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Peters Änderungsvorschläge eingepflegt
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Martin Thoma
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10 changed files with 168 additions and 155 deletions
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@ -13,19 +13,24 @@
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\textbf{Lösung:}
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\[P =
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\begin{pmatrix}
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0 & 1 & 0 \\
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1 & 0 & 0 \\
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0 & 0 & 1
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\end{pmatrix}\]
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durch scharfes hinsehen.
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Nun $L, R$ berechnen:
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\begin{align}
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&\begin{gmatrix}[p]
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\begin{align*}
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&
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&
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A^{(0)} &= \begin{gmatrix}[p]
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3 & 15 & 13 \\
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6 & 6 & 6 \\
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2 & 8 & 19
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\rowops
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\swap{0}{1}
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\end{gmatrix}
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&\\
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P^{(1)} &= \begin{pmatrix}
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0 & 1 & 0\\
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1 & 0 & 0\\
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0 & 0 & 1
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\end{pmatrix},
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&
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A^{(1)} &= \begin{gmatrix}[p]
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6 & 6 & 6 \\
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3 & 15 & 13 \\
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2 & 8 & 19
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@ -33,128 +38,48 @@ Nun $L, R$ berechnen:
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\add[\cdot (-\frac{1}{2})]{0}{1}
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\add[\cdot (-\frac{1}{3})]{0}{2}
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\end{gmatrix}
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\\
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= \begin{pmatrix}
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1 & 0 & 0 \\
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-\frac{1}{2} & 1 & 0 \\
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-\frac{1}{3} & 0 & 1
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\end{pmatrix} \cdot
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&\begin{gmatrix}[p]
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&\\
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L^{(2)} &= \begin{pmatrix}
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1 & 0 & 0\\
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-\frac{1}{2} & 1 & 0\\
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-\frac{1}{3} & 0 & 1
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\end{pmatrix},
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&
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A^{(2)} &= \begin{gmatrix}[p]
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6 & 6 & 6 \\
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0 & 12 & 10 \\
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0 & 6 & 17
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\rowops
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\add[\cdot (-\frac{1}{2})]{1}{2}
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\end{gmatrix}
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\\
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= \begin{pmatrix}
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & -\frac{1}{2} & 1
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\end{pmatrix} \cdot
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\begin{pmatrix}
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1 & 0 & 0 \\
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-\frac{1}{2} & 1 & 0 \\
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-\frac{1}{3} & 0 & 1
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\end{pmatrix} \cdot
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&\begin{gmatrix}[p]
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&\\
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L^{(3)} &= \begin{pmatrix}
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1 & 0 & 0\\
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0 & 1 & 0\\
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0 & -\frac{1}{2} & 1
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\end{pmatrix},
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&
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A^{(3)} &= \begin{gmatrix}[p]
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6 & 6 & 6 \\
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0 & 12 & 10 \\
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0 & 0 & 12
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\colops
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\add[\cdot (-1)]{0}{1}
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\add[\cdot (-1)]{0}{2}
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\end{gmatrix}
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\\
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= \begin{pmatrix}
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1 & 0 & 0 \\
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-\frac{1}{2} & 1 & 0 \\
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-\frac{1}{12} & - \frac{1}{2} & 1
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\end{pmatrix} \cdot
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&\begin{gmatrix}[p]
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6 & 0 & 0 \\
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0 & 12 & 10 \\
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0 & 0 & 12
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\colops
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\add[\cdot (-\frac{10}{12})]{1}{2}
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\end{gmatrix}
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\cdot
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\begin{pmatrix}
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1 & -1 & -1 \\
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0 & 1 & 0 \\
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0 & 0 & 1
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\end{align*}
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Es gilt:
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\begin{align}
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L^{(3)} \cdot L^{(2)} \cdot \underbrace{P^{(1)}}_{=: P} \cdot A^{0} &= \underbrace{A^{(3)}}_{=: R}\\
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\Leftrightarrow P A &= (L^{(3)} \cdot L^{(2)})^{-1} \cdot R \\
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\Rightarrow L &= (L^{(3)} \cdot L^{(2)})^{-1}\\
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&= \begin{pmatrix}
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1 & 0 & 0\\
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\frac{1}{2} & 1 & 0\\
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\frac{1}{3} & \frac{1}{2} & 1
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\end{pmatrix}
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\\
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= \begin{pmatrix}
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1 & 0 & 0 \\
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-\frac{1}{2} & 1 & 0 \\
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-\frac{1}{12} & - \frac{1}{2} & 1
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\end{pmatrix} \cdot
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&\begin{gmatrix}[p]
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6 & 0 & 0 \\
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0 & 12 & 0 \\
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0 & 0 & 12
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\colops
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\mult{0}{\cdot \frac{1}{6}}
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\mult{1}{\cdot \frac{1}{12}}
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\mult{2}{\cdot \frac{1}{12}}
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\end{gmatrix}
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\cdot
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\begin{pmatrix}
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1 & -1 & -1 \\
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0 & 1 & 0 \\
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0 & 0 & 1
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\end{pmatrix}
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\cdot
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\begin{pmatrix}
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1 & 0 & 0 \\
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0 & 1 & -\frac{10}{12} \\
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0 & 0 & 1
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\end{pmatrix}
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\\
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= \begin{pmatrix}
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1 & 0 & 0 \\
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-\frac{1}{2} & 1 & 0 \\
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-\frac{1}{12} & - \frac{1}{2} & 1
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\end{pmatrix} \cdot
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&\begin{gmatrix}[p]
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1
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\end{gmatrix}
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\cdot
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\begin{pmatrix}
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1 & -1 & -\frac{1}{6} \\
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0 & 1 & -\frac{5}{6} \\
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0 & 0 & 1
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\end{pmatrix}
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\cdot
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\begin{pmatrix}
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\frac{1}{6} & 0 & 0 \\
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0 & \frac{1}{12} & 0 \\
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0 & 0 & \frac{1}{12}
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\end{pmatrix}
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\\
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= \underbrace{\begin{pmatrix}
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1 & 0 & 0 \\
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-\frac{1}{2} & 1 & 0 \\
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-\frac{1}{12} & - \frac{1}{2} & 1
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\end{pmatrix}}_L \cdot
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&\begin{gmatrix}[p]
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1
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\end{gmatrix}
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\cdot \underbrace{\frac{1}{72}
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\begin{pmatrix}
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12 & -6 & -1 \\
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0 & 6 & -5 \\
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0 & 0 & 6
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\end{pmatrix}}_R
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\end{align}
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ACHTUNG: Ich habe mich irgendwo verrechnet!
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Siehe \href{http://www.wolframalpha.com/input/?i=%7B%7B1%2C0%2C0%7D%2C%7B-1%2F2%2C1%2C0%7D%2C%7B-1%2F12%2C-1%2F2%2C1%7D%7D*%7B%7B12%2C-6%2C-1%7D%2C%7B0%2C6%2C-5%7D%2C%7B0%2C0%2C6%7D%7D}{WolframAlpha}
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Nun gilt: $P A = L R = A^{(1)}$ (Kontrolle mit \href{http://www.wolframalpha.com/input/?i=%7B%7B1%2C0%2C0%7D%2C%7B0.5%2C1%2C0%7D%2C%7B1%2F3%2C0.5%2C1%7D%7D*%7B%7B6%2C6%2C6%7D%2C%7B0%2C12%2C10%7D%2C%7B0%2C0%2C12%7D%7D}{Wolfram|Alpha})
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\subsection*{Teilaufgabe b}
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@ -182,7 +107,7 @@ Falls $A$ symmetrisch ist, gilt:
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& \Leftrightarrow \text{es gibt eine Cholesky-Zerlegung $A=GG^T$ mit $G$ ist reguläre untere Dreiecksmatrix}\\
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\end{align*}
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Mit dem Hauptminor-Kriterium gilt:
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\subsubsection*{Lösung 1: Hauptminor-Kriterium}
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\begin{align}
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\det(A_1) &= 9 > 0\\
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@ -193,3 +118,15 @@ Mit dem Hauptminor-Kriterium gilt:
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\end{vmatrix} = 9 - 16 < 0\\
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&\Rightarrow \text{$A$ ist nicht positiv definit}
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\end{align}
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\subsubsection*{Lösung 2: Cholesky-Zerlegung}
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\begin{align}
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l_{11} &= \sqrt{a_{11}} = 3\\
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l_{21} &= \frac{a_{21}}{l_{11}} = \frac{4}{3}\\
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l_{31} &= \frac{a_{31}}{l_{11}} = 4\\
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l_{22} &= \sqrt{a_{21} - {l_{21}}^2} = \frac{2 \sqrt{5}}{3}\\
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\dots
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\end{align}
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ACHTUNG: Noch nicht fertig! Irgendwo muss was negatives unter einer
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Wurzel kommen!
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@ -7,43 +7,35 @@ wobei $L$ eine invertierbare, untere Dreiecksmatrix ist.
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Geben Sie die Formel zur Berechnung von $y_i$ an.
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\textbf{Lösung:} TODO! %TODO!
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\textbf{Lösung:}
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\[y_i = \frac{b_i - \sum_{k=i}^{i-1} l_{ik} \cdot y_k}{l_{ii}}\]
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\subsection*{Teilaufgabe b}
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\[Ax = b ? PAx = Pb ? LRx = Pb \]
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Pseudocode:
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\begin{algorithm}[H]
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\begin{algorithmic}
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\Require $p \in \mathbb{P}, a \in \mathbb{Z}, p \geq 3$
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\Procedure{CalculateLegendre}{$a$, $p$}
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\If{$a \geq p$ or $a < 0$}\Comment{rule (III)}
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\State \Return $\Call{CalculateLegendre}{a \mod p, p}$ \Comment{now: $a \in [0, \dots, p-1]$}
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\ElsIf{$a == 0$ or $a == 1$}
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\State \Return $a$ \Comment{now: $a \in [2, \dots, p-1]$}
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\ElsIf{$a == 2$} \Comment{rule (VII)}
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\If{$p \equiv \pm 1 \mod 8$}
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\State \Return 1
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\Else
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\State \Return -1
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\EndIf \Comment{now: $a \in [3, \dots, p-1]$}
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\ElsIf{$a == p-1$} \Comment{rule (VI)}
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\If{$p \equiv 1 \mod 4$}
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\State \Return 1
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\Else
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\State \Return -1
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\EndIf \Comment{now: $a \in [3, \dots, p-2]$}
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\ElsIf{!$\Call{isPrime}{a}$} \Comment{rule (II)}
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\State $p_1, p_2, \dots, p_n \gets \Call{Factorize}{a}$
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\State \Return $\prod_{i=1}^n \Call{CalculateLegendre}{p_i, p}$
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\Else \Comment{now: $a \in \mathbb{P}, \sqrt{p-2} \geq a \geq 3$}
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\If{$\frac{p-1}{2} \equiv 0 \mod 2$ or $\frac{a-1}{2} \equiv 0 \mod 2$}
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\State \Return $\Call{CalculateLegendre}{p, a}$
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\Else
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\State \Return $(-1) \cdot \Call{CalculateLegendre}{p, a}$
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\EndIf
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\EndIf
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\Require Matrix $A$, Vektor $b$
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\Procedure{CalculateLegendre}{$A$, $b$}
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\State $P, L, R \gets \Call{LRZerlegung}{A}$
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\State $b^* \gets Pb$
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\State $c \gets \Call{VorwärtsSubstitution}{L, b^*}$
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\State $x \gets \Call{RückwärtsSubstitution}{R, c}$
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\State \Return $x$
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\EndProcedure
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\end{algorithmic}
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\caption{Calculate Legendre symbol}
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\label{alg:calculateLegendreSymbol}
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\caption{Calculate TODO}
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\label{alg:TODO}
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\end{algorithm}
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\subsection*{Teilaufgabe b}
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\subsection*{Teilaufgabe c}
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Der Gesamtaufwand ist:
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\begin{itemize}
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\item LR-Zerlegung, $\frac{1}{3}n^3 - \frac{1}{3} n^2$
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\item Vektormultiplikation, $2n$
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\item Vorwärtssubstitution, $\frac{1}{2} n^2$
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\item Rückwärtssubstitution, $\frac{1}{2} n^2$
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\end{itemize}
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@ -1 +1,20 @@
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\section*{Aufgabe 3}
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\[f' (x,y) = \begin{pmatrix}
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3 & \cos y\\
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3 x^2 & e^y
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\end{pmatrix}\]
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Und jetzt die Berechnung
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\[f'(x, y) \cdot (x_0, y_0) = f(x,y)\]
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LR-Zerlegung für $f'(x, y)$:
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\begin{align}
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L &= \begin{pmatrix}1 &0 \\ \frac{1}{9} & 1\end{pmatrix}\\
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R &= \begin{pmatrix}3 & \cos y \\ 0 & e^y - x^2 \cos y\end{pmatrix}\\
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P &= I_2\\
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-f ( \frac{-1}{3}, 0) &= \begin{pmatrix} 2\\ -\frac{1}{27}\end{pmatrix}\\
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c &= \begin{pmatrix} 2\\ \frac{7}{27} \end{pmatrix}\\
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(x_1, y_1) &= \begin{pmatrix} \frac{5}{3}\\ \frac{7}{27}\end{pmatrix}
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\end{align}
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@ -1 +1,7 @@
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\section*{Aufgabe 5}
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\subsection*{Teilaufgabe a}
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Die Ordnung $n$ einer Quadraturformel gibt an, dass diese Polynome
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bis zum Grad $\leq n - 1$ exakt löst.
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\subsection*{Teilaufgabe b}
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\subsection*{Teilaufgabe c}
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