Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)
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- Subject: [mg67374] Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)
- From: "Ray Koopman" <koopman at sfu.ca>
- Date: Wed, 21 Jun 2006 02:12:25 -0400 (EDT)
- References: <e7589k$l5d$1@smc.vnet.net><e784ee$foq$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Let r = v / u^2, where
u = Integral[ f[x] , {x, a, b} ] / (b-a), and
v = Integral[ f[x]^2, {x, a, b} ] / (b-a).
Then v = u^2 + Integral[ (f[x]-u)^2, {x, a, b} ] / (b-a),
and hence r >= 1.
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.