Infinite product representations of theta functions
Let f(z) = \prod_{n=1}^{\infty} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz}). Each of these two products converge absolutely and uniformly in any bounded domain of values of z. Hence f(z) is analytic throughout the finite part of the z plane. The zeros of f(z) are simple zeros at the points z=\frac{2n+1}{2}\pi \tau i + m\pi i where m,n\in \mathbb{Z}. So, the zeros of f(z) coincide with the zeros of \vartheta_4(z). Therefore, the function \frac{\vartheta_4(z)}{f(z)} has no poles or zeros in the finite part of the plane. Also, it is easy to see that f(z+\pi) = f(z) and \begin{aligned} f(z+\pi \tau) &= \prod_{n=1}^\infty (1-q^{2n+1}e^{2iz})(1-q^{2n-3}e^{-2iz}) \\ &= f(z) \frac{1-q^{-1}e^{-2iz}}{1-qe^{2iz}} \\ &= -q^{-1}e^{-2iz} f(z) \end{aligned} This is, \frac{\vartheta_4(z)}{f(z)} is a doubly periodic function with periods \pi,\pi\tau and has no poles and zeros. By section 20.12 of [2], it is simply a constant, say G. We have \vartheta_4(z) = G \prod_{n=1}^{\infty} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz})\quad (1) Incrementing z by the half periods \frac{\pi}{2}, \frac{\pi \tau}{2} and \frac{\pi +\pi\tau}{2}, we get the product representations for the other theta functions: \begin{aligned} \vartheta_3(z) &= G \prod_{n=1}^{\infty} (1+q^{2n-1}e^{2iz})(1+q^{2n-1}e^{-2iz})\quad (2) \\ \vartheta_1(z) &= 2G q^{\frac{1}{4}}\sin(z) \prod_{n=1}^{\infty} (1-q^{2n}e^{2iz})(1-q^{2n}e^{-2iz}) \quad (3) \\ \vartheta_2(z) &= 2G q^{\frac{1}{4}}\cos(z) \prod_{n=1}^{\infty} (1+q^{2n}e^{2iz})(1+q^{2n}e^{-2iz}) \quad (4) \end{aligned} Now, all that left is to find the constant G. From identity (3), it easy to see that \vartheta_1'(0) = 2G q^{\frac{1}{4}}\prod_{n=1}^{\infty} (1-q^{2n})^2 Now, using the identity \vartheta_1'(0)=\vartheta_2(0)\vartheta_3(0)\vartheta_4(0), we get: \begin{aligned} 2G q^{\frac{1}{4}}\prod_{n=1}^{\infty} (1-q^{2n})^2 &= \left\{2G q^{\frac{1}{4}}\prod_{n=1}^{\infty}(1+q^{2n})^2 \right\} \times \left\{G\prod_{n=1}^{\infty}(1+q^{2n-1})^2 \right\} \\ &\quad \times \left\{G \prod_{n=1}^{\infty}(1-q^{2n-1})^2 \right\} \\ \implies G &= \prod_{n=1}^\infty (1-q^{2n}) \end{aligned} Note that the rearrangements are justified since all products converge absolutely. Finally, we have \begin{aligned} \vartheta_1(z) &= 2 q^{\frac{1}{4}}\sin(z) \prod_{n=1}^{\infty} (1-q^{2n})(1-2q^{2n} \cos(2z)+ q^{4n}) \quad (5) \\ \vartheta_2(z) &= 2 q^{\frac{1}{4}}\cos(z) \prod_{n=1}^{\infty} (1-q^{2n})(1+2q^{2n} \cos(2z) + q^{4n}) \quad (6) \\ \vartheta_3(z) &= \prod_{n=1}^{\infty} (1-q^{2n})(1+2q^{2n-1} \cos(2z) + q^{4n-2}) \quad (7) \\ \vartheta_4(z) &= \prod_{n=1}^{\infty} (1-q^{2n})(1-2q^{2n-1} \cos(2z) + q^{4n-2}) \quad (8) \end{aligned}
Derivatives of Ratios of Theta Functions
Consider the function \phi(z) = \frac{\vartheta_1'(z)\vartheta_4(z) - \vartheta_1(z)\vartheta_4'(z)}{\vartheta_2(z)\vartheta_3(z)}. Now, \phi(z) is doubly periodic with periods \pi and \frac{\pi\tau}{2}. Relative to these periods, the only poles of \phi(z) are at points congruent to \pi\over 2. It follows that \phi(z) is a constant. Letting z\to 0, we see that \phi(z)=\vartheta_4^2(0). It is therefore established that \frac{d}{dz}\left(\frac{\vartheta_1(z)}{\vartheta_4(z)} \right)=\vartheta_4^2(0) \frac{\vartheta_2(z)\vartheta_3(z)}{\vartheta_4^2(z)} \quad (9) Two other identities of this kind can be derived using the same technique: \begin{aligned} \frac{d}{dz}\left(\frac{\vartheta_2(z)}{\vartheta_4(z)} \right) &=-\vartheta_3^2(0) \frac{\vartheta_1(z)\vartheta_3(z)}{\vartheta_4^2(z)} \quad (10) \\ \frac{d}{dz}\left(\frac{\vartheta_3(z)}{\vartheta_4(z)} \right) &=-\vartheta_2^2(0) \frac{\vartheta_1(z)\vartheta_2(z)}{\vartheta_4^2(z)} \quad (11) \end{aligned} By writing \xi = \frac{\vartheta_1(z)}{\vartheta_4(z)} and making use of identities (13) and (14) from part 1, we obtain: \left(\frac{d\xi}{dz}\right)^2 = (\vartheta_2^2(0) - \xi^2 \vartheta_3^2(0))(\vartheta_3^2(0)-\xi^2 \vartheta_2^2(0)) Using the change of variables y = \frac{\vartheta_3(0)}{\vartheta_2(0)}\xi and u = \vartheta_3^2(0) z, the above differential equation can be written as: \left(\frac{dy}{du}\right)^2 = (1-y^2)(1-k^2 y^2) where k = \frac{\vartheta_2^2(0)}{\vartheta_3^2(0)}. This differential equation has the particular solution: y = \frac{\vartheta_3(0)}{\vartheta_2(0)} \frac{\vartheta_1(u \vartheta_3^{-2}(0))}{\vartheta_4(u \vartheta_3^{-2}(0))} We can write y as a function of u and k as y = \text{sn}(u,k) or simply \text{sn}(u). Clearly, \text{sn}(u,k) has periods 2\pi \vartheta^2_3(0) and \pi \tau \vartheta_3^2(0). It has two simple poles congruent to \frac{1}{2}\pi\tau \vartheta_3^2(0) and \pi \vartheta_3^2(0) + \frac{1}{2}\pi\tau \vartheta_3^2(0). Also, it is easy to see that \int_0^1 \frac{dy}{\sqrt{(1-y^2)(1-k^2 y^2)}} = \text{sn}^{-1}(1) = \frac{\pi}{2}\vartheta_3^2(0) The integral on the left is denoted by complete elliptic integral of the first kind and is denoted by K(k).
References
- Derek F. Lawden (1989). Elliptic Functions and Applications. Springer-Verlag New York
- E. T. Whittaker, G. N. Watson (1927). A Course of Modern Analysis. Cambridge University Press