[Update] Comparision of 5 methods of interpolation to continuous tetration

[sheldonison]

I simply asked if I am correct with my 10 step way to do kneser.

I do not ask for the Riemann mapping so I guess its more of a proof question.

But actually I'd say its a " construction " question.

Is the Kneser proof/solution constructed in the 10 steps I posted or is one or more steps wrong ?

Tommy, the sequence is good, but the Riemann mapping step 5) has a lot of sub-steps. For me, the biggest complexity hurdle in Kneser's construction, besides the fact that I don't have a formal math degree, is taking the \( \alpha(z) \) Abel function, from Schroeder function, of the real axis, which is after the step where he generates the chi-star, but still one or two steps before the Riemann mapping. I don't read German, so have no idea how he proved the infinite region is simply connected, and I wouldn't know how to do so, since the region is increasingly recursively complex.

Here is the rough Abel function of the real axis, showing the repeating pattern; here \( \Im(z)=\frac{1}{sexp(3.5)}\approx 10^{-78} \). Kneser multiplies this repeating pattern by \( 2\pi i \) and then takes the exponent of that; \( \exp(2\pi i z) \). That is the contour that gets wrapped around a unit circle for the Riemann mapping.
   

Here we zoom in on one of the singularities, where sexp(z)=0. The singularity gets ever more complicated as we super-exponentially approach zero. Here, I show what the contour looks like if \( \Im(z)=\frac{1}{\text{sexp}(8.5)} \).
   

My algorithm has a mathematical description, as well as pari-gp code. I don't want to side track too much, but it does something different but equivalent to generate \( \text{sexp}(z)=\alpha^{-1}(z+\theta(z)) \), via a 1-cyclic mapping from the inverse Abel function, \( \alpha^{-1}(z) \), as well as an sexp(z) Taylor series representation at the real axis. This is because the \( \theta(z) \) function has a singularity at the real axis, so adequate convergence is not possible with a reasonable number of terms. So my algorithm actually has to iteratively generate two different equivalent representations of sexp(z).
- Sheldon