M F Hasler <mhasler@dsi972.fr> wrote:

Brad Klee <bradklee@gmail.com> wrote:

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a = (a+b+c)/3 b = [(a*b)^(1/2)+(b*c)^(1/2)+(c*d)^(1/2)]/3 c = (a*b*c)^(1/3)

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Is this the best generalization of the agm to three (or more) variables? Does it even converge?

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I checked that: agm3[1,2,3] = 1.8932121023002878019777... converges

I think the most natural generalisation is to take as "update" for the k-th variable the k-th root of the k-th symmetric polynomial in the variables, normalized by the number of terms. For 3 that would mean

a = (a+b+c)/3 b = [( a*b + b*c + c*a )/3 ]^(1/2) c = (a*b*c)^(1/3)

It gives AGM(1,2,3) = 1.9099262335408153...

Let's compromise by taking the AGM of Klee's 1.8932121023002878019777 and Hasler's 1.9099262335408153, which is about 1.9015599859028125. :-) How about a fully recursive definition with no visible addition, multiplication, powers, or roots: a = agm(agm(a,(b,c)) b = agm(agm(b,(a,c)) c = agm(agm(c,(a,b)) (As I assume Klee and Hasler did, I'm not replacing any of a,b, and c until all three assignments are done.) With 1,2,3 that converges to 1.9091504422214686... For four terms: a = agm(agm(a,b),agm(c,d)) b = agm(agm(a,c),agm(b,d)) c = agm(agm(a,d),agm(b,c)) then use my above agm3 on those three. (I thought they might immediately be equal, but no.) For 1,2,3,4 that converges to 2.3587709092584177... Is there a natural way to extend this approach to 5 or more terms? Of course it's required that it give the same answer if the terms are permuted.