Entry 158

In the previous entry, given the Jacobi theta functions $\vartheta_n(0,q)$ with nome $q = e^{\pi i\tau}$, modular lambda function $\lambda(\tau)$, and complete elliptic integral of the first kind $K(k) = \tfrac{\pi}2\, {_2F_1}\big(\tfrac12,\tfrac12,1,k^2)$ we proposed three identities, one as,

$$\left(\frac{\vartheta_2(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}}\right)^4 \overset{\color{red}?}=\lambda$$ where $\lambda = \lambda(\tau)$. Using the known definitions of $\vartheta_n(0,q)$ and $\lambda(\tau)$, the proposed identity implies 

$$\frac{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}{\eta^2(\tau)}=\left(\frac{\eta^2(\tau)}{\eta(\tfrac{\tau}2)\,\eta(2\tau)}\right)^4$$

For appropriate complex quadratics such as $\tau=\sqrt{-n}$, then the RHS is a radical. But if the numerator of the LHS has a closed-form (implied in this list), then this also gives a closed-form for the denominator $\eta^2(\tau)$ such as

$$\begin{align}\eta^2(\sqrt{-1}) &=\frac{\Gamma^2(\tfrac14)}{(2\pi)^{3/2}}\frac1{2^{1/2}}\\ \eta^2(\sqrt{-2}) &=\frac{\Gamma(\tfrac18)\Gamma(\tfrac38)}{(2\pi)^{3/2}}\frac1{2^{5/4}}\\ \eta^2(\sqrt{-3}) &=\frac{\Gamma^3(\tfrac13)}{(2\pi)^{2}}\frac{3^{1/4}}{2^{2/3}}\end{align}$$

and so on. Since an eta quotient is involved, then $\eta(\sqrt{-d})$ for $d$ with class number $1$ will behave quite orderly and will be discussed in the next entry.