Suppose $f:I\to \mathbb{R}$, where $I$ is an open subset of $\mathbb{R}$, is a smooth function on $I$, $f\in C^{\infty}(I)$. Let $x_0\in I$.

Def. We say that $f$ is analytic on $x_0$ if the Taylor series of $f$ centered at $x=x_0$, ie $$\sum_{n=0}^{+\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n,$$ converge to $f$ in a neighborhood of $x_0$.

My doubt is the following: suppose we know that $f$ is analytic on $x_0$. Then, necessary, $$\sum_{n=0}^{+\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\qquad [1]$$ has radius of convergence $R>0$. From power series theory we know then certainly $[1]$ converges (punctually) on $(x_0-R,x_0+R)$. Does $[1]$ need to converge to $f$ in the whole $(x_0-R,x_0+R)$, or we just need that $[1]$ converge to $f$ in some neighborhood $A$ of $x_0$, $A\subset (x_0-R,x_0+R)$, $A\subset I$, in order to respect the definition of analytic function?

Thanks a lot in advance.


By definition, a neighbourhood of $x_0$ is a set $U$ such that there exists an open set $V$ such that $x_0\in V\subset U$. By definition, that $V$ is open means there is some $\epsilon>0$ such that $(x_0-\epsilon,x_0+\epsilon)\subset V$.

Hence, if $f$ is analytic at $x_0$, then there exists $\epsilon>0$ such that the Taylor series of $f$ at $x_0$ converges to $f$ in $(x_0-\epsilon,x_0+\epsilon)$. However, $\epsilon$ does not have to be the radius of convergence $R$ of the Taylor series. It could still be that the Taylor series converges to something different from $f$ for values in $(x_0-R,x_0+R)\setminus(x_0-\epsilon,x_0+\epsilon)$.

EDIT: For example $|x|$ is analytic at $x_0=1$ with $\epsilon=1$, however the radius of convergence of the Taylor series is infinite.

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  • $\begingroup$ Great. Thanks a lot! That's very clear, sir. $\endgroup$ – eleguitar Dec 15 '18 at 16:30

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