Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I)

[bo198214]

In Chaos Pro, is there a way to find coordinates [x,y] of an interesting region/point? So far I could only use zooming the are which was fine for e^pi/2, e^-e, e^(1/e) , e^pi/2 which are rather distinct points with interestingly different behaviour of iteration z_0=I, z=pixel^z in their neighborouhood, ( I have placed the pictures in the thread) but that is very time consuming if points are less obvious and e.g. off real axis. Imaginary axis is almost invisible.

Unfortunately I dont know either. Perhaps one can apply some tickmarks via the formula, if real and imaginary part are integer then draw a red dot or something like that. But it should be possible with intrinsic methods in ChaosPro. If you use Fractal Explorer the complex coordinates of the current mouse position are shown in the status bar, however fractal explorer is not that sophisticated as ChaosPro and magnitudes slower.

If you consider the limit \( a \) of \( z_{n+1} = b^{z_n} \) then it is clear that \( b^a=a \). However the sequence would only converge to an attracting fixed point and the only attracting fixed point of \( b^z \) is the lower real fixed point, which exists only for \( e^{-e}<b<e^{1/e} \).

Perhaps any finite z on top of iterations of a^(a^(a^.....z) will lead to h(a) via different trajectories.

No, it does not converge for all starting values \( z_0 \), for example \( \lim_{n\to\infty} z_n = \infty \) for \( z_0>4 \) and \( b=\sqrt{2} \).
But if it converges then indeed \( \lim_{n\to\infty} z_n = h(b) \).

And would mapping the trajectory of this convergence give some additional information about the point compared to just computing it step by step on real axis?

*shrug*