I'd like to show that the second-order term in the Taylor polynomial for $f$ centered at $(a,a)$ is equal to $$\frac{1}{2}f_{xx}(a,a)(x-y)^2.$$ By $C^2$ I mean that all of the first and second order partial deriatives exist and are continuous.
I first looked at the form of the term, which is $$\frac{1}{2}f_{xx}(a,a)x^2 + \frac{1}{2}f_{yy}(a,a)y^2 + f_{xy}(a,a)xy.$$ Since I have $f(x,y) = f(y,x)$, differentiating with respect to the first variable gives me $f_x(x,y) = f_y(y,x)$ and differentiating again gives me $f_{xx}(x,y) = f_{yy}(y,x)$, so taking $x = y = a$ I get $f_{xx}(a,a) = f_{yy}(a,a)$. Then the term becomes $$\frac{1}{2}f_{xx}(a,a)(x^2 + y^2) + f_{xy}(a,a)xy.$$
I think what I need to do is show that $f_{xy}(a,a) = -f_{xx}(a,a)$, at which point the second derivative factors out and $x^2 + y^2 - 2xy$ can be simplified to $(x-y)^2$. However, it's not immediately clear to me how I should do this. Specifically, where does the negative sign come into play?
Thanks!
– Marcus Emilsson Nov 17 '14 at 10:14