The Magic of Theta Functions: Contd.

In the previous post we studied some interesting properties of theta functions which were used to relate them to AGM and thereby to elliptic integrals. We will continue to explore further in this direction and start with a remarkable property of theta function $\theta_{3}(q)$.

Transformation formula for Theta Functions

Let $0 < q < 1$, so that $\log q$ is negative and we can write $q = e^{-\pi s},\, s > 0$. The transformation formula for $\theta_{3}(q)$ relates $\theta_{3}(q) = \theta_{3}(e^{-\pi s})$ with $\theta_{3}(e^{-\pi / s})$ in a very simple and beautiful way as follows: $$\sqrt{s}\,\theta_{3}(e^{-\pi s}) = \theta_{3}(e^{-\pi / s})\tag{1}$$ The usefulness of the above transformation formula for numerical calculations is quite obvious. When $s$ is near zero, so that $q$ is very near to 1, the series of $\theta_{3}(q)$ converges very slowly. However, in this case $1 / s$ is large and so $q^{\prime} = e^{-\pi / s}$ is quite small and then the series for $\theta_{3}(q^{\prime})$ converges very very fast. By the above transformation formula we can then calculate $\theta_{3}(q)$ very efficiently when $q$ is near $1$.

The proof of the transformation formula is based on a result from theory of Fourier series called "Poisson Summation Formula" which we will state below without proof (however we will indicate lines of proof which is not difficult altogether):

Poisson Summation Formula: Let $f$ be a non-negative function defined on $\mathbb{R}$ such that it is increasing in $(-\infty, 0]$ and decreasing in $[0, \infty)$. Also assume that the improper Riemann Integral $\int_{-\infty}^{\infty}f(x)\,dx$ exists. Then for all $x \in \mathbb{R}$ we have $$\sum_{n\,=\,-\infty}^{\infty}\frac{f(n + x\,+) + f(n + x\,-)}{2} = \sum_{k\,=\,-\infty}^{\infty}e^{2\pi ikx}\int_{-\infty}^{\infty}f(t)e^{-2\pi ikt}\,\,dt$$ and both the series are absolutely convergent.

The proof of the above theorem can be provided by considering the function $$F(x) = \sum_{n = -\infty}^{\infty}f(n + x)$$ and noting that this function is periodic with period 1. The result is then obtained by finding the Fourier series for $F(x)$.

We will apply the Poisson formula for the function $f(x) = e^{-s\pi x^{2}}$ where $s > 0$. Clearly the function $f$ satisfies all the conditions needed for applying Poisson formula. We need to calculate the integral in the Poisson formula: \begin{align} \int_{-\infty}^{\infty}f(t)e^{-2\pi ikt}\,\,dt &= \int_{-\infty}^{\infty}e^{-s\pi t^{2} - 2\pi ikt}\,\,dt\notag\\ &= \int_{-\infty}^{\infty}e^{-s\pi t^{2}}(\cos(- 2\pi kt) + \imath \sin(-2\pi kt)) \,\,dt\notag\\ &= \int_{-\infty}^{\infty}e^{-s\pi t^{2}}\cos(2\pi kt)\,\,dt - \imath \int_{-\infty}^{\infty}e^{-s\pi t^{2}}\sin(2\pi kt) \,\,dt\notag\\ &= 2\int_{0}^{\infty}e^{-s\pi t^{2}}\cos(2\pi kt)\,\,dt\notag \end{align} On substituing $x = \sqrt{s\pi}\,t$ the above integral reduces to $$\frac{2}{\sqrt{s\pi}}\int_{0}^{\infty}e^{-x^{2}}\cos\left(2\sqrt{\frac{\pi}{s}}\,kx\right)\,dx = \frac{2}{\sqrt{s\pi}}F\left(\sqrt{\frac{\pi}{s}}k\right)$$ where $$F(y) = \int_{0}^{\infty}e^{-x^{2}}\cos(2xy)\,dx$$ Integration by parts gives us \begin{align} F(y) &= \left(e^{-x^{2}}\frac{\sin(2xy)}{2y}\right)_{x = 0}^{x = \infty} - \int_{0}^{\infty}e^{-x^{2}}(-2x)\frac{\sin(2xy)}{2y}\,dx\notag\\ &= \frac{1}{y}\int_{0}^{\infty}xe^{-x^{2}}\sin(2xy)\,dx\notag \end{align} And differentiating the integral for $F(y)$ we get \begin{align} F'(y) &= \int_{0}^{\infty}e^{-x^{2}}(-2x)\sin(2xy)\,dx\notag\\ &= -2\int_{0}^{\infty}xe^{-x^{2}}\sin(2xy)\,dx = -2yF(y)\text{ (from above)}\notag \end{align} so that $F'(y) + 2yF(y) = 0$. Noting that $F(0) = \sqrt{\pi} / 2$ we solve the differential equation for $F(y)$ to get $$F(y) = \frac{\sqrt{\pi}}{2}e^{-y^{2}}$$ It follows that the integral we started with is given by $$\int_{-\infty}^{\infty}f(t)e^{-2\pi ikt}\,\,dt = \frac{1}{\sqrt{s}}e^{-\pi k^{2} / s}$$
The Poisson Summation Formula thus gives us $$\sum_{n\,=\,-\infty}^{\infty}e^{-\pi s (n + x)^{2}} = \frac{1}{\sqrt{s}}\sum_{k\,=\,-\infty}^{\infty}e^{2\pi ikx}e^{-(\pi / s)k^{2}}\tag{2}$$ Putting $x = 0$ in above we get the desired transformation formula for $\theta_{3}$. Putting $x = 1 / 2$ we get $$\sqrt{s}\,\theta_{2}(e^{-\pi s}) = \theta_{4}(e^{-\pi / s})\tag{3}$$ Replacing $s$ by $1 / s$ in above we get $$\sqrt{s}\,\theta_{4}(e^{-\pi s}) = \theta_{2}(e^{-\pi / s})\tag{4}$$ Diving $(3)$ by $(1)$ we get $$\frac{\theta_{2}(e^{-\pi s})}{\theta_{3}(e^{-\pi s})} = \frac{\theta_{4}(e^{-\pi / s})}{\theta_{3}(e^{-\pi / s})}$$ and thus we get in terms of the parameter $k$ $$k(e^{-\pi s}) = k^{\prime}(e^{-\pi / s})\tag{5}$$ and replacing $s$ by $1 / s$ we get $$k(e^{-\pi / s}) = k^{\prime}(e^{-\pi s})\tag{6}$$ Thus the moduli $k(e^{-\pi s})$ and $k(e^{-\pi / s})$ are complementary and we can write the AGM formula as: $$M(1, k(e^{-\pi s})) = \theta^{-2}_{3}(e^{-\pi / s})\tag{7}$$ and $$M(1, k'(e^{-\pi s})) = \theta^{-2}_{3}(e^{-\pi s})\tag{8}$$ Dividing $(8)$ by $(7)$ we get $$\pi\,\frac{M(1, k'(e^{-\pi s}))}{M(1, k(e^{-\pi s}))} = \pi\,\frac{\theta^{2}_{3}(e^{-\pi / s})}{\theta^{2}_{3}(e^{-\pi s})} = \pi s = -\log q$$ (using $(1)$ above) i.e. $$\pi\,\frac{M(1, k'(q))}{M(1, k(q))} = -\log q\tag{9}$$ From this equation we can easily calculate $q$ given the value of modulus $k$. Recasting the above equation in form of the complete elliptic integrals we get $$\pi\,\frac{K'(k)}{K(k)} = -\log q\tag{10}$$ This is more commonly stated as $q = e^{-\pi K'/K}$. That last result is very intriguing and I should elaborate further on its importance. We start with a number $k$ such that $0 < k < 1$ and then define the positive constants: \begin{align} k' &= \sqrt{1 - k^{2}}\notag\\ K &= K(k) = \int_{0}^{\pi / 2 }\frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}}\notag\\ K' &= K(k') = \int_{0}^{\pi / 2 }\frac{d\theta}{\sqrt{1 - k'^{2}\sin^{2}\theta}}\notag \end{align} where the latter constants $K$ and $K'$ can be expressed as hypergeometric series (expanding the integrand as infinite binomial series and integrating term by term) as follows: \begin{align} K &= \frac{\pi}{2}\,F\left(\frac{1}{2}, \frac{1}{2}; 1; k^{2}\right) = \frac{\pi}{2}\sum_{n\,=\,0}^{\infty} \frac{1^{2}\cdot 3^{2}\cdot 5^{2} \cdots (2n - 1)^{2}}{2^{2n}(n!)^{2}}\,k^{2n}\notag\\ K' &= \frac{\pi}{2}\,F\left(\frac{1}{2}, \frac{1}{2}; 1;1 - k^{2}\right) = \frac{\pi}{2}\sum_{n\,=\,0}^{\infty} \frac{1^{2}\cdot 3^{2}\cdot 5^{2} \cdots (2n - 1)^{2}}{2^{2n}(n!)^{2}}\,(1 - k^{2})^{n}\notag \end{align} It is a surprising feat that the above formula can be inverted and we can obtain $k$ given $K$ and $K'$ as follows: \begin{align} q &= e^{-\pi K' / K}\notag\\ k &= \frac{\theta_{2}^{2}(q)}{\theta_{3}^{2}(q)} = \frac{4(q^{1/4} + q^{9/4} + q^{25/4} + \cdots)^{2}}{(1 + 2q + 2q^{4} + 2q^{9} + \cdots)^{2}}\notag \end{align} The inversion formula to obtain $k$ above seems quite mysterious given the expressions for $K$ and $K'$ and our treatment has been a bit artificial in the sense that we defined theta functions out of the blue and then proved their connections with the elliptic integrals.. The way theta functions actually came into being historically is quite different and involves the study of the elliptic functions which are inverses of elliptic integrals in the way circular function $y = \sin x$ is the inverse of the integral (which we might call a circular integral): $$x = \int_{0}^{y} \frac{dt}{\sqrt{1 - t^{2}}}$$ We will turn to the elliptic functions in the next post and will see a lot many properties of theta functions in their full generality.

Note: The material for this and the previous post was derived from Borwein's famous book "Pi and the AGM".