Re: Re: Infrequent Mathematica User

*To*: mathgroup at smc.vnet.net*Subject*: [mg47302] Re: [mg47244] Re: Infrequent Mathematica User*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Fri, 2 Apr 2004 03:31:58 -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> <200403310759.CAA21116@smc.vnet.net> <6D84CC78-8405-11D8-94C5-00039311C1CC@mimuw.edu.pl> <p06020446bc92846fe4d8@[130.95.156.21]> <opr5s9r0sxiz9bcq@smtp.cox-internet.com>*Sender*: owner-wri-mathgroup at wolfram.com

What I meant was, e.g. n = 3; Array[a, {n}] . Array[1 & , {n}] <= Norm[Array[a, {n}]]* Norm[Array[1 & , {n}]] a[1] + a[2] + a[3] <= Sqrt[3]* Sqrt[Abs[a[1]]^2 + Abs[a[2]]^2 + Abs[a[3]]^2] Andrzej Chiba, Japan http://www.mimuw.edu.pl/~akoz/ On 2 Apr 2004, at 13:08, DrBob wrote: > 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] >>> >>> -? >> >> 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/ > >