07/01/2014, 02:25 PM
(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.
- Sheldon