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It is known that hypergeometric functions are closely related to the formula of $\pi$ given by Ramanujan. Trying to master the proof given by the Borwein brothers, I got two identities:

$$\mathrm{A}:(1+k^2)\left(\frac{2K(k)}{\pi}\right)^2={}_3 F_2\left(\begin{matrix}1/4& 3/4& 1/2\\1& 1\end{matrix};\frac{16k^2(1-k^2)^2}{(1+k^2)^4}\right)$$

with $0\leq k\leq\sqrt{2}-1$ and

$$\mathrm{B}:(\frac{1+k^2}{2})\left(\frac{2K^{\prime}(k)}{\pi}\right)^2={}_3 F_2\left(\begin{matrix}1/4& 3/4& 1/2\\1& 1\end{matrix};\frac{16k^2(1-k^2)^2}{(1+k^2)^4}\right)$$

with $\sqrt{2}-1\leq k\leq\sqrt{2}+1$, where $K$ and $K^\prime=K(\sqrt{1-k^2})$ are complete elliptic integrals of the first kind, $F$ is generalized hypergeometric function.

Question: Does A imply B(or B imply A)?

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zy_
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  • As written, these can't both be true. Since the RHS of each are the same, both being true amounts to $$(1+k^2)\left(\frac{2K(k)}{\pi}\right)^2=\frac{1+k^2}{2}\left(\frac{2K'(k)}{\pi}\right)^2$$ which simplifies to $K'(k)=\frac{1}{\sqrt{2}}K(k)$. But this would mean that $K'(k)$ is an exponential function in $k$, which is false. – Semiclassical Nov 02 '14 at 13:57
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    Eh, $K^{\prime}$ does not represent the derivative of $K$ in elliptic integral theory. It is $K(\sqrt{1-k^2})$ – zy_ Nov 02 '14 at 14:30
  • Ah. I'd remembered the $k'$ convention for the compl. modulus but forgotten that it also applied to $K'(k)$ as the compl. elliptic integral. Though, does that actually help? The complementary elliptic integral isn't proportional to the usual one. (The case of $k=1/\sqrt{2}\implies K(k)=K(k)/\sqrt{2}$ is a particularly evident counter-example.) – Semiclassical Nov 02 '14 at 15:39
  • Hmm. A and B cannot be simultaneously true for the same k. I am wondering whether a transformation formula exists. RHS has a singular point at $\sqrt{2}-1$, and I need such a formula for analytic continuation, and I hope this would help. – zy_ Nov 02 '14 at 15:50
  • Given that they both can't be true for all $k$, perhaps you should edit your question include the steps you used to obtain them? That would clarify the context and hopefully show where things went bad. – Semiclassical Nov 02 '14 at 15:51
  • I use Clausen's formula and linear transformation formula for the deduction, and things might go bad for the hypergeometric functions are multi-valued around its singularities. A numerical computation on Mathematica suggest that A is true for $0<k<\sqrt{2}-1$ while B is true for $k>\sqrt{2}-1$. – zy_ Nov 02 '14 at 15:58
  • Let us continue this discussion in chat. – Semiclassical Nov 02 '14 at 16:03

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