02/21/2016, 01:38 PM
(This post was last modified: 02/21/2016, 01:48 PM by sheldonison.)
(02/20/2016, 11:48 PM)marraco Wrote: I do not understand if you speak of "half iterate", because your code uses it internally, or you say that I'm looking for the Taylor series of \( \\[15pt]
{^{0.5}x} \) (I'm looking for \( \\[15pt]
{^{1.5}x} \)).
I updated my post#17; I graphed the sexp(-0.5) which is not real valued for b=0.9; the graph used the simpler Schroeder function from the attracting fixed point instead of the more complicated Kneser solution, but both look nearly identical, and the problem is the function is not real valued at the real axis. So if you start with b=1.1, and rotate around 180 degrees to b=0.9, you get 0.90455-0.2937i; that is using the Schroeder function inverse Abel solution which isn't part of fatou.gp. Sexp(-0.5)=-0.90415-0.2939i. This is a problem because the Schroeder function Abel/inverse Abel solution is real valued for bases>1. So there's a complicated singularity at b=1. I don't claim to understand that singularity at b=1 yet, but I understand it well enough to know that the Taylor series won't converge, because it is a different function rotating clockwise around the singularity at b=1 than rotating counter clockwise around the singularity at b=1. The difference is fatal to getting a converging Taylor series.
To better understand the nature of the singularity, you will have to generate the Schroeder function, both for b=1.1 and b=0.9, and generate sexp(-0.5) for both cases yourself, and see that it is not real valued. My code actually works with the Schroeder function and its inverse for \( z \mapsto \exp(z)+k \), which perhaps muddies the water; \( k = \ln(\ln(z))+1 \), even though there is always simple linear transform using sexpforminvabel(z), connecting the two. The fact that fatou.gp works with the bipolar Kneser tetration using both fixed points muddies the water even more, since Kneser tetration is no longer real valued for real bases<eta either; it has an even more complicated singularity at b=1, than the very complicated Schroeder function singularity at b=1. That is why I graphed the Schroeder function solution, even though both look visually identical.
So yes, the problem is you are looking for an analytic solution, where I don't think there is any simple form for an analytic solution for the sexp(-0.5), rotating around in a circle around b=1. Sorry.
- Sheldon
