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Is there a complex power series $\sum_{n=0}^\infty a_n z^n$, with a finite radius of convergence $0<r<\infty$, that admits an analytic continuation to a disc $\{z\in \mathbb{C} : |z|<R\}$ with strictly bigger radius $R>r$? (The analytic continuation should be defined everywhere on the bigger disc.)

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  • $\begingroup$ Given a function $f(z)$, when talking about radius of convergence one has to specify the point $z_0$ around which this convergence takes place. Oftentimes, this is assumed that $z_0 = 0$ since if not, one looks at function $g(z) = f(z+z_0)$. But this is important distinction since, for example, for function $f(z) = \frac{1}{1+z}$ radius of convergence at $0$ is $1$ but at $100$ (i.e. $100 + 0)i$ it is $99$. $\endgroup$
    – Salcio
    Commented Dec 23, 2022 at 16:50
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    $\begingroup$ Since I wrote sum a_n z^n I assumed that it is implicitly clear that I was talking about a series of powers of z, hence the center of the disk of convergence is z=0 $\endgroup$ Commented Dec 23, 2022 at 18:56
  • $\begingroup$ I was not being critical about your exposition. Just making a point that at different location radius of convergent may be different. $\endgroup$
    – Salcio
    Commented Dec 23, 2022 at 22:05

2 Answers 2

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No: suppose there is some holomorphic function $f$ defined on some open disk with radius $r'>r$ which agrees with $\sum a_n z^n$ on $|z| < r$. Then by the Taylor expansions for analytic functions, $f$ equals its Taylor series $\sum c_n z^n$ on $|z| < r'$ (here $c_n = f^{(n)}(0) /n!$).

But by assumption $f = \sum c_n z^n$ agrees with $\sum a_n z^n$ on $|z| < r$. It follows that $a_n = c_n$ for all $n$ (otherwise if $a_n \neq c_n$ for some $n$, they would have different $n'$th derivatives at zero).

But now using the formula for the radius of convergence, we get the contradiction $$ r = ({\lim \sup}_{n \to \infty} \sqrt[n]{|a_n|})^{-1} = ({\lim \sup}_{n \to \infty} \sqrt[n]{|c_n|})^{-1} \geq r'. $$

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  • $\begingroup$ Sorry, but I don't understand why that answers my question. This just says that the given power series does not converge outside of |z|<r. Why could there not be other power series (not centered at zero) that converge outside of the disc |z|<r and whose values agree with the given one inside |z|<r, and these cover a disc strictly bigger? $\endgroup$ Commented Dec 23, 2022 at 13:24
  • $\begingroup$ @MichaelBächtold I edited my answer, but I'm not sure this is enough.. I'm thinking about it for a bit longer. $\endgroup$
    – Steven
    Commented Dec 23, 2022 at 14:00
  • $\begingroup$ @MichaelBächtold I think now I have a valid proof. It uses Taylor's theorem for holomorphic functions, but this can be found in many books. $\endgroup$
    – Steven
    Commented Dec 23, 2022 at 14:23
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    $\begingroup$ Right, silly me. Should have just looked up that theorem $\endgroup$ Commented Dec 23, 2022 at 19:02
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No.

Suppose $f(z)=\sum_{n=0}^\infty a_n z^n$ whenever $|z|<r$,

and also that $f(z)$ is holomorphic on a bigger disc $|z|<R$.

It is a theorem that if $f(z)$ is holomorphic on the disc $|z|<R$,

there is a Taylor series $\sum_{n=0}^\infty b_n z^n$ that is convergent on $|z|<R$ and that is equal to $f(z)$.

Now, $a_n=\dfrac{f^{(n)}(0)}{n!}=b_n$.

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  • $\begingroup$ See my comment to the other answer. I would have the same question with your answer. $\endgroup$ Commented Dec 23, 2022 at 13:25
  • $\begingroup$ @MichaelBächtold the answer has been edited for clarity $\endgroup$ Commented Dec 23, 2022 at 15:08

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