Tetration series for integer exponent. Can you find the pattern?
#20
(02/21/2016, 01:38 PM)sheldonison Wrote: 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.

(02/21/2016, 01:38 PM)sheldonison Wrote: So if you start with b=1.1, and rotate around 180 degrees to b=0.9,
You mean a circle centered at 1, with radius .1?

I don't need that. I need center at 1.1, with radius .01:

[Image: 0DhETMJ.png?1]

(02/21/2016, 01:38 PM)sheldonison Wrote: 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.
Ok. The taylor series for \( \\[15pt]

{^{x}b} \) won't converge on 1, but \( \\[15pt]

{^{1.5}x} \) is a different function, and we know that it exists and converge for \( \\[15pt]

{^{n}x} \;\, n\in \mathbb{N} \)
I have the result, but I do not yet know how to get it.
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