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I want to prove that $$\frac{4}{\pi^2}K^2=\frac{1}{1+k^2}\sum_{n=0}^{\infty} \frac{(4n)!}{n!^4}\left(\frac{k(k^2-1)}{4(k^2+1)^2}\right)^{2n}$$

See notation reference here Different approach to the sum $\sum_{n=0}^{\infty}\frac{(4n)!}{n!^4} \frac{1}{4^{4n}}$

By the integral representation $$V(x)=\frac{4}{\pi^2}\int_0^1 \frac{K(xt^2)}{\sqrt{1-t^2}}\, dt$$

I am trying this approach, to prove that $$K^2=\frac{1}{1+k^2}\int_0^1 \frac{1}{\sqrt{1-t^2}}K\left(\frac{4k(k^2-1)t^2}{(k^2+1)^2}\right)\, dt$$

I tried switching integrals, and now I'm stuck

Any help is appreciated

  • I suspect you mean $K^2(k)$ where you have $K^2$ for both left-hand sides of the displays. – Eric Towers Jun 29 '24 at 03:08
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    I might start with squaring https://dlmf.nist.gov/15.9.E24 then mash together the product of ${}_2 F_1$s with https://dlmf.nist.gov/15.16.E3 . The general shapes seem to come together in the ballpark of where you want to go. – Eric Towers Jun 29 '24 at 03:37
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    This is consequence of the identity $${}{2}F{1}(a, b; a - b + 1; z) = (1 - z)^{1 - 2b}(1 + z)^{2b - a - 1},{2}F{1}\left[\frac{a - 2b + 1}{2}, \frac{a - 2b + 2}{2}; a - b + 1; \frac{4z}{(1 + z)^{2}}\right]$$ with $a=b=1/2,z=k^2$ followed by an application of Clausen's formula. The identity can be proved by showing that both sides satisfy same differential equation and are equal at $z=0$. – Paramanand Singh Jul 06 '24 at 14:44
  • @ParamanandSingh I will also have to use the Quadratic transformation $x \mapsto 4x(1-x)$. Regardless, thank you for your help –  Jul 21 '24 at 11:57
  • Yes you are right. I forgot to mention that quadratic thing. – Paramanand Singh Jul 21 '24 at 15:18
  • @ParamanandSingh I would like to point out that we don't actually have to use the mentioned formula. We just need to prove $$(1+z)^{2b-1-a} {}_2F_1\left[\frac{a-2b+1}{2}, \frac{a-2b+2}{2}; a-b+1 ;\frac{4z}{(1+z)^2}\right]={}_2F_1(a-2b+1,1-b;1+a-b;z)$$ Which is straight forward by expanding the left hand side as two taylor series and use Cauchy product.

    By that we can just let $a=b=\frac{1}{2}$

     –  Sep 28 '24 at 01:20

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As @ParamanandSingh commented. We will use the formula

$${}_2F_1\left({a},{b};{a-b+1};{z}\right)=(1-z)^{1-2b}(1+z)^{2b-1-a}{}_2F_1\left({\frac{a-2b+1}{2}},{\frac{a-2b+2}{2}};{a-b+1};{\frac{4z}{(1+z)^2}}\right)$$

The proof of the above formula could be seen in this math overflow post. Let $a=\frac{1}{2}$ and $b=\frac{1}{2}$, $z=k^2$, we immediately get

$$\frac{2}{\pi} K(k)=\frac{1}{\sqrt{1+k^2}} {}_2F_1 \left(\frac{1}{4},\frac{3}{4};1;\frac{4k^2}{(1+k^2)^2}\right)$$

While the quadratic transformation $x\mapsto 4x(1-x)$ gives

$${}_2F_1\left(a,b;\frac{a+b+1}{2};x\right)={}_2F_1\left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};4x(1-x)\right)$$

let $a=\frac{1}{4}$, $b=\frac{3}{4}$: $${}_2F_1\left(\frac{1}{4},\frac{3}{4};1;x\right)={}_2F_1\left(\frac{1}{8},\frac{3}{8};1;4x(1-x)\right)$$ But $V(z)={}_3F_2\left(\frac{1}{4},\frac{1}{2},\frac{3}{4};1,1;z^2\right)$ and by Clausen formula: $$V(z)={}_2F_1^2\left(\frac{1}{8},\frac{3}{8};1;z^2\right)$$

Letting $x=\frac{4k^2}{(1+k^2)^2}$ gives $$\frac{2}{\pi}\sqrt{1+k^2}K(k)=\sqrt{V\left(\frac{4k(1-k^2)}{(1+k^2)}\right)}$$ Which is the desired result.

Also we note that the result is only valid for $|k|\leq \sqrt{2}-1$ due to bijectivity.