Modular lambda function: Difference between revisions

Revision as of 19:57, 9 December 2021

Revision as of 20:01, 9 December 2021

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It is the [[Square (algebra)|square]] of the [[Jacobi modulus]],<ref name=C108>Chandrasekharan (1985) p.108</ref> that is, <math>\lambda(\tau)=k^2(\tau)</math>. In terms of the [[Dedekind eta function]] <math>\eta(\tau)</math> and [[theta function]]s,<ref name=C108/>

:<math> \lambda(\tau) = \Bigg(\frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\eta^2(2\tau)}{\eta^3(\tau)}\Bigg)^8 = \frac{16}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^8 + 16} =\frac{\theta_2^4(0,\tau)}{\theta_3^4(0,\tau)} </math>

and,

:<math> \frac{1}{\big(\lambda(\tau)\big)^{1/4}}-\big(\lambda(\tau)\big)^{1/4} = \frac{1}{2}\left(\frac{\eta(\tfrac{\tau}{4})}{\eta(\tau)}\right)^4 = 2\,\frac{\theta_4^2(0,\tfrac{\tau}{2})}{\theta_2^2(0,\tfrac{\tau}{2})}</math>

where<ref name=C63>Chandrasekharan (1985) p.63</ref>

:<math>\theta_2(0,\tau) =\sum_{n=-\infty}^\infty e^{\pi i\tau (n+1/2)^2}</math>

:<math>\theta_3(0,\tau) = \sum_{n=-\infty}^\infty e^{\pi i\tau n^2}</math>

:<math>\theta_4(0,\tau) = \sum_{n=-\infty}^\infty (-1)^n e^{\pi i\tau n^2} </math>

In terms of the half-periods of [[Weierstrass's elliptic functions]], let <math>[\omega_1,\omega_2]</math> be a [[fundamental pair of periods]] with <math>\tau=\frac{\omega_2}{\omega_1}</math>.

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The values of λ*(x) can be computed as follows:

:<math>\lambda^*(x) = \frac{\theta^2_2(0,i\sqrt{x})}{\theta^2_3(0,i\sqrt{x})} </math>

:<math>\lambda^*(x) = \left[\sum_{a=-\infty}^\infty\exp[-(a+1/2)^2\pi\sqrt{x}]\right]^2\left[\sum_{a=-\infty}^\infty\exp(-a^2\pi\sqrt{x})\right]^{-2} </math>