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I would like to evalue the absolute value of the theta function with unit nome, i.e. when $|q|=1$ or equivelantly when $\tau\in\mathbb R$. The theta function is expressed as

$$\vartheta(z;q=e^{\pi i \tau})=\sum_{n=-\infty}^\infty e^{\pi i n^2 \tau}e^{2\pi i n z}=\sum_{n=-\infty}^\infty q^{n^2}e^{2\pi i n z}$$

I understand from a comment on this post that for certain values this can be rewritten in terms of a dirac comb. I.e. for $q$ an $m$-th root of unity (equivelantly $\tau$ is a rational number, and $m$ is the denominator of its simplest form) the expression can be expressed as

$$\sum_{k=0}^{m-1} q^{k^2}\sum_n e^{2i\pi (nm+k) x} =\sum_{k=0}^{m-1} q^{k^2} e^{2i\pi k x} \sum_l \delta(mx-l)$$

And therefore the absolute value of the theta function for $q$ an $m$-th root of unity can be written as

$$|\vartheta(z;q=e^{\pi i \tau})|=\left|\sum_{k=0}^{m-1} q^{k^2} e^{2i\pi k x}\right|\sum_l \delta(mx-l)$$

Is it possible to find an expression of $|\vartheta(z;q=e^{\pi i \tau})|$ in terms of delta functions (or to say it is equal to $0$ or a constant) when $\tau$ is irrational?

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  • $\begingroup$ Do you understand what is a periodic distribution and its Fourier series? It doesn't make much sense to ask for its absolute value, unless you have any good reason for it. $\endgroup$ – reuns May 10 at 19:59
  • $\begingroup$ @reuns it is for a proof of a theorem in computer science. It would suffice to know something like "the absolute value diverges for all values (except at x=n)" for example $\endgroup$ – Cameron May 10 at 20:22
  • $\begingroup$ I don't see what you mean. Do you? $\endgroup$ – reuns May 10 at 20:26
  • $\begingroup$ @reuns For the proof I am working on, the only information required is whether the modulus of the function diverges, or if it is 0. In the case of rational numbers this is straight-forward, as you gave in the linked comment. I would like to know whether it is possible to write a similar expression for irrational $tau$, if we discard information about phase and/or even magnitude. I would also be happy with the answer that this is not possible, given an explanation of why. $\endgroup$ – Cameron May 10 at 20:34

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