On May 9, 2:53 pm, Ralf Bader <ba...@nefkom.net> wrote: > Virgil wrote: > > In article > > <26149057-ce0c-4da2-895e-a7efe7ca5...@h1g2000vbx.googlegroups.com>, > > WM <mueck...@rz.fh-augsburg.de> wrote: > > >> On 9 Mai, 21:36, Virgil <vir...@ligriv.com> wrote: > >> > In article > >> > <b15d6323-e22a-4963-9519-e7f9e948e...@q8g2000vbl.googlegroups.com>, > > >> > 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.