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Why the following is true?

Corollary 1.4: An analytic function has at least one singularity on its circle of convergence.

That is, if f can be extended analytically from $B(p,R)$ to $B(p,R′)$, then the radius of convergence is at least $R′$. So there must be some obstruction to making such an extension,

This does not look like a proof. it seems that the convergent domain must be open. But is it clear? Why it can't convergent on a closed disc?

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Hmmm...what about the case of $f(z)=\exp(-z^{-2})$, expanded in a Taylor series about the point $z=1$? Is $z=0$ defined to be a singularity of $f$, even though $f$ is in some sense perfectly well behaved there, and is in fact $C^\infty$ there? –  Ben Crowell Feb 9 '12 at 4:14
Yes, $z=0$ is an essential singularity of $f$. The restriction of the function to $\mathbb R$ is well-behaved there, but on $\mathbb C$ it is most certainly not well behaved. –  Robert Israel Feb 9 '12 at 6:41
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3 Answers

up vote 3 down vote accepted

I believe that the sentence starting "That is..." is not intended to be a proof of Corollary 1.4 but rather a clarification of the statement. Let me try to restate what is being claimed:

Suppose $f(z)$ is defined and holomorphic on (at least) an open disk of radius $R > 0$ centered at $z_0 \in \mathbb{C}$. Then the radius of convergence of the Taylor series expansion of $f$ at $z_0$ is at least $R$.

This is true, and indeed it is a very standard fact in elementary complex analysis. At this point in my career it's been so long since I've studied this material (and I've never yet taught it...) that my brain has reserved a very small amount of space for remembering why these things are true. I consulted it, and it said "Use Cauchy's estimates," but then that's what it says for most results in this area! Any complex analysis text will do better.

The statement should not be interpreted to mean that if a Taylor series expansion of a function $f$ about a point $z_0$ has radius $R$, then there is necessarily at least one $z$ with $|z-z_0| = R$ such that the Taylor series diverges at $z$. Gerry Myerson's answer gives a simple example where that does not happen: $\sum_{n=1}^{\infty} \frac{z^n}{n^2}$. Nevertheless, there is a sense in which the associated function $f$ has a singularity on the boundary: $f$ cannot be analytically continued to any connected open set containing (the open unit disk and) the point $1$. This has to do with the branch cut in the dilogarithm function $\operatorname{Li}_2(z)$: see e.g. this wonderful article by Don Zagier for lots more information about $\operatorname{Li}_2(z)$.

Added: Okay, I looked at the document in question. Indeed, Corollary 1.4 as I have stated it follows immediately from Theorem 1.3, so what follows the statement is a clarification, not a proof. (And I agree that the unclarified statement of Corollary 1.4 is rather confusing...)

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The radius of convergence is the distance from the center to the nearest singularity (and if there's no singularity, then the radius of convergence is infinite). That's one of my favorite propositions in complex analysis (OK, I'm simple-minded${}\;\ldots\dots$).

Your last paragraph puzzles me. If I'm not mistaken, the question of which points on the boundary of the region of convergence are points at which the series converges, is fairly thorny in general. It had never occurred to me that the region of convergence must be open, nor that it must be closed. I'd have guessed that in some cases, it's neither.

However: If there's no singularity within a distance $R^{\;\prime}>R$ of the point $p$, then the function expanded in power of $z-p$ has radius of convergence at least $R^{\;\prime}$. That is a consequence of the proof that if a function is holomorphic (i.e. complex-differentiable) at $p$, then it's analytic at (i.e. represented by a convergent near) $p$.

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Thanks Michael. My question is: Why can't the domain of convergent be a close disc, say $\bar{B}(0, 1)$ all points outside the disc is divergent. This does not contradict to the definition of the radius of convergent, since the divergent on an open set, there is no 'nearest singularity. –  Sun Feb 9 '12 at 4:07
If it's convergent in a closed disk, then the closure of the set of locations of singularities does not include any point on that disk. Since the boundary of the disk is compact, this implies the closure of the set of locations of singularities is bounded away from the boundary. That means there's a bigger disk with no singularities in its interior. Therefore the radius of convergence would be bigger than the radius of that closed disk. –  Michael Hardy Feb 9 '12 at 16:37
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The series $f(z)=\sum_{n=1}^{\infty}z^n/n^2$ converges on the closed disk of radius 1, so I find Corollary 1.4 puzzling.

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