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Jun 23, 2011 at 1:58 comment added Junkie In reality, after further thought, his answer is the "right way around", for you get these growth rates on the coefficients via (after) knowing that the function is holomorphic on the upper half plane, or unit disc. This follows here, as the denominator of $\eta(q^8)\eta(q^4)^{16}/\eta(q^2)^{24}$ is $\Delta(q^2)$ which has no zeros. But the method of Rademacher is still nice to know, and it applies widely (partitions are another example).
Jun 22, 2011 at 12:27 vote accept Ariyan Javanpeykar
Jun 22, 2011 at 12:27 comment added Ariyan Javanpeykar Thank you very much! I'll accept S. Carnahan's answer below and thank you for your answer again (which was posted before his). Thnx!
Jun 22, 2011 at 11:32 comment added Junkie The classical reference for growth of coefficients is Rademacher with the $j$-invariant via the method of Farey arcs jstor.org/stable/2371313 , and Brisebarre and Philibert perso.ens-lyon.fr/nicolas.brisebarre/Publi/fujijrms.pdf don't even bother to replicate the proof when noting that it holds for all modular functions of weight 0 for the full modular group (Theorem 5.1). I suspect varying it to your case should not yield tumultuous difficulties.
Jun 22, 2011 at 11:11 comment added Junkie As with many modular computations, it appears you can bound $a_j$ by a measure like $j^{\sqrt j}$, so that indeed for any $R<1$ the series will converge, by comparison $\sum_j j^{\sqrt j}e^{2\pi i\tau j}$.
Jun 22, 2011 at 11:07 answer added S. Carnahan timeline score: 3
Jun 22, 2011 at 10:48 comment added Ariyan Javanpeykar Thank you for your comment. It didn't occur to me to search Sloane's database.
Jun 22, 2011 at 10:41 comment added Junkie Sloane gives the coefficients at oeis.org/A115977 including a definition in terms of $\eta$ as 16 times the coefficient of $\eta(q)^8\eta(q^4)^{16}/\eta(q^2)^{24}$. This can yield their growth, and I suspect it is an exercise. I don't know a reference. If you want exact bounds, rather than asymptotics then it is harder, but for convergence radius, this is more than abides.
Jun 22, 2011 at 10:24 history asked Ariyan Javanpeykar CC BY-SA 3.0