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Topic: Matheology � 258
Replies: 53   Last Post: May 11, 2013 10:07 PM

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 ross.finlayson@gmail.com Posts: 2,720 Registered: 2/15/09
Re: Matheology 258
Posted: May 11, 2013 6:55 PM

On May 9, 2:53 pm, Ralf Bader <ba...@nefkom.net> wrote:
> Virgil wrote:
> > In article
> >  WM <mueck...@rz.fh-augsburg.de> wrote:

>
> >> On 9 Mai, 21:36, Virgil <vir...@ligriv.com> wrote:
> >> > In article

>
> >> >  WM <mueck...@rz.fh-augsburg.de> wrote:
> >> > > > WM <mueck...@rz.fh-augsburg.de> wrote:
> >> > > > For all n: f(n) = 1 , lim_n-->oo f(n) = 1
> >> > > > This is required for correctly calculating differential quotients
> >> > > > in analysis. (Just this morning I explained that in class.)

>
> >> > How is
> >> >    "For all n: f(n) = 1 , lim_n-->oo f(n) = 1"
> >> > needed  to calculate  the differential quotient of f(x) = e^x at x =
> >> > pi?

>
> >> It is necessary to calculate the differential quotient of functions
> >> like f(x) = ax + b.

>
> > It is not at all necessary, as many calculus texts manage quite nicely
> > to find the differential quotients of such linear functions without it.

>
> >> > It can ONLY be of any use in correctly calculating differential
> >> > quotients in the rare cases in which the difference quotients at a
> >> > point are all equal regardless of the differences in x.

>
> >> So it is. But even these "rare cases" belong to mathematics and have
> >> to be solved correctly.

>
> > I do not know of any such rare cases that cannot be solved much more
> > simply by ordinary difference quotiens,

>
> >> > I.e., when the delta-y over delta-x ratio is constant, as in linear
> >> > functions.

>
> >> > So apparently WM never gets anywhere beyond the derivatives of linear
> >> > functions.

>
> >> That is not an admissible conclussion.
>
> > Why not? Your sequential arguments are like using sledgehammers to crack
> > eggs.

>
> Not really. They are the way you go if you unwind the definitions.
>

> > Give y = f(x) = a*x + b for all real x, and son point x = x_0
> > The difference quotient from x = x_0 to x = x_0 + h, for any h =\= 0 is
> >    [f(x_0 + h) - f(x_0)]/[(x_0 + h) - (x_0)] =
> >    [a*(x_0 + h) + b - a*x_0 - b]/h =
> >    [a*h]/h =
> >    a.
> > Since this is true for all non-zero real h,
> > it is also the limit as h -> 0.
> > No sequences needed.

>
> > Is that too difficult for your students, or just too difficult for you?
>
> Well, in your above computation, you still need to take lim_(x->x_0), and
> the final a is to be seen as a function with constant value a, for
> lim_(x->x_0) a to make sense. And the limit of that function at x_0 is taken
> through sequences, and that boils down to something remotely resembling
> Mückenheim's crap. One time at least you (resp. students) should see that
> and how the machinery works before starting to be sloppy. Or imagine
> programming a CAS system doing this stuff. You will need to distinguish the
> number a from the constant function with value a, and probably you will
> need to think a little bit to get things straight at this point.
> Mückenheim-style crap like
>  For all n: f(n) = 1 , lim_n-->oo f(n) = 1
>  is necessary to calculate the differential quotient of functions
>  like f(x) = ax + b.
> will make a mess out of your CAS. It does make some sense that the f(.)
> notation is used for functions, and the index notation (x_i) is used for
> sequences, but Mückenheim proudly screws everything up.
>
> But this is just the first level of current Mückenheim crap. This strawman
> about differentiating linear functions was brought in to start a new round
> of Mückenheim's idiotic digit shuffling procedure allegedly proving that in
> what Mückenheim believes to be set theory, "For all n: f(n) = 1 ,
> lim_n-->oo f(n) = 0".

Let f_d(n) = n/d, 0 <= n <= d, d e N.

lim n->d f_d(n) = 1, 1 e ran(f)

f is strictly increasing and constant monotone, for all d.

Then, as d->oo, each f(n), as an expansion, would at least start with .
000..., but the elements of ran(f) are strictly increasing for
increasing n, for all d, and 1 e ran(f) for all d.

To see a corresponding example, integrate: S_0^1 1 dx. The
differential vanishes as b-a -> 0 but the sum over them = 1.

Or: does not the square have area?

Regards,

Ross Finlayson