Recently, I read one paper titled [Modular equations and approximations to π](http://ramanujan.sirinudi.org/Volumes/published/ram06.pdf) 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.



How to prove some identities about infinite product?
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.


