# Algebraic relations between certain infinite products

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.]

This work is in the public domain in the United States because it was published before January 1, 1923.

The author died in 1920, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 80 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works.