01/26/2010, 01:22 PM
(This post was last modified: 01/28/2010, 11:25 AM by sheldonison.)

(01/26/2010, 06:24 AM)mike3 Wrote: Ah. So how would you use the given equation to calculate ?Update: fixed notation to use the Abel function, which is the inverse of the super-exponential developed from the fixed point "c".

Well, again, I was sort of trying to follow the graphs that Jay made, and Henryk's description of Kneser's approach, and it was not quite accessible to me. It appears they were iterating the logarithm on a unit length, [0..1], which is the first step in the limit towards the inverse super-exponential, or the Abel function, . The sequence of logarithms converges towards the fixed point, so the logical next step is the equation for , or the inverse super-exponential developed from the complex fixed point c. And the equation I got for is

Of course, I'm assuming that is the entire sexp_e developed from the fixed point, developed from the approximation for small (large negative) values of z, , which has complex values at the real axis. But I don't think that matches what Kneser is doing, but I think it must be closely related. Then I graphed the contour of , and I was wondering if that was indeed the contour that Kneser Reimann mapped, . But I couldn't tell from Jay's posts, or Henryk's post. It seemed more complicated than that. Anyway, If you already had the sexp_e to start with than

The 1-cyclic function is . Plugging into the equation for .

Of course, we don't know what sexp_e is, so we start with the contour generated over a unit length from [0..1]. A linear approximation for sexp_e is probably a reasonbly starting point, since it has a continuous first derivative. Then do a Riemann mapping of the contour of of the unit length, to a unit circle. After the Riemann mapping, unroll the the unit circle, divide by 2*pi, and we're back to a unit length. Only now is a 1-cyclic function that can be expressed as the unique sum of a complex fourier series exponentially decaying to a constant as i increases. I gather the computational difficulty comes in actually doing the Riemann mapping...

- Shel