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Re: Re: Infrequent Mathematica User


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
>



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