06/10/2011, 01:43 PM
(This post was last modified: 06/10/2011, 05:25 PM by sheldonison.)

(06/10/2011, 08:48 AM)bo198214 Wrote:For clarity, this is the integral, to generate the individual a_n terms for the Fourier series approximation, generated from the sexp(z) approximation.(08/08/2010, 07:14 PM)sheldonison Wrote: Theta(z) has a singularity at all integer values of n. Theta(z) is represented by an infinite sequence of fourier terms. The fourier series for theta(z) can be developed from any arbitrary unit length on the real axis of sexp(z), where z>-2. Only terms with positive values of n are included, and all terms a_n for negative values of n are zero.

I have to ask here again. "Only terms with positive values of n are included.":

Why did you put that restriction?

Is there inherent reason, or is it just that you think this is the most natural and simplest way?

, Kneser Riemann mapping approximation, for

Because sexp(z) here, is only an approximation, including terms with negative values of n would mean that theta(z) would not decay as z goes to . It would also mean that the theta(z) function would only be defined at one and only one value of , since an infinite Fourier series typically only converges where the series is sampled. For this algorithm, I have chosen to sample at imag(z)=0.12i, see Jan 11th 2011 post. By throwing out the a_n terms, with n<0, we have a function which is defined for , with singularities at integers. The singularities are because the sexp(z) approximation function we are generating the Fourier series of is only approximate. So is only an approximately 1-cyclic function. Since the solution we are ultimately looking for has a_n terms with negative values of n all zero, this allows the iterated sequence of functions to converge. Again, I think the picture in the previous post I linked to helps a lot. After doing the Fourier analysis, the function on this circle is used to generate the next sexp(z) approximation, as described in the algorithm, where f(x) is the previous sexp(z) approximation. The sexp(z) would also require a full Laurent series to converge on the unit circle (and then only converge on the unit circle), but we throw out the coefficients for the z^-n terms.

If we had the exact sexp(z) function, and we did an exact infinite Fourier analysis at the real axis, or at any other value of , then all of the a_n terms with negative n would be zero, and the infinite sequence of positive a_n terms gives a function which converges for >=0, as long as z is not an integer.

- Sheldon