Theta Functions and Jacobi Elliptic Functions

While theta functions have applications in number theory and the heat equation, they are also used as helper functions in proving properties of Jacobi elliptic functions. The latter is what we will do in this post. This post will be followed by others to derive various properties of the Jacobi elliptic functions in order to evaluate integrals of them and eventually elliptic integrals.

We begin with the basic definitions of the theta functions. Note that notation for the theta functions is not standard.

Theta Functions

\(z\) = argument
\(\tau\) = lattice parameter, \(\mathfrak{I}(\tau) \gt 0\)
\(q = \mathrm{e}^{i\pi \tau}\) = nome, \(|q| \lt 1\)

a. Exponential forms
\begin{equation}
\theta_{1}(z|\tau) = \theta_{1}(z,q) = -i\sum_{n=-\infty}^{\infty} (-1)^{n} q^{(n+1/2)^{2}} \mathrm{e}^{i(2n+1)z} \\
\theta_{2}(z|\tau) = \theta_{2}(z,q) = \sum_{n=-\infty}^{\infty} q^{(n+1/2)^{2}} \mathrm{e}^{i(2n+1)z} \\
\theta_{3}(z|\tau) = \theta_{3}(z,q) = \sum_{n=-\infty}^{\infty} q^{n^{2}} \mathrm{e}^{i2nz} \\
\theta_{4}(z|\tau) = \theta_{4}(z,q) = \sum_{n=-\infty}^{\infty} (-1)^{n} q^{n^{2}} \mathrm{e}^{i2nz}
\end{equation}

b. Trigonometric forms
\begin{equation}
\theta_{1}(z|\tau) = \theta_{1}(z,q) = 2\sum_{n=0}^{\infty} (-1)^{n} q^{(n+1/2)^{2}} \sin[(2n+1)z] \\
\theta_{2}(z|\tau) = \theta_{2}(z,q) = 2\sum_{n=0}^{\infty} q^{(n+1/2)^{2}} \cos[(2n+1)z] \\
\theta_{3}(z|\tau) = \theta_{3}(z,q) = 1 + 2\sum_{n=1}^{\infty} q^{n^{2}} \cos(2nz) \\
\theta_{4}(z|\tau) = \theta_{4}(z,q) = 1 + 2\sum_{n=1}^{\infty} (-1)^{n} q^{n^{2}} \cos(2nz)
\end{equation}

Jacobi Elliptic Functions as Functions of Theta Functions

The modulus \(k\):
\begin{equation}
\sqrt{k} = \frac{\theta_{2}(0,q)}{\theta_{3}(0,q)}
\end{equation}
The complementary modulus \(k^{\prime}\):
\begin{equation}
\sqrt{k^{\prime}} = \frac{\theta_{4}(0,q)}{\theta_{3}(0,q)}
\end{equation}
3 of the 12 Jacobi elliptic functions:
\begin{align}
\mathrm{sn}(u,k) &= \frac{\theta_{3}(0,q)\theta_{1}(z,q)}{\theta_{2}(0,q)\theta_{4}(z,q)} \\
\mathrm{cn}(u,k) &= \frac{\theta_{4}(0,q)\theta_{2}(z,q)}{\theta_{2}(0,q)\theta_{4}(z,q)} \\
\mathrm{dn}(u,k) &= \frac{\theta_{4}(0,q)\theta_{3}(z,q)}{\theta_{3}(0,q)\theta_{4}(z,q)}
\end{align}
where \(z\) is the argument of the theta functions and \(u\) is the argument of the Jacobi elliptic functions. They are related by the following equation:
\begin{equation}
z=\frac{u}{\theta_{3}^{2}(0,q)}
\end{equation}

Proofs of Basic Properties of Jacobi Elliptic Functions

We assume all properties of theta functions as given. For proofs of these properties, see references 2 and 4 below.

1. First “Pythagorean” Identity
\begin{align}
k^{2} + {k^{\prime}}^{2} &=
\frac{\theta_{2}^{4}(0,q)}{\theta_{3}^{4}(0,q)} + \frac{\theta_{4}^{4}(0,q)}{\theta_{3}^{4}(0,q)}
= \frac{\theta_{2}^{4}(0,q) + \theta_{4}^{4}(0,q)}{\theta_{3}^{4}(0,q)} \\
&= \frac{\theta_{3}^{4}(0,q)}{\theta_{3}^{4}(0,q)} = 1
\end{align}
We used this identity in the last step.

2. Second “Pythagorean” Identity
\begin{align}
\mathrm{cn}^{2}(u,k) + \mathrm{sn}^{2}(u,k)
&= \frac{\theta_{4}^{2}(0,q)\theta_{2}^{2}(z,q)}{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} +
\frac{\theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)}{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} \\
&= \frac{\theta_{4}^{2}(0,q)\theta_{2}^{2}(z,q) + \theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)}
{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} \\
&= \frac{[\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q) \,-\, \theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)] + \theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)}{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} = 1
\end{align}
We used this identity in the last step.

3. Third “Pythagorean” Identity
\begin{align}
\mathrm{dn}^{2}(u,k) + k^{2}\mathrm{sn}^{2}(u,k)
&= \frac{\theta_{4}^{2}(0,q)\theta_{3}^{2}(z,q)}{\theta_{3}^{2}(0,q)\theta_{4}^{2}(z,q)}
+ \frac{\theta_{2}^{4}(0,q)}{\theta_{3}^{4}(0,q)}\,
\frac{\theta_{3}^{2}(0,q)\theta_{1}^{2}(z,q)}{\theta_{2}^{2}(0,q)\theta_{4}^{2}(z,q)} \\
&= \frac{\theta_{4}^{2}(0,q)\theta_{3}^{2}(z,q) + \theta_{2}^{2}(0,q)\theta_{1}^{2}(z,q)}
{\theta_{3}^{2}(0,q)\theta_{4}^{2}(z,q)} \\
&= \frac{[\theta_{3}^{2}(0,q)\theta_{4}^{2}(z,q) \,-\, \theta_{2}^{2}(0,q)\theta_{1}^{2}(z,q)]
\,+\, \theta_{2}^{2}(0,q)\theta_{1}^{2}(z,q)}
{\theta_{3}^{2}(0,q)\theta_{4}^{2}(z,q)} = 1
\end{align}
We used this identity in the last step.

Derivations of Derivatives of the 3 Basic Jacobi Elliptic Functions

1.
\begin{align}
\frac{\partial \,\mathrm{sn}(u,k)}{\partial u}
&= \frac{\theta_{3}(0,q)}{\theta_{2}(0,q)} \frac{\partial}{\partial z} \left(\frac{\theta_{1}(z,q)}{\theta_{4}(z,q)} \right) \frac{\partial z}{\partial u} \\
&= \frac{\theta_{3}(0,q)}{\theta_{2}(0,q)} \, \frac{\theta_{4}^{2}(0,q)\theta_{2}(z,q)\theta_{3}(z,q)}{\theta_{4}^{2}(z,q)} \, \frac{1}{\theta_{3}^{2}(0,q)} \\
&= \left[\frac{\theta_{4}(0,q)\theta_{2}(z,q)}{\theta_{2}(0,q)\theta_{4}(z,q)}\right] \,
\left[\frac{\theta_{4}(0,q)\theta_{3}(z,q)}{\theta_{3}(0,q)\theta_{4}(z,q)}\right]
= \mathrm{cn}(u,k)\mathrm{dn}(u,k)
\end{align}
The derivative of the ratio of theta functions in the first line was obtained from Gradshteyn and Ryzhik 8.199(2).1.

2.
\begin{equation}
\frac{\partial}{\partial u} [\mathrm{cn}^{2}(u,k) + \mathrm{sn}^{2}(u,k) = 1]
\end{equation}
\begin{equation}
2\mathrm{cn}(u,k)\frac{\partial}{\partial u}\mathrm{cn}(u,k) + 2\mathrm{sn}(u,k)\frac{\partial}{\partial u}\mathrm{sn}(u,k) = 0
\end{equation}
\begin{equation}
\frac{\partial}{\partial u}\mathrm{cn}(u,k) = -\frac{\mathrm{sn}(u,k)}{\mathrm{cn}(u,k)}\frac{\partial}{\partial u}\mathrm{sn}(u,k) = -\mathrm{sn}(u,k)\mathrm{dn}(u,k)
\end{equation}

3.
\begin{equation}
\frac{\partial}{\partial u} [\mathrm{dn}^{2}(u,k) + k^{2}\mathrm{sn}^{2}(u,k) = 1]
\end{equation}
\begin{equation}
2\mathrm{dn}(u,k)\frac{\partial}{\partial u}\mathrm{dn}(u,k) + 2k^{2}\mathrm{sn}(u,k)\frac{\partial}{\partial u}\mathrm{sn}(u,k) = 0
\end{equation}
\begin{equation}
\frac{\partial}{\partial u}\mathrm{dn}(u,k) = -k^{2}\frac{\mathrm{sn}(u,k)}{\mathrm{dn}(u,k)}\frac{\partial}{\partial u}\mathrm{sn}(u,k) = -k^{2}\mathrm{sn}(u,k)\mathrm{cn}(u,k)
\end{equation}

References

1. NIST Digital Library of Mathematical Functions
2. A Course of Modern Analysis – Whittaker and Watson
3. Gradshteyn and Ryzhik’s Table of Integrals, Series, and Products, 8th edition
4. Lectures on the Theory of Elliptic Functions – Hancock

Legendre-type relations for generalized complete elliptic integrals by Shingo Takeuchi

Legendre’s relation for the complete elliptic integrals of the first and second kinds is generalized. The proof depends on an application of the generalized trigonometric functions and is alternative to the proof for Elliott’s identity.

The entire paper is available here.

Evaluate the Integral \(\int_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}} \)

How to evalute
\begin{equation}
\int\limits_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}
\end{equation}
was a question posed at Mathematics Stack Exchange. Here is my solution

The integral is of the form of the polynomial under the radical having 1 real and 2 complex roots and the lower limit of integration equal to the real root. So we have
\begin{equation}
\int\limits_{a}^{x} \left( (t-a)[(t-r)^{2} + s^{2}] \right)^{-1/2}\, dt
= \frac{1}{\sqrt{m}} \mathrm{cn}^{-1}\left(\frac{m-(t-a)}{m+(t-a)},k \right)
\end{equation}
where
\begin{equation}
m^{2} = (r – a)^{2} + s^{2} \quad \mathrm{and} \quad k^{2} = \frac{1}{2} – \frac{a-r}{2m}
\end{equation}
\(\mathrm{cn}^{-1}z\) is the inverse of one of the Jacobi elliptic functions and \(k\) is the modulus.

To factor the polynomial inside of the radical, we note that the roots are the cube roots of 1, thus
\begin{align}
t^{3}-1 &= (t-t_{1})(t-t_{2})(t-t_{3}) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]-i\frac{\sqrt{3}}{2} \right) \left(\Big[t+\frac{1}{2}\Big]+i\frac{\sqrt{3}}{2} \right) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]^{2} + \frac{3}{4}\right)
\end{align}

We now have
\begin{equation}
a = 1 \quad r = -\frac{1}{2} \quad s = \frac{\sqrt{3}}{2} \quad m^{2} = 3 \quad k^{2} = \frac{1}{2} – \frac{3}{4\sqrt{3}}
\end{equation}
and thus
\begin{equation}
\int\limits_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}
= 3^{-1/4} \mathrm{cn}^{-1}\left(\frac{\sqrt{3} + 1 – x}{\sqrt{3} – 1 + x},\sqrt{\frac{1}{2} – \frac{3}{4\sqrt{3}}} \right)
\end{equation}