5

The series $\displaystyle\sum \dfrac{z^n}{n^2}$ converges for $\lvert z\rvert<1$ by the ratio test, meaning that the dilogarithm function $\text{Li}_2(z),$ which is equal to the series $\displaystyle\sum \dfrac{z^n}{n^2}$ when it converges, is certainly analytic on $\lvert z\rvert<1$.

Similarly, by the ratio test, the series diverges for $\lvert z\rvert>1$. And we know that for a meromorphic function, the radius of convergence is always the distance from the center to the nearest singularity(wikipedia). Hence we should conclude that this function has a pole somewhere on $\lvert z\rvert=1$?

On the other hand, at the point $z=1$ it converges to $\pi^2/6$ (see Basel problem), it must also converge at the point $z=-1$, and it is stated on this question by Ben that the series is convergent on the whole circle $\lvert z\rvert=1$ (this confused me and prompted the question), and on a comment to an answer to a related question, one is reminded by user 23rd that if a function is analytic on a closed disk, then it is analytic on an open disk of larger radius. So we should conclude that this series is convergent for some $z$ with $\lvert z\rvert>1$? For the whole complex plane? In fact, this Wolfram alpha page does claim that the function is analytic on all of $\mathbb{C}$ (if I'm reading it correctly; it's very terse).

Actually that second related question (singularity of analytic continuation of $f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2}$) already contains the answer to my question: the dilogarithm is analytic on $\mathbb{C}\setminus [1,\infty)$. But I can't understand how that answer is consistent with the other remarks, and the Wolfram page. How is this situation reconciled? Could I get an explanation that's a little more detailed than what's already there?

Community
  • 1
ziggurism
  • 16,756
  • " if a function is analytic on a closed disk, then it is analytic on an open disk of larger radius" But we don't know this series is analytic on ${|z| \le 1}.$ – zhw. Nov 04 '15 at 19:11
  • @zhw.: doesn't convergence at a point imply analyticity? – ziggurism Nov 04 '15 at 19:59
  • No. In fact $(\sum x^n/n^2 - \pi^2/6)/(x-1) \to \infty$ as $x\to 1^-.$ – zhw. Nov 04 '15 at 20:09
  • 1
    @zhw.: right. It is convergent in an open neighborhood of a point which is equivalent to differentiable. We have convergence at $z=1$, but no neighborhood thereof – ziggurism Nov 04 '15 at 20:52
  • By the way, the dilogarithm function isn't meromorphic, because of the branch point and branch cut. – tparker Sep 17 '18 at 17:37

1 Answers1

3

The alternative definition (easily proven equivalent to the one you give for $\lvert z \rvert < 1$) $$ \operatorname{Li}_2 (z) = \int_0^z \frac{\log{(1-t)}}{t} \, dt, $$ where the integral can be taken to be along the ray joining $0$ to $z$ for $z \notin (1,\infty) $, shows that the dilogarithm has a branch point at $z=1$ (since the logarithm inside the integral has one): you can see that no extension across the cut is possible by computing the derivative across the cut, for example.

As we see here, it is possible for a power series to converge on its whole circle of convergence (since the series is absolutely convergent for $\lvert z \rvert \leqslant 1$), but for no larger $\lvert z \rvert$; in this case, because the branch cut "gets in the way" of defining an analytic extension to a larger convex domain.

Chappers
  • 67,606
  • So a statement like "the radius of convergence is always the distance to the nearest pole" is not correct and this function is a counterexample because there is no pole at $z=1$. It should be "radius of convergence is distance to nearest pole or branch point"? That's what the linked Wikipedia article means by "singularity"? – ziggurism Nov 04 '15 at 21:15
  • Yes, although the terminology is likely not universal. Probably the best way to think about a singularity is a point $a$ such that there is no open disk containing $a$ on which the function is analytic. This includes poles, essential singularities and branch points, which are essentially the only possible ways to lose analyticity at a point (if the point is isolated, it's one of the first two, else it's a branch point). – Chappers Nov 05 '15 at 01:40
  • 1
    I guess this question has little to do with special functions like the dilogarithm. If you want a series to exhibit the phenomenon that a branch point can end the radius of convergence instead of a pole, the series for $\sqrt{1-z}$ behaves in exactly the same way, though I found that series harder to work with. – ziggurism Nov 05 '15 at 04:27
  • 1
    @ziggurism Actually isolated singularities and branch points is rather an exceptional way to lose analyticity. The generic one would correspond to emergence of natural boundaries, such as the circle $|z|=1$ for the series $\sum_{n}z^{n^2}$. – Start wearing purple Nov 11 '15 at 11:53
  • @Startwearingpurple: can you elaborate on that idea? What is a natural boundary? Boundary between convergent and divergent domains? Are there some kind of singularities on this boundary? I don't recognize the series you quoted – ziggurism Nov 13 '15 at 02:05
  • Mathematica evaluates the sum as $\dfrac{1}{2}(1+\theta_3(0,x)))$. https://en.wikipedia.org/wiki/Theta_function – ziggurism Nov 14 '15 at 16:55
  • @ziggurism There are several types of nonisolated singularities other than branch points. Roughly speaking, a cluster point is a limit of an infinite sequence of isolated singularities, and a natural boundary is a "continuous line of singularities". – tparker Apr 20 '18 at 00:06
  • @tparker so is the singularity in the dilogarithm a "continuous line of singularities" or a branch point? I thought it was the latter... – ziggurism Apr 20 '18 at 17:50
  • @ziggurism The singularity in the dilogarithm function is a logarithmic branch point. I was just pointing out that your claim that "the radius of convergence is the distance to the nearest pole or branch point" isn't true in general (although it's true for the dilogarithm); the radius of convergence is actually the distance to the nearest singularity, which (unlike what Chappers said) could be a pole, an essential singularity, a branch point, a cluster point, or a natural boundary. – tparker Apr 20 '18 at 18:28
  • @tparker I see. Thank you for the clarification. – ziggurism Apr 20 '18 at 18:32