Re: Re: Infrequent Mathematica User

*To*: mathgroup at smc.vnet.net*Subject*: [mg47299] Re: [mg47244] Re: Infrequent Mathematica User*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Fri, 2 Apr 2004 03:31:50 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

On 2/4/04, Andrzej Kozlowski wrote: >Actually one can use Paul's argument to prove >the following stronger statement: > >Sum[(Subscript[x,i]/(1 + Sum[Subscript[x,j]^2, {j, i}]))^2, {i, n}] < 1 > >for every positive integer n. > >It is easy to see that this implies the >inequality in the original problem (use >Schwarz's inequality). MathWorld only gives the integral form at http://mathworld.wolfram.com/SchwarzsInequality.html The required form of the inequality is at http://mathworld.wolfram.com/CauchysInequality.html >Moreover, the proof is easier since the inductive step is now trivial. Nice. >In addition, the inequality leads to some >intriguing observations and also to what looks >like a bug in Limit (?) Actually, I think the problem is with Series. I've submitted a bug report. >The inequality implies that the sums, considered >as functions on the real line, are bounded and >attain their maxima. So it is natural to >consider the functions f[n] (obtained by setting >all the Subscript[x,i] = Subscript[x,j)] > >f[n_][x_] := NSum[(x/(i*x^2 + 1))^2, {i, 1, n}] > >It is interesting to look at: > >plots = Table[ > Plot[f[n][x], {x, -1, 1}, DisplayFunction -> Identity], {n, 1, 10}]; > >Show[plots, DisplayFunction -> $DisplayFunction] > >The f[n] of course also bounded by 1 and so in the limit we have the function: > >f[x_] = Sum[(x/(i*x^2 + 1))^2, {i, 1, Infinity}] > >PolyGamma[1, (x^2 + 1)/x^2]/x^2 > >which also ought to be bounded bu 1. > >Plotting the graph of this, e.g. > >Plot[f[x], {x, -0.1, 0.1}] > >shows a maximum value 1 at 0 (where the function >is not defined), however Mathematica seems to >give the wrong limit: > >Limit[f[x],x->0] > >-? Series also gives an incorrect result (I think Limit is using this). Cheers, Paul