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Re: Re: Infrequent Mathematica User
*To*: mathgroup at smc.vnet.net
*Subject*: [mg47295] Re: [mg47244] Re: Infrequent Mathematica User
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Fri, 2 Apr 2004 03:31:37 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
To get interesting pictures of f[n] showing the way they converge to f
one should plot the graphs for fairly large values of n, e.g.:
plots = Table[
Plot[f[n][x], {x, -1, 1}, DisplayFunction -> Identity], {n, 100,
1000,100}];
Andrzej
On 2 Apr 2004, at 02:53, 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). Moreover, the proof is easier
> since the inductive step is now trivial.In addition, the inequality
> leads to some intriguing observations and also to what looks like a
> bug in Limit (?)
>
> 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]
>
> -°
>
> ?
>
> Andrzej Kozlowski
> Chiba, Japan
> http://www.mimuw.edu.pl/~akoz/
>
>
> On 31 Mar 2004, at 16:59, Bobby R. Treat wrote:
>
>> OK, as I suspected, there's apparently no standard inequality such as
>> the Cauchy Sum Inequality that applies (at least, not directly). There
>> couldn't be, because they're all "tight" for every n and all can be
>> made strict equalities for some choice of the variables, while neither
>> is true for the problem inequality, which is never an equality and is
>> tight only in the limit as n->Infinity.
>>
>> Meanwhile, the "proof" at the link is a proof only if we blindly trust
>> Reduce. (I doubt that we always can.) I really liked the
>> "straightforward" comment before the last use of Reduce. I had no
>> trouble replacing the first use with an actual proof, but the second
>> is trickier.
>>
>> Bobby
>>
>> Paul Abbott <paul at physics.uwa.edu.au> wrote in message
>> news:<c4bdvq$6vn$1 at smc.vnet.net>...
>>> In article <c4390g$en3$1 at smc.vnet.net>,
>>> drbob at bigfoot.com (Bobby R. Treat) wrote:
>>>
>>>> If you actually see a way to apply one of the standard inequalities
>>>> to
>>>> this sum, please share. Nothing obvious springs to mind.
>>>
>>> For a proof see
>>>
>>>
>>> ftp://physics.uwa.edu.au/pub/Mathematica/MathGroup/
>>> InequalityProof.nb
>>>
>>> Cheers,
>>> Paul
>>>
>>>
>>>> Paul Abbott <paul at physics.uwa.edu.au> wrote in message
>>>> news:<c3ueie$9ti$1 at smc.vnet.net>...
>>>>> Jim Dars wrote:
>>>>>
>>>>>> A Math NG posed the problem:
>>>>>>
>>>>>> Let x1,x2,...,xn be real numbers. Prove
>>>>>> x1/(1+x1^2) + x2/(1+x1^2+x2^2) +...+ xn/(1+x1^2+...+xn^2) <
>>>>>> sqrt(n)
>>>>>
>>>>> To prove this, I would do a search for inequalites, e.g,
>>>>>
>>>>> http://mathworld.wolfram.com/ChebyshevSumInequality.html
>>>>>
>>>>> Also, see
>>>>>
>>>>> Hardy, G. H.; Littlewood, J. E.; and P?lya, G.
>>>>> Inequalities, 2nd ed. Cambridge, England:
>>>>> Cambridge University Press, pp. 43-44, 1988.
>>>>>
>>>>> To investigate using Mathematica, I like to use subscripted
>>>>> variables:
>>>>>
>>>>> s[n_] := Sum[Subscript[x,i]/(1 + Sum[Subscript[x,j]^2, {j,
>>>>> i}]), {i, n}]
>>>>>
>>>>> If you enter this expression into Mathematica and
>>>>> do Cell | Convert to StandardForm (or
>>>>> TraditionalForm) you will get a nicely formatted
>>>>> expression for the n-th left-hand side of the
>>>>> inequality.
>>>>>
>>>>> Note that you can prove the inequality on a
>>>>> case-by-case basis using CylindricalDecomposition
>>>>> (in Version 5.0):
>>>>>
>>>>> CylindricalDecomposition[s[1] > 1, {Subscript[x, 1]}]
>>>>>
>>>>> CylindricalDecomposition[s[2] > Sqrt[2], {Subscript[x, 1],
>>>>> Subscript[x,
>>>>> 2]}]
>>>>>
>>>>> and so on. This may not seem convincing, but see
>>>>> what happens if the you change the inequality:
>>>>>
>>>>> CylindricalDecomposition[s[2] > 1/2, {Subscript[x, 1],
>>>>> Subscript[x, 2]}]
>>>>>
>>>>>> To get a feel for the problem, and maybe spark
>>>>>> an idea, I hoped to look at some few early
>>>>>> maximum values. However, these proved difficult
>>>>>> to come by.
>>>>>
>>>>> NMaximize is the way to go:
>>>>>
>>>>> Table[NMaximize[s[n], Table[{Subscript[x, i], -5, 5}, {i, n}]],
>>>>> {n, 6}]
>>>>>
>>>>> Cheers,
>>>>> Paul
>>>>
>>
>>
>>
>
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