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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?

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    Taylor expansion best aproximate function at the point $c$ by polynomial, but if our begin function is polynomial it is easy to see the best aproximation of polynomial is polynomial (equal degree) – 2017-01-22
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    So the approximation is _exact_? – 2017-01-22
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    Yes, look this video - https://www.youtube.com/watch?v=19x213y_uk4 – 2017-01-22
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    The approximation is exact if we go to degree 3 or higher. In degrees 0, 1 and 2, however, it is _not_ exact. – 2017-01-22

1 Answers 1

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As commented, in case of a polynomial the approximation to degree $\;n\ge3\;$ is exact:

$$f(x)=f(x_0)+\frac{f'(x_0)(x-x_0)}{1!}+\frac{f''(x_0)(x-x_0)^2}{2!}+\frac{f'''(x_0)(x-x_0)^3}{3!}=$$

$$=-\pi\sqrt\pi-\pi-\sqrt\pi-1+(3\pi+2\sqrt\pi+1)(x+\sqrt\pi)+\frac{(-6\sqrt\pi-2)(x+\sqrt\pi)^2}2++\frac{6(x+\sqrt\pi)^3}6$$