Let $f : \mathbb R^2 → \mathbb R$ be a function. Assume that the partial derivatives of f of order less or equal to two at $(0,0)$ exist. Let $g$ be second degree polynomial such that its first order derivatives (as well as its second order derivatives) are equal to the first order derivatives (and the second order derivatives) respectively of $f$ at $(0, 0)$ and such that $g(0, 0) = f(0, 0)$.
Show that $g$ is equal to the second order Taylor polynomial of $f$ about $(0, 0)$.
My attempt:
The 2nd order Taylor polynomial of $f$ at $(0,0)$ is given by:
$P_2(x,y)=f(0,0)+f_x(0,0)(x-0)+f_y(0,0)(y-0)+\dfrac 12\left[f_{xx}(0,0)(x-0)^2+2f_{xy}(0,0)(x-0)(y-0)+f_{yy}(0,0)(y-0)^2\right]$
But we are given that
$f(0,0) = g(0,0)$,
$f_x(0,0) = g_x(0,0)$, $f_y(0,0) = g_y(0,0)$, $f_{xx}(0,0) = g_{xx}(0,0)$, $f_{yy}(0,0) = g_{yy}(0,0)$, and $f_{xy}(0,0) = g_{xy}(0,0)$
So, replacing all these in $P_2(x,y)$, I get: $P_2(x,y) = g(0,0)+g_x(0,0)(x-0)+g_y(0,0)(y-0)+\dfrac 12\left[g_{xx}(0,0)(x-0)^2+2g_{xy}(0,0)(x-0)(y-0)+g_{yy}(0,0)(y-0)^2\right]$
Hence, indeed, $g$ = 2nd order Taylor polynomial of $f$ about $(0,0)$.
Is my attempt correct?