<s>[INST] Expanding the Polynomial Using Taylor Series? Expand the polynomial
$$f(x) = x^3-2x^2-3x+5$$
In power of $(x-2)$

This might be a simple question but all we have to do to solve this question is to expand using the Taylor series with $x =2 $ , as opposed to the Maclaurin at $x=0$.
$f(x) = f(a) + \frac{f'(a)(x-a)^1}{1!}+\frac{f''(a)(x-a)^2}{2!}+ \cdot \cdot \cdot$
$f(2)= -1+1(x-2)+\frac{8(x-2)^2}{2!}+\frac{6(x-2)^3}{3!}$
$f(2) = -1 +(x-2) +4(x-2)^2+(x-2)^3$
So in the end this would be the solution correct?
 [/INST] Yes that's correct.
Moreover for a polynomial of degree $n$, the Taylor series of order $n$ is exact (because all the remaining derivatives are nulls), thus you have:
$$
f(x)= -1+(x-2)+4 (x-2)^2+(x-2)^3
$$
(no need for the $O\left((x-2)^4\right)$ term here, this is a true equality)
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