Taylor-Formel/Quadratische Approximation und Hesse-Form/Aufgabe/Lösung

Es ist einerseits

{}{\begin{aligned}\sum _{r\in \mathbb {N} ^{n},\,\vert {\,r\,}\vert =2}{\frac {1}{r!}}D^{r}f(P)\cdot v^{r}&=\sum _{r=(0,\ldots ,0,1,0,\ldots ,0,1,0,\ldots ,0)}{\frac {1}{r!}}D^{r}f(P)\cdot v^{r}+\sum _{r=(0,\ldots ,0,2,0,\ldots ,0)}{\frac {1}{r!}}D^{r}f(P)\cdot v^{r}\\&=\sum _{1\leq i<j\leq n}D_{i}D_{j}f(P)\cdot v_{i}v_{j}+{\frac {1}{2}}\sum _{1\leq i\leq n}D_{i}D_{i}f(P)\cdot v_{i}^{2}.\end{aligned}}

Andererseits ist

{}{\begin{aligned}\operatorname {Hess} _{P}\,f(v,v)&=\left(v_{1},\,\ldots ,\,v_{n}\right){\left({\left(D_{i}D_{j}f(P)\right)}_{i,j}\right)}{\begin{pmatrix}v_{1}\\\vdots \\v_{n}\end{pmatrix}}\\&=\left(v_{1},\,\ldots ,\,v_{n}\right){\begin{pmatrix}\sum _{j=1}^{n}D_{1}D_{j}f(P)v_{j}\\\sum _{j=1}^{n}D_{2}D_{j}f(P)v_{j}\\\vdots \\\sum _{j=1}^{n}D_{n}D_{j}f(P)v_{j}\end{pmatrix}}\\&=\sum _{i=1}^{n}{\left(\sum _{j=1}^{n}D_{i}D_{j}f(P)v_{j}\right)}v_{i}\\&=\sum _{(i,j)}D_{i}D_{j}f(P)v_{i}v_{j}\\&=\sum _{(i,j),\,i\neq j}D_{i}D_{j}f(P)v_{i}v_{j}+\sum _{(i,i)}D_{i}D_{i}f(P)v_{i}v_{i}\\&=\sum _{(i,j),\,i<j}2D_{i}D_{j}f(P)v_{i}v_{j}+\sum _{i=1}^{n}D_{i}D_{i}f(P)v_{i}^{2}.\end{aligned}}

Mit Hinzunahme des Faktors

{}{\frac {1}{2}}

stimmen die beiden Ausdrücke überein.