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Find the second order Taylor expansion about the point (1,-2) of the function $f(x,y) = (x^2 + y)e^{xy}$.

I begin by computing the matrix of partial derivatives of f.

$Df(x,y)=(2xe^{xy}+e^{xy}y(x^2+y),e^{xy}+e^{xy}x(x^2+y))$

From this I compute the Hessian matrix

$\begin{pmatrix} 2e^{xy}x+e^{xy}x^4+e^{xy}x^2y&e^{xy}xy^2+2e^{xy}y+3e^{xy}x^2+e^{xy}x^3y \\ 2e^{xy}y+e^{xy}xy^2+e^{xy}x^3y+3e^{xy}x^2& 2e^{xy}x+e^{xy}x^4+e^{xy}x^2y \end{pmatrix}$

It's already getting complicated. I need to put point (1,-2) and I'll get not so nice numbers.

Then I evaluate at the point $(1,-2)$ and find $Df(1,-2)=$? and $H(1,-2)=$? Now I need to put these together to compute the degree 2 Taylor polynomial, but I don't have them.

My professor told me that I'm doing it on wrong way, that I should use $e^{\alpha+\beta x+\gamma y...}=e^\alpha e^{\beta x+\gamma y...}$ before expanding the degree 2 Taylor polynomial and that I need to consider only terms up to order two!!!

Obviously there is easier method of doing it, for example using hint of my professor, but I don't know how to do it.

If someone could help me, I would be really thankful.

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  • $\begingroup$ The numbers aren't too bad. Factor out a $e^{xy}$, and then plug in $(1, -2)$ into everything. The thing about only going up to order $2$ makes sense, but the thing about $e^{\alpha+\beta x+\gamma y}$ makes no sense to me. Maybe someone else knows. $\endgroup$ – JasonM Jun 12 '16 at 21:18
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Hint:

It's rather simple conceptually (the computation details are another matter…).

Set $x=1+u$, $y=-2+v$. We have \begin{align*}\mathrm e^{xy}&=\mathrm e^{-2u+v+uv}=\mathrm e^{2}\mathrm e^{-2u+v}\mathrm e^{uv}\\&=\mathrm e^{2}\Bigl(1-2u+v+\frac12(-2u+v)^2+o\bigl(\lVert(u,v)\rVert^2\bigr)\bigl(1+uv+o\bigl(\lVert(u,v)\rVert^2\bigr)\bigr)\\ &=\mathrm e^{2}\Bigl(1-2u+v+2u^2-uv+\frac12v^2+o\bigl(\lVert(u,v)\rVert^2\bigr)\Bigr) \end{align*} There remains to do the same for the first factor (which is its own Taylor polynomial) and multiply the polynomials, truncating the result at order $2$.

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