Re: Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)
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- Subject: [mg67391] 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 02:13:12 -0400 (EDT)
- References: <e7589k$l5d$1@smc.vnet.net> <200606200614.CAA15901@smc.vnet.net> <858BE938-5559-4439-A8F5-02B27E9A1B2B@mimuw.edu.pl>
- Sender: owner-wri-mathgroup at wolfram.com
I forgot to add that your condition: 0 < f(x) <= 1 is not really
needed, it can be relaxed to f[x]>0 or even to a more general one.
his sis clear from the proof I gave and the fact that x/y<= 1 is
equaivalent to x<=y provided y is >0. But one can also see it
independently of the proof. Clearly that bound f[x]<=1 does not
matter, since the ratio
(t - a)*Integrate[f[x]^2, {x, a, t}]/Integrate[f[x], {x, a, t}]^2
will not change if you replace f[x] by k f[x] for any k. The
condition 0<f[x] is only needed to ensure Integrate[f[x], {x, a, t}]
^2 is not 0. Other than that it also does not need, since
Integrate[f[x], {x, a, t}]^2 <=Integrate[Abs[f[x]], {x, a, t}]^2
so if the result is true for f[x]>=0 then it it is also true for any f.
Andrzej Kozlowski
On 21 Jun 2006, at 01:08, Andrzej Kozlowski wrote:
> 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
- Re: Help: ratio of integral of f(x)^2 to square of integral of f(x)