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Tagged with elliptic-integrals hypergeometric-functions
2 questions
2
votes
2
answers
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How did Ramanujan find $\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=\frac{288\sqrt{5}}{5\pi}?$
The formula
$$\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=\frac{288\sqrt{5}}{5\pi}$$
(in older notation) appears as eq. 38 in Ramanujan's paper Modular equations ...
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5
votes
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answer
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What is the surface area of the finite part of the Cayley nodal cubic surface?
The Cayley nodal surface is defined by the equation $x^2+y^2+z^2-2xyz=1$. The finite part of the surface is the tetrahedral part bounded by the 4 nodes $(1,1,1)$, $(1,-1,-1)$, $(-1,1,-1)$, $(-1,-1,1)$....
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