06/25/2014, 08:21 AM
(This post was last modified: 06/25/2014, 08:25 AM by sheldonison.)
(06/25/2014, 06:10 AM)mike3 Wrote: Hmm. It might work. However the function will not be complex-periodic, and therefore will not have a Fourier/exponential series expansion. It needs the periodicity in order to work.This one seems to converge pretty poorly; much slower than ideal. If you take \( f(\text{sexp}(18\pi i))<\approx 10^{-32{ \), and then wrap it around a unit circle, and then represent it with a Laurent series, then a 1000 term series from z^-500 to z^500 will be accurate to a little better than 10^-25. That's reasonably accurate, but it is also a very large number of Taylor/Laurent series terms.
For comparison, the fast converging functions are like exp(-x^2), which is a perfect Gaussian. If you wrap it around a unit circle, so that f(-1)~-10^-32, then such a perfect Gaussian would require a little less than a 100 term Laurent series to be accurate to 32 decimal digits. So this function, needs 10x more terms and even then, it is less accurate -- not at all promising.
Some background: I'm experimenting with Fourier/Laurent convolutions to calculate arbitrarily large Tetration Taylor series coefficients to arbitrary accuracy, and it works really well. But now I realize that it works best when the envelope function behaves like a Gaussian, which is often the case. But a Gaussian envelope converges to zero faster than this f(sexp(z*I)) function.
- Sheldon

