Is it true that for any natural number $n$, that $n$-th order Taylor polynomial of any polynomial at $\vec0$ is just itself ? (If $n\ge$degree of the polynomial)
For example if $f(x_1,\dots,x_m)$ is a polynomial, then $n$-th order Taylor polynomial of $f$ at $\vec 0$ is just $f$ itself ? i.e. $f(x_1,\dots,x_m)=4x_1^2x_2+5x_3^3x_5+7x_1^3+2x_4,n\ge4$ etc.
JUst repeating the comments: Yes, it's true, but as I recall, the proof is pretty tedious. In the one-dimensional case, I know that Spivak's Calculus has a proof (somewhere in Chapter 20, on Taylor polynomials). In the multivariable case, I know that Henri Cartan's Differential Calculus text (Chapter 7) contains a proof. Anyway, just try it out in the 2-variable case by writing out the taylor polynomial for small n. You should see how to generalize.