\[P(x)\]
of finite degree always returns the polynomial \[P(x)\]
.Example:
\[P(x)=x^2\]
as a Taylor series at \[x=1\]
.\[\frac{d (P(x))}{dx}=2x, \; \frac{d^2 (P(x))}{dx^2}=2 \]
and higher derivatives all equal 0.The Taylor series is
\[\begin{equation} \begin{aligned} T(x) &= P(1)+ \frac{d(P(x))}{dx}|_{x=1}(x-1)+ \frac{1}{2!} \frac{d^2(P(x))}{dx^2}|_{x=1}(x-1)^2 \\ &= 1 + 2(x-1)+ (x-1)^2 \\ &= 1+2x-2+x^2-2x+1 \\ &= x^2 \end{aligned} \end{equation}\]
.This is because a polynomial is uniquely determined by it's coefficients, so if
\[P(x_= \sum a_nx^n = \sum b_n x^n\]
then \[a_n=b_n\]
for all \[n\]
.