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I have a simple problem: I need to evaluate the limit $x\rightarrow 1$ of the Jacobi Theta function 2

$$\Theta_2(m,x)=2x^{1/4}\sum_{k=0}^\infty x^{k(k+1)}\cos((2k+1)m)$$

when $m=0$, that to say

$$\Theta_2(0,x)=2x^{1/4}\sum_{k=0}^\infty x^{k(k+1)}$$

I guess that the problem is a simple application of the geometric series, but I do not get the right convergence.

Francesco Turci
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1 Answers1

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The problem is not that simple. Well, it's not very complicated neither, but one needs to use Jacobi's imaginary transformation. Introducing a new variable $\tau$ by $x=e^{i\pi \tau}$, we obtain $$\vartheta_2(0|\tau)=\left(-i\tau\right)^{-\frac12}\vartheta_4\left(0|\tau'\right),\qquad \tau'=-\tau^{-1}.$$ Now as $x\to 1$, we have $\tau\to i0$, $\tau'\to i\infty$ and $x'\to 0$, and therefore the asymptotics is given by $$\vartheta_2\left(0,x\to 1\right)\sim \left(-i\tau\right)^{-\frac12}=\left(-\frac{\ln x}{\pi}\right)^{-\frac12}\sim \left(\frac{1- x}{\pi}\right)^{-\frac12}.$$

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  • Thank you! I really thought that the series were the answer. I now wonder: there is no general result for the $\sum_{k=0}^\infty x^{p(k)}$ where p(k) is a polynomial with real coefficients? – Francesco Turci May 14 '15 at 12:51
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    @FrancescoTurci There is indeed some room for generalization. The idea is to use Poisson summation formula. – Start wearing purple May 14 '15 at 14:23