Re: Gabriel's AFD - Thread 8th March 2005

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 03/10/05 

Date: Thu, 10 Mar 2005 10:23:30 -0600

On 9 Mar 2005 08:20:18 -0800, "Jason" <logamath@yahoo.com> wrote:

>> Huh? If what I said is right, which of course it is, then your
>> _proof_ of the theorem says that
>>
>> f(x+w) - f(x) = f'(x).
>> That's not right.
>
>No. That's not what the theorem says.

I didn't say that that was what the theorem says.
I said that that followed from a certain step in
your supposed proof of the theorem.

You don't seem to have noticed, by the way: the theorem
turns out to be simply false, so you can stop worrying about
how to prove it.

>> If the proof involves "positional derivatives" but you can't tell
>> me what a positional derivative is then the proof is not right.
>
>A *positional derivative* exists everywhere on the interval [x;x+w).
>It is not possible to calculate positional derivatives. A derivative
>becomes *positional* once n and t coincide, I think what this means is
>that although it exists at that point (i.e. _x+ws/n_), the difference
>quotient can no longer be evaluated since
>
> f(x+2w/n) - f(x+w/n)
> --------------------
> w/n
>
>contains infinite values and the *subtractand* (in the numerator) is
>indeterminate. If you could evaluate the above assuming w/n is zero,
>all the quotients would evaluate to zero *always* since f(x)-f(x)=0.
>With this in mind, we do not try to evaluate these derivatives but
>rather the entire sum which is reducible.
>
>Jason Wells

************************

David C. Ullrich