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Given a meromorphic function on $\mathbb{C}$, is the radius of convergence in a regular point exactly the distance to the closest pole?

As Robert Israel points out in his answer, that this is of course an upper bound by the Cauchy-Hadamard principle.

Theo Buehler in the comments gives a refernce for the non obvious direction: Remmert, Theory of complex functions, Chapter 7, §3, p.210ff (p. 164ff of my old German edition). Look for Cauchy-Taylor.

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I can not even see why the radius $R = \infty$, if $f$ is entire. –  plusepsilon.de Jun 22 '11 at 13:13
    
There is a chapter in Wilf's book generatingfunctionology about the relation between the asymptotics of the coefficients of a power series and the nature of the singularities on the boundary of the circle of convergence of the function. –  GEdgar Jun 22 '11 at 13:52
    
Dear GEdgar, you certainly mean e.g. Fatou's theorem or other summability conditions on the $a_n$'s. This comment might better fit to this question of mine: math.stackexchange.com/questions/46898/…, but I do not see how this should relate. –  plusepsilon.de Jun 22 '11 at 15:15
    
I suggest that you take a look at Remmert's book I recommended in another question of yours (or any other book on complex analysis that deserves its title). This is really basic material that is covered everywhere and I find it a bit too much to ask for a proof here. –  t.b. Jun 22 '11 at 16:16
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As I said, it's in any book on complex analysis. E.g. Remmert, Theory of complex functions, Chapter 7, §3, p.210ff (p. 164ff of my old German edition). Look for Cauchy-Taylor. –  t.b. Jun 24 '11 at 7:18

1 Answer 1

up vote 13 down vote accepted

Yes, it is (that should be "pole", not "pol"). If $r$ is the distance from $z_0$ to the closest pole, the function is analytic in $\{z: |z - z_0| < r\}$, so the radius of convergence is at least $r$, but it can't be more than $r$ because $|f(z)| \to \infty$ as $z$ approaches that pole.

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This is, of course, the correct answer. But one might want to give some reminder or reference to the proof, as this is certainly not obvious (e.g. it fails over $\mathbb{R}$ and even more dramatically over $\mathbb{C}_p$, where the principle of analytic continuation is violated). I seem to recall that this has something to do with the Cauchy estimates? (Well, it's a safe guess....) –  Pete L. Clark Jun 22 '11 at 7:34
    
Cauchy's Estimate is one way but more typically directly using Cauchy's formula and then expanding the $1/(z-w)$ term in $$f(w)=1/2\pi i\int_C f(z)/(z-w) dz$$. –  Steve Jun 22 '11 at 12:31
    
The basic theorem, that any function analytic in an open disk has a Taylor series (about the centre of the disk) that converges in that disk, should appear in any introductory Complex Variables text. Caution: the result is often quoted as something like "the radius of convergence is the distance from the centre to the nearest singularity". The problem with this is that if you start with a given function $f$, the sum of the Taylor series agrees with $f$ in a neighbourhood of the centre, but might be different at some points within the circle of convergence (see next comment) –  Robert Israel Jun 24 '11 at 17:56
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For example, consider $f(z) = \sqrt{z}$ (using the principal branch, which has a branch cut on the negative real axis) and $a = -1+i$. The largest disk around $a$ on which $f$ is analytic has radius 1, but the radius of convergence of the Taylor series is $\sqrt{2}$. The sum of the Taylor series is a branch of $\sqrt{z}$, but not the principal branch. –  Robert Israel Jun 24 '11 at 18:00
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@Libertron That is indeed what it means. The radius of convergence $\ge r$ because the function is analytic in $\{z: |z - z_0| < r\}$, and $\le r$ because $|f(z)| \to \infty$ as $z$ approaches the pole, so you conclude it is exactly $r$. –  Robert Israel Jul 21 at 5:48

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