So, I was asked to find the third-degree Taylor expansion for the polynomial: $$g(x,y)=x^2+xy+y^2\,\,\mathrm{at}\,(a,b)=(1,2)$$
Oddly enough, the third-degree Taylor expansion was the function itself. Was this a mere coincidence, because I thought the Taylor expansion of any degree was the function itself around $(a,b)=(0,0)$, i.e. a Maclaurin series?
So, does this result make sense?
On
Note that a polynomial is uniquely determined by its value, and its derivatives at a given point.
Taylor polynomials are polynomials which approximate a given function.
For a Taylor polynomial of degree $n$ the requirements are that the function and the polynomial have the same value, the same first derivative, the second derivative, up to the $n_{th}$ derivatives at the given point.
If it happens that your function is a polynomial of $n_{th}$ degree, then the Taylor polynomial is identical to your function because there is only one such polynomial of degree $n.$
In case of functions of more than one variable, the partial derivatives up to the $n_{th}$ partials are the same and that implies uniqueness.
As you know $g$ is a polynomial (in several variables) so up to certain degree of derivatives it must vanish, so at the end the Taylor expansion is itself, and here, third degree of derivatives makes the $g$ vanishing.