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.