1}{4}} \right) \cdot 1,\left( { - \frac{1}{4}} \right) \cdot 2,\left( { - \frac{1}{4}} \right) \cdot 4} \right) = \underline{\underline {\left( { - \frac{1}{4}, - \frac{1}{2}, - 1} \right)}} $ \\
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$ \underline{\underline { - 5 \cdot \vec c}} = - 5 \cdot \left( {1,1,1} \right) = \underline{\underline {\left( { - 5, - 5, - 5} \right)}} $
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\item[b.]$ \underline{\underline {\left| {\vec a} \right|}} = \left| {\left( {3, - 1,4} \right)} \right| = \sqrt {3^2 + \left( { - 1} \right)^2 + 2^2 } = \sqrt {9 + 1 + 4} = \underline{\underline {\sqrt {14} }} \cong 3.741657387 \\
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\underline{\underline {\left| {\vec b} \right|}} = \left| {\left( {1,2,4} \right)} \right| = \sqrt {1^2 + 2^2 + 4^2 } = \sqrt {1 + 4 + 16} = \underline{\underline {\sqrt {21} }} \cong 4.582575695 \\
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\underline{\underline {\left| {\vec c} \right|}} = \left| {\left( {1,1,1} \right)} \right| = \sqrt {1^2 + 1^2 + 1^2 } = \underline{\underline {\sqrt 3 }} \cong 1.732050808 $
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\item[c.] Sei $ \alpha = \alpha \left( {\vec a,\vec b} \right)$, dann gilt $0 \le \alpha \le \pi$ und
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$\cos \left( \alpha \right) = \frac{{\left\langle {\vec a,\vec b} \right\rangle }}{{\left| {\vec a} \right| \cdot \left| {\vec b} \right|}}\mathop = \limits_{vgl.b.} \frac{{\left\langle {\left( {3, - 1,2} \right),\left( {1,2,4} \right)} \right\rangle }}{{\sqrt {14} \cdot \sqrt {21} }} = \frac{{3 \cdot 1 + \left( { - 1} \right) \cdot 2 + 2 \cdot 4}}{{7 \cdot \sqrt 2 \cdot \sqrt 3 }} = \frac{{3 - 2 + 8}}{{7 \cdot \sqrt 2 \cdot \sqrt 3 }} = \frac{9}{{7 \cdot \sqrt 2 \cdot \sqrt 3 }} $\\
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Der Taschenrechner liefert : $\underline{\underline {\alpha \cong 1.018209678}}\entspricht 58.33911721^ \circ $
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\item[d.]$ \underline{\underline{\vec a \times \vec b }}= \left( {3, - 1,2} \right) \times \left( {1,2,4} \right) = \left( {\left( { - 1} \right) \cdot 4 - 2 \cdot 2,2 \cdot 1 - 4 \cdot 3,3 \cdot 2 - 1 \cdot \left( { - 1} \right)} \right) = \left( { - 4 - 4,2 - 12,6 + 1} \right) = \underline{\underline {\left( { - 8, - 10,7} \right)}}$ \\
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Der gesuchte Flächeninhalt $F$ ist $F = \left| {\vec a \times \vec b} \right| = \sqrt {\left( { - 8} \right)^2 + \left( { - 10} \right)^2 + 7^2 } = \sqrt {64 + 100 + 49} = \underline{\underline {\sqrt {213} }}$ $ \left( { \cong 14.59451952} \right)$
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\item[e.]$ \left[ {\vec a,\vec b,\vec c} \right] = \left\langle {\vec a \times \vec b,\vec c} \right\rangle \mathop = \limits_{vgl.d.} \left\langle {\left( { - 8, - 10,7} \right),\left( {1,1,1} \right)} \right\rangle = - 8 - 10 + 7 = \underline{\underline { - 11}} $ \\
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Das gesuchte Volumen ist $\underline{\underline V} = \left| {\left[ {\vec a,\vec b,\vec c} \right]} \right| = \left| { - 11} \right| = \underline{\underline {11}}$
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\end{list}
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\end{document}
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@@ -0,0 +1,96 @@
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\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}
|
||||
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
|
||||
\newtheorem{algorithm}[theorem]{Algorithm}
|
||||
\newtheorem{axiom}[theorem]{Axiom}
|
||||
\newtheorem{case}[theorem]{Case}
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||||
\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}
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||||
\newtheorem{proposition}[theorem]{Proposition}
|
||||
\newtheorem{remark}[theorem]{Remark}
|
||||
\newtheorem{solution}[theorem]{Solution}
|
||||
\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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||||
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||||
\item[b.] Berechnen Sie $\widetilde{x}$.
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||||
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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:
|
||||
|
||||
\begin{description}
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||||
\item[a.] $\exp(z)=%
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%TCIMACRO{\dsum \limits_{k=0}^{\infty}}%
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||||
%BeginExpansion
|
||||
{\displaystyle\sum\limits_{k=0}^{\infty}}
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||||
%EndExpansion
|
||||
\frac{1}{k!}z^{k}$ \ $\Longrightarrow$ \ \ $x=%
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%TCIMACRO{\dsum \limits_{k=0}^{\infty}}%
|
||||
%BeginExpansion
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||||
{\displaystyle\sum\limits_{k=0}^{\infty}}
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||||
%EndExpansion
|
||||
\frac{1}{k!}(-1)^{k}=\exp(-1)=\underline{\underline{\frac{1}{e}}}$
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||||
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||||
\item[b.] $\widetilde{x}=%
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||||
%TCIMACRO{\dsum \limits_{k=0}^{4}}%
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||||
%BeginExpansion
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||||
{\displaystyle\sum\limits_{k=0}^{4}}
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||||
%EndExpansion
|
||||
\frac{1}{k!}(-1)^{k}=\allowbreak1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}%
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||||
=\frac{12-4+1}{24}=\allowbreak\frac{9}{24}=\frac{3}{8}=\allowbreak
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||||
\underline{\underline{0.375}}\,$
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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 Höchstfehler von $\widetilde{x}$
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||||
\end{description}
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||||
|
||||
|
||||
\end{document}
|
||||
@@ -0,0 +1,173 @@
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\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}
|
||||
%TCIDATA{Version=5.50.0.2953}
|
||||
%TCIDATA{CSTFile=40 LaTeX article.cst}
|
||||
%TCIDATA{Created=Saturday, December 30, 2006 12:55:45}
|
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%TCIDATA{LastRevised=Friday, January 05, 2007 12:09:26}
|
||||
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
|
||||
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
|
||||
%TCIDATA{BibliographyScheme=Manual}
|
||||
%TCIDATA{<META NAME="DocumentShell" CONTENT="Standard LaTeX\Blank - Standard LaTeX Article">}
|
||||
%TCIDATA{ComputeGeneralSettings=0,10,6,0,0,0,0}
|
||||
%BeginMSIPreambleData
|
||||
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
|
||||
%EndMSIPreambleData
|
||||
\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}
|
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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 l\"{o}sbare 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(
|
||||
\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}
|
||||
\item[a.] 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.
|
||||
|
||||
\item[b.] Zeigen Sie, da\ss \ das Gesamtschrittverfahren konvergiert.
|
||||
|
||||
\item[c.] F\"{u}hren Sie eine Apeoteriori-Fehlerabsch\"{a}tzung f\"{u}r
|
||||
$\overrightarrow{x}^{\left( 3\right) }$\ durch.
|
||||
|
||||
\item[d.] F\"{u}hren Sie eine Apriori-Fehlerabsch\"{a}tzung f\"{u}r
|
||||
$\overrightarrow{x}^{\left( 10\right) }$\ durch.
|
||||
\end{itemize}
|
||||
|
||||
L\"{o}sung:
|
||||
|
||||
\begin{itemize}
|
||||
\item[a.] Rechenvorschriften:\newline$x_{1}^{\left( Z\right) }=\frac{1}%
|
||||
{4}\left( 2+x_{2}^{\left( Z-1\right) }+x_{4}^{\left( Z-1\right) }\right)
|
||||
=\frac{1}{2}+\frac{1}{4}x_{2}^{\left( Z-1\right) }+\frac{1}{4}x_{4}^{\left(
|
||||
Z-1\right) }$\newline$x_{2}^{\left( Z\right) }=\frac{1}{4}\left(
|
||||
1+x_{1}^{\left( Z-1\right) }+x_{3}^{\left( Z-1\right) }+x_{5}^{\left(
|
||||
Z-1\right) }\right) =\frac{1}{4}+\frac{1}{4}x_{1}^{\left( Z-1\right)
|
||||
}+\frac{1}{4}x_{3}^{\left( Z-1\right) }+\frac{1}{4}x_{5}^{\left(
|
||||
Z-1\right) }$\newline$x_{3}^{\left( Z\right) }=\frac{1}{4}\left(
|
||||
2+x_{2}^{\left( Z-1\right) }+x_{6}^{\left( Z-1\right) }\right) =\frac
|
||||
{1}{2}+\frac{1}{4}x_{2}^{\left( Z-1\right) }+\frac{1}{4}x_{6}^{\left(
|
||||
Z-1\right) }$\newline$x_{4}^{\left( Z\right) }=\frac{1}{4}\left(
|
||||
2+x_{1}^{\left( Z-1\right) }+x_{5}^{\left( Z-1\right) }\right) =\frac
|
||||
{1}{2}+\frac{1}{4}x_{1}^{\left( Z-1\right) }+\frac{1}{4}x_{5}^{\left(
|
||||
Z-1\right) }$\newline$x_{5}^{\left( Z\right) }=\frac{1}{4}\left(
|
||||
1+x_{2}^{\left( Z-1\right) }+x_{4}^{\left( Z-1\right) }+x_{6}^{\left(
|
||||
Z-1\right) }\right) =\frac{1}{4}+\frac{1}{4}x_{2}^{\left( Z-1\right)
|
||||
}+\frac{1}{4}x_{4}^{\left( Z-1\right) }+\frac{1}{4}x_{6}^{\left(
|
||||
Z-1\right) }$\newline$x_{6}^{\left( Z\right) }=\frac{1}{4}\left(
|
||||
2+x_{3}^{\left( Z-1\right) }+x_{5}^{\left( Z-1\right) }\right) =\frac
|
||||
{1}{2}+\frac{1}{4}x_{3}^{\left( Z-1\right) }+\frac{1}{4}x_{5}^{\left(
|
||||
Z-1\right) }$\newline%
|
||||
\begin{tabular}
|
||||
[c]{l||llllll}%
|
||||
z & $x_{1}^{\left( Z\right) }$ & $x_{2}^{\left( Z\right) }$ &
|
||||
$x_{3}^{\left( Z\right) }$ & $x_{4}^{\left( Z\right) }$ & $x_{5}^{\left(
|
||||
Z\right) }$ & $x_{6}^{\left( Z\right) }$\\\hline\hline
|
||||
1 & $0.5$ & $0.25$ & $0.5$ & $0.5$ & $0.25$ & $0.5$\\
|
||||
2 & $0.6875$ & $0.5625$ & $0.6875$ & $0.6875$ & $0.5625$ & $0.6875$\\
|
||||
3 & $0.8125$ & $0.734325$ & $0.8125$ & $0.8125$ & $0.734375$ & $0.8125$%
|
||||
\end{tabular}
|
||||
\newline
|
||||
|
||||
$\overrightarrow{x}=\overrightarrow{x}^{\left( 3\right) }=\left(
|
||||
0.8125;\text{ }0.734325;\text{ }0.8125;\text{ }0.8125;\text{ }0.734375;\text{
|
||||
}0.8125\right) $
|
||||
|
||||
\item[b.] Berechnung der Kontraktionszahl $\lambda$\newline\newline$%
|
||||
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq1}}^{6}}%
|
||||
%BeginExpansion
|
||||
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq1}}^{6}}
|
||||
%EndExpansion
|
||||
\left\vert \frac{a_{1j}}{a_{11}}\right\vert =0.5;$ \ \ $%
|
||||
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq2}}^{6}}%
|
||||
%BeginExpansion
|
||||
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq2}}^{6}}
|
||||
%EndExpansion
|
||||
\left\vert \frac{a_{2j}}{a_{22}}\right\vert =0.75;$ \ \ $%
|
||||
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq3}}^{6}}%
|
||||
%BeginExpansion
|
||||
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq3}}^{6}}
|
||||
%EndExpansion
|
||||
\left\vert \frac{a_{3j}}{a_{33}}\right\vert =0.5;$\newline\newline$%
|
||||
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq4}}^{6}}%
|
||||
%BeginExpansion
|
||||
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq4}}^{6}}
|
||||
%EndExpansion
|
||||
\left\vert \frac{a_{4j}}{a_{44}}\right\vert =0.5;$ \ \ $%
|
||||
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq5}}^{6}}%
|
||||
%BeginExpansion
|
||||
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq5}}^{6}}
|
||||
%EndExpansion
|
||||
\left\vert \frac{a_{5j}}{a_{55}}\right\vert =0.75;$ \ \ $%
|
||||
%TCIMACRO{\dsum \limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq6}}^{6}}%
|
||||
%BeginExpansion
|
||||
{\displaystyle\sum\limits_{\genfrac{.}{.}{0pt}{1}{j=1}{j\neq6}}^{6}}
|
||||
%EndExpansion
|
||||
\left\vert \frac{a_{6j}}{a_{66}}\right\vert =0.5;$\newline\newline%
|
||||
$\lambda=\max\{0.5;0.75\}=\underline{\underline{0.75}}$\newline$\lambda
|
||||
<1\Longrightarrow$ \underline{das Gesamtschrittverfahren konvergiert} f\"{u}r
|
||||
jeden Startvektor
|
||||
|
||||
\item[c.] $\underset{i=1,...,6}{\max}\left\vert x_{i}^{\left( 3\right)
|
||||
}-x_{i}\right\vert \leq\frac{\lambda}{1-\lambda}$ \ \ \ $\underset
|
||||
{i=1,...,6}{\max}\left\vert x_{i}^{\left( 3\right) }-x_{i}^{\left(
|
||||
2\right) }\right\vert =\frac{0.75}{0.25}\cdot0.171875=\allowbreak
|
||||
\underline{\underline{0.515\,625}}\,$
|
||||
|
||||
\item[d.] $\underset{i=1,...,6}{\max}\left\vert x_{i}^{\left( 10\right)
|
||||
}-x_{i}\right\vert \leq\frac{\lambda^{10}}{1-\lambda}$ \ \ $\underset
|
||||
{i=1,...,6}{\max}\left\vert x_{i}^{\left( 1\right) }-x_{i}^{\left(
|
||||
0\right) }\right\vert =\frac{0.75^{10}}{0.25}\cdot0.5=\underline{\underline{\allowbreak
|
||||
0.112\,627\,029\,\allowbreak4}}$
|
||||
\end{itemize}
|
||||
|
||||
|
||||
\end{document}
|
||||
@@ -0,0 +1,73 @@
|
||||
|
||||
\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}
|
||||
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|
||||
$a=(3,4)$$b=(10,5)$$c=a-\frac{1}{2}b$$c$$|a|$$|b|$$|c|$$(a,c)$$(a,b)$$c=a-\frac{1}{2}b=(3,4)-\frac{1}{2}(10,5)=(3,4)-(10\frac{1}{2},5\frac{1}{2})=(3,4)-(5,\frac{5}{2})=(3-5,4-\frac{5}{2})=(-2,\frac{3}{2})$$|a|=|(3,4)|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$$|b|=|(10,6)|=\sqrt{10^2+5^2}=\sqrt{100+25}=\sqrt{125}=\sqrt{525}=5\sqrt{5}11.18033988$$|c|=|(-2,\frac{3}{2})|=\sqrt{(-2)^2+(\frac{3}{2})^2}=\sqrt{4+\frac{9}{4}}=\sqrt{\frac{16+9}{4}}=\sqrt{\frac{25}{4}}=\frac{5}{2}$$=(a,c)0$$()=\frac{a,c}{|a||c|}%TCIMACRO{{=}}%%BeginExpansion{=}%EndExpansion_vgl.b.\frac{(3,4),(-2,\frac{3}{2})}{5\frac{5}{2}}=\frac{3(-2)+4\frac{3}{2}}{\frac{25}{2}}=\frac{-6+6}{\frac{25}{2}}=0=\frac{}{2}(90^)$$=(a,b)$$0$$()=\frac{a,b}{|a||b|}%TCIMACRO{{=}}%%BeginExpansion{=}%EndExpansion_vgl.b.\frac{(3,4),(10,5)}{55\sqrt{5}}=\frac{310+45}{25\sqrt{5}}=\frac{50}{25\sqrt{5}}=\frac{2}{\sqrt{5}}$$=^-1(\frac{2}{\sqrt{5}})=%TCIMACRO{{arccos}}%%BeginExpansion{arccos}%EndExpansion(\frac{2}{\sqrt{5}})0.463647609(26.565051177078^)$
|
||||
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|
||||
\section{Aufgabe 1}
|
||||
|
||||
Gegeben sind: \hspace{10mm}$\vec a = \left( {3,4} \right)$,\hspace{5mm}$\vec b = \left( {10,5} \right)$,\hspace{5mm}$ \vec c =
|
||||
\vec a - \frac{1} {2}\vec b$
|
||||
|
||||
\begin{list}{\ding{42}}
|
||||
{\setlength{\topsep}{0.3cm}
|
||||
\setlength{\itemsep}{0.3cm}
|
||||
\setlength{\leftmargin}{6mm}
|
||||
\setlength{\labelwidth}{4mm}
|
||||
\setlength{\parsep}{2mm}
|
||||
\setlength{\labelsep}{2mm}
|
||||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||||
\item[a.]Bestimmen Sie $\vec c$ durch Zeichnung und Rechnung!
|
||||
\item[b.]Bestimmen Sie $\left|\vec a\right|$, $\left|\vec b\right|$, $\left|\vec c\right|$.
|
||||
\item[c.]Bestimmen Sie den Öffnungswinkel $\alpha \left( {\vec a,\vec c} \right)$ und $\alpha \left( {\vec a,\vec b} \right)$.
|
||||
\end{list}
|
||||
|
||||
\subsection{Lösung}
|
||||
\subsubsection{Zeichnung}
|
||||
\begin{list}{\ding{42}}
|
||||
{\setlength{\topsep}{0.3cm}
|
||||
\setlength{\itemsep}{0.3cm}
|
||||
\setlength{\leftmargin}{6mm}
|
||||
\setlength{\labelwidth}{4mm}
|
||||
\setlength{\parsep}{2mm}
|
||||
\setlength{\labelsep}{2mm}
|
||||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||||
\item[a.]\includegraphics{Loesung001}
|
||||
\end{list}
|
||||
|
||||
\subsubsection{Rechnung}
|
||||
\begin{list}{\ding{42}}
|
||||
{\setlength{\topsep}{0.3cm}
|
||||
\setlength{\itemsep}{0.3cm}
|
||||
\setlength{\leftmargin}{6mm}
|
||||
\setlength{\labelwidth}{4mm}
|
||||
\setlength{\parsep}{2mm}
|
||||
\setlength{\labelsep}{2mm}
|
||||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||||
\item[a.]$\underline{\underline{\vec c}} = \vec a - \frac{1} {2}\vec b = \left( {3,4} \right) - \frac{1} {2}\left( {10,5}
|
||||
\right) = \left( {3,4} \right) - \left( {10 \cdot \frac{1} {2},5 \cdot \frac{1} {2}} \right) = \left( {3,4} \right) - \left({5,\frac{5}{2}} \right) =\left( {3 - 5,4 - \frac{5} {2}} \right) = \underline{\underline {\left( { - 2,\frac{3} {2}} \right)}}$
|
||||
\item[b.]$\underline{\underline {\left| {\vec a} \right|}} = \left| {\left( {3,4} \right)} \right| = \sqrt {3^2 + 4^2 } = \sqrt {9 +
|
||||
16} = \sqrt {25} = \underline{\underline 5}$\\ \\%
|
||||
$\underline{\underline {\left| {\vec b} \right|}} = \left| {\left( {10,6} \right)} \right| = \sqrt {10^2 + 5^2 } = \sqrt
|
||||
{100 + 25} = \sqrt {125} = \sqrt {5 \cdot 25} = 5\sqrt 5 \cong \underline{\underline {11.18033988}}$\\%
|
||||
$\underline{\underline {\left| {\vec c} \right|}} = \left| {\left( { - 2,\frac{3}{2}} \right)} \right| = \sqrt {\left( { - 2} \right)^2 + \left( {\frac{3}{2}} \right)^2 } = \sqrt {4 + \frac{9}{4}} = \sqrt {\frac{{16 + 9}}{4}} = \sqrt {\frac{{25}}{4}} = \underline{\underline {\frac{5}{2}}} $
|
||||
\item[c.]$\alpha = \alpha \left( {\vec a,\vec c} \right)\text{, }0 \le \alpha \le \pi$\\
|
||||
\vspace{2mm}
|
||||
\hspace{-1.5mm}$\cos \left( \alpha \right) = \frac{{\left\langle {\vec a,\vec c} \right\rangle }}{{\left| {\vec a} \right| \cdot \left| {\vec c} \right|}}\mathop = \limits_{vgl.b.} \frac{{\left\langle {\left( {3,4} \right),\left( { - 2,\frac{3}{2}} \right)} \right\rangle }}{{5 \cdot \frac{5}{2}}} = \frac{{3 \cdot \left( { - 2} \right) + 4 \cdot \frac{3}{2}}}{{\frac{{25}}{2}}} = \frac{{ - 6 + 6}}{{\frac{{25}}{2}}} = 0 \Rightarrow \underline{\underline {\alpha = \frac{\pi }{2}}} \left( { \entspricht 90^ \circ } \right)$
|
||||
|
||||
$\beta = \alpha \left( {\vec a,\vec b} \right)$, $0 \le \beta \le \pi$\\
|
||||
\vspace{2mm}
|
||||
\hspace{-1.5mm}$\cos \left( \beta \right) = \frac{{\left\langle {\vec a,\vec b} \right\rangle }}{{\left| {\vec a} \right| \cdot \left| {\vec b} \right|}}\mathop = \limits_{vgl.b.}\frac{{\left\langle {\left( {3,4} \right),\left( {10,5} \right)} \right\rangle }}{{5 \cdot 5 \cdot \sqrt 5 }} = \frac{{3 \cdot 10 + 4 \cdot 5}}{{25 \cdot \sqrt 5 }} = \frac{{50}}{{25 \cdot \sqrt 5 }} = \frac{2}{{\sqrt 5 }} $.\\
|
||||
Es folgt:\\
|
||||
$\beta = \cos^{-1} \left(\frac{2}{\sqrt 5}\right)= \arccos\left(\frac{2}{\sqrt 5}\right) \cong \underline{\underline{0.463647609}} \left( \entspricht 26.565051177078^\circ \right)$
|
||||
|
||||
\end{list}
|
||||
@@ -0,0 +1,62 @@
|
||||
\section{Aufgabe 1}
|
||||
|
||||
Gegeben sind: \hspace{10mm}$\vec a = \left( {3,4} \right)$,\hspace{5mm}$\vec b = \left( {10,5} \right)$,\hspace{5mm}$ \vec c =
|
||||
\vec a - \frac{1} {2}\vec b$
|
||||
|
||||
\begin{list}{\ding{42}}
|
||||
{\setlength{\topsep}{0.5cm}
|
||||
\setlength{\itemsep}{0.5cm}
|
||||
\setlength{\leftmargin}{6mm}
|
||||
\setlength{\labelwidth}{4mm}
|
||||
\setlength{\parsep}{2mm}
|
||||
\setlength{\labelsep}{2mm}
|
||||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||||
\item[a.]Bestimmen Sie $\vec c$ durch Zeichnung und Rechnung!
|
||||
\item[b.]Bestimmen Sie $\left|\vec a\right|$, $\left|\vec b\right|$, $\left|\vec c\right|$.
|
||||
\item[c.]Bestimmen Sie den Öffnungswinkel $\alpha \left( {\vec a,\vec c} \right)$ und $\alpha \left( {\vec a,\vec b} \right)$.
|
||||
\end{list}
|
||||
|
||||
\subsection{Lösung}
|
||||
\subsubsection{Zeichnung}
|
||||
\begin{list}{\ding{42}}
|
||||
{\setlength{\topsep}{0.5cm}
|
||||
\setlength{\itemsep}{0.5cm}
|
||||
\setlength{\leftmargin}{6mm}
|
||||
\setlength{\labelwidth}{4mm}
|
||||
\setlength{\parsep}{2mm}
|
||||
\setlength{\labelsep}{2mm}
|
||||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||||
\item[a.]\includegraphics{Loesung0001}
|
||||
\end{list}
|
||||
|
||||
\subsubsection{Rechnung}
|
||||
\begin{list}{\ding{42}}
|
||||
{\setlength{\topsep}{0.5cm}
|
||||
\setlength{\itemsep}{0.5cm}
|
||||
\setlength{\leftmargin}{6mm}
|
||||
\setlength{\labelwidth}{4mm}
|
||||
\setlength{\parsep}{2mm}
|
||||
\setlength{\labelsep}{2mm}
|
||||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||||
\item[a.]$\underline{\underline{\vec c}} = \vec a - \frac{1} {2}\vec b = \left( {3,4} \right) - \frac{1} {2}\left( {10,5}
|
||||
\right) = \left( {3,4} \right) - \left( {10 \cdot \frac{1} {2},5 \cdot \frac{1} {2}} \right) = \left( {3,4} \right) - \left({5,\frac{5}{2}} \right) =\left( {3 - 5,4 - \frac{5} {2}} \right) = \underline{\underline {\left( { - 2,\frac{3} {2}} \right)}}$
|
||||
\item[b.]$\underline{\underline {\left| {\vec a} \right|}} = \left| {\left( {3,4} \right)} \right| = \sqrt {3^2 + 4^2 } = \sqrt {9 +
|
||||
16} = \sqrt {25} = \underline{\underline 5}$\\ \\%
|
||||
$\underline{\underline {\left| {\vec b} \right|}} = \left| {\left( {10,6} \right)} \right| = \sqrt {10^2 + 5^2 } = \sqrt
|
||||
{100 + 25} = \sqrt {125} = \sqrt {5 \cdot 25} = 5\sqrt 5 \cong \underline{\underline {11.18033988}}$\\%
|
||||
$\underline{\underline {\left| {\vec c} \right|}} = \left| {\left( { - 2,\frac{3}{2}} \right)} \right| = \sqrt {\left( { - 2} \right)^2 + \left( {\frac{3}{2}} \right)^2 } = \sqrt {4 + \frac{9}{4}} = \sqrt {\frac{{16 + 9}}{4}} = \sqrt {\frac{{25}}{4}} = \underline{\underline {\frac{5}{2}}} $
|
||||
\item[c.]$\alpha = \alpha \left( {\vec a,\vec c} \right)\text{, }0 \le \alpha \le \pi$\\
|
||||
\vspace{2mm}
|
||||
\hspace{-1.5mm}$\cos \left( \alpha \right) = \frac{{\left\langle {\vec a,\vec c} \right\rangle }}{{\left| {\vec a} \right| \cdot \left| {\vec c} \right|}}\mathop = \limits_{vgl.b.} \frac{{\left\langle {\left( {3,4} \right),\left( { - 2,\frac{3}{2}} \right)} \right\rangle }}{{5 \cdot \frac{5}{2}}} = \frac{{3 \cdot \left( { - 2} \right) + 4 \cdot \frac{3}{2}}}{{\frac{{25}}{2}}} = \frac{{ - 6 + 6}}{{\frac{{25}}{2}}} = 0 \Rightarrow \underline{\underline {\alpha = \frac{\pi }{2}}} \left( { \entspricht 90^ \circ } \right)$
|
||||
\\
|
||||
\vspace{5mm}
|
||||
|
||||
$\beta = \alpha \left( {\vec a,\vec b} \right)$, $0 \le \beta \le \pi$\\
|
||||
\vspace{2mm}
|
||||
\hspace{-1.5mm}$\cos \left( \beta \right) = \frac{{\left\langle {\vec a,\vec b} \right\rangle }}{{\left| {\vec a} \right| \cdot \left| {\vec b} \right|}}\mathop = \limits_{vgl.b.}\frac{{\left\langle {\left( {3,4} \right),\left( {10,5} \right)} \right\rangle }}{{5 \cdot 5 \cdot \sqrt 5 }} = \frac{{3 \cdot 10 + 4 \cdot 5}}{{25 \cdot \sqrt 5 }} = \frac{{50}}{{25 \cdot \sqrt 5 }} = \frac{2}{{\sqrt 5 }} $.\\
|
||||
Es folgt:\\
|
||||
$\beta \cos^{-1} \left(\frac{2}{sqrt 5}\right) \cong $
|
||||
|
||||
|
||||
|
||||
\end{list}
|
||||
Binary file not shown.
@@ -0,0 +1,30 @@
|
||||
\section{Aufgabe 2}
|
||||
Welche Gegenkraft $\vec F$ hebt die folgenden vier Einzelkräfte, die an einem Massepunkt angreifen, in der Wirkung
|
||||
auf?\\
|
||||
|
||||
\hspace{10mm}$ \vec F_1 = \left( {200N,110N} \right)$\hspace{10mm}$\vec F_2 = \left( { - 10N,30N} \right)$
|
||||
|
||||
\hspace{10mm}$ \vec F_3 = \left( {40N,85N} \right)$\hspace{14mm}$\vec F_4 = \left( { - 30N, - 50N} \right)$\\ \\
|
||||
Von welchem Betrag ist $\vec{F}$? Unter welchem Winkel greifen $\vec{F_1}$ und $\vec{F_2}$ den Massepunkt an?
|
||||
\subsection{Lösung}
|
||||
$ \overrightarrow F = - \left( {\overrightarrow {F_1 } + \overrightarrow {F_2 } + \overrightarrow {F_3 } + \overrightarrow {F_4 } } \right)\\
|
||||
= - \left( {200N - 10N + 40N - 30N,110N + 30N + 85N - 50N} \right)\\
|
||||
= - \left( {200N,175N} \right) \\
|
||||
= \underline{\underline {\left( { - 200N, - 175N} \right)}}$\\
|
||||
|
||||
$\underline{\underline {\overrightarrow F }} = \sqrt { - \left( {200} \right)^2 + \left( { - 175} \right)^2 } N =\sqrt {\left( {8 \cdot 25} \right)^2 + \left( {7 \cdot 25} \right)^2 } N =25\cdot \sqrt{113}N \entspricht \underline{\underline{265.7536453 N}} $\\
|
||||
\\
|
||||
Der gesuchte Winkel sei $\alpha \text{, } 0\leq\alpha\leq \pi$.
|
||||
\\
|
||||
Dann gilt: $\cos\left(\alpha \right)=\frac{{\left\langle {\overrightarrow {F_1 } ,\overrightarrow {F_2 } } \right\rangle }}{{\left| {\overrightarrow {F_1 } } \right| \cdot \left| {\overrightarrow {F_2 } } \right|}}$. ${\Huge \otimes}$
|
||||
|
||||
$ \left\langle {\vec F_1 ,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
|
||||
\over F} _2 } \right\rangle = \left\langle {\left( {200N,110N} \right),\left( { - 10N,30N} \right)} \right\rangle = - 2000N^2 + 3300N^2 = 1300N^2 $
|
||||
|
||||
$\left| {\vec F_1 } \right| = \sqrt {\left( {200} \right)^2 + \left( {110} \right)^2 } N \cong 228.2542442N$
|
||||
|
||||
$\left| {\vec F_2 } \right| = \sqrt {\left( { - 10} \right)^2 + \left( {30} \right)^2 } N \cong 31.6227766N$\\
|
||||
|
||||
Das Einsetzen in ${\Huge \otimes}$ ergibt:
|
||||
|
||||
$\cos \left( \alpha \right) \cong 0.1801044696 \Rightarrow \arccos \left( \alpha \right) \cong \underline{\underline{1.38970367}} \left( {{\rm 79}{\rm .62415508}^ \circ } \right) $
|
||||
@@ -0,0 +1,21 @@
|
||||
\section{Aufgabe 2}
|
||||
Welche Gegenkraft $\vec F$ hebt die folgenden vier Einzelkräfte, die an einem Massepunkt angreifen, in der Wirkung
|
||||
auf?\\
|
||||
|
||||
\hspace{10mm}$ \vec F_1 = \left( {200N,110N} \right)$\hspace{10mm}$\vec F_2 = \left( { - 10N,30N} \right)$
|
||||
|
||||
\hspace{10mm}$ \vec F_3 = \left( {40N,85N} \right)$\hspace{14mm}$\vec F_4 = \left( { - 30N, - 50N} \right)$\\ \\
|
||||
Von welchem Betrag ist $\vec{F}$? Unter welchem Winkel greifen $\vec{F_1}$ und $\vec{F_2}$ den Massepunkt an?
|
||||
\subsection{Lösung}
|
||||
$ \overrightarrow F = - \left( {\overrightarrow {F_1 } + \overrightarrow {F_2 } + \overrightarrow {F_3 } + \overrightarrow {F_4 } } \right)\\
|
||||
= - \left( {200N - 10N + 40N - 30N,110N + 30N + 85N - 50N} \right)\\
|
||||
= - \left( {200N,175N} \right) \\
|
||||
= \underline{\underline {\left( { - 200N, - 175N} \right)}}$\\
|
||||
|
||||
$\underline{\underline {\overrightarrow F }} = \sqrt { - \left( {200} \right)^2 + \left( { - 175} \right)^2 } N =\sqrt {\left( {8 \cdot 25} \right)^2 + \left( {7 \cdot 25} \right)^2 } N =25\cdot \sqrt{113}N \entspricht \underline{\underline{265.7536453 N}} $\\
|
||||
\\
|
||||
Der gesuchte Winkel sei $\alpha \text{, } 0\leq\alpha\leq \pi$.
|
||||
\\
|
||||
Dann gilt: $\cos\left(\alpha \right)=\frac{{\left\langle {\overrightarrow {F_1 } ,\overrightarrow {F_2 } } \right\rangle }}{{\left| {\overrightarrow {F_1 } } \right| \cdot \left| {\overrightarrow {F_2 } } \right|}}$. ${\Huge\ast}$
|
||||
|
||||
|
||||
Binary file not shown.
@@ -0,0 +1,44 @@
|
||||
\section{Aufgabe 3}
|
||||
Gegeben sind: $\vec{a}=\left(3,-1,2\right)\text{, }\vec{b}=\left(1,2,4\right)\text{, }\vec{c}=\left(1,1,1\right)$.
|
||||
|
||||
Berechnen Sie:
|
||||
\begin{list}{\ding{42}}
|
||||
{\setlength{\topsep}{0.3cm}
|
||||
\setlength{\itemsep}{0.3cm}
|
||||
\setlength{\leftmargin}{6mm}
|
||||
\setlength{\labelwidth}{4mm}
|
||||
\setlength{\parsep}{1mm}
|
||||
\setlength{\labelsep}{2mm}
|
||||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||||
\item[a.]$\vec{a}+\vec{b}\text{, }\vec{a}-\vec{b}\text{, }4\cdot\vec{a}\text{, }-\frac{1}{4}\cdot\vec{b}\text{, }-5\cdot\vec{c}$,
|
||||
\item[b.]$\left|\vec{a}\right|\text{, }\left|\vec{b}\right|\text{, }\left|\vec{c}\right|$,
|
||||
\item[c.]$\alpha\left(\vec{a},\vec{b}\right)$,
|
||||
\item[d.]$\vec{a}\times\vec{b}$ und den Flächeninhalt des von $\vec{a}$ und $\vec{b}$ aufgespannten Parallelogramms,
|
||||
\item[e.]$\left[\vec{a},\vec{b},\vec{c}\right]$ und das Volumen des von $\vec{a},\vec{b}$ und $\vec{c}$ aufgespannten Spats.
|
||||
\end{list}
|
||||
\subsection{Lösung}
|
||||
\begin{list}{\ding{42}}
|
||||
{\setlength{\topsep}{0.3cm}
|
||||
\setlength{\itemsep}{0.3cm}
|
||||
\setlength{\leftmargin}{6mm}
|
||||
\setlength{\labelwidth}{4mm}
|
||||
\setlength{\parsep}{1mm}
|
||||
\setlength{\labelsep}{2mm}
|
||||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||||
\item[a.]$\underline{\underline{\vec{a}+\vec{b}}}=\left(3,-1,2\right)+\left(1,2,4\right)=\left(3+1,-1+2,2+4\right)=\underline{\underline{\left(4,1,6\right)}}$ \\
|
||||
$\underline{\underline{\vec{a}-\vec{b}}}=\left(3,-1,2\right)-\left(1,2,4\right)=\left(3-1,-1-2,2-4\right)=\underline{\underline{\left(2,-3,-2\right)}}$ \\
|
||||
$\underline{\underline {4 \cdot \vec a}} = 4 \cdot \left( {3. - 1,2} \right) = \left( {4 \cdot 3,4 \cdot \left( { - 1} \right),4 \cdot 2} \right) = \underline{\underline {\left( {12, - 4,8} \right)}} $\\
|
||||
$ \underline{\underline { - \frac{1}{4} \cdot \vec b}} = - \frac{1}{4} \cdot \left( {1,2,4} \right) = \left( {\left( { - \frac{1}{4}} \right) \cdot 1,\left( { - \frac{1}{4}} \right) \cdot 2,\left( { - \frac{1}{4}} \right) \cdot 4} \right) = \underline{\underline {\left( { - \frac{1}{4}, - \frac{1}{2}, - 1} \right)}} $ \\
|
||||
$ \underline{\underline { - 5 \cdot \vec c}} = - 5 \cdot \left( {1,1,1} \right) = \underline{\underline {\left( { - 5, - 5, - 5} \right)}} $
|
||||
\item[b.]$ \underline{\underline {\left| {\vec a} \right|}} = \left| {\left( {3, - 1,4} \right)} \right| = \sqrt {3^2 + \left( { - 1} \right)^2 + 2^2 } = \sqrt {9 + 1 + 4} = \underline{\underline {\sqrt {14} }} \cong 3.741657387 \\
|
||||
\underline{\underline {\left| {\vec b} \right|}} = \left| {\left( {1,2,4} \right)} \right| = \sqrt {1^2 + 2^2 + 4^2 } = \sqrt {1 + 4 + 16} = \underline{\underline {\sqrt {21} }} \cong 4.582575695 \\
|
||||
\underline{\underline {\left| {\vec c} \right|}} = \left| {\left( {1,1,1} \right)} \right| = \sqrt {1^2 + 1^2 + 1^2 } = \underline{\underline {\sqrt 3 }} \cong 1.732050808 $
|
||||
\item[c.] Sei $ \alpha = \alpha \left( {\vec a,\vec b} \right)$, dann gilt $0 \le \alpha \le \pi$ und
|
||||
$\cos \left( \alpha \right) = \frac{{\left\langle {\vec a,\vec b} \right\rangle }}{{\left| {\vec a} \right| \cdot \left| {\vec b} \right|}}\mathop = \limits_{vgl.b.} \frac{{\left\langle {\left( {3, - 1,2} \right),\left( {1,2,4} \right)} \right\rangle }}{{\sqrt {14} \cdot \sqrt {21} }} = \frac{{3 \cdot 1 + \left( { - 1} \right) \cdot 2 + 2 \cdot 4}}{{7 \cdot \sqrt 2 \cdot \sqrt 3 }} = \frac{{3 - 2 + 8}}{{7 \cdot \sqrt 2 \cdot \sqrt 3 }} = \frac{9}{{7 \cdot \sqrt 2 \cdot \sqrt 3 }} $\\
|
||||
Der Taschenrechner liefert : $\underline{\underline {\alpha \cong 1.018209678}}\entspricht 58.33911721^ \circ $
|
||||
\item[d.]$ \underline{\underline{\vec a \times \vec b }}= \left( {3, - 1,2} \right) \times \left( {1,2,4} \right) = \left( {\left( { - 1} \right) \cdot 4 - 2 \cdot 2,2 \cdot 1 - 4 \cdot 3,3 \cdot 2 - 1 \cdot \left( { - 1} \right)} \right) = \left( { - 4 - 4,2 - 12,6 + 1} \right) = \underline{\underline {\left( { - 8, - 10,7} \right)}}$ \\
|
||||
Der gesuchte Flächeninhalt $F$ ist $F = \left| {\vec a \times \vec b} \right| = \sqrt {\left( { - 8} \right)^2 + \left( { - 10} \right)^2 + 7^2 } = \sqrt {64 + 100 + 49} = \underline{\underline {\sqrt {213} }}$ $ \left( { \cong 14.59451952} \right)$
|
||||
\item[e.]$ \left[ {\vec a,\vec b,\vec c} \right] = \left\langle {\vec a \times \vec b,\vec c} \right\rangle \mathop = \limits_{vgl.d.} \left\langle {\left( { - 8, - 10,7} \right),\left( {1,1,1} \right)} \right\rangle = - 8 - 10 + 7 = \underline{\underline { - 11}} $ \\
|
||||
Das gesuchte Volumen ist $\underline{\underline V} = \left| {\left[ {\vec a,\vec b,\vec c} \right]} \right| = \left| { - 11} \right| = \underline{\underline {11}}$
|
||||
\end{list}
|
||||
|
||||
Binary file not shown.
@@ -0,0 +1,9 @@
|
||||
\section{Aufgabe 4}
|
||||
Die Kraft $\vec{F}$ mit $\left|\vec{F}\right|=85N$ verschiebt einen Massenpunkt um eine Strecke $\vec s$ mit $\left|\vec{s}\right|=32m$; dabei wird eine Arbeit von $W=1360J$ verrichtet.
|
||||
Unter welchen Winkel greift die Kraft an?
|
||||
|
||||
\subsection{Lösung}
|
||||
Gesucht ist $\alpha = \alpha\left(\vec{F},\vec{a}\right)$. \\
|
||||
Es gilt $0\leq\alpha\leq\pi$ und $\cos\left(\alpha\right)=\frac{\left\langle \vec{F},\vec{s}\right\rangle}{\left| \vec{F} \right| \cdot \left| \vec{s} \right|}=\frac{W}{\left| \vec{F} \right| \cdot \left| \vec{s} \right|}=\frac{1360J}{32 \cdot 85 N\cdot m}=\frac{1360}{2720}=\frac{1}{2}$. \\
|
||||
Es folgt: $\underline{\underline{\alpha=\frac{\pi}{3}}}$ in Bogenmaß bzw. $\underline{\underline{\alpha = 60^ \circ }}$
|
||||
im Gradmaß.
|
||||
Binary file not shown.
@@ -0,0 +1,87 @@
|
||||
\section{Aufgabe 5}
|
||||
Gegeben sind die folgenden Matrizen:\\
|
||||
$A = \left( {\begin{array}{*{20}c}
|
||||
1 & 0 & 2 \\
|
||||
0 & 3 & 0 \\
|
||||
\end{array}} \right) \text{, } B = \left( {\begin{array}{*{20}c}
|
||||
4 & 5 \\
|
||||
0 & 0 \\
|
||||
\end{array}} \right) \text{, } C = \left( {\begin{array}{*{20}c}
|
||||
0 & 1 \\
|
||||
0 & { - 1} \end{array}} \right) \text{, } D = \left( {\begin{array}{*{20}c}
|
||||
1 & 3 \\
|
||||
0 & 2 \\
|
||||
\end{array}} \right) \text{, } F = \left( {\begin{array}{*{20}c}
|
||||
1 & { - \frac{3}{2}} \\
|
||||
0 & {\frac{1}{2}} \\
|
||||
\end{array}} \right) $
|
||||
\begin{list}{\ding{42}}
|
||||
{\setlength{\topsep}{3mm}
|
||||
\setlength{\itemsep}{3mm}
|
||||
\setlength{\leftmargin}{6mm}
|
||||
\setlength{\labelwidth}{4mm}
|
||||
\setlength{\parsep}{2mm}
|
||||
\setlength{\labelsep}{2mm}
|
||||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||||
\item[a.]Berechnen Sie die folgenden Matrizen bzw. begründen Sie das Nichterklärtsein:\\
|
||||
$A+B\text{, }B+C\text{, }C-D\text{, }4\cdot F$
|
||||
\item[b.]Berechnen Sie die folgenden Matrizen bzw. begründen Sie das Nichterklärtsein:\\
|
||||
$B\cdot A\text{, }A\cdot B\text{, }B\cdot C\text{, }C\cdot B$\\
|
||||
\item[c.]Berechnen Sie $D\cdot F$ und $F\cdot D$. Folgern Sie, daß $D$ invertierbar ist und geben Sie $D^{-1}$ an.
|
||||
\item[d.]Berechnen Sie $A^t$ und $B^t$.
|
||||
\end{list}
|
||||
|
||||
\subsection{Lösung}
|
||||
\begin{list}{\ding{42}}
|
||||
{\setlength{\topsep}{3mm}
|
||||
\setlength{\itemsep}{3mm}
|
||||
\setlength{\leftmargin}{6mm}
|
||||
\setlength{\labelwidth}{4mm}
|
||||
\setlength{\parsep}{2mm}
|
||||
\setlength{\labelsep}{2mm}
|
||||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||||
\item[a.]\underline{\underline{$A+B$ ist nicht erklärt}}, da $A$ und $B$ unterschiedliches Format haben.\\
|
||||
$ \underline{\underline {B + C}} = \left( {\begin{array}{cc}
|
||||
4 & 5 \\
|
||||
0 & 0 \\
|
||||
\end{array}} \right) + \left( {\begin{array}{cc}
|
||||
0 & 1 \\
|
||||
0 & { - 1} \\
|
||||
\end{array}} \right) = \left( {\begin{array}{cc}
|
||||
{4 + 0} & {5 + 1} \\
|
||||
{0 + 0} & {0 + \left( { - 9} \right)} \\
|
||||
\end{array}} \right) = \underline{\underline {\left( {\begin{array}{cc}
|
||||
4 & 6 \\
|
||||
0 & { - 1} \\
|
||||
\end{array}} \right)}} $ \\
|
||||
|
||||
$\underline{\underline {C - D}} = \left( {\begin{array}{cc}
|
||||
0 & 1 \\
|
||||
0 & { - 1} \\
|
||||
\end{array}} \right) - \left( {\begin{array}{cc}
|
||||
1 & 3 \\
|
||||
0 & 2 \\
|
||||
\end{array}} \right) = \left( {\begin{array}{cc}
|
||||
{0 - 1} & {1 - 3} \\
|
||||
{0 - 0} & { - 1 - 2} \\
|
||||
\end{array}} \right) = \underline{\underline {\left( {\begin{array}{cc}
|
||||
{ - 1} & { - 2} \\
|
||||
0 & { - 3} \\
|
||||
\end{array}} \right)}} $\\
|
||||
|
||||
$\underline{\underline {4 \cdot F}} = 4 \cdot \left( {\begin{array}{cc}
|
||||
1 & { - \frac{3}{2}} \\
|
||||
0 & {\frac{1}{2}} \\
|
||||
\end{array}} \right) = \left( {\begin{array}{cc}
|
||||
{4 \cdot 1} & {4 \cdot \left( { - \frac{3}{2}} \right)} \\
|
||||
{4 \cdot 0} & {4 \cdot \frac{1}{2}} \\
|
||||
\end{array}} \right) = \underline{\underline {\left( {\begin{array}{cc}
|
||||
4 & { - 6} \\
|
||||
0 & 2 \\
|
||||
\end{array}} \right)}} $
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\end{list}
|
||||
Binary file not shown.
@@ -0,0 +1,53 @@
|
||||
\section{Aufgabe 6}
|
||||
%\begin{floatingfigure}[l]{72mm}
|
||||
%\begin{wrapfigure}[1]{l}{72mm}
|
||||
% \includegraphics{Aufgabe006}
|
||||
%\end{wrapfigure}
|
||||
%\end{floatingfigure}
|
||||
|
||||
\begin{figure}[htbp]
|
||||
\begin{minipage}{0.5\textwidth}
|
||||
\centering
|
||||
\includegraphics{Aufgabe006}
|
||||
\end{minipage}\hfill
|
||||
\begin{minipage}{0.48\textwidth}
|
||||
Gegeben ist der folgende Vierpol:\\
|
||||
\begin{list}{\ding{42}}
|
||||
{\setlength{\topsep}{0.3cm}
|
||||
\setlength{\itemsep}{0.3cm}
|
||||
\setlength{\leftmargin}{6mm}
|
||||
\setlength{\labelwidth}{4mm}
|
||||
\setlength{\parsep}{2mm}
|
||||
\setlength{\labelsep}{2mm}
|
||||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||||
\item[a.]Bestimmen Sie eine Matrix\\$A$ mit $
|
||||
A \cdot \left( {\begin{array}{*{20}c}
|
||||
{I_1 } \\
|
||||
{I_2 } \\
|
||||
\end{array}} \right) = \left( {\begin{array}{*{20}c}
|
||||
{U_1 } \\
|
||||
{U_2 } \\
|
||||
\end{array}} \right)$
|
||||
($A$ heißt Widerstandsmatrix)
|
||||
\item[b.]Berechnen Sie $U_1$ und $U_2$ mit Hilfe von a. für folgende Werte:
|
||||
$R_1 = 10\Omega\text{, }R_2 = 20\Omega \text{, }I_1 = 0.5A\text{, }I_2 = 2A.$
|
||||
\end{list}
|
||||
\end{minipage}
|
||||
\end{figure}
|
||||
|
||||
|
||||
|
||||
\subsection{Lösung}
|
||||
Vorab:
|
||||
\begin{list}{\ding{42}}
|
||||
{\setlength{\topsep}{0.2cm}
|
||||
\setlength{\itemsep}{0.1cm}
|
||||
\setlength{\leftmargin}{6mm}
|
||||
\setlength{\labelwidth}{4mm}
|
||||
\setlength{\parsep}{2mm}
|
||||
\setlength{\labelsep}{2mm}
|
||||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||||
\item[1)]$I_1+I_2-I=0$
|
||||
\item[2)]$U_1-R_2I-R_1I_1=0$
|
||||
\item[1)]$U_2-R_2T-R_1I_2=0$
|
||||
\end{list}
|
||||
Binary file not shown.
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 7}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
Binary file not shown.
@@ -0,0 +1,2 @@
|
||||
\section{Aufgabe 8}
|
||||
xyz
|
||||
@@ -0,0 +1,2 @@
|
||||
\section{Aufgabe 9}
|
||||
xyz
|
||||
@@ -0,0 +1,43 @@
|
||||
\section{Aufgabe 10}
|
||||
%\begin{floatingfigure}[l]{72mm}
|
||||
%\includegraphics{Aufgabe010}
|
||||
%\end{floatingfigure}
|
||||
|
||||
%\begin{wrapfigure}[5]{l}{75mm}
|
||||
% \includegraphics{Aufgabe010}
|
||||
%\end{wrapfigure}
|
||||
|
||||
\begin{figure}[htbp]
|
||||
\begin{minipage}{0.5\textwidth}
|
||||
\centering
|
||||
\includegraphics{Aufgabe010}
|
||||
\end{minipage}\hfill
|
||||
\begin{minipage}{0.48\textwidth}
|
||||
Gegeben ist die folgende Schaltung:\\
|
||||
wobei $U_2=20V\text{, }U_1=-10V\text{, }R_1=6\Omega\text{, }R_2=2\Omega\text{, }R_3=3\Omega\text{.}$\\
|
||||
Stellen Sie ein lineares Gleichungssystem zur Bestimmung von $I_1\text{, }I_2\text{, }I_3$ in Matrixform auf.\\ \\
|
||||
Lösen Sie es mit dem Gaußalgorithmus.
|
||||
\end{minipage}
|
||||
\end{figure}
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
%\parpic[l]{\framebox{\includegraphics{Aufgabe010}}}
|
||||
|
||||
\subsection{Lösung}
|
||||
\begin{list}{\ding{42}}
|
||||
{\setlength{\topsep}{0.2cm}
|
||||
\setlength{\itemsep}{0.1cm}
|
||||
\setlength{\leftmargin}{6mm}
|
||||
\setlength{\labelwidth}{4mm}
|
||||
\setlength{\parsep}{2mm}
|
||||
\setlength{\labelsep}{2mm}
|
||||
\renewcommand{\makelabel}[1]{\textbf{#1}}}
|
||||
\item[Knotengleichung A:]$I_1+I_2-I_3=0$
|
||||
\item[Maschengleichung I:]sdfsdfsdfsdfsd
|
||||
\item[Maschengleichung II:]fsdfsd
|
||||
\end{list}
|
||||
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 11}
|
||||
\includegraphics{Aufgabe011}
|
||||
xyz
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 12}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 13}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,2 @@
|
||||
\section{Aufgabe 14}
|
||||
xyz
|
||||
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 15}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,2 @@
|
||||
\section{Aufgabe 16}
|
||||
xyz
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 17}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 18}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,5 @@
|
||||
\section{Aufgabe 19}
|
||||
\includegraphics{Aufgabe019}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 20}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 21}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 22}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
Binary file not shown.
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 23}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 24}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 25}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 26}
|
||||
\includegraphics{Aufgabe026}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 27}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 28}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 29}
|
||||
\includegraphics{Aufgabe029}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 30}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 31}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,2 @@
|
||||
\section{Aufgabe 32}
|
||||
xyz
|
||||
@@ -0,0 +1,2 @@
|
||||
\section{Aufgabe 33}
|
||||
xyz
|
||||
@@ -0,0 +1,2 @@
|
||||
\section{Aufgabe 34}
|
||||
xyz
|
||||
@@ -0,0 +1,2 @@
|
||||
\section{Aufgabe 35}
|
||||
xyz
|
||||
@@ -0,0 +1,2 @@
|
||||
\section{Aufgabe 36}
|
||||
xyz
|
||||
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 37}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 38}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 39}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,3 @@
|
||||
\section{Aufgabe 40}
|
||||
xyz
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 41}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 42}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 43}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 44}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 45}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,5 @@
|
||||
\section{Aufgabe 46}
|
||||
xyz
|
||||
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 47}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 48}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 49}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 50}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 51}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 52}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 53}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 54}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
@@ -0,0 +1,3 @@
|
||||
xœãf``õsñ÷ñ Ñ30bdòA˜ ˆ… @J3±ñ `Æ!i
|
||||
’d„
|
||||
€‰ Ñ1Å¥IÅ©É%™ùyÕ>ߊKóÒk@ê?/Ÿ
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 55}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
Binary file not shown.
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 56}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
Binary file not shown.
@@ -0,0 +1,2 @@
|
||||
\section{Aufgabe 57}
|
||||
xyz
|
||||
@@ -0,0 +1,2 @@
|
||||
xœãf``õsñ÷ñ Ñ30bdòaX�„¡4HÌØä?0#‰
|
||||
"Éššƒd�å /
|
||||
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 58}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
Binary file not shown.
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 59}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
Binary file not shown.
@@ -0,0 +1,4 @@
|
||||
\section{Aufgabe 60}
|
||||
xyz
|
||||
|
||||
\subsection{Lösung}
|
||||
Some files were not shown because too many files have changed in this diff Show More
Reference in New Issue
Block a user