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Suppose I have a function $f$ that is real analytic at $x = 0$, that means there exists $\delta > 0$ such that $$ f(x) = \sum_{n = 0}^{\infty} c_n x^n $$ for all $|x| \leq \delta$, for some real coefficients $c_n$. Now it follows that the $c_n$ are necessarily the Taylor series coefficients of $f$ at $0$. Suppose the radius of convergence of the Taylor series of $f$ is $R >0$, that is the series converge absolutely for $|x|<R$.

Does it then follow that $$ f(x) = \sum_{n = 0}^{\infty} c_n x^n $$ for all $|x|<R$? I think it should but I wasn't too sure... any comments are appreciated!

Edit. What I am wondering is if the radius of convergence of the Taylor series is $R$, then what condition on $f$ (it has to be more than just being analytic at $0$, thanks to the comment below) ensures that the equality (the function equals the Taylor series) holds for all $|x|< R$?

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    $\begingroup$ No, without any additional assumption on $f$, why should this follow? Consider $f$ which is defined through $f(x) = \sin(x) $ for $|x|< \delta$ and $f(x) = \sin{\delta}$ for $x\ge \delta$, $f(x) = - \sin(\delta)$ for $x\le -\delta$. This is analytic near $x=0$, but, in general, only continuous. I guess what you intended to ask is whether $g(x) := \sum_{n=0}^\infty c_nx^n$ is analytic for $|x|<R$... $\endgroup$
    – Thomas
    Oct 10, 2021 at 16:15
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    $\begingroup$ ...or whether the equality holds for $f$ if $f$ is analytic for $|x|<R$. $\endgroup$
    – Thomas
    Oct 10, 2021 at 16:21
  • $\begingroup$ Thank you for the comments. Let me edit the question. $\endgroup$
    – Johnny T.
    Oct 10, 2021 at 19:29

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You need some further condtions on $f$. If $f$ is only known to be analytic at $0$, then beyond $[x-\delta, x+\delta]$, it can do anything. It doesn't even have to be continuous. However, if $f$ is real analytic everywhere on $|x|<R$ (which is obviously sufficient for $f(x)$ = Taylor series inside the disk) is sufficient.

At eath point, use the power series to extend the definition of $f$ to be a complex analytic germ around that point, so we can have lots of pieces of complex analytic function which conicide with each other on at least an interval on $\mathbb R$ (as they have all to agree with $f$ on $|x|<R$), and from complex analysis, we know they must then all agree with each other everywhere (up to monodromy, but there is no topological obstruction in this simple setup). Therefore we have a well-defined complex analytic function on the complex disc $|x|<R$ which extends $f(x)$, and for complex analytic funciton, the function is always equal to its power series within the convergence radius.

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