(05/22/2023, 05:55 AM)Daniel Wrote: Has anyone studied the iterates of \( \frac{1}{e^{-1/z}} \)?

\[\frac{1}{e^{-1/z}}=e^{1/z}\]

\[\frac{1}{e^{-\frac{1}{e^{-1/z}}}}=e^{e^{1/z}}\]

\[\frac{1}{e^{-\frac{1}{e^{-\frac{1}{e^{-1/z}}}}}}=e^{e^{e^{1/z}}}\]

Im sorry but iterating exp(1/z) I get

\[e^{1/z}\]

\[e^{e^{-1/z}}\]

\[e^{e^{-e^{-1/z}}}\]

etc

So I would like to point out :

1) your result is wrong.

2) the fixpoint at oo of exp(x) is not really an analytic fixpoint. In fact exp(oo) takes on any value, even 0.

exp(x) - x = 0 does not even have a fixpoint at positive infinity in the sense that lim exp(x) - x > 0.

3) This looks alot like using base -e ( MINUS e ! ).

This base is thus below the sheltron region rather than above.

iterating exp(1/z) =?=

exp^[s]( exp_{-e}^[f(s)]( (-1)^g(s) 1/z ) )

for some f(s) and g(s) ??? hmm

4) Maybe we should go backwards :

iterate ln(z)^(-1)

( the inverse of exp(1/z) )

or the general case ln(z)^(t)

Notice ln(z)^(-1) has the fixpoint

ln(z)^(-1) = z

=>

ln(z) = 1/z

=>

z ln(z) = 1

z= exp( LambertW(1) )

( about 1.7632.. )

---

Now if we differentiate the fixpoint of 1/ln(z) we get

1/( z ln(z)^2 )

plugging in the fixpoint exp(W(1) ) we get

1/ ( exp(W(1)) W(1)^2 )

=

1/( 1 * W(1) )

=

W(1)^(-1)

=

also exp(W(1)) !!

( about 1.7632.. )

so the fixpoint has its own derivative cool.

***

other bases or log(x)^2 could also be considered.

But maybe you do not want to go there ?

Regards

tommy1729