Re: Re: What kind of math problem is this?

*To*: mathgroup at smc.vnet.net*Subject*: [mg21045] Re: [mg20950] Re: [mg20896] What kind of math problem is this?*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Sun, 12 Dec 1999 23:51:12 -0500 (EST)*References*: <199911200607.BAA04739@smc.vnet.net> <199912010650.BAA07636@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Daniel Lichtblau wrote: > > Seth Chandler wrote: > > > > This may be a little off topic, but I was hoping someone here could help. > > > > Let r and f be real numbers between 0 and 1 inclusive. How does one > > determine if there is a function u that satisfies this relation and, if so, > > what the function is. > > > > InverseFunction[u][f u[x] + (1 - f)u[y]] == (1 - r)(f x + (1 - f)y) + r x > > > > [ ... ] > > The answer depends to some extent on what exactly you are willing to > use. Before going into more detail, I'll reformulate your problem a bit. > You require a function u[x], invertible on some domain of interest, such > that > > f*u[x] + (1 - f)*u[y] == u[(1 - r)*(f* x + (1 - f)*y) + r*x] > > By letting f+r-f*r == k, we may rewrite the rhs as > > u[k*x + (1-k)*y] > > The case where k==f is trivial (just take u[x]==x) and it arises when > either r==0 or f==1. We'll assume neither of these occurs. > > One can show there is no differentiable function u[x] that satisfies > this functional equation. To prove this, assume u is differentiable at x > and chose y = x + dx/(1-k). We then have > > u[x+dx] == f*u[x] + (1-f)*u[x+dx/(1-k)] > > Invoking differentiability at x we obtain > > u[x] + u'[x]*dx + o(dx) == > f*u[x] + (1-f)*(u[x]+u'[x]*dx/(1-k)+o(dx/(1-k))) > > where in both places o(...) vanishes faster than dx (think of dx as > small, i.e. we take a limit as it goes to zero). Rewriting the rhs we > have > > u[x] + (1-f)/(1-k)*u'[x]*dx + (1-f)/(1-k)*o(dx/(1-k)) > > As the small-o terms vanish faster than the linear terms, sufficiently > close to x these cannot be equal because the linear terms in dx > disagree. > > I have not successfully convinced myself whether or not there can be a > nowhere differentiable function u[x] that does what you want. But > already we realize there will be no useful computable u[x] (how many > nowhere-differentiable functions can you compute?). > [ ... ] Okay, I can do a bit better on this score. Call a function u[x] valid if it is invertible and satisfies the functional equation u[k*x + (1-k)*y] == m*u[x] + (1 - m)*u[y] where we assume k!=m. I'll indicate why u[x] cannot be continuous. Say it is. First note that invertibility and continuity together imply u[x] must be (strictly) monotonic. Clearly we can subtract a constant and still satisfy the functional equation (and retain continuity and invertibility), hence we may assume u[0] = 0. Then we know u[1] != 0 (or it would not be invertible). As we can clearly multiply by a nonzero scalar without invalidating the function, we may furthermore assume u[1] = 1. Letting x = 1/k and y = 0 we get 1 == u[1] == m*u[1/k] so u[1/k] == 1/m. We let x = 1/k^2, y = 0 and similarly obtain u[1/k^2] == 1/m^2. Likewise we find u[1/k^s] == 1/m^s for positive integer exponents s, and the same idea works for negative integer exponents as well. Now take x = 0 and y = 1/(1-k) to get u[1/(1-k)] == 1/(1-m). Taking y = 1/(1-k)^t similarly gives u[1/(1-k)^t] == 1/(1-m)^t for integer exponents t. Right away we see that NO function u[x] will work if k is 1/2. This is because u[1/2] == u[1/(1-1/2)] == 1/(1-m) != 1/m == u[1/2]. (This really did not require an assumption that u[x] be continuous, you just need to modify the argument a bit.) In the general case just note that we have two sequences in the x-u[x] plane {k^s,m^s} and {(1-k)^t,(1-m)^t} that change at different rates. Perhaps it is easier to see this by taking logarithms. The sequences now lie on lines, the slope of the first is s*Log[m] / (s*Log[k]) == Log[m]/Log[k] and likewise the slope of the second is Log[1-m]/Log[1-k]. These slopes cannot be equal by assumption that m!=k. From this one can conclude that eventually there will be a point {x1,y1} = {s1*Log[k],s1*Log[m]} on the first line, and a corresponding point {x2,y2} = {t1*Log[1-k],t1*Log[1-m]} on the second, such that x1<x2 but y1>y2 or vice versa (depending on which slope is larger). This implies that u[x] must violate monotonicity. I believe there cannot be a nonconstant function u[x] satisfying the basic functional equation, but so far have been unable to prove that (even assuming invertibility). Daniel Lichtblau Wolfram Research

**References**:**Re: What kind of math problem is this?***From:*Daniel Lichtblau <danl@wolfram.com>

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