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@@ -18,7 +18,7 @@ |
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Es ist |
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\begin{salign*} |
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F(x) &= \frac{1}{2} \sum_{i,j=1}^{n} a_{ij}x_j x_i - \sum_{i=1}^{n} b_i x_i |
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\intertext{Damit folgt} |
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\intertext{Da $A$ symmetrisch, gilt $a_{ij} = a_{ji}$. Damit folgt} |
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\frac{\partial F}{\partial x_i} &= \frac{1}{2} \left(2 \sum_{j=1, i\neq j}^{n} a_{ij}x_j |
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+ 2 a_{ii} x_i \right) - b_i \\ |
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&= \sum_{j=1}^{n} a_{ij} x_j - b_i \\ |
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@@ -183,10 +183,10 @@ |
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\[ |
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\frac{\partial F}{\partial x_j} = \sum_{i=1}^{n} 2 f_i(x) \frac{\partial f_i}{\partial x_j} |
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= 2 \left( f(x), (J_f(x))_{j-\text{te Spalte}} \right)_2 |
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\implies \nabla F = 2 J_f(x^{k}) f(x^{k}) |
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\implies \nabla F = 2 J_f(x^{k})^{T} f(x^{k}) |
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.\] Damit folgt |
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\[ |
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x^{k+1} = x^{k} - 2 \alpha_{opt} J_f(x^{k}) f(x^{k}) |
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x^{k+1} = x^{k} - 2 \alpha_{opt} J_f(x^{k})^{T} f(x^{k}) |
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.\] |
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\item Newton-Verfahren: |
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\[ |
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