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.

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

1 Answer 1


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 $2^{-1/4}$.

  • $\begingroup$ I'm not quite familiar with Ramanujan class invariants. Could you please provide more detail? $\endgroup$
    – Wane
    May 16, 2021 at 3:52
  • 1
    $\begingroup$ @Wane: I have updated the answer with some details. But a full understanding requires a working knowledge of elliptic integrals, theta functions. $\endgroup$
    – Paramanand Singh
    May 16, 2021 at 4:22
  • $\begingroup$ Very nice solution. $\endgroup$ May 16, 2021 at 5:42
  • $\begingroup$ 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)? $\endgroup$
    – Wane
    May 16, 2021 at 7:01
  • 1
    $\begingroup$ @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. $\endgroup$
    – Paramanand Singh
    May 16, 2021 at 8:35

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