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

• To: mathgroup at smc.vnet.net
• Subject: [mg67386] Re: [mg67364] Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Wed, 21 Jun 2006 02:13:03 -0400 (EDT)
• References: <e7589k$l5d$1@smc.vnet.net> <200606200614.CAA15901@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

First of all this is has nothing to do with Mathematica and you do  
not even  seem to know the proper Mathematica syntax for Integrate (I  
wonder how this got past the moderator, twice). To make it more  
acceptable I will try to use Mathematica in the proof as much as  
possible and I do hope I am not solving your homework!



Clearly, what you want to show is equivalent to proving that the  
function:

g[t_] := (t - a)*Integrate[f[x]^2, {x, a, t}] - Integrate[f[x], {x,  
a, t}]^2

is >=0 for all real t>=a. This is trivially true for t=a. Since g is  
differentiable we can compute D[g[t],t].  Using Mathematica we obtain:


D[g[t], t]


(t - a)*f[t]^2 - 2*Integrate[f[x], {x, a, t}]*f[t] + Integrate[f[x] 
^2, {x, a, t}]


Let's re-arrange this by hand. The above is equal to:


(t - a)*f[t]^2 + Integrate[f[x]^2-2*f[x]*f[t]+f[t]^2, {x, a, t}]-  
Integrate[f[t]^2, {x, a, t}]

that is

((t - a)*f[t]^2- Integrate[f[t]^2, {x, a, t}]) + Integrate[f[x]^2-2*f 
[x]*f[t]+f[t]^2, {x, a, t}]

that is

(t - a)*f[t]^2- Integrate[f[t]^2, {x, a, t}]) + Integrate[(f[x]-f[t]) 
^2, {x, a, t}]

However:


Simplify[(t - a)*f[t]^2 - Integrate[f[t]^2, {x, a, t}]]

0

so we are left with


Integrate[(f[x]-f[t])^2, {x, a, t}]

But the integral of a non-negative function over an interval is non- 
negative. Hence we have shown that D[g[t],t]>=0 and g[a]=0, which  
implies what we wished to prove.

Andrzej Kozlowski

Tokyo, Japan




On 20 Jun 2006, at 15:14, 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.
>

• References:
  • Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)
    • From: ronnen.levinson@gmail.com