Theta function identities associated with Ramanujan's modular equations of degree 15 (Q601197)

The complete elliptic integral of the first kind \(K(k)\) is defined by \[ K(k)=\int_{0}^{\pi/2}\frac{\text{d}\phi}{\sqrt{1-k^2\sin^{2}\phi}}, \] where \(0<k<1\) is called \textit{the modulus} of \(K\); \(k'=\sqrt{1-k^2}\) is called the complementary modulus. Let \(K\), \(K'\), \(L\), \(L'\) denote complete elliptic integrals of the first kind with the moduli \(k\), \(k'\), \(l\) and \(l'\), respectively. \textit{A modular equation} of degree \(n\) is a relation between the moduli \(k\) and \(l\) which is implied by \(n\frac{K'}{K}=\frac{L'}{L}\). For example, \(n=2\) gives \(l^2(1+k)^2=4k\). Let \(K\), \(K'\), \(L_{1}\), \(L'_{1}\), \(L_{2}\), \(L'_{2}\), \(L_{3}\) and \(L'_{3}\) denote complete elliptic integrals of the first kind corresponding, in pairs, to moduli \(k=\sqrt{\alpha}\), \(l=\sqrt{\beta}\), \(\sqrt{\gamma}\), \(\sqrt{\delta}\) and their complementary moduli, respectively. Suppose that the following equalities hold: \[ n_{1}\frac{K'}{K}=\frac{L'_{1}}{L_{1}},\quad n_{2}\frac{K'}{K}=\frac{L'_{2}}{L_{2}},\quad n_{3}\frac{K'}{K}=\frac{L'_{3}}{L_{3}}, \] where \(n_{3}=n_{1}n_{2}\). Then a mixed modular equation of composite degree \(n_{3}\) is a relation between moduli \(\sqrt{\alpha}\), \(\sqrt{\beta}\), \(\sqrt{\gamma}\), \(\sqrt{\delta}\), induced by the above equalities. The current paper deals with the case \(n_{1}=3\), \(n_{2}=5\). S. Ramanujan stated, and B. Berndt proved several modular equations of degree \(15\). Using these modular equations and a method of parametrizations, B. Berndt derived several theta identities. In the paper under review the authors take the opposite direction of investigations. They derive independent, alternative proofs of many of these theta function identities, obtaining some new ones. These identities yield afterwards modular equations. For example, they prove two new modular equations of composite degree \(15\): \[ \begin{aligned} (\beta\gamma)^{1/8}-(\alpha\delta)^{1/8}=&1-(\alpha\beta\gamma\delta)^{1/8}-[(1-\alpha)(1-\beta)(1-\gamma)(1-\delta)]^{1/8}\\ =&[\alpha\delta(1-\beta)(1-\gamma)]^{1/8}+[\beta\gamma(1-\alpha)(1-\delta)]^{1/8}.\end{aligned} \]