complex iteration (complex "height")

[bo198214]

lower fixed point \( a=\Omega=0.567143.. \)
we get the base by \( b={1/e}=\Omega^{1/\Omega} \)
if \( z={1/\Omega} \) then

Hm, perhaps I have to add that the formula given is designed for strictly increasing functions with an attracting fixed point. However if you set \( b<1 \) then the function \( b^x \) is strictly decreasing.
It was discussed somewhere on this forum already that it most likely can not have real-valued fractional iterations there (real valued only at integer iterations).

I dont know how this formula behaves in this case (whether it converges), at least the fixed point is still attractive, so try your luck! Would be interesting whether this supports my conjecture of complex values in the range of \( e^{-e}<b<1 \).

\( {1/e}[4]{1/\Omega}=\lim_{n\to\infty} \log_{1/e}^{\circ n}(\Omega*2 -1*\exp_{1/e}^{\circ n}(1)) \)

At this point I do not know what to do with limits -how do You apply n to iterating logarithms, and at the same time to iteration of exponentiation of 1/e inside it?

It just means that if you set \( n \) high enough you get quite close to the "right" value. And we discussed already how \( f^{\circ n}(x) \) is defined. For example \( \exp_b^{\circ 3}(x)=b^{b^{b^x}} \).

If we could move limit inside, than:

\( \lim_{n\to\infty} \exp_{1/e}^{\circ n}(1))=h({1/e)}=\Omega \)

And \( 2*\Omega-\Omega=\Omega \)

\( \log_{1/e}(\Omega)= \Omega \)

Yeah, but unfortunately its not possible to move the limit inside!
Thats the strange beauty of this formula.

If we take the same but:

\( z={-1/\Omega} \)

And then I dont want to think about negative bases, perhaps first finish with bases \( e^{-1}<b<1 \)!