Do analytic function and power series agree whenever the power series converges?

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a function which is analytic at $x=0$. Then $$f(x)=\sum_{n=0}^\infty a_nx^n$$ for all $x$ in a neighborhood of $0$. Let $R$ be the radius of convergence of $\sum_{n=0}^\infty a_nx^n$. Then obviously $R>0$. My question is: Is it possible that there is a point $p\in(-R,R)$ such that $$f(p)\neq\sum_{n=0}^\infty a_np^n \ ?$$

Answer 1

Yes. Let $$f(x)=\begin{cases}\sin x\quad&\mbox{if}\ \ |x|<\pi,\\ 0\quad&\mbox{if}\ \ |x|\geq\pi.\end{cases}$$ Then $$f(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$ in a neighborhood of $x=0$, but $$f(3\pi/2)=0\neq-1=(3\pi/2)-\frac{(3\pi/2)^3}{3!}+\frac{(3\pi/2)^5}{5!}-\cdots.$$