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From Rademacher's book (Topics in Analytic Number Theory) I'm using the functional equation of $\vartheta_2(0|\tau) = 2\sum_{m=0}^\infty q^{\left(m+\frac12\right)^2} = \vartheta_2(\tau)$ and the fact that it vanishes at the cusps $\tau = \frac{a}{b}$, $a$ odd, $b$ even.

The order of vanishing at infinity is just $\frac14$. I'd like to find the order of vanishing at the cusp $\tau = \frac12$ (no particular reason for this choice). Using say $A = \left(\begin{array}{lr} 1 & 0 \\ 2 & 1 \end{array}\right) \in \Gamma_0(2)$ and plugging into the functional equation I get $$\vartheta_2\left(\frac{\tau}{2\tau+1}\right) = i e^{-\pi i/4} \sqrt{-i(2\tau+1)}\vartheta_2(\tau)$$ As I understand it I'd want to plug $\tau = i\infty$ into this, which would give me $\vartheta_2\left(\frac12\right)$ in terms of $\vartheta_2(i\infty) = q^\frac{1}{4}+...$ However, I don't know how to handle the $\sqrt{2\tau+1}$ term if I was to put $\tau = i\infty$, so I feel like there's a big hole in my understanding.

The equation from Rademacher is given as $$\vartheta_2\left(\frac{\upsilon}{c\tau+d}\bigg|\frac{a\tau+b}{c\tau+d}\right) = i^{3-d-(a+1)/2}e^{\pi i (dc+ab-ac-1)/4}\sqrt{\frac{c\tau+d}{i}}e^{\pi i c\upsilon^2/(c\tau+d)}\vartheta_{1-c, 1-d}(\upsilon|\tau)$$ on page 182, for $c>0$, $a$ odd.

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  • $\begingroup$ If $f(\tau) = f(\tau+a)$ is meromorphic on $Im(\tau) > 0$ then $g(q) = f(\frac{a\log q}{2i \pi })$ is well-defined and meromorphic on $0 < |q| < 1$. So you can apply the usual tools of complex analysis : if $q=0$ is an isolated singularity of $g(q)$, you get a Taylor series for $g$ and a $q$ expansion for $f$ valid "at the cusp $i\infty$" i.e. for $Im(\tau)$ large enough. $\endgroup$
    – reuns
    Dec 23, 2016 at 3:48

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The convention when working with modular forms $f(\tau)$ of weight $k$ is for the order of vanishing at a cusp $a/c$ to mean the order of vanishing of $(c\tau+d)^{-k} f\left(\frac{a\tau+b}{c\tau+d}\right)$ as a function of $q = e^{2 \pi i \tau}$. One also speaks of the value of a modular form at a cusp as the constant term in the $q$-expansion. This is a bit different than the usual notion of plugging in points, or examining Taylor expansions that one usually things about in complex analysis.

For example, for the regular Jacobi theta function $\theta(\tau) = \sum_{n \in \mathbb{Z}} q^{n^{2}}$, one has $\theta(-1/\tau) = (-i \tau)^{1/2} \theta(\tau)$. One says that $\theta(\tau)$ is holomorphic and non-vanshing at the cusp at zero because $\tau^{-1/2} \theta(-1/\tau)$ has an expansion in integral powers of $q$ with a nonzero constant term. However, for $r$ real and positive, we have $\theta(ir) \to \infty$ as $r \to 0$, as is fairly clear from the $q$-expansion.

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  • $\begingroup$ So instead I just want to consider $$(2\tau+1)^{-\frac{1}{2}}\vartheta_2\left(\frac{\tau}{2\tau+1}\right) = i e^{-\pi i/2} \vartheta_2(\tau) = i e^{-\pi i/2}(q^\frac14+...)$$ So the order of vanishing at $\tau=\frac12$ is also just $\frac14$? $\endgroup$
    – HmmmBeee
    Dec 20, 2016 at 15:09
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    $\begingroup$ Yes, that's correct. $\endgroup$
    – Jeremy Rouse
    Dec 20, 2016 at 15:42

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