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Empty
 one year ago
AGM / Elliptic integral relationship proof from here:
https://en.wikipedia.org/wiki/Arithmetic%E2%80%93geometric_mean#Proof_of_the_integralform_expression
Empty
 one year ago
AGM / Elliptic integral relationship proof from here: https://en.wikipedia.org/wiki/Arithmetic%E2%80%93geometric_mean#Proof_of_the_integralform_expression

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Empty
 one year ago
Best ResponseYou've already chosen the best response.0\[I(x,y) = \int\limits_0^{\pi/2} \frac{d \theta}{x^2 \cos^2 \theta + y^2 \sin^2 \theta}\] Is there a geometric reason for the substitution of \[\sin \theta = \frac{ 2x \sin \phi}{(x+y)+(xy) \sin^2 \phi}\] which allows us to see that: \[I(x,y) = I(\frac{1}{2} (x+y),\sqrt{xy})\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1x,y are real number ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I believe in general they are, although I am curious about when x and y are complex.

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 one year ago
Best ResponseYou've already chosen the best response.0Overall I just wanna understand how in the world Gauss was able to come to this relationship between the elliptic integral and AGM honestly haha. The AGM basically is the closed form of an elliptic integral it seems, which I am shocked no one has mentioned this powerful function before.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I forgot the square root sign here: \[I(x,y) = \int\limits_0^{\pi/2} \frac{d \theta}{\sqrt{x^2 \cos^2 \theta + y^2 \sin^2 \theta}}\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1to solve ,we can do \[\int\limits \frac{ d \theta }{ x^2\cos^2\theta+y^2\sin^2\theta }=\\\int\limits \frac{ d \theta }{ 1+\frac{y^2\sin^2\theta}{x^2\cos^2\theta} }\frac{1}{x^2\cos^2\theta}=\\\frac{1}{x^2} \int\limits \frac{ d \theta }{ 1+(\frac{y}{x}\tan \theta)^2 }(1+\tan^2\theta)=\\ u=\frac{y}{x}\tan \theta \\=\frac{1}{x^2}*\frac{x}{y} \arctan(\frac{y}{x}\tan \theta)\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, sorry about that, it wasn't this integral to begin with, I had written it wrong by accident.

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 one year ago
Best ResponseYou've already chosen the best response.0Also, I am not really interested in solving the integral, since it's not solvable.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0You want to show I(a,b) = I(am(a,b), gm(a,b)) using some geometry and relate it to gauss ?

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1yes do you check? \[I(\frac{x+y}{2},\sqrt{xy})\] maybe in writing ,and arranging find the trick !

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Well I just worked it out, but it just seems like a lot of terrible algebra sadly! @amoodarya I guess I would like to see some sort of motivation for why there is any relationship between repeating this process: \[a_{n+1} = \frac{a_n+b_n}{2}\]\[b_{n+1}=\sqrt{a_nb_n}\] and the integral in a way. How did Gauss know to look here or think this? I can follow through the algebra to show that it's true, but I want to know how ellipses and these averages are related, this seems very strange and fascinating!

Empty
 one year ago
Best ResponseYou've already chosen the best response.0It appears as if someone has explained it here, http://paramanands.blogspot.com/2009/08/arithmeticgeometricmeanofgauss.html#.VZ0JZPm6fIU I can't seem to follow his reasoning on my own, if anyone wants to try to understand this article with me step by step, speak up! Otherwise I'll just sorta take a break from this haha. I don't even know where elliptic integrals really come from other than in calculating arc length or why they are important at all.
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