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?

Thanks in advance.

  • 2
    $\begingroup$ 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 $a$. $\endgroup$
    – reuns
    Commented Nov 5, 2020 at 12:21
  • $\begingroup$ Thanks for the explanation $\endgroup$
    – JoseAf
    Commented Nov 5, 2020 at 19:02

1 Answer 1


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 $1$.

This situation where the domain is restricted (so is not "well-chosen") is the only possibility where the radius of Taylor series can be strictly larger than the one of the largest disk. Otherwise you will have an equality.


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