Saturday, May 24, 2025

Entry 88

The modular lambda function \(\lambda(\tau)\) $$\lambda(\tau) = \left(\frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)}\right)^8$$ discussed in the previous post solves, among other things, $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)}  = -\tau\sqrt{-1}$$ For example, let \(\tau=\sqrt{-2}\) so $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)}  = \sqrt{2}$$

and the Mathematica command ModularLambda[tau] yields a real number equal to \(\lambda\big(\tau) = (1-\sqrt2)^2\). Other \(\tau=\sqrt{n}\) can be found in Mathworld's list. But we can use more general complex \(\tau\). For example, let \(\tau=\frac{1+\sqrt{-2}}2\) which is no longer in the list. So $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-\lambda(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)} = {-\left(\tfrac{1+\sqrt{-2}}2\right)} \sqrt{-1}$$ and we find the complex number, $$\lambda(\tau) = 4\big({-1}+\sqrt2\big)^3\left(4+\sqrt{2(1-5\sqrt2)}\right) \approx 1.1370849 + 0.9905592 i$$ which is a root of quartic with two real roots and two complex roots. And so on for other complex quadratic irrationals \(\tau\).

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