RE: [math-fun] elliptic mean
<< Must a Mean associate? -- Rich Well, from ((x+y)/2 + z)/2 = x/4 + y/4/ + z/2 we see that even the arithmetic mean fails to associate. On the other hand, I have in the past worked with mean associates (though that was not obligatory). --Dan
<<=RCS << Must a Mean associate? -- Rich
=Dan ...the arithmetic mean fails to associate. Good point. I once had to explain this to a school teacher who was averaging test scores this way.
But what is the Official Definition of "mean" anyway? Looking at mathworld, it would seem that we must have x M y = y M x x min y <= x M y <= x max y ax M ay = a(x M y) Is that it?
On Fri, 29 Apr 2005, Marc LeBrun wrote:
But what is the Official Definition of "mean" anyway? Looking at mathworld, it would seem that we must have
Kermit Sigmon wrote several papers on topological "means". According to him a mean is a commutative idempotent topological groupoid. Of course, no associativity is assumed. He also considered what he called n-means, which have n-ary operations, symmetric in all arguments and satisfying f(x,x,...x) = x for all x. Mostly his papers were concerned with topological matters. Here are some of the titles: Sigmon, Kermit Cancellative medial means are arithmetic. Duke Math. J. 37 1970 Sigmon, Kermit Acyclicity of compact means. Michigan Math. J. 16 1969 111--115. Sigmon, Kermit Medial topological groupoids. Aequationes Math. 1 1968 217--234. Sigmon, Kermit On the existence of a mean on certain continua. Fund. Math. 63 1968 311--319. Sigmon, Kermit A note on means in Peano continua. Aequationes Math. 1 1968 no. 1-2, 85--86.
On Friday 29 April 2005 19:48, Daniel Asimov wrote:
<< Must a Mean associate? -- Rich
Well, from ((x+y)/2 + z)/2 = x/4 + y/4/ + z/2 we see that even the arithmetic mean fails to associate.
Let (A,<) be a totally ordered set. Define a "mean" on A to be a function m : AxA -> A such that m(x,x) = x for all x m(x,y) = m(y,x) for all x,y m(x,--) is monotone nondecreasing for all x. Suppose m is an associative mean. Then, in particular, m(x,m(x,y)) = m(m(x,x),y) = m(x,y). That is: if z = m(x,y) for some y then m(x,z) = z. Let's say that m is "archimedean" if, for any x and z, there exists y such that z = m(x,y). There is no associative archimedean mean on a set with more than one element; for if m is such a mean then the condition in the previous paragraph is vacuous and we have m(x,z)=z unconditionally, and likewise m(x,z)=x unconditionally, and so x=z unconditionally. Not all means are archimedean. For instance, min and max are nonarchimedean means; they are also associative. Here's another: let A = {0,1,2} and let m(0,0)=0, m(2,2)=2, m(x,y)=1 otherwise. Then m is associative. I suspect associativity is a very stringent condition; it also seems an unnatural one. (And not the one Gosper was actually pointing out the failure of, as Guy has noted.) -- g
On Sat, 30 Apr 2005, Gareth McCaughan wrote:
On Friday 29 April 2005 19:48, Daniel Asimov wrote:
<< Must a Mean associate? -- Rich
...
Not all means are archimedean. For instance, min and max are nonarchimedean means; they are also associative. Here's another: let A = {0,1,2} and let m(0,0)=0, m(2,2)=2, m(x,y)=1 otherwise. Then m is associative.
Here's another example. Take m(x,y) to be the "simplest" number in the closed interval [x..y] (or [y..x] for x>y), with "simplest" as in Conway's surreal numbers. This means m(x,x) = x, and for x<y, m(x,y) is the unique integer in [x..y] of smallest absolute value, or, if there are no integers in the interval, then the unique diadic rational with the smallest denominator. This mean, however, fails to satisfy the identity m(ax,ay) = a m(x,y). David Moulton
On Monday 02 May 2005 18:06, David P. Moulton wrote: [me:]
Not all means are archimedean. For instance, min and max are nonarchimedean means; they are also associative. Here's another: let A = {0,1,2} and let m(0,0)=0, m(2,2)=2, m(x,y)=1 otherwise. Then m is associative.
[David:]
Here's another example. Take m(x,y) to be the "simplest" number in the closed interval [x..y] (or [y..x] for x>y), with "simplest" as in Conway's surreal numbers. This means m(x,x) = x, and for x<y, m(x,y) is the unique integer in [x..y] of smallest absolute value, or, if there are no integers in the interval, then the unique diadic rational with the smallest denominator. This mean, however, fails to satisfy the identity m(ax,ay) = a m(x,y).
Lovely! More generally: let << be a total order on R (or on whatever set our mean is supposed to be defined on) and let m(x,y) be the <<-smallest element of [x,y]. (Put whatever restrictions are needed on << to ensure that there is one.) Then m is an associative mean. Min and max arise when << is either < or >. My 0,1,2 example arises when << ranks 1 below 0 and 2. David's example is where << compares Conway simplicities. These are all the associative means. Let m be an associative mean. Say that x << y iff m(x,y)=x. If x << y << z then m(x,z) = m(m(x,y),z) = m(x,m(y,z)) = m(x,y) = x, so << is transitive. Clearly x << y << x implies x=y. Hence << is a partial order. Now suppose m(x,z) = y and x <= t <= z. Then I claim y << t. For m(y,t) <= m(y,z) = m(m(x,z),z) = m(x,m(z,z)) = m(x,z) = y and m(y,t) >= m(x,y) = m(x,m(x,z)) = m(m(x,x),z) = m(x,z) = y and hence m(y,t) = y. Thus m(x,z) is the <<-minimal element of [x,z]. Thus far, << isn't a total order, but we can break ties however we like (say, via <) and it won't change any of the above. -- g
So, to every associative mean there corresponds a partial order << such that every interval contains a unique <<-minimal element, which the mean picks out. What does this tell us about *homogeneous* associative means, i.e., ones on R+ for which m(ta,tb) = t.m(a,b)? (Note 1. For a partial order, "minimal" is ambiguous. In this instance I mean that the element is actually <<-smaller than everything else in the interval. I think the term is usually used the other way.) (Note 2. We really want to ask the following more general question, but I don't know how to answer it. We have a total order (A,<) and a group G of its automorphisms; what associative means are there that are invariant under G?) We must have a << b => ta << tb. Another way of saying this: whether a << b depends only on b/a. Let S = { x : 1 << x } and T = { x : x << 1 }; then T = 1/S in the obvious sense by symmetry, 1 is in neither, and transitivity says that both are closed under multiplication. Every interval contains a <<-minimal element, which means that for any t [1,t] contains s such that everything other than 1 in [1/s,t/s] is >> 1. This is impossible unless one endpoint of the interval is 1; so for any t either everything in (1,t] is >> 1 or everything in [1/t,1) is >> 1. That is, either all of (1,oo) is >> 1 or all of (0,1) is >> 1. That is, the only homogeneous associative means on R+ are min and max. I suspect one can adapt this to the slightly less special case of the question in Note 2 where G=A and < is compatible with the group operation, but my brain's full of cotton wool right now so I shan't bother. -- g
Ok, you've aroused my curiosity. What means _do_ associate? "Mean": "statistical norm or average value". Mean of 2 variables would then somehow incorporate both distributions to produce a norm or average value. Maximum & minimum associate -- they are lattice operations. Presumably, an associating mean would also be a lattice op? At 11:48 AM 4/29/2005, Daniel Asimov wrote:
<< Must a Mean associate? -- Rich
Well, from ((x+y)/2 + z)/2 = x/4 + y/4/ + z/2 we see that even the arithmetic mean fails to associate.
On the other hand, I have in the past worked with mean associates (though that was not obligatory).
--Dan
participants (6)
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Daniel Asimov -
David P. Moulton -
Edwin Clark -
Gareth McCaughan -
Henry Baker -
Marc LeBrun