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From: jsa@alexandria (Jon S Anthony)
Subject: Re: OO, C++, and something much better!
Date: 1997/02/25
Message-ID: <JSA.97Feb24193207@alexandria>#1/1
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Distribution: world
References: <JSA.97Jan16141937@alexandria>
 <5de62l$f13$1@goanna.cs.rmit.edu.au>
Organization: PSI Public Usenet Link
Newsgroups: 
 comp.lang.c++,comp.lang.smalltalk,comp.lang.eiffel,comp.lang.ada,comp.object
Date: 1997-02-25T00:00:00+00:00
List-Id: <comp.lang.ada>


In article <JSA.97Feb21195346@alexandria> jsa@alexandria (Jon S Anthony) writes:

I wrote the following utter rubbish:

>  Cf = set of continuous functions on R
> 
>  ': Cf -> Cf, where f' is the derivative of f

Total _obvious_ BS.  The canonical counter example: |x|.  Fortunately
Brian Rogoff was nice enough to hit me in the head a couple of times
in order to, shall we say, realign my thinking with my saying.

What I was actually thinking was:

   ': Df -> Cf, f' = derivative of f, Df = differentiable functions (*)

Of course, in this context, ' is no longer an operator, so its a bad
example anyway.


I suppose the "right" example here would have been:

   I : Cf -> Cf, I(f) = anti derivative of f with constant c (aka
                        indefinite integral)

This is basically a quick corollary of the FTC.

  proof: since f is continuous, by the FTC, f has an anti
  derivative F with constant term c, i.e., F' = f, so F is obviously
  differentiable => F is continous.


(*) Which claims: if f is differentiable (at a) then f' is continuous
    (at a)

For those who are now wisely skeptical,

proof:

    f'(a+h) = lim  [f((a+h)+k) - f(a+h)]/k
              k->0

   lim  f'(a+h) = lim  lim [f((a+h)+k) - f(a+h)]/k
   h->0           h->0 k->0

                = lim  [lim f((a+k)+h) - lim f(a+h)]/k
                  k->0  h->0             h->0

   since f is differentiable, f is continuous and by f's continuity
   the above is

                lim  [f(a+k) - f(a)]/k
                k->0

                = f'(a)

/Jon
-- 
Jon Anthony
Organon Motives, Inc.
Belmont, MA 02178
617.484.3383
jsa@organon.com