Re: John Gabriel's Theorem Revisited.

From: Yan Zhang (yanzhang_at_fas.harvard.edu)
Date: 02/18/05 

Date: Fri, 18 Feb 2005 18:07:46 -0500

David C. Ullrich wrote:
> On Thu, 17 Feb 2005 16:08:35 -0500, Yan Zhang
> <yanzhang@fas.harvard.edu> wrote:
>
>
>>Jason wrote:
>>
>>>If we redefine the derivative at a point using Gabriel's
>>>limit/quotient:
>>>
>>>Definition:
>>>
>>>The derivative is the limit of the finite difference quotient:
>
>
> [i]
>
>
>>> f(x + w/n) - f(x)
>>> f'(x) = Lim ------------------
>>> n->Infinity w/n
>>>
>>>
>>>If this limit is zero or undefined at a point, we say the function f is
>>>not differentiable at the point (x,f(x)).
>>
>>How is this any different from
>
>
> [ii[
>
>
>> f(x + e) - f(x)
>>f'(x) = Lim ---------------
>> e->0 e
>>?
>
>
> It's very different.
>
> First, [i] makes no sense as stated, because as you sort of point
> out below, the right side of [i] defines a function of two variables,
> x and w - [i] would make sense if it were written
>
> f(x + w/n) - f(x)
> f'(x,w) = Lim ------------------
> n->Infinity w/n
>
> To make [i] into a sensible definition we need to add a few
> English words, saying that f is differentiable if there is
> a number f'(x) such that [i] holds for all w <> 0.
>
> That's probably just a problem with writing style, no doubt
> whoever proposed [i] as a new and improved definition meant
> that it should hold for all w <> 0. But worse, even with
> that clarification, [i] is simply not the same as [ii].
>
> An example showing that the two are different is a teensy
> bit complicated. Choose a sequence of number r_j such that
> r_j > 0, r_j -> 0, and r/j/r_k is irrational for all j, k
> with j <> k. Define f by f(r_j) = 1 and f(x) = 0 for other
> x.
>
> Then f is not even continuous at 0, so it's certainly
> not differentiable there; the sequence r_j shows that
> [ii] does not hold. But definition [i] says that
> f'(0) = 0! This is because given w <> 0 the fact
> that r_j/r_k is irrational shows that f(w/n) = 0
> except for at most one value of n.

Right, but as Mr. Ulrich argued with Mr. Wells, and in the other branch
of this topic where Mr. Wade was involved, I believe it was agreed that
there was the extra condition that f be continuous on [x, x+w],
on which I think they are equivalent.

[snip]

>>I don't think what we are doing is saying a number "IS" an infinitesimal
>>and it "exists" or not. The word "infinitesimal" comes up when we take
>>limits like we have to.
>
>
> Actually it doesn't come up even then, or it needn't. There are
> no infinitesmals involved in any of the standard definitions for
> standard analysis.

I think what I was trying to say is we are not using "infinitesimal" in
the formal definition anywhere in this argument, rather getting involved
into confusion on an "infinitesimal" as a number or a limit going to 0.
It was not worded correctly - in either case, I think talking about
the/an "infinitesimal" has no relation to the matter at hand.

-Yan