How to prove the infinite product expression of this Jacobian elliptic $\operatorname{sn}$ function?

Question

How to prove the following infinite product expression?

\begin{align*} \operatorname{sn}(x,k)=\tanh \left(\frac{\pi\,\!x}{2 K(k')}\right)\prod\limits_{n=1}^{+\infty\,}\frac{ \tanh \left(\frac{\pi\left(2K(k) n-x\right)}{2K(k')}\right) \tanh \left(\frac{\pi \left(2K(k) n+x\right)}{2K(k')} \right)} {\tanh^2\left(\frac{\pi\,\!K(k)}{K(k')}\left(n-\frac{1}{2}\right)\right)} \end{align*}

Are there similar infinite product expressions for other Jacobian elliptic functions?

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NumberForm[JacobiSN[0.01, 0.2^2], 15]
NumberForm[
 Tanh[(\[Pi]*x)/(2 EllipticK[1 - k^2])]*
   NProduct[(
     Tanh[(\[Pi] (2 EllipticK[k^2] n - x))/(
       2 EllipticK[1 - k^2])] Tanh[(\[Pi] (2 EllipticK[k^2] n + x))/(
       2 EllipticK[1 - k^2])])/
     Tanh[(\[Pi]*EllipticK[k^2])/
        EllipticK[1 - k^2]*(n - 1/2)]^2 /. {k -> 0.2, x -> 0.01}, {n, 
     1, 20}, WorkingPrecision -> 15] /. {k -> 0.2, x -> 0.01}, 15]

Output

0.009999826667967988 + 0.0*I
0.00999982666796799

It is important to note that in Mathematica's default function, the second parameter of the Jacobian elliptic function sn is $m=k^2$.

Answer 1

Let's note the standard product formulas \begin{align} \text{sn}(u, k) &= \dfrac{2q^{1/4}}{\sqrt{k}}\sin\left(\dfrac{\pi u}{2K}\right)\prod_{n = 1}^{\infty}\dfrac{1 - 2q^{2n}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n}}{1 - 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}\tag{1}\\ \text{cn}(u, k) &=\dfrac{2q^{1/4}\sqrt{k'}}{\sqrt{k}}\cos\left(\dfrac{\pi u}{2K}\right)\prod_{n = 1}^{\infty}\dfrac{1 + 2q^{2n}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n}}{1 - 2q^{2n - 1}\cos\left(\dfrac{\pi u}{K}\right) + q^{4n - 2}}\tag{2} \end{align} where $q=\exp(-\pi K'/K) $.

From these we get the product for $\text{sc} (u, k) $ as $$\text{sc} (u, k) =\frac{1}{\sqrt{k'}}\tan\left(\frac{\pi u} {2K}\right)\prod_{n=1}^{\infty}\dfrac{1-2q^{2n}\cos\left(\dfrac{\pi u} {K} \right) +q^{4n}}{1+2q^{2n}\cos\left(\dfrac{\pi u} {K} \right) +q^{4n}}\tag{3}$$ Let's further observe Jacobi imaginary transformation $$\text{sn} (u, k) =-i\, \text{sc} (iu, k') $$ and then using $(3)$ we have the following product for $\text{sn} (u, k) $ $$\text{sn} (u, k) =\frac{1}{\sqrt{k}}\tanh\left(\frac{\pi u} {2K'}\right)\prod_{n=1}^{\infty}\dfrac{1-2q'^{2n}\cosh\left(\dfrac{\pi u} {K'} \right) +q'^{4n}}{1+2q'^{2n}\cosh\left(\dfrac{\pi u} {K'} \right) +q'^{4n}}$$ where $q'=\exp(-\pi K/K') $.

Next we need to observe that $$\frac{ 1-q'^{2n}\cosh(\pi u/K') +q'^{4n}}{1+q'^{2n}\cosh(\pi u/K') +q'^{4n}}=\frac{1-q'^{2n}\exp(-\pi u/K')}{1+q'^{2n}\exp(-\pi u/K')}\cdot \frac{1-q'^{2n}\exp(\pi u/K')} {1+q'^{2n}\exp(\pi u/K')} $$ and the right hand side can be written as $$\tanh\left(\frac{n\pi K} {K'} - \frac{\pi u} {2K'}\right)\tanh\left(\frac{n\pi K} {K'} +\frac{\pi u} {2K'}\right)$$ and hence we have the expansion $$\text{sn} (u, k) =\frac{1}{\sqrt{k}}\tanh\left(\frac{\pi u} {2K'}\right)\prod_{n=1}^{\infty}\tanh\left(\frac{n\pi K} {K'} - \frac{\pi u} {2K'}\right)\tanh\left(\frac{n\pi K} {K'} +\frac{\pi u} {2K'}\right)\tag{4}$$ Further note that we have the product expansion $$k'=\frac{\vartheta_4^2(q)}{\vartheta_3^2(q)}=\prod_{n=1}^{\infty}\left(\frac{1-q^{2n-1}}{1+q^{2n-1}}\right)^4$$ which leads to $$\sqrt{k} =\frac{\vartheta_4(q')}{\vartheta_3(q')}=\prod_{n=1}^{\infty}\left(\frac{1-q'^{2n-1}}{1+q'^{2n-1}}\right)^2=\prod_{n=1}^{\infty}\tanh^2\left(\frac{(2n-1)\pi K}{2K'}\right)\tag{5}$$ Using $(4),(5)$ we get the desired product expansion in question.

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For each elliptic function of the arguments $u, k$ we can apply Jacobi imaginary transformation and write it as a function of arguments $iu, k'$ and use their standard product representations to get expansions in terms of $\pi u/K'$.

Answer 2

https://en.wikipedia.org/wiki/Jacobi_elliptic_functions#Definition_in_terms_of_Jacobi_theta_functions gives expressions for the Jacobi elliptic functions as products of theta functions, and expressions for the theta functions as infinite products. E.g., $$ {\rm cn}(u;\,k)={\theta_{01}(q)\theta_{11}(\zeta;\,q)\over\theta_{11}(q)\theta_{01}(\zeta;\,q)} $$ where $$ \theta_{01}(v;\,w)=\prod_{n=1}^{\infty}(1-w^{2n})(1-2\cos(2v)w^{2n-1}+w^{4n-2}) $$ and $\theta_{01}(q)=\theta_{01}(0;\ q)$, and some equally complicated formula for $\theta_{11}$.

Answer 3

The first question is

How to prove the following infinite product expression? \begin{align*} \operatorname{sn}(x,k)=\tanh \left(\frac{\pi\,\!x}{2 K(k')}\right)\prod\limits_{n=1}^{+\infty\,}\frac{ \tanh \left(\frac{\pi\left(2K(k) n-x\right)}{2K(k')}\right) \tanh \left(\frac{\pi \left(2K(k) n+x\right)}{2K(k')} \right)} {\tanh^2\left(\frac{\pi\,\!K(k)}{K(k')}\left(n-\frac{1}{2}\right)\right)} \end{align*}

I don't know of any reference to such an expansion using hypberbolic tangent. The closest using just sine and cosine is from Gradshteyn and Ryzhik, Table of Integrals, Series, and Products, section $\bf{8.146.23}$ or other sources (where $\,q=e^{-\pi K(k')/K(k)}$)

$$ \text{sn} u = \frac{2\sqrt[4]{q}}{\sqrt{k}} \sin \frac{\pi u}{2K(k)} \prod_{n=1}^\infty \frac{1-2q^{2n}\cos \frac{\pi u}{K(k)}+q^{4n}} {1-2q^{2n-1}\cos \frac{\pi u}{K(k)}+q^{4n-2}}. $$

The key idea is that elliptic functions are doubly periodic with equal number of zeros and poles (up to multiplicity) in a period parallelogram. For the elliptic function $\,\text{sn}(z|m)\,$ there are two zeros at $\,z=0\,$ and $\,z=2K(m)\,$ and two poles at $\,z=iK'(m)\,$ and $\,z=2K(m)+iK'(m).\,$ The two periods are $\,4K(m)\,$ and $\,2iK'(m).\,$ The zeros and poles alternate along vertical lines of the period rectangles. As $\,m\to 1^{-}\,$ the real period $\,4K(m)\to\infty\,$ and $\,\text{sn}(z|m)\to \tanh(z).\,$ The function $\,\tanh(z)\,$ has zeros at $\,n\pi i\,$ and poles at $\,(n-\frac12)\pi i.\,$ Thus, It has an infinite product expansion as

$$ \tanh(z) = z \prod_{n=1}^\infty \frac{z^2 + (n\pi)^2} {z^2 + ((n-1/2)\pi)^2}\left(\frac{n}{n-1/2}\right)^{-2}. \tag1 $$

Thus, to a first approximation

$$ \text{sn}(z) := \text{sn}(z|m) \approx \tanh \left(\frac{z\,\pi}{2K'(m)}\right) \tag2 $$

and missing the rest of its zeros and poles.

Define the function

$$ T(z) := T(z|m) = \tanh\left(\frac{z\,\pi K(m)}{2K'(m)}\right). \tag3 $$

The infinite product expansion of sn is

$$ \text{sn}(zK(m)) = T(z)\prod_{n=1}^\infty \frac{T(2n-z)T(2n+z)}{T(2n-1)^2}. \tag4 $$

A quick check using Wolfram Mathematica:

z = 13/100; m = 21/100; M = 7; DP = 15;
K0 = EllipticK[m]; K1 = EllipticK[1-m];
T[z_] := Tanh[z*Pi/2*K0/K1];
N[{JacobiSN[z*K0, m], T[z]*Product[
   T[2*n - z]*T[2*n + z] /
   T[2*n - 1]^2, {n, 1, M}]}, DP] //NumberForm
(* {0.214382745916949, 0.214382745916948} *)

Note that in this expansion, the zeros and poles match which implies the two sides of the equation differ only by a constant factor. I don't yet have a proof that the constant factor is $1$ or of the identity

$$ k^{1/4} = m^{1/8} = \prod_{n=1}^\infty T(2n-1). \tag5 $$

The second question is

Are there similar infinite product expressions for other Jacobian elliptic functions?

The answer is yes. It is known that $$ \text{sc}(z,k) = -i\,\text{sn}(i\,z,k'), \quad \tanh(z) = -i\,\tan(i\,z) \tag6 $$

which means that a slightly modified version of the infinite product expansion for sn becomes the infinite product expansion for sc using tan instead of tanh.

I think there may be an expansion for $\text{sd}(z,k)$ but I will have to do some work for that.