(03/06/2013, 11:57 PM)sheldonison Wrote: Can it have a Tetration superfunction with sexp(-1)=0, sexp(0)=1, sexp(1)=b, sexp(2)=b^b? This is a base on the shellthron boundary, with one fixed point \( L=\frac{1}{e} \) where \( b^L=L \) and
\( b={(\frac{1}{e})}^e\approx 0.065988 \)
This base has a rationally indifferent multiplier=-1, which has a pseudo period=2. Because the multiplier is an exact root of unity, rather than an irrational root of unity, this base doesn't have a Schroeder function.
\( b^{L+x}\approx L-x+1.35914x^2-1.23151x^3+... \)
Iterating this function starting with z=1 doesn't lead to behavior that seem much like Tetration. I was curious if it was possible to generate an sexp(z) solution given that I think there is only one fixed point, and that the fixed is rationally indifferent, in this case with a pseudo period of 2.
As I understand it, the abel function for this function, would be generated from b^b^z. If you iterate the b^z function twice, b^b^z would be a parabolic case with an Abel function for each of the four leaves on the Leau-Fatou flower. But I don't know how you can combine these four Abel functions together into a single inverse abel function for b^z. Note in the equation below, that there is no x^2 term.
\( b^{b^{L+x}}\approx L+x-1.2315x^3+0.836897x^4+... \)
At the real axis starting close to the fixed point and iterating twice always gets you a little bit closer to the fixed point, but the convergence is very slow. For example, starting with z=0, iterating z=b^z oscillates towards the fixed point.
I generated some results and graphs for a complex base on the shellthron boundary with a rationally indifferent fixed point with a multiplier of \( \exp(\frac{2\pi i}{5}) \), with a pseudo period=5, which also lacks a Schroeder function because the multiplier is a 5th root of unity; the other fixed point is repelling. See post #10 of this thread, http://math.eretrandre.org/tetrationforu...hp?tid=729 This result was generated via the conjectured merged complex tetration solution, using both fixed points of the base. It seems that if there is a solution for other rationally indifferent roots of unity on the Shell Thron boundary, than there could be a solution for this base as well.
In the upper half of the complex plane, such a solution for base (1/e)^e might only be well behaved near the imaginary axis, exponentially converging towards the fixed point as \( \Im(z) \) increases much like the conjectured solution with pseudo period=5. Such a solution would become increasingly chaotic as real(z) increased or decreased. I have no idea what the solution could look like in the lower half of the complex plane, since I think there is only one fixed point.
- Sheldon
Hmm. We can track the two "principal" fixed points in the plane as they go from, say, base 2, to this base via a non-real contour from above (since I think \( \left(\frac{1}{e}\right)^e \) lies on the branch cut from \( (-oo, e^{1/e}] \), and we "conventionally" define such functions to be "continuous from above" (if the cut goes to the right, then it is "continuous from below").). The upper fixed point should be \( L_{+} \approx 0.36787944117144232159552377016146086745 \) and the lower \( L_{-} \approx -0.19574575248807635792808172595109672367 + 1.6911999209105686636060727529631245323*I \), assuming my method of repeated Newton method + slowly inching the base toward \( \left(\frac{1}{e}\right)^e \) through a complex plane contour worked. \( L_{+} \) is neutral, as you already know, but \( L_{-} \) is not, and we should be able to generate the regular superfunction there.
The interesting part is that \( L_{-} \) is not a fixed point on the principal branch of the base-\( \left(\frac{1}{e}\right)^e \) logarithm. Instead it is fixed on the branch with \( k = -1 \), i.e. \( \frac{\log(z) - 2\pi i}{-e} \), where \( \log \) is the principal branch of the natural log.
The superfunction generated from the "lower" fixed point looks like this:
(scale is from -10 to +10 on both axes).
So I'd guess that it should look like that in the lower half-plane.
Not sure what to do about the upper half-plane, though.
