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I need help, I'm looking at Elliptic integrals and why you can use AGM to compute it. I got the point why, but I didn't understand some steps where Gauss make a proof on it. He came with new variable $\varphi'$, where

$\sin(\varphi) = \frac{2a\sin^2(\varphi')}{(a+b)+(a-b)\sin^2(\varphi')}$,

then you need to make $\cos^2(\varphi) = 1 -\sin^2(\varphi) $, and that's where the struggle begins, so you got like

$\cos^2(\varphi) = \frac{(a+b)^2+2(a^2-b^2)\sin^2(\varphi')+(a-b)^2\sin^4(\varphi') - 4a^2\sin^4(\varphi')}{[(a+b)+(a-b)\sin^2(\varphi')]^2}\qquad(1)$

now you need to put in $ a_1 = \frac{a+b}{2}, b_1=\sqrt{a\cdot b} $ so you get

$\cos^2(\varphi) =\frac{4a_1^2-4(2a_1^2-b_1^2)\sin^2(\varphi')+4(a_1^2-b_1^2)\sin^4(\varphi')}{[(a+b)+(a-b)\sin^2(\varphi')]^2}=\frac{4a_1^2\cos^4(\varphi')+4b_1^2\sin^2(\varphi')\cos^2(\varphi')}{[(a+b)+(a-b)\sin^2(\varphi')]^2} \qquad (2)$

Soo, can be someone so kind and show me how I put in $a_1,b_1$, into $(1)$ to get $(2)$ please, I just don't see it and want to get it clear.

Thank you for you time and have a nice day.

J. W. Tanner
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MathIsFun
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1 Answers1

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The substitution is C&P'ed incorrectly. The correct one is $$\sin\varphi=\frac{2a\color{red}{\sin\varphi'}}{(a+b)+(a-b)\sin^2\varphi'}.$$ (Don't forget to take the $d\varphi$ part into account too.)

metamorphy
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    Thanks man, I didn't notice, that I take the bad substitution. I write it on paper and try to make it with wrong one. I got it now.

    Thank you, very much.

     – MathIsFun Feb 28 '20 at 22:57