# Definition of analytic function at a point

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?

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.