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Re: Matheology � 300
Posted:
Jul 30, 2013 3:10 PM
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> > In article > > > <e5226cfa-5122-4b21-98c8-f1c504bb1748@googlegroups.com > > > >, > > 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
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