Entry 104

Given \(q = e^{2\pi i \tau}\) and the \(q\)-continued fraction discussed in Entry 95. $$\begin{align}\quad\text{K}_4(q) &= \frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\\ & = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+q+\cfrac{q^2} {1+q^2+\cfrac{q^3} {1+q^3+\ddots }}}}\end{align}$$

and compare it to the modular lambda function $$\lambda(\tau) = \left(\frac{\sqrt2\,\eta(\tfrac12\tau)\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8\quad$$ therefore the two are related by $$\text{K}_4(\tau) = \big(\lambda(2\tau)\big)^{1/8}$$ Mathematica can calculate \(\lambda(\tau)\) as ModularLambda[tau] and there is list of exact values in Mathworld which also implies exact values for \(\text{K}_4(\tau)\). But we will give a nice consistent form when \(d = 4m\) has class number \(h(-d)=2\) for \(m=5,13,37\) as $$\begin{align}\lambda\big(\sqrt{-5}\big) &= \frac12-\sqrt{\left(\tfrac{{-1}+\sqrt{5}}2\right)^3}\\ \lambda\big(\sqrt{-13}\big) &= \frac12-3\sqrt{\left(\tfrac{{-3}+\sqrt{13}}2\right)^3}\\ \lambda\big(\sqrt{-37}\big) &= \frac12-21\sqrt{\left({-6}+\sqrt{37}\right)^3}\end{align}$$ as well as the complex values $$\begin{align}\frac1{\lambda\left(\frac{1+\sqrt{-5}}2\right)} &= \frac12-\sqrt{\left(\tfrac{{-1}-\sqrt{5}}2\right)^3}\\ \frac1{\lambda\left(\frac{1+\sqrt{-13}}2\right)} &= \frac12-3\sqrt{\left(\tfrac{{-3}-\sqrt{13}}2\right)^3}\\ \frac1{\lambda\left(\frac{1+\sqrt{-37}}2\right)} &= \frac12-21\sqrt{\left({-6}-\sqrt{37}\right)^3}\end{align}$$