I want to compute second order Taylor polynomial at point $p(1,1)$ for function: $$ f(x,y)=xe^{-xy^2} $$ I can compute gradient and hessian matrix for required derivatives of the function so forming Taylor polynomial will be possible.
$$ \nabla f(1,1) = \begin{bmatrix} 0 \\ -2e^{-1} \end{bmatrix}$$
$$ H(1,1)= \begin{bmatrix} -e^{-1} & -2e^{-1} \\ 2e^{-1} & 2e^{-1} \end{bmatrix} $$
Second degree Taylor polynomial should be
$$ p_2(x,y)=f(1,1)+f_x(1,1)(x-1)+f_y(1,1)(y-1)+\frac{1}{2!} \left[\begin{array}sf_{xx}(1,1)(x-1)^2+2f_{xy}(1,1)(x-1)(y-1)+f_{yy}(1,1)(y-1)^2 \end{array}\right] $$
Now if I simply plug in values I've computed.
$$ p_2(x,y)=e^{-1}+0(x-1)-2e^{-1}(y-1)+\frac{1}{2!}\left[\begin{array} a -e^{-1}(x-1)^2+2(-2e^{-1})(x-1)(y-1)+2e^{-1}(y-1)^2\end{array}\right] $$
since $f_x(1,1)=0$ we can remove $f_x(1,1)(x-1) $
$$ p_2(x,y)=e^{-1}-2e^{-1}(y-1)+\frac{1}{2!}\left[\begin{array} a -e^{-1}(x-1)^2+2(-2e^{-1})(x-1)(y-1)+2e^{-1}(y-1)^2\end{array}\right] $$
This was interpretation of formula $$ f(a+h)\approx \sum_{j=0}^{m}\frac{(h\cdot \nabla)^jf(a)}{j!}. $$ I am not 100% sure I interpreted the idea right but this seems correct to me. If someone can verify this is right / show that there is flaw that would be highly appreciated.