Benutzer:Digamma/Taylor-Formel

Es soll die Funktion

$f:\mathbb {R} ^{2}\to \mathbb {R} ,~x\mapsto \exp(x_{1}-x_{2})\cdot \log(1-x_{2})$

mit $x=(x_{1},x_{2})\in \mathbb {R} ^{2}$ um den Punkt $a=(a_{1},a_{2})=(1,0)\in \mathbb {R} ^{2}$ entwickelt werden.

In diesem Beispiel soll die Funktion bis zum zweiten Grad entwickelt werden. Es gilt also $k=2$. Wegen $|j|\leq k$ müssen, gemäß der Multiindexschreibweise die Tupel $(0,0)$, $(1,0)$, $(0,1)$, $(2,0)$, $(1,1)$ und $(0,2)$ berücksichtigt werden.

Die partiellen Ableitungen der Funktion lauten:

${\frac {\partial f}{\partial x_{1}}}(a)=\left[\exp(x_{1}-x_{2})\cdot \log(1-x_{2})\right]_{x=(1,0)}=0$

${\frac {\partial f}{\partial x_{2}}}(a)=\left[-\exp(x_{1}-x_{2})\cdot \left(\log(1-x_{2})+{\frac {1}{1-x_{2}}}\right)\right]_{x=(1,0)}=-e$

${\frac {\partial ^{2}f}{\partial x_{1}^{2}}}(a)=\left[\exp(x_{1}-x_{2})\cdot \log(1-x_{2})\right]_{x=(1,0)}=0$

${\frac {\partial ^{2}f}{\partial x_{1}\partial x_{2}}}(a)=\left[-\exp(x_{1}-x_{2})\cdot \left(\log(1-x_{2})+{\frac {1}{1-x_{2}}}\right)\right]_{x=(1,0)}=-e$

${\frac {\partial ^{2}f}{\partial x_{2}}}(a)=\left[\exp(x_{1}-x_{2})\left(\log(1-x_{2})+{\frac {2}{1-x_{2}}}-{\frac {1}{(1-x_{2})^{2}}}\right)\right]_{x=(1,0)}=e$

Es folgt mit der mehrdimensionalen Taylor-Formel:

${\begin{aligned}f(x)\approx &f(a)+{\frac {1}{1!}}{\frac {\partial f}{\partial x_{1}}}(a)~(x_{1}-a_{1})+{\frac {1}{1!}}{\frac {\partial f}{\partial x_{2}}}(a)~(x_{2}-a_{2})\\&+{\frac {1}{2!}}{\frac {\partial ^{2}f}{\partial x_{1}^{2}}}(a)~(x_{1}-a_{1})^{2}+{\frac {1}{1!1!}}{\frac {\partial ^{2}f}{\partial x_{1}\partial x_{2}}}(a)~(x_{1}-a_{1})(x_{2}-a_{2})\\&+{\frac {1}{2!}}{\frac {\partial ^{2}f}{\partial x_{2}^{2}}}(a)~(x_{2}-a_{2})^{2}\\&=0+0-e(x_{2}-0)+0-e(x_{1}-1)(x_{2}-0)+{\frac {1}{2}}e(x_{2}-0)^{2}\\&=-x_{1}x_{2}e+{\frac {1}{2}}x_{2}^{2}e\end{aligned}}$

Eine alternative Darstellung der Taylorentwicklung basiert nicht auf der Multiindexdarstellung, sondern auf einer Matizendarstellung der Taylor-Formel. So gilt auch:

${\begin{aligned}f(x)&\approx f(a)+J_{f}(a)(x-a)+{\frac {1}{2}}(x-a)^{T}H_{f}(a)(x-a)\\&=f(a)+{\begin{pmatrix}{\frac {\partial f}{\partial x_{1}}}(a)&{\frac {\partial f}{\partial x_{2}}}(a)\end{pmatrix}}{\begin{pmatrix}x_{1}-a_{1}\\x_{2}-a_{2}\end{pmatrix}}\\&\qquad +{\frac {1}{2}}{\begin{pmatrix}x_{1}-a_{1}&x_{2}-a_{2}\end{pmatrix}}{\begin{pmatrix}{\frac {\partial ^{2}f}{\partial x_{1}^{2}}}(a)&{\frac {\partial ^{2}f}{\partial x_{2}\partial x_{1}}}(a)\\{\frac {\partial ^{2}f}{\partial x_{1}\partial x_{2}}}(a)&{\frac {\partial ^{2}f}{\partial x_{2}^{2}}}(a)\end{pmatrix}}{\begin{pmatrix}x_{1}-a_{1}\\x_{2}-a_{2}\end{pmatrix}}\\&=0+{\begin{pmatrix}0&-e\end{pmatrix}}{\begin{pmatrix}x_{1}-1\\x_{2}\end{pmatrix}}+{\frac {1}{2}}{\begin{pmatrix}x_{1}-1&x_{2}\end{pmatrix}}{\begin{pmatrix}0&-e\\-e&e\end{pmatrix}}{\begin{pmatrix}x_{1}-1\\x_{2}\end{pmatrix}}\\&=-x_{1}x_{2}e+{\frac {1}{2}}x_{2}^{2}e\end{aligned}}$

mit der Jacobi-Matrix $J_{f}$ und der Hesse-Matrix $H_{f}$.