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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: [mg57367] Re: Solve or Reduce on a monstrosity of an expresssion (and a prize!)
  • From: Andrzej Kozlowski <andrzej at akikoz.net>
  • Date: Wed, 25 May 2005 06:03:20 -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> <458D701E-37FA-425F-89C4-52A5628E22CF@akikoz.net> <Pine.LNX.4.58.0505241736010.24188@boston.eecs.umich.edu> <1D91709F-1E4D-4B98-95F5-695F7BD65577@akikoz.net> <Pine.LNX.4.58.0505242049410.3527@boston.eecs.umich.edu>
  • Sender: owner-wri-mathgroup at wolfram.com

On 25 May 2005, at 10:00, Daniel Reeves wrote:

> wow, this is huge progress.  medium prize, at the least, is cinched :)
>
> eager to hear how you establish that the derivative has <= 2 roots...

First, I only think I could probably prove this, but that does not  
mean I actually have a rigorous proof. I have a sketch of a possible  
proof that I will describe, but it may not work. What is worse, even  
if it is essentially correct, it is probably the easy part of the  
problem. Checking the two "simple identities" for every integer may  
turn out to be awfully hard, perhaps as hard as Fermat's Last Theorem?
So I don't think that my contribution so far deserves any prize :-(

Anyway, this is my idea for a argument proving that D[f[x,n]]==0 has  
only two roots, for every n. For each complex number n, think of the   
expression D[f[x,n]] as an element of  the field Al of "algebraics"  
over the complex numbers, that is the algebraic closure of the field C 
[x] of rational functions in one variable with coefficients in the  
complex numbers C. This is not just an algebraic object but also a  
topological space. Each element f[x] of Al can be thought as a  
function on C given by a radical expression like the one in your  
problem. We know that for each f[x] there is a polynomial p[x] such  
that all the roots of f[x]==0 are also roots of p[x]==0. As Daniel  
Lichtblau pointed out in a recent posting the function
First[GroebnerBasis[f[x],x]]  gives such a polynomial. I think one  
can probably make sure that the polynomial is monic, that is, its  
highest degree term is 1. So consider the mapping f[x]->First 
[GroebnerBasis[f[x],x]] from the topological space Al to the space of  
all polynomials in one variable. Now, here comes the point I am most  
uncertain about: I think this mapping can be made continuous. In  
other words, if we change our radical expression f[x] only slightly  
than the corresponding polynomial will only change slightly. In  
particular, it's degree will not jump. If that is so  we are nearly  
done. We need only to check only one more thing, which is that any  
expression D[f[x,n],x] can be connected to D[f[x,m],x] by a  
continuous path. Looking at the monstrous expression it seems to me  
that it is so, anyway, as long as we keep n>1 .  But since we can  
compute the polynomial for a number of integers n and we know that it  
has degree 2, it would seem to follow that it must always have degree 2.

To really be sure the argument is correct would require very good  
understanding of what the GrobenerBasis algorithm really does in such  
situations and I have only a vague understanding. In particular I  
have never seen anywhere the issue of continuity being discussed, but  
then I have not looked for it.
But one must remember even if we could prove this we could still be  
almost as far from proving the full statement as we were at the start.

Andrzej Kozlowski





>
>
> --- \/   FROM Andrzej Kozlowski AT 05.05.25 09:16 (Tomorrow)   \/ ---
>
>
>> I am now pretty sure that I could now prove the general result
>> provided that I could establish two "simple" facts, which are that:
>>
>>
>> FullSimplify[D[f[x, n], x] /. x -> (n - 1)/n,
>>    Element[n,Integers] && n > 2]
>>
>> FullSimplify[D[f[x, n], x] /.
>>     x -> (-n^2 - 2*n - 1)/((n - 1)*n),
>>    Element[n,Integers] && n > 2]
>>
>> are both zero. In other words, I think I can prove (there are some
>> details that I would have to check but I am pretty sure they are
>> fine)   that the derivative can have no more than two roots, so if
>> the above are the roots everything is done. But unfortunately after
>> 24 hours Matheamtica has not returned any answer to the first of the
>> above. I have not even tried the second.
>> Perhaps a more specialized algebra program for this sort of thing
>> might do better?
>>
>> Andrzej Kozlowski
>>
>
> -- 
> http://ai.eecs.umich.edu/people/dreeves  - -  google://"Daniel Reeves"
>
> "I have enough money to last me the rest of my life, unless I
> buy something." -- Jackie Mason
>
>


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