(07/01/2014, 02:25 PM)sheldonison Wrote:(07/01/2014, 10:07 AM)mike3 Wrote: Bummer... did a more accurate measurement on the convergence -- looks like 100 coefficients in each direction will only give 15-16 digits of accuracy.
So, I tried generating a Laurent series for \( f=\exp((\frac{z-1}{z+1})^2) \), wrapped around a unit circle, with similar results for the Laurent series. A Fourier/Laurent series on a unit circle will have the coefficients eventually decay according to the nearest singularity in the complex plane, |a_n| = r^n, or |r|^-n, depending on the radius of that singularity. If the function is not analytic on the unit circle, then the first derivative with the discontinuity determines the rate of decay of the coefficients. I looked for an equation online, something like a_n~n^-(k+1), but couldn't find it. Anyway, in this case, we have a Gaussian, and I'm pretty sure that all of the derivatives are continuous on the unit circle -- and all of the derivatives go to zero at z=-1, where we have the singularity. Hence, the coefficients apparently decay at an almost analytic rate, but not quite, but decay faster than n^-k, where k is any integer.
Have you tried using lim n-> oo (1+x/n)^n ~ exp(x) ?
Sorry if this is a bad idea.
Another idea Is using the parallelogram contour integration of G(z).
The parallelogram should be paralel to the pseudoperiod of sexp(z).
Since G(iz) = 0 in the real limit z -> +oo we get that the integral around this paralellogram of G(z) = 0 ( since analytic).
And thus this integral reduces to 2 integrals over a line with slope equal to the speudoperiod.
Now by a staircase nesting of contour integrals we can set up the equations for the points on this line. Though that is complicated.
This relates to my recent idea and thread about pseudo double-periodicity for sexp.
regards
tommy1729

