Entry 96

Level 5. Given $q = e^{2\pi i \tau}$, Ramanujan's theta function $f(a,b)$, its one-parameter version $f(-q) = f(-q,-q^2)$, and the q-Pochhammer symbol. We have the Rogers-Ramanujan sum-product identities,

$$\begin{align}H(q) &= \frac{f(-q,-q^4)}{f(-q)} \,=\, \sum_{n=0}^\infty \frac {q^{n^2+n}} {(q;q)_n} = \prod_{n=1}^\infty\frac{1}{(1-q^{5n-2})(1-q^{5n-3})}\\ G(q) &=  \frac{f(-q^2,-q^3)}{f(-q)} = \sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n}  = \prod_{n=1}^\infty\frac{1}{(1-q^{5n-1})(1-q^{5n-4})}\end{align}$$ Their ratio is the Rogers-Ramanujan continued fraction

$$\begin{align}R(q) &= \frac{a}{b} = \frac{q^{11/60} H(q)}{q^{-1/60}\,G(q)} = q^{1/5}\prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-2})}{(1-q^{5n-2})(1-q^{5n-3})}\\ &=\cfrac{q^{1/5}} {1+\cfrac{q} {1+\cfrac{q^2} {1+\cfrac{q^3} {1+\ddots}}}}\end{align}$$ Let $a=q^{11/60}H(q)$ and $b= q^{-1/60}\,G(q)$. Then $(a, b)$ for appropriate $\tau$ are actually radicals. Proof in Entry 97.