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

```

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