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Topic: Matheology § 300
Replies: 152   Last Post: Aug 1, 2013 6:47 AM

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Re: Matheology � 300
Posted: Jul 30, 2013 3:10 PM

> > In article
> >

>

> > >,
> > mueckenh@rz.fh-augsburg.de wrote:
> >

> > > On Saturday, 29 June 2013 18:23:47 UTC+2,
> Michael
> > Klemm wrote:
> > > > "Michael Klemm" <m_f_klemm@t-online.de> wrote
> in
> > message
> > > > news:kqmhs1\$1ih\$1@solani.org... > The velocity
> of
> > the
> > > > convergence/divergence is given by the
> equation
> > The velocity of the
> > > > convergence/divergence is given by the
> equation
> > f(n+1) - f(oo) =
> > > > -a[f(oo) - f(n)].
> >
> > Regards Michael

> > >
> > > where f(oo) = lim{n-->oo}f(n)
> > >
> > > Regards, WM

> >
> > Not necessarily.
> >
> > So WM is wrong, again, as uusal!!!
> >
> > One need not have f(oo) = lim_(n -> oo) f(n) at

> all,
> > as long as one has
> > a defined value for f(oo) and also for f(n) for

> each
> > n in |N..
> >
> > "Continuity" of f at x = oo is not required for

> f(n)
> > - f(oo) to make
> > sense for every n in |N..
> > --
> >
> >

>
> Nonsense.
>
> Let f(0) = N.
> Then it's easy to see that:
> f(n) = a^n.N + b.(1-a^n)/(1-a)
> And lim(n->oo) f(n) = b/(1-a), but only for |a| <= 1
> Symbolically we may write then: f(oo) = b/(1-a)
> For |a| > 1 the limit and hence f(oo) does not
> exist.
> As every beginning student in calculus knows.
> Micheal Klemm's "argument" is matheology.
>
> Han de Bruijn

Sorry, I meant a < 1 instead of a <= 1.

Han de Bruijn

Date Subject Author
6/26/13 mueckenh@rz.fh-augsburg.de
6/26/13 Virgil
6/26/13 YBM
6/26/13 Scott Berg
6/27/13 mueckenh@rz.fh-augsburg.de
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