(06/10/2022, 09:56 AM)Catullus Wrote: On the topic of the behavior of tetration. How slow does (e^-e)^^x converge anyway? as n becomes larger and larger (e^-e)^^n approaches e^-1 which is approximately .368. even (e^-e)^^16384~.375. Fun fact: 1/((e^-e)^^16384-e^-1) ~ 141.404. 141.404 ~ sqrt(2)*100.
Also maybe there should be a symbol for e^-e. (Also called the iterated exponential constant.) Like how there is a symbol for e^e^-1.
I've always thought we should write \(\eta_-\) for \(e^{-e}\), as it acts kinda like the polar opposite of \(\eta\).
As to how fast the tetration converges, it is slower than any exponential, and additionally
\[
\sum_{n=0}^\infty \eta_-\uparrow\uparrow n - e^{-1}\,\,\text{Does not converge}\\
\]
Whereas for every value \(b\) in the Shellthron region we have exponential convergence to the fixed point.
So just hazarding a guess, \(\eta_-\uparrow\uparrow n = e^{-1} + O(1/n)\).
Probably converges slower though, by the amount your showing. I know there is a way to get the correct O term. It's something to do with the petals. I believe there are two petals about \(e^{-e}\), so if memory serves right, it should be something more like \(O(\sqrt{n})\). And this aligns with the estimate you have above. So I'd bet it's something like:
\[
\eta_-\uparrow\uparrow n = e^{-1} + O(1/\sqrt{n})
\]
EDIT:
Found it in Milnor. If \(f(z) = e^{2\pi i/k}z + O(z^2)\), then in the attracting petals (there will be \(k\)), the function \(f^{\circ n}(z) = O(1/\sqrt[k]{n})\) So yes, an initial estimate would be \(O(\frac{1}{\sqrt{n}})\).
