Re: Solve or Reduce on a monstrosity of an expresssion (and a prize!)

*To*: mathgroup at smc.vnet.net*Subject*: [mg57296] Re: [mg57278] Solve or Reduce on a monstrosity of an expresssion (and a prize!)*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Tue, 24 May 2005 05:12:33 -0400 (EDT)*References*: <200505230620.CAA04045@smc.vnet.net> <EEA0FBBD-9C31-419A-8D0B-7C73AE4DA32E@akikoz.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 23 May 2005, at 21:17, Andrzej Kozlowski wrote: > On 23 May 2005, at 15:20, Daniel Reeves wrote: > > >> >> Mathemahomies, >> I have a beast of a function (though continuously differentiable) >> that I >> need to prove is strictly decreasing in a certain range (which I >> *know* it >> is just from plotting it). Every combination I can think of of >> Reduce and >> Solve and Simplify with assumptions leaves Mathematica spinning >> its wheels >> indefinitely. >> >> Do you have any ideas for cajoling Mathematica into crunching through >> this? >> >> Here's the function: >> >> f[x_,n_] := 9/2/c[x,n]^2*(n+1)b[x,n]^2 (x-d[x,n])(x-x*d[x,n]+d[x,n]+ >> d[x,n]^2+n (d[x,n]-1) (x+d[x,n])) >> >> where >> >> a[x_,n_] := 9*(n+1)^2 + Sqrt[3(n+1)^3 (x^2 (n-1) + 27(n+1))]; >> >> b[x_,n_] := (a[x,n](n-1) x^2)^(1/3); >> >> c[x_,n_] := -3^(2/3) x^2 (n^2-1) + 3^(1/3)(x^2(n^2-1) (9 + 9n + >> Sqrt[3(n+1) (x^2(n-1) + 27(n+1))]))^(2/3); >> >> d[x_,n_] := c[x,n] / (3 b[x,n] (n+1)); >> >> >> Show that f[x,n] is strictly decreasing for x in (0,(n-1)/n) for all >> integers n >= 2. >> >> Note that the limit of f[x,n] as x->0 is (n-1)/(2(n+1)) > 0 >> and f[(n-1)/n,n] == 0. So it would suffice to show that f' has no >> roots >> in (0,(n-1)/n). >> >> >> PS: I have a cool prize for information leading to a solution! >> (whether or >> not it actually involves Mathematica) >> >> -- >> http://ai.eecs.umich.edu/people/dreeves - - google://"Daniel >> Reeves" >> >> Sowmya: Is this guy a mathematician? >> Terence: Worse, an economist. At least mathematicians are honest >> about >> their disdain for the real world. >> >> >> > > Unfortunately I can only manage the most trivial case, n=2, which I > do not think deserves any prize but it seems to show that if anyone > does get the prize it will be well deserved. > > We evaluate your definitions and then set > > n = 2; > > Even in this simplest of cases trying to use Solve directly > > > Solve[D[f[x,2],x]==0,x] > > $Aborted > > takes for ever, so we resort to Groebner basis (actually what I > just called "a hack" in another discussion ;-)) > > We first evaluate > > z = FullSimplify[D[f[x, 2], x] /. x^2 + 81 -> u^2, u > 0]; > > And then use GroebnerBasis: > > > GroebnerBasis[{x^2 + 81 - u^2, z}, {x}, {u}] > > {4*x^2 + 16*x - 9} > > This we can actually solve (even without Mathematica): > > > r=Solve[4*x^2 + 16*x - 9 == 0, x] > > {{x -> -(9/2)}, {x -> 1/2}} > > we can check that the derivative of f[x,2] does indeed vanish at > these points: > > > D[f[x,2],x]/.r//FullSimplify > > {0,0} > > Assuming that Groebner basis is correctly implemented this also > shows that f'[x,2] has no roots in the interval (0,1/2) so we are > done. > > > Unfortunately the same method with n=3 already seems to take for > ever. For small n>2 it may be possible perhaps to combine in some > way numerical methods with Groebner basis to show similarly that D[f > [x,n],x] has no roots in the interval (0,(n-1)/n) but obviously > this won't work for a general n. At the moment the general problem > looks pretty intractable to me, so much that I am almost willing to > offer an additional prize to whoever manages to solve it (but I > won't make any offers before seeing other people's attempts ;-)) > > Andrzej Kozlowski > > I still have not got a proof, but thinking about the above proof for the case n=2 some conjectures come to my mind which might just possible point tothe the way to the general proof. The main idea is Daniel Reeves' one: to show that D[f[x,n],x] is not zero on (0,(n-1)/ 2). If we look at the above argument for n=2 we see that it amounts the fact that D[f[x,n],x]==0 has just two rots, one of which is at (n-1)/n and the other is negative. One is tempted to conjecture that the same is true for any n: the derivative has just two roots, one is at (n-1)/n and the other is negative (I have no good guess at this time as to where it might be, in the n=2 example it is at -9/2). At least for one part of the conjecture there is some confirmation: D[f[x,3],x]/.x->2/3//FullSimplify 0 FullSimplify[D[f[x, 4], x] /. x -> (4 - 1)/4] 0 although again the general n seems to take for ever. There may also be a non-computational proof that the equation can only have two roots and that one of them must be negative. I will think about it more when I have a little more free time but at the moment i view these only as long shot conjectures. Andrzej

**References**:**Solve or Reduce on a monstrosity of an expresssion (and a prize!)***From:*Daniel Reeves <dreeves@umich.edu>