11/24/2008, 06:23 AM
(This post was last modified: 11/26/2008, 02:09 AM by Kouznetsov.)
Henryk, I am still with your deduction. I hope to confirm it soon. While, it seems to me that you make the slog more basic than sexp, and I begin to like it. If we calculate few tens of derivatives of sexp at minus unity, invert the series and calculate some tens of terms of the Tailor expansion for slog around zero, it will be precize and fast representation. If close to the branch point, we can use the asymptotic expansion, which is even faster. We may get the same 14 significant digits with only a hundred of flops. (Now it takes 10^4 flops per value). Then, we can evaluate tetration as inverse of slog, and it will be also fast and precize representation. We get two orders of magnitude in the speed of evaluation of sexp and slog. With these representations, we can evaluate the holomorphic pentation in the similar maner, define sslog and so on. Do you plan to create the pentationforum?
P.S. Henryk, I hope to have the simplification of your proof. It leads to the linear integral equation of Fredholm of 2d kind with the smooth kernel; smooth Fredholm-2 always has one solution, and the only one.
I need to verify it with the numerical implementation; then I post it together with figures. The resulting slog can be much faster than sexp I have for now; it can be so fast as other special functions already implemented in the conventional software (C++, Fortran, Mathematica, etc.).
P.S. Henryk, I hope to have the simplification of your proof. It leads to the linear integral equation of Fredholm of 2d kind with the smooth kernel; smooth Fredholm-2 always has one solution, and the only one.
I need to verify it with the numerical implementation; then I post it together with figures. The resulting slog can be much faster than sexp I have for now; it can be so fast as other special functions already implemented in the conventional software (C++, Fortran, Mathematica, etc.).
