Suppse the power series $ \sum_{n=0}^\infty a_n (x-a )^n$ has positive radius of convergence $R$ and thus defines a real analytic fuction $f$ on $(a-R,a+R).$If $x_0$ is a point with $|x_0-a|<R,$ Let $r=R-|x_0-a|,$then we can certainly define a power series
$$\sum_{n=0}^\infty \frac{f^n(x_0 )}{n!}(x-x_0)^n,x\in(x_0-r,x_0+r).$$
Prove: $$f(x)= \sum_{n=0}^\infty \frac{f^n(x_0 )}{n!}(x-x_0)^n , x\in(x_0-r, x_0+r).$$
The following is my way of thinking:
Using Taylor's Theorem,Let $x$ is a fixed point in $(x_0-r, x_0+r),$then for every $n\in\mathbb{N^{+}},$we have $$f(x)-\sum_{k=0}^n \frac{f^k(x_0 )}{k!}(x-x_0)^k=\frac{f^{n+1}({\xi }_{n})}{(n+1)!}(x-x_0)^{n+1},$$
where ${\xi }_{n}$ is a ponit between $x_0$ and $x$ .$$\left|f(x)-\sum_{k=0}^n \frac{f^k(x_0 )}{k!}(x-x_0)^k \right|\leq \frac{|f^{n+1}({\xi }_{n})|}{(n+1)!}|x-x_0|^{n+1}\leq |f^{n+1}({\xi }_{n})|\frac{r^{n+1}}{(n+1)!},$$
Since that $\lim_{n\rightarrow \infty}\frac{r^{n+1}}{(n+1)!}=0,$
If we have a positive constant $M$, such that $|f^{n+1}({\xi }_{n})|\leq M,(n=1,2,...).$We can get the answer!
Until now I have failed to find this $M$,I have a hunch on that $M$ should be exist.
Any help would be greatly appreciated!