Alternate solution of tetration for "convergent" bases discovered

[sheldonison]

@sheldonison:

Um, I'm not comparing base \( 2.33 + 1.28i \) to regular, rather base \( 1.33 + 1.28i \) to its regular iteration at the real attracting fixed point (so-called "regular tetration").

Mike,

Thanks for the clarifications. Again, love those png color plots, and I would really appreciate it if you have a link to how to create them ..... (ideally from pari-gp).

Now, onto B=1.33+1.28i. The fixed repelling point is
L1 = -1.13500 - 1.98958i
with a corresponding period of -3.6051+2.4470i.
\( \text{period}(B)=2Pi*I/(L*\log(B)+\log(\log(B))) \)

The fixed attracting point (which would be your third graph, developed by iterating log(log(log(z))) starting from near the attracting fixed point) is
L2 = 0.57937+0.64723i
with a corresponding period of 3.58745 - 0.32994i

This matches your third graph, which has approximately five iterations over 20 unit lengths.

Now your second graph might appear to be a combination of these two periods, with a period corresponding to L2 on the left, and a period corresponding to L1 on the right. This seems difficult to imagine to me update, seems like it might, (or might not) work just fine, but it will take one or two posts to explain. Suppose the regular super function(z) is developed from the repelling fixed point by regular iteration, from L1. RegularSuper(z) is entire. Call your function f(z).

Since your function approximates the periodicity of the Regular Super function on the right, would this equation hold, where \( \theta(z) \) is a 1-cyclic function?
\( f(z)=\text{RegularSuper}(z+\theta(z)) \)

edit question Mike, where are the singularities in your second graph? Are they organized in one line?
- Sheldon