Re: John Gabriel's Average Tangent Theorem

From: Jason (logamath_at_yahoo.com)
Date: 02/16/05 

Date: 16 Feb 2005 08:38:30 -0800

David,
Sorry. There is one thing I noticed: Instead of saying differentiable,
he should have said continuous. My apology. Otherwise I see no problem
with the definition.

Besides there is nothing wrong with the math and this is what you need
to look at!

Jason

Jason wrote:
> > >The DERIVATIVE of f(x) denoted by f'(x) is the gradient of the
> tangent
> > >to the curve f(x) at the point [x; f(x)] provided f(x) is
> > >differentiable over [x;x+w].
> > >So what's your problem with this?
>
> > Is this his _definition_ of the derivative, as oppposed to just
> > a statement about derivatives? If it's a statement about
derivatives
> > then it's wrong because a few words are used incorrectly, and it's
> > also curiously weak. But if it's supposed to be the _definition_
> > of the derivative, as I gather it is, then it's simply incoherent,
> > exactly as Wade said/
>
> This is his definition of the derivative - that's what his page says.
> You have made quite a few statements all of which I disagree with
you:
>
> - You say words are used incorrectly. What words are used
incorrectly?
> I see no problem with his wording.
> - You say if it is his definition, then it's incoherent. What is
> incoherent about it? I do not see any incoherence.
>
> > The smallest problem is the use of the word "gradient". In
> > fact the derivative of a function is the _slope_ of the
> > tangent - the tangent is a line, and lines have slopes,
> > lines do not have gradients.
>
> There is no difference between the word gradient and slope. They mean
> exactly the same thing in this case. And you are incorrect about
lines
> not having gradients. A tangent can be a line and it does have a
> gradient/slope. There is no problem with his wording here. Frankly,
> what you are writing sounds incoherent to me.
>
> > A big problem is defining the derivative as the slope
> > ("gradient") of the tangent. The usual definition of
> > tangent line is the line passing through the point
> > (x,f(x)) with slope equal to f'(x). Unless he has
> > previously given a coherent definition of "tangent"
> > that does not mention derivatives the definition
> > is circular.
>
> Again you are talking a whole load of rubbish here David. Just read
> this paragraph again and you will realize that you have said nothing
> but
> incoherent rubbish. What are you trying to say? You use the word
> incoherent too easily: there is decidely nothing incoherent about
John
> Gabriel's English and what he states makes perfect sense. Perhaps you
> should be looking more at the mathematics rather than trying to shoot
> it all done on what you perceive to be incoherent. The mathematics
> speaks
> for itself. Please look at this and direct your comments at the
> mathematics. I have no problem with Gabriel's use of the English
> language.
>
> > The "provided f is differentiable" is also circular,
> > since "f is differentiable" means "f has a derivative".
> > This makes the definition incoherent: It says that
> > _if_ f has a derivative then the derivative is so
> > and so. If we don't know what the word "derivative"
> > means then this "definition" does not tell us what
> > it means, because we cannot understand the definition
> > until _after_ we know what the word "derivative" means.
>
> Gabriel states through his mathematics what the derivative means.
> Simply
> put, his formula allows one to calculate any derivative from first
> principles. There are no circular definitions or otherwise.
>
> > (Another less important point is the "provided f is
> > differentiable over [x;x+w]". If this were a statement
> > about derivatives, as opposed to being the _definition_,
> > then it would be sort of correct, but requiring that
> > f be differentiable over [x;x+w] is very strange,
> > since all that is needed is that f be differentiable
> > at x.)
>
> No. It must be differentiable over [x;x+w] because if we are
> approximating the derivative as we are approaching from the right,
i.e.
> x+w, all those derivatives must exist for that interval, no matter
how
> small the interval. For if these do not exist, then we cannot use any
> formula to approximate or find a derivative through first principles
> because we are calculating gradients at each point as the two points
x
> and x+w begin to coincide. So you are very incorrect saying this. If
> what you are saying is true, then the classic definition fails too:
>
> f(x+w) - f(x)
> i.e. f'(x) = Lim ------------- is not possible.
> w->0 w
>
> For any choice of w, we must have x+w differentiable. Did you do
> calculus?
> It would appear to me that you are very confused. What you are saying
> sounds like you are a complete novice?
>
> Jason Wells

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