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Re: Solve or Reduce on a monstrosity of an expresssion (and a prize!)
*To*: mathgroup at smc.vnet.net
*Subject*: [mg57294] Re: [mg57278] Solve or Reduce on a monstrosity of an expresssion (and a prize!)
*From*: Andrzej Kozlowski <andrzej at akikoz.net>
*Date*: Tue, 24 May 2005 05:12:31 -0400 (EDT)
*References*: <200505230620.CAA04045@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
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
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