Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)
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- Subject: [mg67399] Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)
- From: dh <dh at metrohm.ch>
- Date: Thu, 22 Jun 2006 06:21:10 -0400 (EDT)
- References: <e7589k$l5d$1@smc.vnet.net> <e7846p$fmp$1@smc.vnet.net> <e7aoeo$8t5$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Please not that there is a typo in the foregoing msg.: Plus + should
read Times *.
Sorry about this, Daniel
dh wrote:
> Hi Ronnen,
> after you changed the question to (square in the denominator):
> r = (b-a) Integral[ f(x)^2 dx, {x, a, b} ] /
> Integral[ f(x) dx, {x, a, b} ]^2
> your guess is certanly right. The proof is done using Cauch-Schwarz
> inequality. If S is a (real) scalar product, then
> S[f1,f2]^2 <= S[f1,f1]+S[f2,f2]
> Now consider the scalar product: Integrate[f1[x] f2[x],{x,a,b}]:
>
> Integrate[1 f[x],{x,a,b}]^2 = S[1,f]^2 <= S[1,1] + S[f,f] = (b-a) +
> Integrate[f[x]^2,{x,a,b}]
>
> Daniel
>