Re: Re: Infrequent Mathematica User
- To: mathgroup at smc.vnet.net
- Subject: [mg47301] Re: [mg47244] Re: Infrequent Mathematica User
- From: DrBob <drbob at bigfoot.com>
- Date: Fri, 2 Apr 2004 03:31:56 -0500 (EST)
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- Sender: owner-wri-mathgroup at wolfram.com
I think this equality does it very nicely. Not sure how we'd use either of the others. http://mathworld.wolfram.com/ChebyshevSumInequality.html Anyway, thanks for the extra insights, all of you. Bobby On Fri, 2 Apr 2004 10:51:28 +0800, Paul Abbott <paul at physics.uwa.edu.au> wrote: > 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] >> >> -Æ? > > Series also gives an incorrect result (I think Limit is using this). > > Cheers, > Paul > -- Using M2, Opera's revolutionary e-mail client: http://www.opera.com/m2/