[Date Index]
[Thread Index]
[Author Index]
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/
>
>
Prev by Date:
**RE: Constant function Integrate Assumption - More**
Next by Date:
**Re: why the overflow problems when using assigned values a,b,c vs. actual numbers?**
Previous by thread:
**Re: Re: Infrequent Mathematica User**
Next by thread:
**Re: Re: Infrequent Mathematica User**
| |