In his paper "Algebraic Relations between Certain Infinite Products" (Proceedings of the London Mathematical Society, 2, XVIII, 1920) Ramanujan mentioned about the functions $G(q)$ and $H(q)$ which are part of Rogers-Ramanujan Identities: \begin{align} G(q) &= 1 + \frac{q}{1 - q} + \frac{q^{4}}{(1 - q)(1 - q^{2})} + \frac{q^{9}}{(1 - q)(1 - q^{2})(1 - q^{3})} + \cdots\notag\\ &= \frac{1}{(1 - q)(1 - q^{6})(1 - q^{11})\cdots}\times\frac{1}{(1 - q^{4})(1 - q^{9})(1 - q^{14})\cdots}\tag{1}\\ H(q) &= 1 + \frac{q^{2}}{1 - q} + \frac{q^{6}}{(1 - q)(1 - q^{2})} + \frac{q^{12}}{(1 - q)(1 - q^{2})(1 - q^{3})} + \cdots\notag\\ &= \frac{1}{(1 - q^{2})(1 - q^{7})(1 - q^{12})\cdots}\times\frac{1}{(1 - q^{3})(1 - q^{8})(1 - q^{13})\cdots}\tag{2} \end{align} Ramanujan gave a proof of these identities in another paper (which is described in this blog post).

Ramanujan mentions further "I have now found an algebraic relation between $G(q)$ and $H(q)$, viz.": $$H(q)\{G(q)\}^{11} - q^{2}G(q)\{H(q)\}^{11} = 1 + 11q\{G(q)H(q)\}^{6}\tag{3}$$ which is equivalent to $$\frac{1}{R^{5}(q)} - 11 - R^{5}(q) = \frac{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{6}}{q\{(1 - q^{5})(1 - q^{10})(1 - q^{15})\cdots\}^{6}}\tag{4}$$ where $R(q) = q^{1/5}H(q)/G(q)$ and is proved here.

In his characteristic style he ends his paper with another formula without proof. To quote him: "Another noteworthy formula is $$H(q)G(q^{11}) - q^{2}G(q)H(q^{11}) = 1\tag{5}$$ Each of these formulae is the simplest of a large class."

Is there any proof available for the formula $(5)$? Are the formulas belonging to "a large class" as envisaged by Ramanujan discovered and proved in literature? Any references or a proof would be highly appreciated.

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Here's another one,

$$G(q^2)G(q^3)+qH(q^2)H(q^3) = \frac{1}{G(q^6)H(q)-qG(q)H(q^6)}$$

In fact, there are at least forty of these. See Berndt et al's:

Ramanujan's Forty Identities for the Rogers-Ramanujan Cfrac.

  • $\begingroup$ thanks for that paper. It seems to have exactly what i need. $\endgroup$ – Paramanand Singh Dec 7 '14 at 11:45

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