Let's say we have a polynomial function, say
$f(x) = x^3-x^2+x-1$
and we have to calculate both the Taylor expansion in the point $x_0=-\sqrt{\pi}$
up to a generic $n$-th degree and the Lagrange remainder.
Now, what I haven't really well understood is this: after three derivations the derivative will be constantly $0$, right? So this means the Lagrange remainder will be $0$ too and therefore the Taylor approximation from a certain term onwards won't be anymore an approximation but an exact form?
As commented, in case of a polynomial the approximation to degree $\;n\ge3\;$ is exact:
$$f(x)=f(x_0)+\frac{f'(x_0)(x-x_0)}{1!}+\frac{f''(x_0)(x-x_0)^2}{2!}+\frac{f'''(x_0)(x-x_0)^3}{3!}=$$
$$=-\pi\sqrt\pi-\pi-\sqrt\pi-1+(3\pi+2\sqrt\pi+1)(x+\sqrt\pi)+\frac{(-6\sqrt\pi-2)(x+\sqrt\pi)^2}2++\frac{6(x+\sqrt\pi)^3}6$$