a simple iteration -> 1,39865929053... - sci.math

Message from discussion a simple iteration -> 1,39865929053...

hagman

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More options Jul 6, 7:57 pm

Newsgroups: sci.math

From: hagman <goo...@von-eitzen.de>

Date: Sun, 6 Jul 2008 07:57:06 -0700 (PDT)

Local: Sun, Jul 6 2008 7:57 pm

Subject: Re: a simple iteration -> 1,39865929053...

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On 5 Jul., 21:20, amy666 <tommy1...@hotmail.com> wrote:

> ( only considering reals , imput and output )

> A = 1,39865929053...

> f(0) = f(1) = x_0
> f(n) = exp(f(n-1)) - exp(f(n-2))

> ( f(2) is thus always = 0 )

> for some x_0 the sequence f(n) seems to diverge to oo and for others it seems bounded.

> it seems that the sequence for 1 < x_0 < A is bounded.

> and for larger x_0 > A it diverges to oo.

> thus of course the big question is : do we have a closed form for A = 1,39865929053... ???

> its a simple iteration , similar to tetration (especially when it diverges to oo ).

> if we do not have closed form for A , do we at least have an equation for it ?

> perhaps if we also allow tetration-like functions in our equation ?

> such a simple construct , i missed something trivial ??

> for large n ;
> f(n) = exp(f(n-1)) - exp(f(n-2)) =
> exp(f(n-1)) - exp(exp(f(n-3) - exp(f(n-4)))
> etc

> does that help ?

> regards

> tommy1729

How did you arrive at your A?
For me it does not look too obvious that e.g. x_0 = 100 should produce
divergence -> +oo.
After all (with spreadsheet precision), everything stays within the
same up an down pattern
for more than 150 steps before the inevitable happens.
OTOH, this shows that we cannot be sure that some x_0 < A really
produces a bounded (from above) sequence just by numerical evidence of
10000 steps, say.

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