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

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 umumenu@gmail.com Posts: 910 Registered: 3/9/08
Re: Matheology § 300
Posted: Jul 31, 2013 5:20 AM
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> On Saturday, 29 June 2013 13:53:26 UTC+2, Michael
> Klemm wrote:

> >> in real mathematics there is no f(oo). Therefore
> it cannot differ from lim{n-->oo}f(n).
>

> > Counterexample: f(0) arbitrary, f(n+1) = af(n)+b.
> Then f(oo) = f(oo+1) = af(oo)+b, f(oo) = b/(1-a).
> Thus, f(oo) is defined even if the sequence (f(n))_n
> is divergent. Regards Michael
>
> Jerk!
>
> Regards, WM

It's easy to see that:
f(n) = a^n.f(0) + 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

Han de Bruijn

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