Re: Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)

*To*: mathgroup at smc.vnet.net*Subject*: [mg67391] Re: [mg67364] Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Wed, 21 Jun 2006 02:13:12 -0400 (EDT)*References*: <e7589k$l5d$1@smc.vnet.net> <200606200614.CAA15901@smc.vnet.net> <858BE938-5559-4439-A8F5-02B27E9A1B2B@mimuw.edu.pl>*Sender*: owner-wri-mathgroup at wolfram.com

I forgot to add that your condition: 0 < f(x) <= 1 is not really needed, it can be relaxed to f[x]>0 or even to a more general one. his sis clear from the proof I gave and the fact that x/y<= 1 is equaivalent to x<=y provided y is >0. But one can also see it independently of the proof. Clearly that bound f[x]<=1 does not matter, since the ratio (t - a)*Integrate[f[x]^2, {x, a, t}]/Integrate[f[x], {x, a, t}]^2 will not change if you replace f[x] by k f[x] for any k. The condition 0<f[x] is only needed to ensure Integrate[f[x], {x, a, t}] ^2 is not 0. Other than that it also does not need, since Integrate[f[x], {x, a, t}]^2 <=Integrate[Abs[f[x]], {x, a, t}]^2 so if the result is true for f[x]>=0 then it it is also true for any f. Andrzej Kozlowski On 21 Jun 2006, at 01:08, Andrzej Kozlowski wrote: > First of all this is has nothing to do with Mathematica and you do > not even seem to know the proper Mathematica syntax for Integrate > (I wonder how this got past the moderator, twice). To make it more > acceptable I will try to use Mathematica in the proof as much as > possible and I do hope I am not solving your homework! > > > > Clearly, what you want to show is equivalent to proving that the > function: > > g[t_] := (t - a)*Integrate[f[x]^2, {x, a, t}] - Integrate[f[x], {x, > a, t}]^2 > > is >=0 for all real t>=a. This is trivially true for t=a. Since g > is differentiable we can compute D[g[t],t]. Using Mathematica we > obtain: > > > D[g[t], t] > > > (t - a)*f[t]^2 - 2*Integrate[f[x], {x, a, t}]*f[t] + Integrate[f[x] > ^2, {x, a, t}] > > > Let's re-arrange this by hand. The above is equal to: > > > (t - a)*f[t]^2 + Integrate[f[x]^2-2*f[x]*f[t]+f[t]^2, {x, a, t}]- > Integrate[f[t]^2, {x, a, t}] > > that is > > ((t - a)*f[t]^2- Integrate[f[t]^2, {x, a, t}]) + Integrate[f[x] > ^2-2*f[x]*f[t]+f[t]^2, {x, a, t}] > > that is > > (t - a)*f[t]^2- Integrate[f[t]^2, {x, a, t}]) + Integrate[(f[x]-f > [t])^2, {x, a, t}] > > However: > > > Simplify[(t - a)*f[t]^2 - Integrate[f[t]^2, {x, a, t}]] > > 0 > > so we are left with > > > Integrate[(f[x]-f[t])^2, {x, a, t}] > > But the integral of a non-negative function over an interval is non- > negative. Hence we have shown that D[g[t],t]>=0 and g[a]=0, which > implies what we wished to prove. > > Andrzej Kozlowski > > Tokyo, Japan > > > > > On 20 Jun 2006, at 15:14, ronnen.levinson at gmail.com wrote: > >> Hi folks. >> >> Sorry, I omitted a trailing exponent in my definition of r: >> >> r = (b-a) Integral[ f(x)^2 dx, {x, a, b} ] / >> Integral[ f(x) dx, {x, a, b} ]^2 >> >> I hope this correction makes my question clearer. >> >> Thanks, >> >> Ronnen. >> >> ronnen.levinson at gmail.com wrote: >>> Hi. >>> >>> I'm trying to determine whether the following ratio >>> >>> r = (b-a) Integral[ f(x)^2 dx, {x, a, b} ] / >>> Integral[ f(x) dx, {x, a, b} ] >>> >>> is always greater than or equal to one for 0 < f(x) <= 1. All values >>> all real. >>> >>> I've obtained r>=1 for all tested choices of f(x), but seek >>> guidance to >>> find the general answer. >>> >>> Yours truly, >>> >>> Ronnen Levinson. >>> >>> P.S. E-mailed CC:s of posted replies appreciated. >> >

**References**:**Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)***From:*ronnen.levinson@gmail.com