Can analytic function be represented as a taylor series?

Question

Let $D$ be open in $\mathbb{K}$.

Let $f:D\rightarrow \mathbb{K}$ be an analytic function.

Then, $f$ is infinitely differentiable and $\forall x_0\in D$, there exists a neighborhood $N$ of $x_0$ such that $\forall x\in N\cap D, f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n$.

My question is that, is $f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n, \forall x\in D$?

If not, what would be a counterexample?

Answer 1

The function $\frac{1}{x}$ is analytic on $D=\mathbb R\setminus \{0\}$, for each $x_0$, the taylor series for the function only converges for $x\in [x_0-|x_0|, x_0+|x_0|]$.

Answer 2

For $\mathbb{K} = \mathbb{R}$ take $D = \mathbb{R}$ and $f(x) = \frac{1}{1+x^2}$. The Taylor series of $f$ at $x = 0$ is $\sum_{n=1}^{\infty}(-1)^n x^{2n}$, with convergence set $[-1,1)$. The radius of convergence of this series is $1$ and this is due to the fact that the function of complex variable $z\mapsto \frac{1}{1+z^2}$ is not defined at $z= \pm i$.

In general, any real analytic function on $\mathbb{R}$ is the restriction of a unique complex analytic with domain of definition $\tilde D \supset \mathbb{R}$. If $\tilde D$ is not the whole $\mathbb{C}$ then the Taylor series of $f$ at any point $x_0$ will have a radius of convergence equal to the distance from $x_0$ to the complement of $\tilde D$.