The Logic Off Topic list

--- In logic-ot@y..., GAmoore@a... wrote:
> Let me give you an example of the sort of fundamental mistake that 
> Fourier made. 
> 
> If you have sequence of functions {  f_n(x) : n = 1, 2, .....} then it is 
> >obivously< true that
> the you can interchange the integral and the limit :
> 
>         integral( limit ( f_n(x), n -> infinty) , x = 0 to x = infinity) =
>             limit ( integral(   f_n(x) , x = 0 to x = infinity)  , n -> 
> infinty)
> 
> Right? Wrong. Consider this simple example :
>    let  f_n(x) = 1/n for 0 <= x <= n, and f_n(x) = 0 elsewhere.
> 
>   Then the 
>        integral(   f_n(x) , x = 0 to x = infinity)  = 1 for all n, 

I'm sorry but wouldn't the result of this be: limit( ((1/n)*x, x = 0 to x = y), y -> infinity) which is not equal to 1 for all n?

> and therefore
>        limit ( integral(   f_n(x) , x = 0 to x = infinity)  , n -> 
> infinty) = 1 

This result is correct although the previous assumption was not correct.

> 
> On the other hand, 
>    limit ( f_n(x), n -> infinty)  = 0 (the zero function)
>    integral( 0, n -> infinty) , x = 0 to x = infinity) = 0
> 
> Clearly 1 does not equal 0.
> 
> If you want a stronger example, consider the Heaviside function
>    H_n(x) = 1 for x >= n, and 0 for x < n.
> 
> This function has a limit of the zero function also, but each of the 
> integrals are infinity, so your equation above would look like
>     infinity = 0
> Clearly this is false.

Your conclusion is right. But I thought the equation you started with was ONLY true under certain pre-conditions, the one about the limit of the integral is the same as the integral of the limit. One of the pre-conditions being that the functions must 'converge uniformly'( I hope this is the correct term). This means that the limit( abs(f_n -f), n -> infinity) = 0, where abs is the absolute function and f is a function. You introduced a function into the equation that violates that pre-condition; both functions aren't uniformly convergent.
Check your calculus books, man. At least you gave me a jolt to check my college first year calculus stuff again. Thanks, :-).

> This is the kind of thing Fourier did. He assumed these kinds of things 
> could be done by 'pushing symbols around' as we say in mathematics.

Wrong. Fourier's equations satisfy the precondition I mention above. Maybe he didn't write it in his publications this way, but they were valid. Stop dissing Fourier, man!! :-).