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In celebration of Pi Day, I messed around with the following Ramanujan formula:

$$\sum_{n=0}^{\infty} \binom{2 n}{n}^3 \frac{42 n+5}{2^{12 n+4}} = \frac1{\pi} $$

It turns out that, through some identities such as the following:

$$\sum_{n=0}^{\infty} \binom{2 n}{n}^3 x^n = \frac{4 K\left(\frac{1}{2} \left(1-\sqrt{1-64 x}\right)\right)^2}{\pi ^2} $$

we can derive an analytical expression for the original sum. This analytical expression leads me to the following question:

How does one prove the following relation?

$$ 2 \sqrt{7} E(m) K(m) - (2+\sqrt{7}) K(m)^2 = \frac{\pi}{2} $$

where

$$K(m) = \int_0^{\pi/2} \frac{d \phi}{\sqrt{1-m \sin^2{\phi}}} $$ $$E(m) = \int_0^{\pi/2} d\phi \, \sqrt{1-m \sin^2{\phi}} $$

and

$$m = \frac{8-\sqrt{63}}{16} $$

It seems to me that there is some combination of the integrals that lends itself to a massive simplification, but I have yet to have found it. Thus, I pose this to see if any of my fellow integral killers can drudge up any insights I may have missed.

J. M. ain't a mathematician
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Ron Gordon
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  • Have you already exploited Legendre's relation? https://en.wikipedia.org/wiki/Legendre%27s_relation – Jack D'Aurizio Mar 15 '17 at 17:24
  • Fortunately this is one of the three series for $1/\pi$ for which Ramanujan provided a proof. The method of Ramanujan is explained in my posts http://paramanands.blogspot.com/2012/03/modular-equations-and-approximations-to-pi-part-1.html and http://paramanands.blogspot.com/2012/03/modular-equations-and-approximations-to-pi-part-2.html – Paramanand Singh May 18 '17 at 17:36

1 Answers1

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Gotcha: it is enough to exploit Legendre's relation and the fact that $$ m=\lambda^*(7)^2$$ with $\lambda^*$ being the elliptic lambda function.

Jack D'Aurizio
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    Thank you for the tips. It turns out that we also need the elliptic alpha function as well to complete the proof. http://mathworld.wolfram.com/EllipticAlphaFunction.html – Ron Gordon Mar 16 '17 at 02:14
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    @RonGordon: you are clearly right (and there probably is no need to remark that :) ). I just went to the Wikipedia page related with some particular $K/K'$ ratios and figured out the exact details later, but they are simple to collect once we notice such peculiar identity. Wonderful times we live in, everyone can be Ramanujan for five minutes :D – Jack D'Aurizio Mar 16 '17 at 02:19
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    I just love that I now understand the theory behind one of Ramanujan's insane sums. It demonstrates just how ridiculously far ahead he was. – Ron Gordon Mar 16 '17 at 02:20