Green Eggs and HAM: Tetration for ALL bases, real and complex, now possible?

[sheldonison]

If we let \( G(z) = e^{z^2} \mathrm{tet}(z) \), then .... we can represent the tetration via

\( G(z) = \sum_{n=-\infty}^{\infty} a_n \left(\frac{1 + z}{1 - z}\right)^n \).

where \( a_n \) are the coefficients of the Fourier series for the wrapped unit circle, and the fraction is just the inverse Moebius mapping taking the imaginary axis to the circle. That is, the choice of basis functions for the HAM is given by

\( b_n(z) = \left(\frac{1 + z}{1 - z}\right)^n \).

100 coefficients then gives 32 places accuracy.

Mike, it sounds very promising. I like the Gaussian scaling, and the equations; it seems like a novel promising approach, to combine all of these different ideas with Kouznetsov's Cauchy integral. I'm struggling with the Laurent series on a unit circle, representing G(-i oo) to G(+i oo), where the inside of the circle is either the left or right half of the complex plane. Since the function is represented as a Laurent series, I guess it doesn't matter which is whiich (inside/outside unit circle <=> left/right half complex plane).

So where are the nearest singularities to the unit Laurent series circle? There would be sexp(-2,-3,-4.....) And also sexp(oo). But doesn't real(oo) also get mapped to the unit circle boundary too? I'm also confused about which half of the complex plane is inside the circle, and which half is outside... (1+x)/(1-x)=-2 for x=3. But for the inverse, (x-1)/(1+x), then (-1/3) would map to -2. I think real(oo) is mapped to -1 or 1, depending on whether we use the Mobius or its inverse. So how did the singularity for real(oo), which must be on the unit circle, get cancelled out so that it doesn't mess up the Laurent series? Even if the real(oo) gets cancelled out, sexp(4)~=oo so it may as well be a singularity and sexp(4) is mapped to either x=0.6 or x=-5/3. I'm just trying to figure out the radius of convergence of this approach.