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On the Wolfram webpage, one can find the following identity for the complete elliptic integal of the first kind $K(z)$:

$$K(z) = \frac{2}{1+\sqrt{1-z}} K\Big( \Big( \frac{1-\sqrt{1-z}}{1+\sqrt{1-z}} \Big)^2 \Big)$$

See http://functions.wolfram.com/EllipticIntegrals/EllipticK/17/01/0007/.

Does anyone have a good reference for this (I checked the DLMF and Abramowitz/Stegun, but did not find anything). Or is there a simple proof that I am missing?

Thanks in advance!

Stephan Wagner
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  • Have you looked at Borwein/Borwein's Pi and the AGM? – Angina Seng May 24 '18 at 19:04
  • Thanks for the hint, I will check. BTW, I can actually prove it: one shows that the two sides satisfy the same second-order DE (substituting $\sqrt{1-z} = x$ first helps with calculations) with the same initial values for x=1 resp. z=0. But that does not seem particularly elegant... – Stephan Wagner May 24 '18 at 19:08
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    look at this answer which uses AGM to prove this identity (note: Mathematica is using a different convention for $K(z)$, $K(z) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1 - z\sin^2\theta}}$, the $z$ it uses is really the $k^2$ in other refs) – achille hui May 24 '18 at 19:09
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    @achillehui Why does that not come as a surprise....? – Angina Seng May 24 '18 at 19:11
  • That does settle it, thanks. Somehow this answer escaped me when I was searching through StackExchange. – Stephan Wagner May 24 '18 at 19:11

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