From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: ArcTan and 1/x
Date: 25 Aug 1997 21:33:46 GMT

In article <01bcaf7c$750fb360$02ea77ce@norman>,
Preferred Customer <user@msn.com> wrote:
>I noticed one day that the derivative of Arctan(x) is very similar to the
>derivative of -1/x.  I tried taking the limit of the integral 1/(x^2+a) as
>a approaches 0 from both negative and positive dirrections.  As a definite
>integral it gets confusing but as an indefinate integral the results were
>most suprising.  Please try it and give me any input.

So we let F_b(x) = (1/b) Arctan(x/b) and notice its derivative is 1/(x^2+b^2),
while F_0(x) = -1/x has a derivative of  1/x^2. That is, for any  x,
we see that the derivatives of the  F_b  at x  converge to the derivative
of  F_0.  So?  What that means is that (at each  x)  the slopes of the
F_b  get closer and closer to that of  F_0.  

If it bothers you that
the values of  F_b(x)  themselves are not looking like  F_0(x), relax;
after all, the derivatives of  F_b  and  F_b + C  are the same for any
constant. You could for example have defined

	F_b (x) = (1/b) Arctan(x/b) - (1/b) Arctan(1/b) - 1

which clearly has the same derivative as the original  F_b; it's value at
x=1 is exactly the same as  F_0(1). If you'll graph some of these functions
F_b  now, near  x=1,  you'll see that they are indeed getting closer and
closer to the function  F_0  in that region.

Actually there are some interesting issues here. It's not at all clear
what it means for a sequence of functions  f_n  to "get closer and closer" to
a fixed function  f. Given the most natural definitions, even if the  f_n  
approach  f, it's not true that the derivatives of the f_n  will approach
the derivative of f. (Try e.g.  f_n(x) = (1/n) cos(nx), which approaches
the constant function f(x)=0 pointwise, even though the derivatives
(f_n)'(x) = sin(nx) take values as large as +1.) The fundamental problem
is that differentiation is a limit process, and when you ask
"is the derivative of the limit the same as the limit of the derivatives?",
you're trying to interchange two limit processes, which will in general
give two different answers. (Try letting  m  and  n  go to infinity in
different orders in the expression  (m-n)/(m+n).)

dave