#hyperboliccotangent — Public Fediverse posts
Live and recent posts from across the Fediverse tagged #hyperboliccotangent, aggregated by home.social.
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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
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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
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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
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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
-
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