regular sexp:different fixpoints

[Gottfried]

For me, the entire "bummer" post (same topic), and this paper is really interesting. My personal opinion is on the left side, the "smoothest" superexp is developed from the fixed point of four, and on the right side the "smoothest" superexp is developed from the fixed point of two. In between, they're both a little bit "less smooth".

Hi Sheldon -

well, it may look so, and we may go with it. But I'm not convinced. I'm still thinking, there may be some corrective term in the definition of our fixpoint-dependent functions which let them converge to the same value.
What I've done in continuing the above thoughts was to consider, that the difference on exp_b,2°h(x) and exp_b,4°h(x) also means, that the same value of y must agree to different heights wrt fp2 and fp4. So the sinusoidal form of the above curve should indicate, that at integer iterates the functions synchronize and to the half iterate one function grows faster and to the next half iterate the other one. I've thus plotted this height differences dependent on equidistant height arguments.
Surely we expect periodicity wrt iterates (mod 1), but also the curve seems perfectly sinusoidal. So there seems to be a pair of parameters, the amplitude and a phase-shift to synchronize the two functions, so that we have -like for instance with the zeta-function with positive and negative arguments- a functional relation with different fixpoint-parameter. (The red line is just a mirror of the blue line to make asymmetries better visible)

   
   

There is even one more aspect. If you see the related pictures in the bummer thread, then whatever initial x0-point in exp_b,2°h(x0) they use (Dmitri and Andrew used x0=3, I think), then you'll observe, that the x-axis of the sinusoidal is also bended and not vertically symmetric in the middle.

See a sample plot which shows the same curves of differences when I use for x0 values differeing from 3.
(curves have different x0, x-axis is iteration height from there)

   

I could improve vertical symmetry by changing to another x0 (bold red line); which means, that there are points, where the symmetry is even perfect. I optimized that point manually - but to my surprise, there was one possibility to get a (only near?) perfect selection: that was the mirror of x0=1 into that range 2 < x < 4, giving about 2.93
"Mirror" means here: in the schröder-decomposition of the function determine the height-parameter for m0=schröder(x-t), where x is 1, t the fixpoint and u its logarithm.
We'll get for x=1
1=schröder°-1(m0) +t = schröder^-1( u^0*schröder(1-t))+t
Then
x0= mirror(1) = schröder°-1 ( -m0) +t
meaning mirror(1) = b^^(pi*I)
I think, that was something 2.43, and three integer iterates from that was also my manually found value of 2.934...

So if this procedere finds a "normative" value for the phase of the differences, we need another one for the amplitude to define the relation between the two fixpoint-dependent tetrations.