Entry 105

Let $q = e^{2\pi i\tau}$. For convenience, define the Rogers-Ramanujan continued fraction in terms of $\tau$, hence $R(q) = R(\tau)$. Given the Dedekind eta function $\eta(\tau)$, it is known that $$\frac1{R(\tau)}-R(\tau) = \frac{\eta(\tau/5)}{\eta(5\tau)}+1$$ Thus a quadratic $$R(\tau) = \frac{- \frac{\eta(\tau/5)}{\eta(5\tau)}-1+\sqrt{\left( \frac{\eta(\tau/5)}{\eta(5\tau)}+1\right)^2+4}}2$$ For example, let $\tau = \sqrt{-1}$, then the formula gives us Ramanujan's evaluation, $$R(\sqrt{-1}) = \sqrt[4]5\sqrt{\phi}-\phi = 0.284079\dots$$ with golden ratio $\phi = \frac{1+\sqrt5}2$. The formula also works when $\tau = \frac{1+\sqrt{-n}}2$ though $R(\tau)$ is now negative. Since it contains $q^{1/5}$, one has to be careful with $5$th roots. We give some nice new evaluations for discriminants $d=5m$ with class number $h(-d)=2$ for $m=(3,7,23,47)$ also using the golden ratio $\phi$

$$\begin{align}\frac1{R^5\Big(\tfrac{1+\sqrt{-3/5}}2\Big)} -R^5\Big(\tfrac{1+\sqrt{-3/5}}2\Big)-11 &= -(\sqrt5)^3\phi\\ \frac1{R^5\Big(\tfrac{1+\sqrt{-7/5}}2\Big)} -R^5\Big(\tfrac{1+\sqrt{-7/5}}2\Big)-11 &= -(\sqrt5)^3\phi^3\\ \frac1{R^5\Big(\tfrac{1+\sqrt{-23/5}}2\Big)} -R^5\Big(\tfrac{1+\sqrt{-23/5}}2\Big)-11 &= -(\sqrt5)^3\phi^9\\ \frac1{R^5\Big(\tfrac{1+\sqrt{-47/5}}2\Big)} -R^5\Big(\tfrac{1+\sqrt{-47/5}}2\Big)-11 &= -(\sqrt5)^3\phi^{15}\\ \end{align}$$ Note the $5$th powers $R^5(\tau)$ and how well-behaved their evaluations are.