Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2006
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2006

[Date Index] [Thread Index] [Author Index]

Search the Archive

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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67375] Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)
  • From: dh <dh at metrohm.ch>
  • Date: Wed, 21 Jun 2006 02:12:27 -0400 (EDT)
  • References: <e7589k$l5d$1@smc.vnet.net> <e7846p$fmp$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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


  • Prev by Date: Quaternion problem
  • Next by Date: Problems editing a mathematica package, please help.
  • Previous by thread: Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)
  • Next by thread: Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)