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 $$\sum_{n=-\infty}^{\infty} x ^{n^{2}}y^{2n} = \pi_{m=1}^\infty (1-x^{2m})(1+x^{2m-1}y^2)(1+x^{2m-1}y^-2) $$$$\sum_{n=-\infty}^{\infty} x ^{n^{2}}y^{2n} = \prod_{m=1}^\infty (1-x^{2m})(1+x^{2m-1}y^2)(1+x^{2m-1}y^{-2}) $$
substitute $ x = \frac{1}{R^2}$ , $y= \frac{1}{R}$
$$ 1+ 2\sum_{n=1}^\infty \frac{1}{ R^{2n(n+1)}} = \pi_{n=1}^\infty (1- R^{-4m})(1+R^{-4m})(1+R^{-4m+4}) $$$$ 1+ 2\sum_{n=1}^\infty \frac{1}{ R^{2n(n+1)}} = \prod_{n=1}^\infty (1- R^{-4m})(1+R^{-4m})(1+R^{-4m+4}) $$ how $ \;\;\; (1+R^{-4m})(1+R^{-4m+4}) = (1+R^{-4m})^2 $
If Jacobi's triple product identity is $$\sum_{n=-\infty}^{\infty} x ^{n^{2}}y^n = \pi_{m=1}^\infty (1-x^{2m})(1+x^{2m}y)(1+x^{2m}y^-1) $$$$\sum_{n=-\infty}^{\infty} x ^{n^{2}}y^n = \prod_{m=1}^\infty (1-x^{2m})(1+x^{2m}y)(1+x^{2m}y^{-1}) $$
substitute $ x = \frac{1}{R^2}$ , $y= \frac{1}{R^2}$
$$ 1+ 2\sum_{n=1}^\infty \frac{1}{ R^{2n(n+1)}} = \pi_{m=1}^\infty (1- R^{-4m})(1+R^{-4m-2})(1+R^{-4m+2}) $$$$ 1+ 2\sum_{n=1}^\infty \frac{1}{ R^{2n(n+1)}} = \prod_{m=1}^\infty (1- R^{-4m})(1+R^{-4m-2})(1+R^{-4m+2}) $$
for second equation, subs $ x= \frac{1}{R^2}, y= 1 $
$$ 1+ 2\sum_{n=1}^\infty \frac{1}{ R^{2n^{2}}} = \pi_{m=1}^\infty (1- R^{-4m})(1+R^{-4m})^2 $$$$ 1+ 2\sum_{n=1}^\infty \frac{1}{ R^{2n^{2}}} = \prod_{m=1}^\infty (1- R^{-4m})(1+R^{-4m})^2 $$
