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If I know $f(a)$, $f'(a)$, $f''(a)$ and so on. I can conclude easily $f(x)$ is approximately $f(a) + (x - a) f'(a)$, but how do I further use $f''(a)$ to calculate a better approximation of the curve? I can't seem to find a way to use the second derivative.

Yes I know Taylor's series, but the question is how do I get that using the mean value theorem? Most importantly I want to learn how to apply the mean value theorem, and use it in this situation.

Luca Bressan
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koe
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  • Do you know Taylor expansions ? Around $x=a$, $f(x)=f(a)+(x-a) f'(a)+\frac{1}{2} (x-a)^2 f''(a)+O\left((x-a)^3\right)$ – Claude Leibovici Sep 12 '16 at 09:21
  • Mean value theorem will only give you a certain degree of approximation. There is no better measure than the Taylor series for a function value. – Sarvesh Ravichandran Iyer Sep 12 '16 at 09:32
  • @Claude L then how do i intuitively get that? – koe Sep 12 '16 at 09:37
  • See http://math.stackexchange.com/questions/1267295/quadratic-cubic-etc-approximations-without-the-taylor-series or http://math.stackexchange.com/questions/122766/where-do-the-higher-order-terms-in-taylor-series-come-from?rq=1 –  Sep 25 '16 at 18:46

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