Expression for infinite product

Question

can anyone show me how

$$\displaystyle\frac{4}{R}\displaystyle\Pi_{n=1}^{\infty} \left(\frac{1+R^{-4n}}{1+R^{-4n+2}}\right)^4= \frac{1}{R}\left(1+2 \sum_{n=1}^ {\infty} \frac{1}{R^{2n(n+1)}}\right)^2\left(1+2 \sum_{n=1}^{\infty}\frac{1}{R^{2n^{2}}}\right)^{-2} $$

The above identity is obtained by Komatu [page no, 58] (Y. Komatu, "A coefficient problem for function univalent in an annulus", Kodai math. Rep., 8, 1956, 49-70.)

Jacobi's triple product identity is $$\sum_{n=-\infty}^{\infty} x ^{n^{2}}y^{2n} = \prod_{m=1}^\infty (1-x^{2m})(1+x^{2m-1}y^2)(1+x^{2m-1}y^{-2}) $$

substitute $ x = \frac{1}{R^2}$ , $y= \frac{1}{R}$

$$ 1+ 2\sum_{n=1}^\infty \frac{1}{ R^{2n(n+1)}} = \prod_{n=1}^\infty (1- R^{-4m})(1+R^{-4m})(1+R^{-4m+4}) $$ how $ \;\;\; (1+R^{-4m})(1+R^{-4m+4}) = (1+R^{-4m})^2 $

If Jacobi's triple product identity is $$\sum_{n=-\infty}^{\infty} x ^{n^{2}}y^n = \prod_{m=1}^\infty (1-x^{2m})(1+x^{2m}y)(1+x^{2m}y^{-1}) $$

substitute $ x = \frac{1}{R^2}$ , $y= \frac{1}{R^2}$

$$ 1+ 2\sum_{n=1}^\infty \frac{1}{ R^{2n(n+1)}} = \prod_{m=1}^\infty (1- R^{-4m})(1+R^{-4m-2})(1+R^{-4m+2}) $$

for second equation, subs $ x= \frac{1}{R^2}, y= 1 $

$$ 1+ 2\sum_{n=1}^\infty \frac{1}{ R^{2n^{2}}} = \prod_{m=1}^\infty (1- R^{-4m})(1+R^{-4m})^2 $$

Answer 1

First of all, the original identity has an exponent of $4$ on the left hand side that you have missed. Moreover it is not correct as stated, by simply comparing constant terms of both sides. However it is just a simple typo probably, because the following is true $$\frac{4}{R}\prod_{n=1}^{\infty} \left(\frac{1+R^{-4n}}{1+R^{-4n+2}}\right)^4= \frac{1}{R}\left(2+2 \sum_{n=1}^ {\infty} \frac{1}{R^{2n(n+1)}}\right)^2\left(1+2 \sum_{n=1}^{\infty}\frac{1}{R^{2n^{2}}}\right)^{-2}.$$ To prove it you just have to use some specializations of Jacobi's triple product identity. Indeed it says $$\sum_{n=-\infty}^{\infty} x^{n^2}y^n=\prod_{m=1}^{\infty}(1-x^{2m})(1+x^{2m}y)(1+x^{2m}y^{-1})$$ substituting $x=\frac{1}{R^2}$ and $y=\frac{1}{R^2}$ gives $$2+2\sum_{n=1}^{\infty} \frac{1}{R^{2n(n+1)}}=2\prod_{m=1}^{\infty} \left(1-\frac{1}{R^{4m}}\right)\left(1+\frac{1}{R^{4m}}\right)^2$$ and substituting $x=\frac{1}{R^2}$ and $y=1$ gives $$1+2\sum_{n=1}^{\infty}\frac{1}{R^{2n^2}}=\prod_{m=1}^{\infty}\left(1-\frac{1}{R^{4m}}\right)\left(1+\frac{1}{R^{4m-2}}\right)^2.$$ Komatu's identity follows by dividing the first equation by the second and squaring.

Answer 2

As the other answer pointed out, with the typo fixed, the equation is $$\frac{4}{R}\prod_{n=1}^{\infty} \left(\frac{1+R^{-4n}}{1+R^{-4n+2}}\right)^4= \frac{1}{R}\left(2+2 \sum_{n=1}^ {\infty} \frac{1}{R^{2n(n+1)}}\right)^2\left(1+2 \sum_{n=1}^{\infty}\frac{1}{R^{2n^{2}}}\right)^{-2}.$$ After replacing $R$ with $q^{-1/2}$ we get $$4q^{1/2}\prod_{n=1}^{\infty} \left(\frac{1+q^{2n}}{1+q^{2n-1}}\right)^4 = 4q^{1/2}\left(1+ \sum_{n=1}^ {\infty} q^{n(n+1)}\right)^2 \Big{/} \left(1+2 \sum_{n=1}^{\infty}q^{n^{2} }\right)^2.$$ The left side is $k(q)$ and the right side is $\theta_2(0,q)^2/\theta_3(0,q)^2$ which is a standard equation. See, for instance, DLMF equation 20.9.1 and Abramowitz and Stegun equations 16.38.5 and 16.38.7.