# Is $\prod_{n=0}^\infty \left(1-\frac{1}{\cosh ^2((n+1/2)\pi)}\right)=\frac{1}{\sqrt[4]{2}}$ true?

The infinite product $$\prod_{n=0}^\infty \left(1-\frac{1}{\cosh ^2((n+1/2)\pi)}\right)$$ agrees with $$\frac{1}{\sqrt[4]{2}}$$ to at least 100 decimal places.

The "identity" is reminiscent of $$\sqrt[4]{1-\lambda (i)}=\frac{1}{\sqrt[4]{2}}$$ where $$\lambda$$ is the modular lambda function. I tried to use $$\theta_3(z|\tau)=\theta_3(0,\tau)\prod_{n=0}^\infty \frac{\cos ((n+1/2)\pi\tau+z)\cos ((n+1/2)\pi\tau -z)}{\cos^2 ((n+1/2)\pi \tau)}$$ where $$z,\tau\in\mathbb{C}$$ and $$\operatorname{Im}\tau\gt 0$$.

$$\theta_3(\pi/2|i)$$ leads to $$\prod_{n=0}^\infty \tanh^2 \left(\left(n+\frac{1}{2}\right)\pi\right)$$ and $$\theta_3(i|i)$$ leads to $$\prod_{n=0}^\infty \cosh \left(1-\left(n+\frac{1}{2}\right)\pi\right)\cosh\left(1+\left(n+\frac{1}{2}\right)\pi\right)\operatorname{sech}^2\left(\left(n+\frac{1}{2}\right)\pi\right)$$ but these products don't seem to give the answer.

• Your product can be written as $\prod_{n\geq 0}\tanh^2((n+(1/2))\pi)$. May 16, 2021 at 3:33
• A similar (but somewhat easier) product has been evaluated here. May 16, 2021 at 5:06

If $$q=e^{-\pi}$$ then we can see that your product equals $$\prod_{n\geq 1}\frac{(1-q^{2n-1})^2}{(1+q^{2n-1})^2}$$ which equals $$(g_1/G_1)^2$$ where $$g, G$$ represent Ramanujan class invariants. This is indeed $$2^{-1/4}$$ as $$G_1=1,g_1=2^{-1/8}$$.

A little amount of explanation about Ramanujan class invariants is necessary here.

Let $$k\in(0,1)$$ be the elliptic modulus and define complete elliptic integral of first kind $$K(k) =\int_{0}^{\pi/2}\frac{dx} {\sqrt{1-k^2\sin^2x}}\tag{1}$$ Further we define complementary modulus $$k'=\sqrt{1-k^2}$$ and the expressions $$K(k), K(k')$$ are usually denoted by $$K, K'$$ if the value of $$k$$ is known from context.

A lot of magic is hidden in the elliptic modulus $$k$$ and the value of $$k$$ can be obtained if $$K, K'$$ are given using nome $$q=\exp (-\pi K'/K)$$.

We have $$k=\frac{\vartheta_2^2(q)}{\vartheta_3^2(q)}, k'=\frac{\vartheta_4^2(q)}{\vartheta_3^2(q)}\tag{2}$$ where \begin{align} \vartheta_2(q)&=\sum_{n\in\mathbb {Z}} q^{(n+(1/2))^2}\notag\\ &=2q^{1/4}\prod_{n=1}^{\infty}(1-q^{2n})(1+q^{2n})^2 \tag{3a}\\ \vartheta_3(q)&=\sum_{n\in\mathbb {Z}} q^{n^2}\notag\\ &=\prod_{n=1}^{\infty} (1-q^{2n})(1+q^{2n-1})^2\tag{3b}\\ \vartheta_4(q) &= \vartheta_3 (-q) \tag{3c} \end{align} are theta functions of Jacobi with one parameter being $$0$$. The equality of series and product expressions above is due to Jacobi Triple Product identity.

Ramanujan defined his class invariants $$g_N, G_N$$ using functions $$g, G$$ as \begin{align} g(q) &=2^{-1/4}q^{-1/24}\prod_{n=1}^{\infty} (1-q^{2n-1})\tag{4a}\\ G(q)&=2^{-1/4}q^{-1/24}\prod_{n=1}^{\infty} (1+q^{2n-1})\tag{4b}\\ g_N &=g(\exp(-\pi\sqrt{N}))\tag{4c}\\ G_N &=G(\exp (-\pi\sqrt{N})) \tag{4d} \end{align} where $$N$$ is a positive rational number.

It can be proved using the product expressions for theta functions that $$g(q) =(2k/k'^2)^{-1/12},G(q)=(2kk')^{-1/12}\tag{5}$$ It can also be proved with some effort (say using the theory of modular equations) that if $$N$$ is a positive rational number and $$q=\exp(-\pi\sqrt{N})$$ then the values of $$k, k'$$ are algebraic and hence $$G_N, g_N$$ are also algebraic numbers.

If $$N=1$$ we have $$q=e^{-\pi} =\exp (-\pi K'/K)$$ so that $$K'=K$$ and $$k'=k=1/\sqrt{2}$$ and from $$(5)$$ we get $$g_1=2^{-1/8},G_1=1$$. Using these values the product in question evaluates to $$2^{-1/4}$$.

• I'm not quite familiar with Ramanujan class invariants. Could you please provide more detail?
– Wane
May 16, 2021 at 3:52
• @Wane: I have updated the answer with some details. But a full understanding requires a working knowledge of elliptic integrals, theta functions. May 16, 2021 at 4:22
• Very nice solution. May 16, 2021 at 5:42
• Is it possible to evaluate $\prod_{n=0}^\infty \left(1-\frac{r}{\cosh^2((n+1/2)\pi)}\right)$ for all $r\in\mathbb{Q}$ in this way (are the products even algebraic)?
– Wane
May 16, 2021 at 7:01
• @Wane: I don't know for sure. If it can be expressed in terms of theta functions then one may hope for some evaluation in closed form. In your original question if $\pi$ is replaced by $\pi\sqrt{N}, N>0,N\in\mathbb {Q}$ then the technique of my answer works. May 16, 2021 at 8:35