#lemniscateconstant — Public Fediverse posts
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Let \(\varpi=\dfrac{\Gamma^2\left(\frac14\right)}{2\sqrt{2\pi}}=2.62205755\ldots\) be the lemniscate constant. Then,
\[\Large\displaystyle\sum_{n=1}^\infty\dfrac{1}{\sinh^4(\pi n)}=\dfrac{\varpi^4}{30\pi^4}+\dfrac{1}{3\pi}-\dfrac{11}{90}\]#Series #Sum #InfiniteSum #LemniscateConstant #GammaFunction #Lemniscate #LemniscateOfBernoulli #Bernoulli #Math #Maths #InfiniteSeries #HyperbolicSines
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A lattice sum with a surprising closed form:
\[\Large\displaystyle\sum_{\substack{(m,n)\in\mathbb Z^2\\(m,n)\neq0}}\dfrac{1}{(m+ni)^4}=\dfrac{\varpi^4}{15}=\dfrac{\Gamma^8\left(\frac14\right)}{960\pi^2}\]
#LatticeSum #ClosedForm #LemniscateConstant #Series #Sum #GammaFunction #Pi -
A lattice sum with a surprising closed form:
\[\Large\displaystyle\sum_{\substack{(m,n)\in\mathbb Z^2\\(m,n)\neq0}}\dfrac{1}{(m+ni)^4}=\dfrac{\varpi^4}{15}=\dfrac{\Gamma^8\left(\frac14\right)}{960\pi^2}\]
#LatticeSum #ClosedForm #LemniscateConstant #Series #Sum #GammaFunction #Pi -
An interesting infinite product!
\[\displaystyle\prod_{n=1}^\infty\coth^{(-1)^n}\left(\dfrac{\pi n}{2}\right)=\dfrac{\sqrt\pi}{\sqrt[4]2\sqrt{\varpi}}\]
where \(\varpi=2.62205755\ldots\) is the lemniscate constant (the ratio of the perimeter of Bernoulli's lemniscate to its diameter, analogous to the definition of \(\pi\) for the circle). Both \(\varpi\) and \(\pi\) are proven to be transcendental.
#lemniscate #lemniscateconstant #product #infiniteproduct #HyperbolicFunction #HyperbolicCotangent #Function #pi #maths #math