Re: The definite integral.

From: tetrahedron (jarynth_at_yahoo.com)
Date: 03/16/05 

Date: 16 Mar 2005 02:28:13 -0800

David C. Ullrich wrote
> My first comment is that it seems to me that trying to
> show that ftc can be proved without mvt is really not a good
> idea, because it's going to reinforce Jason's delusions
> about how a proof of ftc using mvt is somehow inappropriate.

No, actually it's possible to prove ftc "without" mvt: all you have to
do is embed the proof of mvt in the proof of ftc and start with the
assupmtions for both. I haven't been following the threads about
Gabriel's theorem very thoroughly, and I'm not sure what his followers
are trying to prove. In addition, Gabriel's site is confusing.

> Now about whether this proof is actually valid. I don't
> think it is. You never stated the hypotheses explicitly,
> but as of
>
> >2) f is differentiable, hence continuous; f' is integrable, hence (by
> >def.) bounded
>

> I take it that you're assuming just that f is differentiable and f'
> is Riemann integrable. I may be missing something, but it also
> appears to me that you need to know that f is "uniformly
> differentiable" for the proof above to work, ie that (modulo
> fiddling about cases where x + h lies outside the domain of f)

You're right, I was too hasty.

>
> (*) (f(x+h) - f(x))/h -> f'(x) _uniformly_ in x.
>
> Of course (*) is equivalent to saying that f' is continuous
> (that (*) implies f' is continuous is just the fact that a
> uniform limit of continuous functions is continuous, while
> the other direction is, eg, clear from mvt...)
>
> If you're not using (*) then I don't see offhand why
> w*sup_s o_s(w/n) tends to 0 as n -> infinity; why is
> that?
>
> >3) The o_s are Landau symbols corresponding to functions different for
> >every value of the index s; you can expand them in f and f' and you
> >can set them to 0 outside dom(f); they hide the parameter x, but you
> >can consider the sum on the left as a unifromly convergent sum of
> >functions of x: because of 2, the o_s are bounded wrt s, even if s is
> >replaced by a continuous parameter.
> >4) equations are true in both senses, so if you dont like the top-down
> >direction you may consider reading the proof bottom-up.
> >
> >Regards
> >
> >
> >
> >"Jason" <logamath@yahoo.com> wrote in message news:<1110866870.134632.127940@g14g2000cwa.googlegroups.com>...
> >> > Well, you could try cracking a calculus text or typing,
> >> > riemann integral definition, at Google.
> >> > I won't give the most general definiton here, but
> >> > here is one that works for continuous f'.
> >> >
> >> >
> >> > x+w w n-1 ws
> >> > Integral f'(t) dt = Lim - SIGMA f'(x + --- )
> >> > x n->oo n s=0 n
> >>
> >> Hell, I am really sorry but this is gabriel's definition. It is
> >> decidedly not a riemann sum! Unfortunately you can't use this because
> >> you have been disagreeing with it all along.
>
>
> ************************
>
> David C. Ullrich

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