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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>
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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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