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Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)
*To*: mathgroup at smc.vnet.net
*Subject*: [mg67388] Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)
*From*: ronnen.levinson at gmail.com
*Date*: Wed, 21 Jun 2006 02:13:06 -0400 (EDT)
*References*: <e7589k$l5d$1@smc.vnet.net><e784ee$foq$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Hi.
I'd like to thank Carl K. Woll and Daniel Huber, who provided solutions
to my typographically corrected question using (a) Lagrange multipliers
and (b) the Cauchy Schwarz inequality, respectively.
The simplest solution (similar to Daniel's) came from a colleague of
mine at Lawrence Berkeley National Lab:
Define mean value operator
bar{g(x)} = Integral[g(x) dx, {x, a, b}] / (b-a).
Then for f(x) real and b != a,
[f(x) - bar{f(x)}]^2 >= 0
Expanding the left hand side,
f(x)^2 - 2 f(x) bar{f(x)} + bar{f(x)}^2 >= 0
Integrating both sides over x from a to b and then dividing by (b-a),
bar{f(x)^2} - 2 bar{f(x)} bar{f(x)} + bar{f(x)}^2 >= 0
bar{f(x)^2} - bar{f(x)}^2 >= 0
and hence
r = bar{f(x)^2} / bar{f(x)}^2 >= 1
which is what I sought to check.
Yours truly,
Ronnen.
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.
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