2
$\begingroup$

In the way of studying an enumerative problem I have found that I have to estimate the Taylor coefficients of functions of the following form. For two polynomials $P(x)$ and $Q(x)$ with $P(0)=Q(0)=1$, I need to study the asymptotics of the Taylor coefficients of

$$\frac{P(x)}{Q(x)}\frac{Q(x^2)}{P(x^2)}\frac{P(x^4)}{Q(x^4)}\dots$$

Is there some reference that could be used in order to see how to deal with these type of problems?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Here's a guess at something to try. Write $R(x) = \frac{P(x)}{Q(x)}$. Your series $F(x)$ satisfies

$$F(x) = \frac{R(x)}{F(x^2)}$$

so taking logarithms we get

$$\log F(x) = \log R(x) - \log F(x^2).$$

Recurrences of roughly this form are studied in VII.5 ("Unlabeled non-plane trees and Polya operators") of Flajolet and Sedgewick's Analytic Combinatorics; the basic idea is to use the recurrence to isolate (probably numerically) the location of the dominant singularity of $F$, then extract asymptotics by studying the behavior of the singularity (edit: or saddle points, see Chapter VIII although I don't think that chapter treats a recurrence like this).

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Mmm..not sure if this works. All schemes in the suggested chapter of Flajolet - Sedgewick are of implicit form, in which one can apply an inversion argument (despite not knowing where the singularity is, then one can get good approximations by numerical methods). In the case I am treating the expression is explicit. For instance, if $P$ and $Q$ only have roots of modulo 1, the singularities are dense in the unit circle, so one needs other type of arguments. $\endgroup$
    – Johnny Cage
    Sep 20, 2022 at 4:49
  • 3
    $\begingroup$ I should've been clearer that this was a guess and I didn't actually know if it would work; will edit to clarify. If the singularities are dense on the unit circle then instead of finding singularities you can try to find saddle points; I have no idea how easy that will be in your case. $\endgroup$
    – Qiaochu Yuan
    Sep 20, 2022 at 4:53
  • $\begingroup$ Really appreciate your feedback and suggestion! $\endgroup$
    – Johnny Cage
    Sep 20, 2022 at 5:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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