05/24/2011, 12:12 AM
(This post was last modified: 05/24/2011, 03:52 AM by sheldonison.)
(05/23/2011, 09:53 PM)bo198214 Wrote:I'll describe what I did to generate the Taylor series for sexp(z). Its not fancy or anything -- its just what worked for me to investigate base eta, with its parabolic convergence. I use pari-gp, to generate an interpolating polynomial. Only I center the polynomial around sexp(100), where the sexp(z) function is already converging towards e, and is fairly well behaved. Then I generate 25 points on either side of sexp(100), using 67 digits of precision.(05/23/2011, 08:42 PM)sheldonison Wrote: Your results matches the Taylor series I get for \( \text{sexp}_{\eta}(z-1) \), developed at -1, where sexp(-1)=0.
Ya but your values are much more accurate as it seems, how do you obtain them?
Now I have a 50 term polynomial, centered around z=100, which seems to have nice convergence properties. The 50th series term of the interpolating polynomial is ~2E-102. The error terms seem to be around 10^-50 close to z=100, and increases to 10^-33 as the radius increases to 20 units. I haven't done a theoretical interpolation error analysis, but I have graphed sexpeta(z)-eta^sexpeta(z-1).
The next step is akin to a discreet version of analytic continuation, where we pick a sample circle. I used a unit circle centered around z=100. I usually pick 200 evenly spaced points around the unit circle, and for each of those points, take the logarithm base eta 100 times. Now we have another unit circle, with 200 sample points, centered around z=0. Use the Cauchy integral to generate a Taylor series from those 200 sample points, and you get a Taylor series, centered around z=0. I have routines that will generate a Taylor series with any number of sample points and sample radius, sort of like a discreet version of analytic continuation.
- Sheldon

