# Taylor Series of analytic function

Why can the radius of convergence of the Taylor series of an analytic function around $$z_0$$ be larger than the largest disk centered at $$z_0$$ of the region of analyticity?

In such situation, could in the region outside of the domain of analyticity the Taylor series not converge to the function?

• Also some singularities might or might not hide behind a branch cut, try with $\log : \Bbb{C}-(-\infty,0]\to \Bbb{C}$ and the Taylor series of $\log z$ and $\frac1{\log z - \log (2-i)-2i\pi}$ at $2+i$, which converge for $|z-(2+i)|<|2+i|$ and $|z-(2+i)|< 2$. For $f$ analytic around $a$, the radius of convergence of its Taylor series at $a$ is the largest $r$ such that there exists $g$ analytic on $|z-a|<r$ agreeing with $f$ on a disk containing $a$. Commented Nov 5, 2020 at 12:21
For example, consider $$f =\exp : \mathbb{D} \rightarrow \mathbb{C}$$ the restriction of the exponential map to $$\mathbb{D} = \lbrace z \in \mathbb{C}, |z|<1 \rbrace$$. The Taylor series of $$f$$ at $$z=0$$ has an infinite radius of convergence, but the largest disk centered around $$0$$ in $$\mathbb{D}$$ has radius $$1$$.