06/07/2010, 12:34 AM
(06/02/2010, 10:56 PM)sheldonison Wrote: After that, do you agree that converting from your f(y) to g(y) by iterating natural logarithms would appear to be exactly the same? Here, the superfunction <=> f(y), and TommySexp <=> g(y). The only reason I threw in the "k" constant was so that the resulting sexp_e(z) can be normalized so that TommySexp_e(0)=1. Otherwise, with the k constant=0, I get TommySexp(0)=0.92715... The value I'm using for k is 0.067838366.
\( \operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} (
\operatorname{superfunc}(z+n+k)) \)
no , i use iterations of exp(x) !
you only use logaritms and the superfunction of 2sinh(x) ...
i use logaritms , superfunction of 2sinh(x) and exp(x).
numerically it might not be a big difference for large values ( since 2sinh(x) and exp(x) are close for large x ) , but for small x or properties it might be a huge difference ... maybe that is why you get 0.92715... ?
i dont know if i need a correcting 'k'.
and btw as for the diff-eq , i made a correction to that post (typo mainly but important ) , maybe it will help.
regards
tommy1729