05/12/2015, 11:45 PM
(05/12/2015, 09:55 PM)tommy1729 Wrote: First i want to say that the equation
Slog(ln(x)) = slog(x) - 1 is sometimes better then
Slog(exp(x)) = slog(x) + 1.
Basically because slog is NOT periodic as the exp suggests.
....
Let the fixpoints be L and L*.
...
How does slog behave around the singularities at L and L* ??
What is slog(-5)? What is the slog(z) as z goes to minus infinity?
slog(-5)=slog(exp(-5))-1=slog(0.0067)-1 = -1.993817....
as z goes to minus infinity slog(z) goes to -2. Since sexp(z) has a fairly straightforward logarithmic singularity at z=-2, \( \text{sexp}(z-2) \approx \log(\text{sexp\,}'(-1)\cdot x) \), then slog(z) as z goes to minus infinity is approximated by the inverse:
\( \lim_{\Re(z) \to -\infty}\; \text{slog}(z) \approx -2 + \frac{\exp(z)}{\text{sexp\,}'(-1)} \)
So if the cutpoints are drawn correctly, then for real(z)<real(L), slog is 2pi i periodic! The question becomes how do you draw the slog cutpoints; which Dimitrii Kouznetsov investigated. The main singularity is at L, L*, which is a complicated singularity since it involves both the singularity of the Koenig solution for the Abel function at the complex fixed point, as well as the perturbation due to the theta mapping. But the Koenig abel function singularity is dominant, and as you circle around the fixed point, you approximately add multiples of the sexp pseudo period which is ~= 4.4470 + 1.0579i. It is only approximate because there is a second singularity at the fixed point of L. if \( \alpha(z) \) is the Koenig Abel function at the fixed point then:
\( \text{slog}(z) = \alpha(z) + \theta(\alpha(z))\;\; \) where theta(z) is a 1-cyclic function which goes to a constant as imag(z) goes to infinity
- Sheldon
