How to prove some identities about infinite product?

Question

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.

Answer 1

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 $RHS=2^{\frac{1}{6}}q^{\frac{1}{24}}(\frac{\theta_2 \theta_4}{\theta^2_3})^{-\frac{1}{6}}=(\frac{2\theta_3^2}{\theta_2\theta_4})^{\frac{1}{6}}q^{\frac{1}{24}}.$

It is enough to prove the following :

$$ \prod(1+q^{2n-1})^2=(\frac{2\theta_3^2}{\theta_2\theta_4})^{\frac{1}{3}}q^{\frac{1}{12}}$$ Put (1),(2),(3) into the above identity, Jacobi triple product Identity is obtained. Hence the result is established.