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+--- Thread: Has anyone studied the iterates of \(\frac{1}{e^{-1/z}}\) (/showthread.php?tid=1753)
Has anyone studied the iterates of \(\frac{1}{e^{-1/z}}\) - Daniel - 05/22/2023
Has anyone studied the iterates of \( \frac{1}{e^{-1/z}} \)?
\[\frac{1}{e^{-1/z}}=e^{1/z}\]
\[\frac{1}{e^{-\frac{1}{e^{-1/z}}}}=e^{e^{1/z}}\]
\[\frac{1}{e^{-\frac{1}{e^{-\frac{1}{e^{-1/z}}}}}}=e^{e^{e^{1/z}}}\]
RE: Has anyone studied the iterates of \(\frac{1}{e^{-1/z}}\) - Daniel - 05/22/2023
I am working on using the exponential function's fixed point at \(\infty\) in an attempt to define tetration \(^z a\) for \(a>e^{1/e}\). Generalizing \(e^{-\frac{1}{z}}\) to \(g(z)=1/f{(\frac{1}{z})}\) where \(f(\infty)=\infty\).
The iterates of \(g(z)\) are,
\[\frac{1}{f\left(\frac{1}{z}\right)},\frac{1}{f\left(f\left(\frac{1}{z}\right)\right)},\frac{1}{f\left(f\left(f\left(\frac{1}{z}\right)\right)\right)},\frac{1}{f\left(f\left(f\left(f\left(\frac{1}{z}\right)\right)\right)\right)}\]
RE: Has anyone studied the iterates of \(\frac{1}{e^{-1/z}}\) - tommy1729 - 05/22/2023
(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