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I met a problem of determining shape of a power series's value domain. The power series $f(z)=\sum a_n z^n$ is with real coefficients and $a_0=0$, $a_1>0$. A positive real number, $R$, is the series's singularity, which means $f(z)$ is divergent if $|z|>R$. Suppose $f(z)$ is convergent at $R$, I am curious what the shape of such power series $f(z)$'s value domain looks like when $|z|<R$. Could it be like a round pancake? Is it possible that there exists a $z$ in $f(z)$'s convergence domain making $f(z)>R$?

Here is an example. $\arctan z$'s Taylor series is $z-z^3/3+z^5/5-\cdots$, $1$ is its singularity. There is no $z$ in series's convergence domain making $\arctan z>1$.

zyynankai
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  • Every power series either converges everywhere, or converges nowhere but its centre $c$; or else there exists $R>0$ such that the series converges when $|z-c|<R$ and diverges when $|z-c|>R$. Other than which points on the circle $|z-c|=R$ are included in the region of convergence, those are the only shapes that a region of convergence can have (point, disk, whole plane). There is a notion of analytic continuation that is very interesting, but that's separate from the question of where a power series can converge. – Greg Martin Jul 14 '22 at 07:01
  • Thanks for your comment. However, I care about the shape of $f(z)$'s VALUE domain, not z-domain. – zyynankai Jul 14 '22 at 07:25
  • Okay, I definitely misread that, sorry. Reading it again, I'm confused by some of the details. What is the relevance of the power series converging at $R$ itself? And the inequality $f(z)>R$ is easy to arrange by just adding a large constant to $a_1$, for example. (Also $\arctan z$ doesn't have a singularity at $1$, but rather at $\pm i$.) – Greg Martin Jul 14 '22 at 17:09

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