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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
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