> 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.
> > 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.
as said for large numbers the iterations quickly starts behaving like tetration.
so if you think about , its logical that for numbers >= e we have divergeance.
on the other hand guessing around 1.398 is not evident.
> 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.
even stranger , the number of iterations we need to arrive at divergeance does not follow simple laws ;
you would expect larger x_0 always diverging faster than smaller x_0 , this is not always true.
