Entry 157

Given the Jacobi theta functions \(\vartheta_n(0,q)\) which traditionally uses the nome \(q = e^{\pi i\tau}\). Define the modular lambda function \(\lambda(\tau)\),

$$\lambda(\tau) = \frac{16}{\left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+16} = \left(\frac{\sqrt2\,\eta(\tau/2)\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 = \left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^4$$

Then we propose the three identities below and for appropriate \(\tau\) such as \(\tau = \sqrt{-d}\) and \(\lambda = \lambda(\tau)\) that the ratios below are radicals,

$$\begin{align}\left(\frac{\vartheta_2(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}}\right)^4 &\overset{\color{red}?}=\lambda\\ \left(\frac{\vartheta_4(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}}\right)^4 &\overset{\color{red}?}=1-\alpha\\ \left(\frac{\vartheta_3(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}}\right)^4 &\overset{\color{red}?}=1\end{align}$$

Adding the first two implies the third. Hence, after removing the common denominator

$$\big(\vartheta_2(0,q)\big)^4+\big(\vartheta_4(0,q)\big)^4 = \big(\vartheta_3(0,q)\big)^4$$

which is known to be true. As eta quotients in the same order above, 

$$\left(\frac{2\eta^2(2\tau)}{\eta(\tau)}\right)^4+\left(\frac{\eta^2\big(\tfrac{\tau}2\big)}{\eta(\tau)}\right)^4 = \left(\frac{\eta^5(\tau)}{\eta^2\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}\right)^4$$

Also, the equalities $$\vartheta_3(0,q) = \sum_{m=-\infty}^\infty q^{m^2} = \frac{\eta^5(\tau)}{\eta^2\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}$$which has a cubic version given in Entry 156.