We have discussed various interesting properties of elliptic functions and related theta functions in detail in previous posts. In particular we discussed that these elliptic functions are doubly periodic. It is only natural then to seek the Fourier series corresponding to these functions. However in this case we will use only the real periods to expand these functions in a Fourier series. It turns out that the Fourier expansions provide us many important identities which can be used in surprisingly many ways to connect to number theory.

Let $ z = \pi u / 2K$ so that the functions $ \text{sn}(u, k), \text{cn}(u, k), \text{dn}(u, k)$ become periodic with periods $ 2\pi, 2\pi, \pi$ respectively when considered as functions of $ z$. Also these are analytic in the strip $ |\Im(z)| < \pi K^{\prime} / 2K = (\pi / 2)\Im(\tau)$. We can then consider the case for $ \text{dn}(u)$ which is an even function and periodic with period $ \pi$ and therefore can be expressed as $$\text{dn}(u) = \sum_{n = -\infty}^{\infty}a_{n}e^{2inz} = \frac{d_{0}}{2} + \sum_{n = 1}^{\infty}d_{n}\cos (2nz)$$ where $ a_{-n} = a_{n}$ so that $$d_{n} = 2a_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi}\text{dn}(u)e^{-2inz}\,dz$$ Calculating the above integral requires the use of residue calculus. We can first look at the integral along the parallelogram (in counterclockwise direction) joining the points $ -\pi, \pi, \pi\tau, -2\pi + \pi\tau$. Clearly the function at hand has poles at $ \pi\tau / 2$ and $ -\pi + \pi\tau / 2$ and at these points the residue is given by the expression $$-i\left(\frac{\pi}{2K}\right)e^{-ni\pi\tau} = -i\left(\frac{\pi}{2K}\right)q^{-n}$$ By residues theorem the integral is given by \begin{align} \int_{C}\text{dn}(u)e^{-2inz}\,dz &= 2\pi i(\text{sum of residues in the parallelogram})\notag\\ &= 2\pi i \cdot -i\left(\frac{\pi}{K}\right)q^{-n} = \left(\frac{2\pi^{2}}{K}\right)q^{-n}\notag \end{align} It is now easy to observe that while evaluating the integral along boundary $ C$ of the parallelogram the integrals on the left and right boundaries cancel each other due to periodicity and the integral on the upper boundary is given by $$\int_{\pi\tau}^{-2\pi + \pi\tau}\text{dn}(u)e^{-2inz}\,dz = -\int_{-2\pi + \pi\tau}^{\pi\tau}\text{dn}(u)e^{-2inz}\,dz$$ Putting $ z = t -\pi + \pi\tau$ the integral reduces to $$e^{-2ni\pi\tau}\int_{-\pi}^{\pi}e^{-2int}\,\text{dn}\left(\frac{2K}{\pi}t\right)\,dt = q^{-2n}\int_{-\pi}^{\pi}\text{dn}(u)e^{-2inz}\,dz$$ It then follows that $$\int_{C}\text{dn}(u)e^{-2inz}\,dz = (1 + q^{-2n})\int_{-\pi}^{\pi}\text{dn}(u)e^{-2inz}\,dz$$ and thus finally $$\int_{-\pi}^{\pi}\text{dn}(u)e^{-2inz}\,dz = \frac{1}{1 + q^{-2n}}\left(\frac{2\pi^{2}}{K}\right)q^{-n}$$ The Fourier coefficient is now given by $$d_{n} = \frac{2\pi}{K}\frac{q^{n}}{1 + q^{2n}}$$ We finally have the Fourier series expansion as $$\text{dn}(u, k) = \frac{\pi}{2K} + \frac{2\pi}{K}\sum_{n = 1}^{\infty}\frac{q^{n}\cos(2nz)}{1 + q^{2n}}$$ Similarly we can establish \begin{align} \text{sn}(u, k) &= \frac{2\pi}{Kk}\sum_{n = 0}^{\infty}\frac{q^{n + 1 / 2}\sin(2n + 1)z}{1 - q^{2n + 1}}\notag\\ \text{cn}(u, k) &= \frac{2\pi}{Kk}\sum_{n = 0}^{\infty}\frac{q^{n + 1 / 2}\cos(2n + 1)z}{1 + q^{2n + 1}}\notag \end{align}

Two corollaries immediately follow from this theorem:

The above two corollaries are more properly known as

### Fourier Series Expansion of Elliptic Functions

Let $ f(z)$ be a periodic function with period $ 2\pi$ and let us also assume that it is analytic in some strip $ a < \Im(z) < b$. Then we can clearly expand it in a Fourier series of the form $$f(z) = \sum_{n = -\infty}^{\infty}c_{n}e^{inz},\,\,\,\,\,\,c_{n} = \frac{1}{2\pi}\int_{z_{0}}^{z_{0} + 2\pi}f(z)e^{-inz}\,dz$$ where the integration has to be done along a path connecting $ z_{0}$ and $ z_{0} + 2\pi$ which lies entirely in the strip $ a < \Im(z) < b$.Let $ z = \pi u / 2K$ so that the functions $ \text{sn}(u, k), \text{cn}(u, k), \text{dn}(u, k)$ become periodic with periods $ 2\pi, 2\pi, \pi$ respectively when considered as functions of $ z$. Also these are analytic in the strip $ |\Im(z)| < \pi K^{\prime} / 2K = (\pi / 2)\Im(\tau)$. We can then consider the case for $ \text{dn}(u)$ which is an even function and periodic with period $ \pi$ and therefore can be expressed as $$\text{dn}(u) = \sum_{n = -\infty}^{\infty}a_{n}e^{2inz} = \frac{d_{0}}{2} + \sum_{n = 1}^{\infty}d_{n}\cos (2nz)$$ where $ a_{-n} = a_{n}$ so that $$d_{n} = 2a_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi}\text{dn}(u)e^{-2inz}\,dz$$ Calculating the above integral requires the use of residue calculus. We can first look at the integral along the parallelogram (in counterclockwise direction) joining the points $ -\pi, \pi, \pi\tau, -2\pi + \pi\tau$. Clearly the function at hand has poles at $ \pi\tau / 2$ and $ -\pi + \pi\tau / 2$ and at these points the residue is given by the expression $$-i\left(\frac{\pi}{2K}\right)e^{-ni\pi\tau} = -i\left(\frac{\pi}{2K}\right)q^{-n}$$ By residues theorem the integral is given by \begin{align} \int_{C}\text{dn}(u)e^{-2inz}\,dz &= 2\pi i(\text{sum of residues in the parallelogram})\notag\\ &= 2\pi i \cdot -i\left(\frac{\pi}{K}\right)q^{-n} = \left(\frac{2\pi^{2}}{K}\right)q^{-n}\notag \end{align} It is now easy to observe that while evaluating the integral along boundary $ C$ of the parallelogram the integrals on the left and right boundaries cancel each other due to periodicity and the integral on the upper boundary is given by $$\int_{\pi\tau}^{-2\pi + \pi\tau}\text{dn}(u)e^{-2inz}\,dz = -\int_{-2\pi + \pi\tau}^{\pi\tau}\text{dn}(u)e^{-2inz}\,dz$$ Putting $ z = t -\pi + \pi\tau$ the integral reduces to $$e^{-2ni\pi\tau}\int_{-\pi}^{\pi}e^{-2int}\,\text{dn}\left(\frac{2K}{\pi}t\right)\,dt = q^{-2n}\int_{-\pi}^{\pi}\text{dn}(u)e^{-2inz}\,dz$$ It then follows that $$\int_{C}\text{dn}(u)e^{-2inz}\,dz = (1 + q^{-2n})\int_{-\pi}^{\pi}\text{dn}(u)e^{-2inz}\,dz$$ and thus finally $$\int_{-\pi}^{\pi}\text{dn}(u)e^{-2inz}\,dz = \frac{1}{1 + q^{-2n}}\left(\frac{2\pi^{2}}{K}\right)q^{-n}$$ The Fourier coefficient is now given by $$d_{n} = \frac{2\pi}{K}\frac{q^{n}}{1 + q^{2n}}$$ We finally have the Fourier series expansion as $$\text{dn}(u, k) = \frac{\pi}{2K} + \frac{2\pi}{K}\sum_{n = 1}^{\infty}\frac{q^{n}\cos(2nz)}{1 + q^{2n}}$$ Similarly we can establish \begin{align} \text{sn}(u, k) &= \frac{2\pi}{Kk}\sum_{n = 0}^{\infty}\frac{q^{n + 1 / 2}\sin(2n + 1)z}{1 - q^{2n + 1}}\notag\\ \text{cn}(u, k) &= \frac{2\pi}{Kk}\sum_{n = 0}^{\infty}\frac{q^{n + 1 / 2}\cos(2n + 1)z}{1 + q^{2n + 1}}\notag \end{align}

### Some Number Theory

We can use the above results to establish some theorems of number theory. To proceed if we put $ u = 0$ in the expansion for $ \text{dn}(u, k)$ we get \begin{align} \frac{2K}{\pi} &= 1 + 4\sum_{n = 1}^{\infty}\frac{q^{n}}{1 + q^{2n}}\notag\\ &= 1 + 4\sum_{n = 1}^{\infty}q^{n}\sum_{m = 0}^{\infty}(-1)^{m}q^{2mn}\notag\\ &= 1 + 4\sum_{n = 1}^{\infty}\sum_{m = 0}^{\infty}(-1)^{m}q^{(2m + 1)n}\notag \end{align} We can now put $$ i = (2m + 1)n,\, d = 2m + 1, \, m = (d - 1) / 2$$ so that \begin{align} \frac{2K}{\pi} &= 1 + 4\sum_{i = 1}^{\infty}\left(\sum_{d \mid i,\, d\,\text{odd}}(-1)^{(d - 1)/2}\right)q^{i}\notag\\ &= 1 + 4\sum_{n = 1}^{\infty}\left(\sum_{d \mid n,\,\,d\,\text{odd}}(-1)^{(d - 1) / 2}\right)q^{n}\notag \end{align} The terms in the inner sum are of the form $ (-1)^{(d - 1) / 2}$ so that the term is $ +1$ if $ d \equiv 1 (\text{mod}\,4)$ and $ -1$ if $ d \equiv 3 (\text{mod}\,4)$. Hence we can write the inner sum as a function $ \delta(n)$ which can be expressed as: $$\delta(n) = \#(\text{divisors of}\, n\,\text{of the form}\, 4i + 1) - \#(\text{divisors of}\,n\,\text{of the form}\,4i + 3)$$ where $ \#$ represents the count of things under its argument. Therefore we have $$\frac{2K}{\pi} = 1 + 4\sum_{n = 1}^{\infty}\delta(n)q^{n}$$ The magic happens when we consider the fact that $ 2K / \pi$ can be expressed in terms of theta functions: $$\frac{2K}{\pi} = \theta_{3}^{2}(q) = \left(\sum_{n = -\infty}^{\infty}q^{n^{2}}\right)^{2} = \sum_{n = 0}^{\infty}r_{2}(n)q^{n} = 1 + \sum_{n = 1}^{\infty}r_{2}(n)q^{n}$$ where $ r_{2}(n)$ is number of ways in which $ n$ can be written as sum of two squares (counting order as well as sign of the integers) and $ r_{2}(0) = 1$. Upon comparing the coeffcients it follows that $ r_{2}(n) = 4\delta(n)$ for all positive integers $ n$. Thus we have the theorem (first established by Jacobi using elliptic function theory):*The number of ways in which a positive integer can be expressed as the sum of two squares is four times the difference between the number of its divisors which are of the form $ 4i + 1$ and the number of its divisors which are of the form $ 4i + 3$.*Two corollaries immediately follow from this theorem:

*A prime number $ p$ of the form $ p = 4i + 1$ can be expressed as the sum of two squares in essentially one way.**A prime number $ p$ of the form $ p = 4i + 3$ can not be expressed as the sum of two squares.*The above two corollaries are more properly known as

*Fermat's theorem on two squares*or sometimes as*Thue's Lemma*. The same technique can be used to prove similar theorems about expressing a number as sum of four squares but the proof is a bit complicated.**Print/PDF Version**
Wow,this is so exciting !! Thank-you for writing!

mathter2357

January 20, 2017 at 9:42 AMDear Mr. Paramanand Singh,

Your calculation of the Fourier coefficient for the elliptic function dn(u,k) is very nice and direct. Could you, please, send me the corresponding calculation of the Fourier coefficient for the function sn(u,k)? Or even a reference (it may already have been developed by some textbook author)?

Thank you in advance,

Fabio M. S. Lima (fabio_msl@yahoo.com)

Institute of Physics, University of Brasilia

Fabio

December 22, 2017 at 1:26 AM@Fabio,

Thanks for your positive feedback. You can see the development of these Fourier Series in the classic text "A Course of Modern Analysis" by Whittaker and Watson. It should be available in any university library. Do refer pages 510 and later.

Regards

Paramanand

Paramanand Singh

December 24, 2017 at 8:46 PM