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Topic: Estimate the integral of the product of two functions
Replies: 2   Last Post: Aug 16, 2012 7:44 PM

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Posts: 85
From: Houston, TX
Registered: 3/9/08
Re: Estimate the integral of the product of two functions
Posted: Aug 16, 2012 7:44 PM
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On Thursday, August 16, 2012 2:34:29 PM UTC-5, nagornyi wrote:
> The functions g(x) and w(x) are defined on [a,b], the first one is bounded, the second one is positive bounded:
> |g(x)|<=M
> 0<=w(x)<=m.
> There are two additional conditions:
> 1. int[a,b] w(x) dx = 1
> 2. int[a,b] g(x) dx = 0
> Question: | int[a,b] g(x) w(x) dx | <= ?
> =========================================
> Using only the first condition, we can write:
> | int[a,b] g(x) w(x) dx | <= |g(x)|*| int[a,b] w(x) dx | <= M.
> Introduction of the second condition should (seems like) improve the estimate. But how?

We consider the case in which m>=1, a=0 and b=1.
Fix t in the interval [1/2/M,1-1/2/M].
Take g(x)=1/(2t) over [0,t) and -1/(2(1-t)) over [t,1]. Then |g(x)|<=M and int[a,b] g(x) dx = 0.
Next, take w(x)=m-(m-1)/t over [0,t) and m over [t,1]. Then int[a,b] w(x) dx = 1.
Finally | int[a,b] g(x) w(x) dx |
=| int[0,t] 1/(2t)(m-(m-1)/t) dx +int[t,1] -1/(2(1-t))m dx |
=| t/(2t)(m-(m-1)/t) +(1-t)(-1/(2(1-t))m) |
so the upper bound on the integral is at least (m-1)*M.

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