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MathGroup Archive 2006

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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: [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
> 


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