sexp for base (1/e)^e ~= 0.0660?

[sheldonison]

I was inspired by Mike's http://math.eretrandre.org/tetrationforu...42#pid6742 recent post to go back and work this some more. Here is an even better multiple term approximation for the superfunction for base=\( e^{-e} \). I would imagine this might even have the mythical infinite form that would converge exactly, that I imagined must exist, since we see tetration solutions for rationally indifferent complex bases.....
superfunction(z) \( \approx \frac{1}{e} + \exp(\pi i z) + \)
\( (0.679570457114761308840071867838 ) \exp(2\pi i z) + \)
\( (-0.615754674910887518935868955048 z) \exp(3\pi i z) + \)
\( (-0.8368973717994861558720220689 z -0.1046121714749357694840027586 )\exp(4\pi i z) + \)
\( (0.5687307295119191570636485542 z^2 -0.3981115106583434099445539880 z) \exp(5\pi i z) + \)
\( (1.030646938212337523757747084 z^2
-0.2834279080083928190333804481 z +
0.01932463009148132857045775787 ) \exp(6\pi i z) + \)
\( (-0.5836643424374061380329675644z^3 +0.9805560952948423118953855083z^2 - 0.3343562875962855162160285621z) \exp(7\pi i z) +... \)

I don't know how to find the equivalent function out on the web. The coefficients are real valued linear combinations of the coeffients for B^L. They were generated in the order they were listed, by iteratively solving the equation requiring that B^superfunction(z)=superfunction(z+1), for each partial term, one at a time. The algebra is messy, but I finally got it the algebra to work. Note that for odd terms, the periodic term must be multiplied by z since \( \exp(3\pi iz)=-\exp(3\pi i(z+1)) \). So all of the coefficients from n=3 upward are much more complex "multiple" term coefficients, multplied by z. Starting with n=5 and n=6, we also need both a z^2 multiplier, and a "z" multiplier. At \( \Im(z)=1i \), each new "multiple term" increases the accuracy by approximately 23x, and at \( \Im(z)=2i \), each new "multiple term" increases accuracy by 500x, so that the equation above at 2i, is accurate to \( 3 \times 10^{-21} \), and at i, it is accurate to \( 2 \times 10^{-11} \)

Even with the formal solution, this equation quickly misbehaves, iterating in either direction, as you increase or decrease the real part of z, but this is also expected, due to the crazy behavior of rationally indifferent fixed points. The more amazing part is that this crazy behavior can be stitched together with a theta mapping (we think), with the other fixed point, and lead to something well behaved in the negative imaginary part of the complex plane too!

I have also generated a similar solution for iterating \( z \mapsto z^2-\frac{3}{4} \), which is also a rationally indifferent 2-periodic point in the Mandelbrot set, whose coefficients turn out to be simpler, and rational valued. For this mapping, the fixed point is -1/2. I've written a pari-gp program that can automatically generate these multiple term coefficients, presumably for arbitrarily large n, though I've only generated them for these two cases. Again, the existence of this type of solution has been puzzling me for a long time, and I would really like to see someone else's work on a similar problem.

Iterating: \( z \mapsto z^2-\frac{3}{4} \),
superfunction(z) \( \approx \frac{-1}{2} + \exp(\pi i z) + \)
\( (\frac{1}{2}) \exp(2\pi i z) + \)
\( (-z) \exp(3\pi i z) + \)
\( (-z + \frac{5}{8})\exp(4\pi i z) + \)
\( (\frac{3z^2}{2} + \frac{-11z}{4} ) \exp(5\pi i z) + \)
\( (2 z^2 +
\frac{-21z}{4} +
\frac{31}{16}) \exp(6\pi i z) + \)
\( (\frac{-5z^3}{2} + 11z^2 - \frac{99z}{8} ) \exp(7\pi i z) + .... \)
- Sheldon