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 > >