The triangle inequality a+b>c transforms to 2b>0, and similar for cyclic permutation. If (a,b,c) are positive lengths, even if they do not satisfy the triangle inequality, they will after one iteration. Thereafter the lengths can be folded into a triangle (or it’s mirror image). Similarly, if (a,b,c) are positive angles, choose angular units where a+b+c=pi. All subsequent iterations preserve the sum, so each triple determines a unique triangle (again, up to reflection). By the limiting property, both sequences converge to equilateral. 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? —Brad
On Sep 18, 2020, at 12:03 AM, Henry Baker <hbaker1@pipeline.com> wrote:
Suppose we have a set of 3 distinct numbers, {a,b,c}.
Then averaging them in pairs preserves distinctness: {(a+b)/2,(b+c)/2,(c+a)/2}.
Proof: assume not, e.g., (a+b)/2=(b+c)/2 => a=c. Contradiction. QED.
So we can go backwards:
{a,b,c} => {a+b-c,b+c-a,c+a-b}
So, {(a+b)/2,(b+c)/2,(c+a)/2} => {(a+b+b+c-c-a)/2=b,(b+c+c+a-a-b)/2=c,(c+a+a+b-b-c)/2=a}
Note also that 'averaging' preserves the sum of the triple (hence its average): (a+b)/2+(b+c)/2+(c+a)/2=a+b+c. Thus, 'going forward' and 'going backwards' preserves this sum.
OK, so now let's iterate this averaging process N times.
I contend that this process makes the numbers more and more 'average'; i.e., all three converge towards the same number: (a+b+c)/3.
Yet, we can still recover our original 3 numbers by running the process backwards N times.
Even better, we can go backwards *further* than our original 3 numbers.
Curiously, if we start with {-1,0,1}, we get {-2^k,0,2^k}, for all k.
Note now that a,b,c need not be 'numbers'; they could be vectors of numbers, so we can play this game in higher dimensions. E.g., we can choose triples of (x,y) pairs to represent triangles, so does iterating them converge towards equilateral triangles?
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