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@ -7,6 +7,103 @@ Die Jacobi-Matrix von $f$ lautet:
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Hierfür wurde in in der ersten Spalte nach $x$ abgeleitet und in der
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zweiten Spalte nach $y$.
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\subsection*{Lösungsvorschlag 1}
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Laut Skript ist eine Iteration gegeben durch:
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\begin{align}
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x_{k+1}&=x_{k}-f'(x_k)^{-1}\cdot f(x_k)
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\end{align}
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Zur praktischen Durchführung Lösen wir
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\[f'(x_0, y_0)\Delta x = -f(x_0,y_0)\]
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mit Hilfe der LR Zerlegung:
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\begin{align}
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%
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f'(x_0,y_0) &= L \cdot R \\
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\Leftrightarrow \begin{pmatrix}
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3 & 1\\
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\frac{1}{3} & 1
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\end{pmatrix}
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&=
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\overbrace{\begin{pmatrix}
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1 & 0\\
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\frac{1}{9} & 1
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\end{pmatrix}}^L \cdot
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\overbrace{\begin{pmatrix}
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3 & 1\\
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0 & \frac{8}{9}
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\end{pmatrix}}^R\\
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%
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L \cdot c &= -f(x_0,y_0) \\
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\Leftrightarrow
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\begin{pmatrix}
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1 & 0\\
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\frac{1}{9} & 1
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\end{pmatrix}
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\cdot c
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&=
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\begin{pmatrix}
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-2\\
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\frac{26}{27}
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\end{pmatrix}\\
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\Rightarrow
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c &= \begin{pmatrix}
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-2\\
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\frac{32}{27}
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\end{pmatrix}\\
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%
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R\cdot \Delta x &= c\\
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\Leftrightarrow
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\begin{pmatrix}
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3 & 1\\
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0 & \frac{8}{9}
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\end{pmatrix}
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\cdot \Delta x &=
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\begin{pmatrix}
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-2\\
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\frac{32}{27}
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\end{pmatrix}\\
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\Rightarrow \Delta x &=
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\begin{pmatrix}
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-\frac{10}{9}\\
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\frac{4}{3}
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\end{pmatrix}
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\end{align}
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Anschließend berechnen wir
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\begin{align}
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\begin{pmatrix}
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x_1\\
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y_1
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\end{pmatrix} &=
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\begin{pmatrix}
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x_0\\
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y_0
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\end{pmatrix}+\Delta x \\
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\Leftrightarrow\begin{pmatrix}
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x_1\\
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y_1
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\end{pmatrix} &=
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\begin{pmatrix}
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\frac{1}{3}\\
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0
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\end{pmatrix} +
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\begin{pmatrix}
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-\frac{10}{9}\\
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\frac{4}{3}
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\end{pmatrix} \\
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\Leftrightarrow\begin{pmatrix}
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x_1\\
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y_1
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\end{pmatrix} &=
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\begin{pmatrix}
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-\frac{7}{9}\\
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\frac{4}{3}
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\end{pmatrix}
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\end{align}
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\subsection*{Lösungsvorschlag 2}
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Und jetzt die Berechnung %TODO: Was ist hiermit gemeint?
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\[f'(x, y) \cdot (x_0, y_0) = f(x,y)\] %TODO: Was ist hiermit gemeint?
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