Algebraic relations between certain infinite products

(Proceedings of the London Mathematical Society, 2, xviii, 1920, Records for 13 March 1919)

It was proved by Prof. L. J. Rogers[1] that

{\displaystyle {\begin{aligned}\scriptstyle {G(x)}&\scriptstyle {=1+{\frac {1}{1-x}}+{\frac {x^{4}}{(1-x)(1-x^{2})}}+{\frac {x^{9}}{(1-x)(1-x^{2})(1-x^{3})}}+\ldots }\\&\scriptstyle {={\frac {1}{(1-x)(1-x^{6})(1-x^{11})}}\ldots \times {\frac {1}{(1-x^{4})(1-x^{9})(1-x^{14})\ldots }},}\end{aligned}}}

and

{\displaystyle {\begin{aligned}\scriptstyle {H(x)}&\scriptstyle {=1+{\frac {x^{2}}{1-x}}+{\frac {x^{6}}{(1-x)(1-x^{2})}}+{\frac {x^{12}}{(1-x)(1-x^{2})(1-x^{3})}}+\ldots }\\&\scriptstyle {={\frac {1}{(1-x^{2})(1-x^{7})(1-x^{12})\ldots }}\times {\frac {1}{(1-x^{3})(1-x^{8})(1-x^{13})\ldots }}.}\end{aligned}}}

Simpler proofs were afterwards found by Prof. Rogers and myself[2].

I have now found an algebraic relation between ${\displaystyle \scriptstyle {G(x)}}$ and ${\displaystyle \scriptstyle {H(x)}}$, viz.:

${\displaystyle \scriptstyle {H(x)\{G(x)\}^{11}-x^{2}G(x)\{H(x)\}^{11}=1+11x\{G(x)H(x)\}^{6}}}$.

Another noteworthy formula is

${\displaystyle \scriptstyle {H(x)G(x^{11})-x^{2}G(x)H(x^{11})=1}}$.

Each of these formulæ is the simplest of a large class.

1. Proc. London Math. Soc., Ser. 1, Vol. xxv, 1894, pp. 318–343.
2. Proc. Camb. Phil. Soc., Vol. xix, 1919, pp. 211–216. A short account of the history of the theorems is given by Mr Hardy in a note attached to this paper. [For Ramanujan's proofs see No. 26 of this volume: those of Rogers, and the note by Hardy referred to, are reproduced in the notes on No. 26 in the Appendix.]