# If it converges, how to show that power series converges to $f(x)$?

I had a very basic question.

Suppose $f(x)$ is a function. And let us say it has a power series :-

$$f(x) = \sum_{n=0}^\infty a_nx^n.$$

Suppose we are operating inside the region of convergence. Then for each value of $x$ in this region the series converges. My question is how do we know that at a given point say $x=x_0$ in this region ,

$f(x_0)$ and $\sum_{n=0}^\infty a_nx_0^n$ take the same value . I mean maybe the series converges but to some value other than $f(x_0)$ . I am somehow unable to see it.

Is it due to its very definition ? As in, we said let $f(x) = \sum_{n=0}^\infty a_nx^n$ and then we went on to find the values of the co-efficients? For ex. in Taylor's series expansion of $e^x$ . So if it has a region of convergence then in that ,the power series has to equal $f(x)$ for all values of $x$.

Please pardon me if this question seems outright silly.

Edit :- I mean the well behaved functions which are analytic . I mean such as $e^x$ and $sinx$ . Not just the ones having a region of convergence.

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A real analytic function agrees by definition with its power series. – Yiorgos S. Smyrlis Feb 22 '14 at 9:38
I learned on the net that a function is analytic iff it is holomorphic. I wished to study a proof for this. I understand holomorphicity involves the Cauchy Riemann conditions. For ex ,how, if we show that e^x satisfies the holomorphic condition it satisfies the analytic condition. By holomorphic I mean a function which is complex differentiable. And by analytic I mean, a function for which power series converges to f(x) . I hope I have these definitions right. Its just that I am unable to see how one implies the other. – ameyask86 Feb 23 '14 at 8:29

Unfortunately, it is not even true.

Take for example $$f(x)=\left\{ \begin{array}{lll} \mathrm{e}^{-1/x^2} & \text{if} & x>0, \\ 0 & \text{otherwise.} \end{array} \right.$$ Then $f^{(n)}(0)=0$, for all $n\in\mathbb N$, and hence the power series $$\sum_{n=0}^\infty f^{(n)}(0)\frac{x^n}{n!},$$ has radius of convergence $r=\infty$. But it does not agree with $f$ is no interval $(-a,a)$!

In the case $f$ is real analytic, it means that $f$ is expressible, locally, as a power series. So $f$ and the power series agree, by definition of real analyticity.

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But it is not analytic function right ? I meant for a function such as e^x . See I am talking about the most well behaved function in this regard such as e^x. – ameyask86 Feb 22 '14 at 9:28
This should be in the comments section. – Fly by Night Feb 22 '14 at 9:30
Ok. And so an analytic function is one for which the series converges to f(x). So if we encounter a function our goal is to show it is analytic and find its region of convergence. – ameyask86 Feb 22 '14 at 9:38
It is rather easy to know whether or not a function is real analytic. Real analytic functions are precisely those which extend as holomorphic in a neighborhood of the real line (or of an interval). – Yiorgos S. Smyrlis Feb 22 '14 at 9:40
I was looking at ways to show a function is analytic. Can you point me to a resource where it is showed how these tests imply that the function and power series agree. – ameyask86 Feb 22 '14 at 9:47

A very similar question has been asked before. Take a look at this post and its answers.

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