In branches of algebra impinging on the enumeration of partitions, one often encounters formulas like

$$\prod_i \left( \frac{1}{1-q^i} \right)^{n_i}$$

for some integers $n_i$. E.g., with $n_i = 1$, this counts partitions, and with $n_i = i$, plane partitions.

Such formulas are usually understood as formal products, but if you tried to take them seriously as analytic functions, you would be thinking about functions with poles at all roots of unity.

Is there a branch of analysis which studies such functions? Does that theory have consequences in combinatorics?

(I don’t know even enough analysis to know what tag to put.)

  • 11
    $\begingroup$ The examples you gave are analytic in the open unit disc with the unit circle as a natural boundary. Not sure what you mean by poles at the roots of unity. Obviously you can use complex analysis/modular forms to give asymptotics etc. Not clear to me what your question is. $\endgroup$ – Lucia Jan 28 '18 at 23:06
  • 3
    $\begingroup$ Hard to say what it'll say without knowing what you want! $\endgroup$ – Lucia Jan 28 '18 at 23:12
  • 3
    $\begingroup$ Maybe one point of entry would be to look at work on mock modular forms -- eg look at write-ups of Zagier or Ono. Ramanujan's original idea of a mock modular form was along your lines of "functions with some known behavior around roots of unity" (which however are not genuine modular forms). $\endgroup$ – Lucia Jan 28 '18 at 23:27
  • 4
    $\begingroup$ Maybe you want to study the Hardy-Littlewood circle method, q.v. This was devised specifically to further the analytic study of $\prod(1-q^i)^{-1}$ and to derive consequences for the partition function. $\endgroup$ – Gerry Myerson Jan 29 '18 at 1:00
  • 1
    $\begingroup$ See arxiv.org/abs/1401.1893 and the references cited there for an application of the circle method to generating functions like these. $\endgroup$ – Ira Gessel Jan 29 '18 at 4:10

One general warning about interpretation as analytic functions: Just because an infinite product formula looks like it tells you where the zeros and poles are, it might not work the way it does for finite products.

For example, we can consider a similar-looking infinite product $$\prod_{i\geq 0}\frac1{1+q^{2^i}} = \frac{1}{1+q+q^2+q^3+\cdots} = 1-q,$$ which looks like it should have poles at all $2^i$-th roots of -1, but in fact it doesn't have any poles in its analytic continuation.

(My apologies if this was already understood -- is there a less-obvious reason for the claim about having poles at all roots of unity?)

One analytic result about such combinatorially-inspired functions is the Polya-Carlson theorem, which gives a surprisingly strong dichotomy: if a power series $f(q) = \sum_{n\geq 0}a_nq^n$ has integer coefficients and radius of convergence 1 (so it defines an analytic function on the open unit disk), then either

a) the function cannot be analytically continued past the unit circle, or

b) the function is rational, of the form $f(q) =\frac{g(q)}{(1-q^m)^n}$ for polynomial $g(q)$ and $m,n\in \mathbb N$.

| cite | improve this answer | |
  • 1
    $\begingroup$ The textbook I linked to might be a good reference for further information, but I have no idea if it's "standard"; it was just the top google result when I searched for that theorem. $\endgroup$ – Harry Richman Feb 6 '18 at 4:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.