From: "denis.feldmann" Subject: Re: limit formula asked Date: Fri, 17 Dec 1999 17:56:07 +0100 Newsgroups: sci.math Keywords: sequence rapidly converging to arithmetic-geometric mean manuel a écrit dans le message <83dnt2$6gj$1@news.si.fct.unl.pt>... >Define 2 sequences xn and yn with: >x1:=a; y1:=b( 0x(n+1):= sqrt(xn*yn) >y(n+1):= (xn+yn)/ 2 > >It is easy to prove that the they converge, and I want a formula depending >on a and b for the value of the limit of ( xn ,a ,b ). > >manuel Just for the fun of being first to answer: if M(a,b) is the limit, pi/M(a,b)= int(1/sqrt((x^2+a^2)(x^2+b^2), x= -infinity..infinity) M(a,b) is the arithmetico-geometric mean; the integral is an elliptic one (no elementary expression); the result is due to Gauss (proof use the change of variable y=x-ab/x (!)) ============================================================================== From: spamless@Nil.nil Subject: Re: limit formula asked Date: 17 Dec 1999 18:53:19 -0500 Newsgroups: sci.math manuel wrote: > Define 2 sequences xn and yn with: > x1:=a; y1:=b( 0 x(n+1):= sqrt(xn*yn) > y(n+1):= (xn+yn)/ 2 > It is easy to prove that the they converge, and I want a formula depending > on a and b for the value of the limit of ( xn ,a ,b ). Well, it is called the arithmetic-geometric mean of (a,b). If a=1, b=SQRT(1-k^2) then the limit is pi/(2*K(k)) where K is the complete elliptic integral of the first kind. ============================================================================== From: spamless@Nil.nil Subject: Re: limit formula asked Date: 17 Dec 1999 18:56:37 -0500 Newsgroups: sci.math manuel wrote: > Define 2 sequences xn and yn with: > x1:=a; y1:=b( 0 x(n+1):= sqrt(xn*yn) > y(n+1):= (xn+yn)/ 2 > It is easy to prove that the they converge, and I want a formula depending > on a and b for the value of the limit of ( xn ,a ,b ). (note, it is easy to see that you can scale so that one of a,b is 1 and the other is smaller than one and the result is just the limit in that case times the scale factor ... e.g. let X=x/a, Y=y/a, b'=b/a to see that X,Y satisfy the the same recursion with X_0=1 now)