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MathGroup Archive 2005

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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57313] Re: Solve or Reduce on a monstrosity of an expresssion (and a prize!)
  • From: Andrzej Kozlowski <andrzej at akikoz.net>
  • Date: Tue, 24 May 2005 05:12:52 -0400 (EDT)
  • References: <200505230620.CAA04045@smc.vnet.net> <EEA0FBBD-9C31-419A-8D0B-7C73AE4DA32E@akikoz.net> <Pine.LNX.4.58.0505231817420.9452@boston.eecs.umich.edu>
  • Sender: owner-wri-mathgroup at wolfram.com

Actually I have now found that


GroebnerBasis[FullSimplify[D[f[x, n], x]], x]
  also works for n other than 2 and confirms my conjecture.  For  
example;


GroebnerBasis[FullSimplify[D[f[x, 3], x]], x]


{-9*x^2 - 18*x + 16, 9*x + 18 - 16/x,
   -18*x - 9*(x^2 + 54) + 502,
   27*6^(1/3)*Sqrt[x^2 + 54]*x + 35*6^(5/6)*x -
    68*6^(1/3)*Sqrt[x^2 + 54] + 350*6^(1/6)*
     (x^2*(Sqrt[6]*Sqrt[x^2 + 54] + 18))^(1/3) -
    140*6^(5/6)}


Solve[-9*x^2 - 18*x + 16 == 0, x]


{{x -> -(8/3)}, {x -> 2/3}}

I suspect that this is actually what FindInstance uses (?)


So here is a new formulation of the conjecture: the roots of the  
equation D[f[x,n],x]==0 are precisely the roots of the quadratic  
equation


x^2 + (n*x)/(n - 1) + (2*x)/(n - 1) + x/((n - 1)*n) +
   x/n - x - n/(n - 1) - 1/(n - 1) + 1/((n - 1)*n) +
   1/((n - 1)*n^2) ==0


Andrzej Kozlowski



On 24 May 2005, at 08:11, Daniel Reeves wrote:

> Ooh ooh!  I just got FindInstance to prove it for any particular n.  
> (Tried
> it for all n up to 75 and it takes about 20 seconds each on my  
> machine and
> doesn't slow down as n increases, though every once in a while  
> there's an
> n that takes 2 minutes).
>
> n = 7;
> FindInstance[FullSimplify[D[f[x,n],x]] == 0 && 0<x<(n-1)/n, {x},  
> Reals]
>   --> returns {} and so QED.
>
> Prize for the general case bigger than ever...
>
>
>
>> 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
>>
>
> -- 
> http://ai.eecs.umich.edu/people/dreeves  - -  google://"Daniel Reeves"
>
> "The good Christian should beware of mathematicians and all those who
> make empty prophecies.  The danger already exists that mathematicians
> have made a covenant with the devil to darken the spirit and confine
> man in the bonds of Hell."
>                 -- St. Augustine
>
>


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