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It is simple to see that the following series converges absolutely and uniformly on $\mathcal{H}$ for all k positive:

$F_{2k}(z) = \sum_{n \in \mathbb{Z}} q^{n^{2k}}$

And this series being a generating function for higher degree forms is like a theta series analogue. But because the Fourier transform of the said function is not so nicely behaved ( the best I could compute required some very bad hyper geometric series), we do not get a ‘Modular form’ like symmetry.

So instead of using the Fourier transform, and then the poisson summation formula to get a symmetry on the Fourier series, is it possible to use some other integral transform, and then use the corresponding eigenfunction series based on that transform to get a ‘nice’ generating function with some transformation properties?

If not, is it possible to study the obstructions that one faces when trying to find the symmetries of a generating function like that?

(p.s. this question arose from a comment by D. Zagier in one of his lectures that every generating function is in some or the other form related to some Modular form, but I suppose we’ll need something more general than just Modular forms to study such functions)

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  • $\begingroup$ I would be surprised if anything interesting about your general $F_{2k}$ is known. The comment of Zagier should be read under certain context, e.g. what did he mean by "every generating function"? I guess what he had in mind was a big picture of Langlands program. $\endgroup$ – WhatsUp Oct 3 '20 at 18:00
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You can try to follow the standard theta argument for general $k$, to get a series expansion, except not one that relates the function to itself.

Consider $$G_{2k}(z) = \sum_{n=1}^{\infty} e^{-n^{2k} z}$$ Take the Mellin transform of both sides $$\int_0^\infty G_{2k}(z)z^{s-1}dz = \Gamma(s) \zeta(2ks)$$ Replace $z$ with $1/z$, and $s$ with $-s$ $$\int_0^\infty G_{2k}(1/z)z^{s-1}dz = \Gamma(-s) \zeta(-2ks))$$ Substituting in the functional equation for $\zeta(s)$ and expand both $\zeta$ and $\sin$ as a sum: $$= \Gamma(-s)\Gamma(1+ks) (2\pi)^{-ks} (i^{ks}-i^{-ks}) \zeta(1+ks)$$ Thus, expanding out $\zeta$ as a sum and taking the inverse Mellin transform, you express $G_{2k}(z)$ as a sum over integers of a function that is an inverse mellin transform of a gamma function production (ie: a Meijer G-function).

When $k=1/2$ the function summed is basically $1/(1-z)$ giving you the expansion of $1/(e^z-1)$ in terms of its poles. When $k=1$ the function summed is $e^{-z^2}$ giving you the functional equation. For other $k$ the function is a hypergemetric function satisfying some nice properties, but not referring back to the original function. There are some nice uses of these sums in proving that functional equations cannot have degree strictly between 1 and 2

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  • $\begingroup$ Thanks for the answer, I’d reached the same conclusion that the hypergeometric series would not refer back to itself, that is why I was wondering if the approach of using the Mellin transform is not the answer to this sort of a series? $\endgroup$ – Sagars Oct 6 '20 at 12:56

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