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.
Best Answer
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.