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Question. Suppose $f : (x_{0}-\epsilon, x_{0}+\epsilon) \rightarrow \mathbb{R}$ is given by power sereis $x \mapsto \sum_{n=0}^{\infty}a_{n}(x - x_{0})^{n}$ where the sum is convergent everywhere in the open interval. Then can we find a holomorphic extension $F : D \subseteq \mathbb{C} \rightarrow \mathbb{C}$ such that $D = \{z \in \mathbb{C} : |z - x_{0}| < \epsilon\}$ and $F|_{(x_{0}-\epsilon, x_{0}+\epsilon)} = f$? If we can't, can we do this on a smaller disc $D' \subseteq D$?

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Yes, we can. From root test we know that $\epsilon\cdot\limsup_{n\to\infty}|a_n|^{1/n}\le 1$. Then the radius of convergence of $\sum_{n=0}^\infty a_n(z-x_0)^n$ for complex $z$ is also no less than $\epsilon$, so we can define $F(z)$ as the sum of this series, which is holomorphic on $D$.

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