Jacobi Theta Function modularity translation quick q

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SUMMARY

The discussion focuses on the Jacobi theta series, specifically the property that ##\theta^{m}(\tau + 1) = \theta^{m}(\tau)##. The series is defined as ##\theta^{m}(\tau) = \sum\limits^{\infty}_{n=0} r_{m}(\tau) q^{n}##, where ##q^{n} = e^{2\pi i n \tau}##. The participant attempts to incorporate an additional factor of ##e^{2\pi i n}## into the series, suggesting a new Fourier coefficient ##r'{m}= r_{m} e^{2\pi i n}##. The conclusion emphasizes the periodicity of the series due to the property that ##e^{i 2n\pi} = 1## for integer values of n.

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Homework Statement



I have the Jacobi theta series: ##\theta^{m}(\tau) = \sum\limits^{\infty}_{n=0} r_{m}(\tau) q^{n} ##,

where ##q^{n} = e^{2\pi i n \tau} ## and I want to show that ##\theta^{m}(\tau + 1) = \theta^{m}(\tau) ##

(dont think its needed but) where ##r_{m} = ## number of ways of writing ##m## as the sum of ##n## squares.

Homework Equations



the above

The Attempt at a Solution


[/B]
so i get an extra ##e^{2\pi i n} ## factor, ## \theta^{m}(\tau) = \sum\limits^{\infty}_{n=0}r_{m}(\tau) q^{n} e^{2\pi i n} ##

I think it should be obvious what to do now, but I don't know what to do next?

Something like defining a new Fourier coefficient, ##r'{m}= r_{m} e^{2\pi i n} ## and then since the sum is to ? but that doesn't seem proper enough?

Many thanks.
 
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Hmm. You know that if n is an integer, e^{i 2n\pi} = 1?
 
stevendaryl said:
Hmm. You know that if n is an integer, e^{i 2n\pi} = 1?
oh my ! thank you ha :)
 

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