I realize now why I was so sure that I'd get more accuracy out of my accelerated solutions.
It turns out I was almost right. You see, I can get hundreds of digits of precision out of my accelerated solutions, though only dozens of digits of accuracy.
Precision
I qualify precision to mean that \( \mathrm{slog}\left(\exp(z)\right)=\mathrm{slog}(z)+1 \).
In this sense, my accelerated 1200 term solution is precise to well over 400 decimal digits at about z=-0.55, with accuracy dropping off exponentially, or linearly if measured in digits, as we move in either direction from that point (because either z or exp(z) is moving away from the origin). By z=-1 or z=0, precision has dropped off to about 170 digits, which is twice what I'd originally hoped for (80 digits).
Edit: If I include the additional terms for the logarithms used to accelerate the solution (i.e., if I use the "residue" and then add the logarithms), then precision is centered at z=0 and is essentially limited by the imprecision of the matrix solver for now.
I don't even see imprecision at 2048 bits until z=-0.398 to the left and z=0.324 to the right. By z=-0.5 to the left, precision is still in excess of 500 digits, and by 0.5 to the right, it's in excess of 450 digits. It's not until about -1.175 and 0.788 that I see precision drop below 100 digits.
As it is, this amount of precision is extremely good. If I could just find a way to get similar accuracy!
Accuracy
I qualify accuracy to mean that the slog of a finite solution, evaluated at non-integers (typically at z=0.5), is equal to the theoretical value of the natural solution of the infinite system (or at least, taking a limit of finite solutions with size going to infinity, assuming such a limit exists). Accuracy is obviously harder to measure, but we can ballpark it, to within a few orders of magnitude, for example.
In this sense, my accelerated 1200-term solution is accurate to about 20-25 digits at z=0.5, which is far less accuracy than I'd hoped for.
It turns out I was almost right. You see, I can get hundreds of digits of precision out of my accelerated solutions, though only dozens of digits of accuracy.
Precision
I qualify precision to mean that \( \mathrm{slog}\left(\exp(z)\right)=\mathrm{slog}(z)+1 \).
In this sense, my accelerated 1200 term solution is precise to well over 400 decimal digits at about z=-0.55, with accuracy dropping off exponentially, or linearly if measured in digits, as we move in either direction from that point (because either z or exp(z) is moving away from the origin). By z=-1 or z=0, precision has dropped off to about 170 digits, which is twice what I'd originally hoped for (80 digits).
Edit: If I include the additional terms for the logarithms used to accelerate the solution (i.e., if I use the "residue" and then add the logarithms), then precision is centered at z=0 and is essentially limited by the imprecision of the matrix solver for now.
I don't even see imprecision at 2048 bits until z=-0.398 to the left and z=0.324 to the right. By z=-0.5 to the left, precision is still in excess of 500 digits, and by 0.5 to the right, it's in excess of 450 digits. It's not until about -1.175 and 0.788 that I see precision drop below 100 digits.
As it is, this amount of precision is extremely good. If I could just find a way to get similar accuracy!
Accuracy
I qualify accuracy to mean that the slog of a finite solution, evaluated at non-integers (typically at z=0.5), is equal to the theoretical value of the natural solution of the infinite system (or at least, taking a limit of finite solutions with size going to infinity, assuming such a limit exists). Accuracy is obviously harder to measure, but we can ballpark it, to within a few orders of magnitude, for example.
In this sense, my accelerated 1200-term solution is accurate to about 20-25 digits at z=0.5, which is far less accuracy than I'd hoped for.
~ Jay Daniel Fox