The Logic Off Topic list

Thoughts from the mind of yoonchinet@..., 08-11-2001:

>--- In logic-ot@y..., GAmoore@a... wrote:
>  >    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?

The graph of the functions Greg describes looks like (in Courier please)


      |
1/n -|=======================
      |
      |
      |
      +----------------------+=======================
                             n

i.e. a horizontal line at height 1/n, from x=0 to x=n, and at height 
0 for x>n.  The integral is simply the surface between the function 
and the x-axis, which is clearly n*1/n = 1.

Your
     limit( ((1/n)*x, x = 0 to x = y), y -> infinity)
is indeed not 1 for all n (in fact, it's infinity for all n>0). 
However, this limit is not the same as the previous integral. 
(1/n)*x is a primitive function for 1/n -- but 1/n is _not_ what's 
being integrated.

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

Correct.  This limit is simply (well, loosely speaking) an infinite 
line at height 1/infinite, which still encloses a surface of 1.


>  > On the other hand,
>>     limit ( f_n(x), n -> infinty)  = 0 (the zero function)
>>     integral( 0, n -> infinty) , x = 0 to x = infinity) = 0

Also correct.


cheers,
HJ
-- 
     Hendrik Jan Veenstra
     email: mailto:h@...
     www:   http://www.ision.nl/users/h/index.html