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MathGroup Archive 2004

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Re: Infrequent Mathematica User

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47245] Re: Infrequent Mathematica User
  • From: drbob at bigfoot.com (Bobby R. Treat)
  • Date: Wed, 31 Mar 2004 02:59:56 -0500 (EST)
  • References: <163.2d33e8ae.2d90396b@aol.com> <001f01c41166$1d4ae340$58868218@we1.client2.attbi.com> <c3ueie$9ti$1@smc.vnet.net> <c4390g$en3$1@smc.vnet.net> <c4bdvq$6vn$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Here's a replacement proof that's more transparent.

For the induction step, it's sufficient to apply a==Sqrt[n-1] to the
statement (which I'll prove) that

a/Sqrt[x[1]^2 + 1] + x[1]^2/(x[1]^2 + 1) < Sqrt[a^2 + 1]

for all real a and x[1]. It's strictly true (trivially) if x[1]==0,
since

x[1]/(x[1]^2 + 1) + a/Sqrt[x[1]^2 + 1] < Sqrt[a^2 + 1] /. x[1] -> 0
a < Sqrt[1 + a^2]

Now it's sufficient to show that the EQUALITY is never true for real
x[1] and a. (I use Solve here, but the two sides can be squared to
form a quadratic equation in a.)

(Simplify[#1, x[1] \[Element] Reals] & )[
  a /. Solve[x[1]/(x[1]^2 + 1) + a/Sqrt[x[1]^2 + 1] == Sqrt[a^2 + 1],
a]]
{-I + 1/(x[1]*Sqrt[1 + x[1]^2]), I + 1/(x[1]*Sqrt[1 + x[1]^2])}

Neither root is real. QED.

Bobby

Paul Abbott <paul at physics.uwa.edu.au> wrote in message news:<c4bdvq$6vn$1 at smc.vnet.net>...
> In article <c4390g$en3$1 at smc.vnet.net>,
>  drbob at bigfoot.com (Bobby R. Treat) wrote:
> 
> > If you actually see a way to apply one of the standard inequalities to
> > this sum, please share. Nothing obvious springs to mind.
> 
> For a proof see
> 
>   ftp://physics.uwa.edu.au/pub/Mathematica/MathGroup/InequalityProof.nb
> 
> Cheers,
> Paul
> 
> 
> > Paul Abbott <paul at physics.uwa.edu.au> wrote in message 
> > news:<c3ueie$9ti$1 at smc.vnet.net>...
> > > Jim Dars wrote:
> > > 
> > > >A Math NG posed the problem:
> > > >
> > > >  Let x1,x2,...,xn be real numbers. Prove
> > > >  x1/(1+x1^2) + x2/(1+x1^2+x2^2) +...+ xn/(1+x1^2+...+xn^2) < sqrt(n)
> > > 
> > > To prove this, I would do a search for inequalites, e.g,
> > > 
> > >   http://mathworld.wolfram.com/ChebyshevSumInequality.html
> > > 
> > > Also, see
> > > 
> > > Hardy, G. H.; Littlewood, J. E.; and Pólya, G. 
> > > Inequalities, 2nd ed. Cambridge, England: 
> > > Cambridge University Press, pp. 43-44, 1988.
> > > 
> > > To investigate using Mathematica, I like to use subscripted variables:
> > > 
> > >    s[n_] := Sum[Subscript[x,i]/(1 + Sum[Subscript[x,j]^2, {j, i}]), {i, n}]
> > > 
> > > If you enter this expression into Mathematica and 
> > > do Cell | Convert to StandardForm (or 
> > > TraditionalForm) you will get a nicely formatted 
> > > expression for the n-th left-hand side of the 
> > > inequality.
> > > 
> > > Note that you can prove the inequality on a 
> > > case-by-case basis using CylindricalDecomposition 
> > > (in Version 5.0):
> > > 
> > >   CylindricalDecomposition[s[1] > 1, {Subscript[x, 1]}]
> > > 
> > >   CylindricalDecomposition[s[2] > Sqrt[2], {Subscript[x, 1], Subscript[x, 
> > >   2]}]
> > > 
> > > and so on. This may not seem convincing, but see 
> > > what happens if the you change the inequality:
> > > 
> > > CylindricalDecomposition[s[2] > 1/2, {Subscript[x, 1], Subscript[x, 2]}]
> > > 
> > > >To get a feel for the problem, and maybe spark 
> > > >an idea, I hoped to look at some few early 
> > > >maximum values.  However, these proved difficult 
> > > >to come by.
> > > 
> > > NMaximize is the way to go:
> > > 
> > >   Table[NMaximize[s[n], Table[{Subscript[x, i], -5, 5}, {i, n}]], {n, 6}]
> > > 
> > > Cheers,
> > > Paul
> >


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