今研究中のやつ - 赤げふの数学

$\displaystyle |Lemniscatesine (x)|=2^{1/4} e^{-\pi /6} \left( \prod_{k=1}^{\infty} \left(1+e^{-4 \pi k}+2e^{-2\pi k} \cosh \pi \right) \right) \sqrt{\cosh \pi -1 }\sqrt{\dfrac{1 -\cos(2\pi x/\varpi )}{\cosh \pi +\cos (2\pi x /\varpi )}} \prod_{n=1}^{\infty} \dfrac{1+e^{-4\pi n} -2e^{-2\pi n} \cos (2\pi x/\varpi )}{\sqrt{\left( 1+e^{-4\pi n} +2e^{-2\pi n} \cosh (\pi )\cos(2\pi x/\varpi ) \right)^2 +4e^{-4\pi n} \sinh^{2} (\pi) \sin^{2} (2\pi x/\varpi ) }}$

$A=\cosh \pi$

$C=e^{-2\pi}$ と置くと

$\displaystyle | Lemniscatesine (\dfrac{\varpi}{2\pi} \arccos z)| =2^{1/4} C^{1/12} \left( \prod_{k=1}^{\infty} \left(1+C^{2k}+2C^{k} A \right) \right) \sqrt{A-1}\sqrt{\dfrac{1 -z}{A +z}} \prod_{n=1}^{\infty} \dfrac{1+C^{2n} -2C^{n} z}{\sqrt{\left( 1+C^{2n} +2C^{n} Az \right)^2 +4C^{2n} (A^{2}-1)(1-z^{2}) }}$