01/19/2019, 02:39 PM
(This post was last modified: 01/20/2019, 12:42 PM by sheldonison.)
jaydfox Wrote:the accuracy you mentioned (76 digits)... I assumed this limitation in accuracy was in your solution. But it finally dawned on me that this limitation was in ... the coefficients I posted.
Correct. I regenerated my slog, and optimized the sample points to get as much accuracy as possible in the \( \theta_R \) mapping to compare Kneser slog and Jay's slog.
There is the limitation that your 800th term is 10^-126, so we need to keep the sample radius as small as possible. Here, the \( \theta_\R \) sample points are with slog(z) where |z|<0.706, which is the radius of the red circle. Sampling on the green dots sexp(0.431i-1.405...0.431i-0.405), we are limited to 35 complex conjugate pairs in the Fourier series, before the Fourier series becomes random noise.
With this Fourier series sampling, it is impossible to theoretically do better than 247 decimal digits since 10^-126*0.706^800~=10^-247. And in fact, for a large chunk of the red circle of radius=0.706, I'm seeing accuracies of about 244 decimal digits with this 35 complex conjugate pair terms in the \( \theta_\R \) mapping. All points within the red circle are accurate to 1.15E-213 with this theta mapping. All points within a unit circle are accurate to within 1.5E-126, limited by the 800 terms in Jay's slog series. It is intuitive to compare the error plot below to the red circle above, and the Fourier series sample points within the red circle. Here is the error plot equation: precision is limited by the number of terms in Jays slog, and the number of terms in the \( \theta_\R \) mapping.
\( \frac{10^{-240}}{\text{JaySlog}(z)-\text{KneserSlog}(z)-\theta_\R(\text{KneserSlog})} \);
The Kneser slog used was accurate to 276 decimal digits, which is the most accurate computation I've ever done. A Fourier transform just on the real axis could have 322 decimal digits of consistency, but then all we would have is a line, instead of an enclosed region in the complex plane.
- Sheldon

