07/02/2014, 10:13 PM
(07/02/2014, 07:56 AM)mike3 Wrote:(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.
What if, perhaps, a better approach here would be to use Gaussians themselves as the set of basis functions? That is, write
\( G(z) = \sum_{n=0}^{\infty} a_n e^{\epsilon_n (z - z_n)^2} \),
where \( \epsilon_n > 0 \) are real parameters, and \( z_n \in i\mathbb{R} \) are imaginary parameters? Now we have three, instead of one, coefficient sequences to deal with, but this might work. We have great freedom to tweak the various parameters (\( \epsilon_n \) and \( z_n \)) so as to optimize the convergence now.
The trouble is doing the arithmetic on these series. Addition is simple enough (assuming same \( \epsilon_n \) and \( z_n \) between the two series), but multiplication seems a little trickier.
A generalized zeta type function of z^2 !?
This is more complicated than a general dirichlet series !!
Whatever you ask me , dont ask for the zero's
regards
tomm1729