We have defined today in class:
A function $f:S\to\mathbb{C}$ (where $S\subseteq \mathbb{C}$ is open) is anayltic at a point $z_0\in \Bbb{C}$ if there exist $R>0$ and a sequence $(a_n)_n\subseteq \Bbb{C}$ s.t. $f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n$ for all $z\in B_R(z_0)$ (the open ball of radius R centered in $z_0$).
My guess is that $f$ would be analytic in this case for every $z_1\in B_R(z_0)$ as well (i.e. has convergent taylor series around $z_1$). For simplicity, let's assume $z_0=0$, $R=1$ and $z_1=\frac{1}{2}$. Probably, now I should be able to view $f$ as $f(z)=\sum_{n=0}^{\infty}a_n(z-\frac{1}{2})^n$ for all $z\in B_\frac{1}{2}(\frac{1}{2})$.
Is this true? Maybe other coefficients? I've tried to do subtitutions with no success.
Thanks!
Yes, that's true. It's a standard theorem about power series that if the radius of convergence of a power series $\sum_{n=0}^\infty a_n(z-z_0)^n$ is $R$, then the function$$\begin{array}{ccc}B_R(z_0)&\longrightarrow&\mathbb{C}\\z&\mapsto&\displaystyle\sum_{n=0}^\infty a_n(z-z_0)^n\end{array}$$is analytic.