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christian
2020-06-26 21:57:05 +02:00
parent 0c6d5e130e
commit 154cef4ed6
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sose2020/num/uebungen/num8.pdf
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sose2020/num/uebungen/num8.tex
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@@ -18,7 +18,7 @@
Es ist Es ist
\begin{salign*} \begin{salign*}
F(x) &= \frac{1}{2} \sum_{i,j=1}^{n} a_{ij}x_j x_i - \sum_{i=1}^{n} b_i x_i F(x) &= \frac{1}{2} \sum_{i,j=1}^{n} a_{ij}x_j x_i - \sum_{i=1}^{n} b_i x_i
\intertext{Damit folgt} \intertext{Da $A$ symmetrisch, gilt $a_{ij} = a_{ji}$. Damit folgt}
\frac{\partial F}{\partial x_i} &= \frac{1}{2} \left(2 \sum_{j=1, i\neq j}^{n} a_{ij}x_j \frac{\partial F}{\partial x_i} &= \frac{1}{2} \left(2 \sum_{j=1, i\neq j}^{n} a_{ij}x_j
+ 2 a_{ii} x_i \right) - b_i \\ + 2 a_{ii} x_i \right) - b_i \\
&= \sum_{j=1}^{n} a_{ij} x_j - b_i \\ &= \sum_{j=1}^{n} a_{ij} x_j - b_i \\
@@ -183,10 +183,10 @@
\[ \[
\frac{\partial F}{\partial x_j} = \sum_{i=1}^{n} 2 f_i(x) \frac{\partial f_i}{\partial x_j} \frac{\partial F}{\partial x_j} = \sum_{i=1}^{n} 2 f_i(x) \frac{\partial f_i}{\partial x_j}
= 2 \left( f(x), (J_f(x))_{j-\text{te Spalte}} \right)_2 = 2 \left( f(x), (J_f(x))_{j-\text{te Spalte}} \right)_2
\implies \nabla F = 2 J_f(x^{k}) f(x^{k}) \implies \nabla F = 2 J_f(x^{k})^{T} f(x^{k})
.\] Damit folgt .\] Damit folgt
\[ \[
x^{k+1} = x^{k} - 2 \alpha_{opt} J_f(x^{k}) f(x^{k}) x^{k+1} = x^{k} - 2 \alpha_{opt} J_f(x^{k})^{T} f(x^{k})
.\] .\]
\item Newton-Verfahren: \item Newton-Verfahren:
\[ \[