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

• To: mathgroup at smc.vnet.net
• Subject: [mg57391] Re: Solve or Reduce on a monstrosity of an expresssion (and a prize!)
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Thu, 26 May 2005 04:31:44 -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> <F016112F-BF6B-4D4E-8A3C-1F38AB3DDC5F@akikoz.net> <Pine.LNX.4.58.0505251303210.3527@boston.eecs.umich.edu>
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```Daniel Reeves wrote:
> yikes.  ok, let's assume that all works.  So it remains to show that both
> of the following expressions of n are zero for all n >= 2.  Strangely,
> FullSimplify can do this for all specific n up to 113 but 114 seems to
> stump it.
>
> here are the expressions:
> [...]

I can point out some aspects to a probabilistic argument. Making it into
a full fledged proof would require figuring out some bounds that offhand
I don't know how to do, then verifying for appropriately many cases.

First you need to show that there are 2 roots for each x, at least when
n>=2. For this it suffices to show it holds for generic n. Since we
cannot seem to do that computationally, instead do it for "random" n.
Calling your expression fp, do, say,

fp7 = Simplify[fp /. n->7]
gb7 = GroebnerBasis[fp7,x]

Now check for a quadratic polynomial (in x) in gb7. If you find one,
that means there are at most 2 solutions when n=7. This can of course be
repeated for other values of n. The point is that, for integers larger
than 2, this computation appears to be feasible. Offhand I think this
idea (of verifying at sufficiently many values of n) suffices, though
there may be a lurking branch cut issue. Note that showing there are two
solutions for n=7 is not quite sufficient in and of itself because 7
might not be generic. In this setting, nongeneric points would lie on a
variety (roots of a polynomial), hence be a set of measure zero.

Verification that x->(n-1)/n and x->-(n+1)^2/(n*(n-1)) are the two
solutions can be done by making the substitutions into fp and verifying,
for sufficiently many n>=2, that these vanish. The reason is that
univariate algebraic functions vanishing at sufficiently many integer
points vanish everywhere that branches don't change (e.g. x-Sqrt[x^2]
vanishes on positive integers and indeed on all positive reals). I think
the branch cut issue goes away when you have the proviso that n>=2,
because for the substitutions above all radicands are positive in that
region. But maybe I'm missing something and there is more to it than this.

I should mention that if you did not know those were the appropriate
substitutions, they might be deduced by finding particular {x,n} pairs
and getting more values via homotopy continuation (basically, setting up
and solving an ODE via NDSolve). Values from this could then be used to
interpolate an implicit relation. This was the gist of some work I
described in a note last week:

http://forums.wolfram.com/mathgroup/archive/2005/May/msg00627.html

Offhand I do not know how well it would work for your particular
algebraic expression, but it strikes me as a plausible approach. (A
better one for this case might be to take an implicit plot of the zero
set of fp and guess the relation).

Daniel Lichtblau
Wolfram Research

```

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