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 >>>> >> >> >> >