eta as branchpoint of tetrational

[sheldonison]

@sheldonison: Thanks for the coeffs. I'd like to see if I could make a color graph. What is the starting point you use for the upper regular superfunction? (i.e. the value at 0)

Hey Mike,
\( \text{Usexp}_{\sqrt{2}}(z) \) is developed from the L=4 fixed repelling point, with \( \text{Usexp}_{\sqrt{2}}(0)=5.767053253764297762019157833944 \).
\( \text{Lsexp}_{\sqrt{2}}(z) \) is developed from the L=2 fixed attracting point, with f(-1)=0, f(0)=1, f(1)=sqrt(2), and the "merger" of the two functions takes place at half the period of the LSexp=8.57i. In practice, one could literally say that for imag(z)<=8.57i Fatou(z)=Lsexp(z), and for imag(z)>=8.57i Fatou(z)=Usexp(z+k), ignoring any of the theta(z) terms, because the two functions are so very nearly identical where they merge.
- Sheldon

That is odd, since in the complex-base case you dug up that thread on, the "merger" looks to occur right along the real axis, just as it does for, say, base \( e \). This makes me suspicious if this function is really the limit of the complex-base function for \( b < \eta \). Unless, of course, the graphs are deceptive. Which is possible, given how "subtle" the curve differences are.

And half the period, or half the average of the two periods?

At half the period of the Lower Superfunction, all three functions, LSexp(z), Usexp(z+k), and Fatou(z) are nearly equal, differing by approximately 1E-25. At the real axis, the Fatou(z) differs from LSexp(z) by 1E-50. At imag(z) =(Uperiod+Lperiod)/2, the average of the two periods, we have Fatou(z) differing from Usexp(z+k) by about 1E-50. I actually have calculated all three of these functions to >100 decimal digits accuracy, to verify my results!

If we had developed the two superfunctions of sqrt(2) on their other real valued section, at half their respective imaginary periods, than the merger between the two functions would have occurred at the real axis. But that's not what comes out of the limit equations, where we get the primary real axis of the two superfunctions. Base e is entirely different, since it involves a Kneser Riemann mapping, as opposed to the much simpler merger for Fatou(z). If you have a Kneser Riemann mapping, then the merger occurs at the real axis, which will have singularities at z=-2,-3,-4.....
- Sheldon