# How to prove some identities about infinite product?

Recently, I read one paper titled Modular equations and approximations to π by Ramanujan, in which there are some formulas for $$q=\pi i \tau$$( where $$\tau=x+yi, y>0$$, hence $$|q|<1)$$ :

$$\prod_{n=1}^\infty\left(1+q^{2n-1}\right)=2^{\frac{1}{6}} q^{\frac{1}{24}}(kk')^{-\frac{1}{12}} ~~~ (1)$$

and $$\prod_{n=1}^\infty\left(1-q^{2n-1}\right)= 2^{\frac{1}{6}} q^{\frac{1}{24}}k^{-\frac{1}{12}}k'^{\frac{1}{6}} ~~~~(2)$$

where $$k=k(\tau)$$ is the Jacobi modulus， $$k^2(\tau)=\lambda(\tau)$$, the elliptic modular function, and $$k'=\sqrt{1-k^2}.$$

The following result can be calculated by Mathematica: $$\left(1+e^{-\pi }\right)\left(1+e^{-3 \pi }\right)\left(1+e^{-5 \pi }\right) \cdots=2^{\frac{1}{4}} e^{-\pi / 24}.$$

But I do not know how to prove these formulas (1) and (2). I would appreciate if someone could give some suggestions.

• Also posted to m.se, math.stackexchange.com/questions/3634037/… without notification to either site, a violation of site norms. – Gerry Myerson Apr 20 at 12:26
• I guess you want to say $q=e^{\pi i \tau}$... – Xarles Apr 20 at 13:16
• Second one is $q^{1/24} \eta(q) / \eta(q^2)$ and first one is $q^{1/24} (\eta(q^2)/ \eta(q) )/ (\eta(q^4)/ \eta(q^2))$ by matching terms in infinite products. So you want to match the eta quotients with the $k$ and $k'$. – Will Sawin Apr 20 at 13:29
• What is $\eta(q)$? – Jacob.Z.Lee Apr 20 at 23:45
• The Dedekind $\eta$-function, most likely. en.wikipedia.org/wiki/Dedekind_eta_function – Gerry Myerson Apr 20 at 23:48

First， by theta function we have $$k=\frac{\theta_2}{\theta_3},k'=\frac{\theta_4}{\theta_3},$$ where $$\theta_2=2q^{\frac{1}{4}} G \prod (1+q^{2n})^2; ~(1)$$ $$\theta_3= G \prod (1+q^{2n-1})^2;(2)$$ $$\theta_4= G \prod (1-q^{2n-1})^2;(3)$$ and $$G= \prod (1-q^{2n})^2.$$
So we have $$RHS=2^{\frac{1}{6}}q^{\frac{1}{24}}(\frac{\theta_2 \theta_4}{\theta^2_3})^{-\frac{1}{6}}=(\frac{2\theta_3^2}{\theta_2\theta_4})^{\frac{1}{6}}q^{\frac{1}{24}}.$$
$$\prod(1+q^{2n-1})^2=(\frac{2\theta_3^2}{\theta_2\theta_4})^{\frac{1}{3}}q^{\frac{1}{12}}$$ Put (1),(2),(3) into the above identity, Jacobi triple product Identity is obtained. Hence the result is established.