From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Wednesday, July 15, 2009 10:29:59 AM Subject: Re: [math-fun] clean 16-th order AGM iteration (Just playing around with the standard definition of AGM, I see that it tends to converge to 15 decimal places almost always within 5 or 6 iterations -- so it already has impressive convergence.) Define an equivalence relation via (x,y) ~ (x',y') whenever AGM(x,y) = AGM(x',y'). Then what do the equivalence classes look like as subsets of (0,oo)^2 ? They're most likely all curves; what are their equations? --Dan Also -- and this is probably well-known -- it seems that the "weighted AGM": a(n+1) = p*a(n) + q*g(n) g(n+1) = a(n)^p * g(n)^q where 0 < p < 1, p+q= 1, always converges for positive a(0),g(0) as well. _____________________________________________________________________ AGM(x,y) is symmetric and homogeneous of degree 1 in x and y, so it is sufficient to look at the function f(x) such that AGM(x,f(x)) = 1. Maple tells me [asympt(GaussAGM(1,x));] that asymptotically AGM(1,x) = (pi/2) (x / log(4x)) + O(1/x). From this it follows that asymptotically f(x) = 4 x exp( -(pi/2) x). Using this expression for f(x), AGM(x,f(x)) is already 0.995 at x=2. -- Gene