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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.

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    $\begingroup$ Also posted to m.se, math.stackexchange.com/questions/3634037/… without notification to either site, a violation of site norms. $\endgroup$ Apr 20, 2020 at 12:26
  • $\begingroup$ I guess you want to say $q=e^{\pi i \tau}$... $\endgroup$
    – Xarles
    Apr 20, 2020 at 13:16
  • $\begingroup$ 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 $k'$. $\endgroup$
    – Will Sawin
    Apr 20, 2020 at 13:29
  • $\begingroup$ What is $\eta(q)$? $\endgroup$ Apr 20, 2020 at 23:45
  • $\begingroup$ The Dedekind $\eta$-function, most likely. en.wikipedia.org/wiki/Dedekind_eta_function $\endgroup$ Apr 20, 2020 at 23:48

1 Answer 1

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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.

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  • $\begingroup$ I think there is exponent of two missing on theta 3 after the RHS, thanks again for posting the proof. $\endgroup$
    – Dabed
    Apr 24, 2020 at 13:49
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    $\begingroup$ Yes, you are right. Thanks. $\endgroup$ Apr 24, 2020 at 13:56

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