Professor asked me to prove why is $$a_n = \frac{f^{(n)} (x_0)}{n!}$$ I know that Taylor series $$f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (x_0)}{n!} (x - x_0)^n$$ and I can see the resemblance with the power series $$f(x) = \sum_{n = 0}^{\infty} a_n (x - x_0)^n$$ but I have no idea how to prove the relation.


closed as unclear what you're asking by Jack, TheGeekGreek, J. M. is a poor mathematician, Siong Thye Goh, kimchi lover Aug 30 '17 at 2:01

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    $\begingroup$ Set the last series equal to $f(x)$? At that point the question comes down to uniqueness of power series, which takes a little work to prove. $\endgroup$ – Ian Aug 29 '17 at 19:48

Let $R$ be the radius of convergence of $(a_n)_n$, then let define $f\colon]x_0-R,x_0+R[\rightarrow\mathbb{R}$ by: $$f(x):=\sum_{n=0}^{+\infty}a_n(x-x_0)^n.$$ Then, since the series defining $f$ converges normally on compacts of $]x_0-R,x_0+R[$, one has: $$f^{(n)}(x_0)=n!a_n.$$ Whence the result.

  • $\begingroup$ Thanks man, it really baffled me $\endgroup$ – user2202368 Aug 29 '17 at 20:20

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