On Fri 18 Sept. 2020, 08:41, Brad Klee <bradklee@gmail.com> wrote :
(...) Another possibility is: 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) Is this the best generalization of the agm to three (or more) variables? Does it even converge?
There was some recent discussion about generalization of AGM to more arguments, and I also found some older one on math-/stackexchange, where I recently answered a question asked more than 8 years ago: https://math.stackexchange.com/questions/442062/to-find-the-limit-of-three-t... . 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*d )/3 ]^(1/2) c = (a*b*c)^(1/3) i.e. slightly different than what you suggested for the 2nd one ("b"), In that stackexchange answer I showed that it converges. There are different proposals for other generalisations, <https://math.stackexchange.com/questions/1794795/arithmetic-geometric-mean-of-3-numbers> which do not give the same result but might be valid and maybe useful in particular situations. I find the above one the most appealing (natural / simple / straightforward) one, though. It gives AGM(1,2,3) = 1.9099262335408153... - Maximilian