(06/04/2014, 03:11 AM)JmsNxn Wrote: ... fully determined by the sequence of numbers \( a_n \) such that \( (^{a_n} e) = n \).
Particularly:
\( \int_0 ^\infty t^{z-1} \sum_{n=0}^\infty a_n (-t)^n/n!\,dt = \G(z)slog(-z) \) for \( 0 < \Re(z) < \Re(L) \)
that's pretty interesting.
I'll have to mull on what we can accomplish with this but a very good idea I'm thinking about is determining these a_n (or a criterion they all satisfy) and seeing what else we can do with them.
FURTHERMORE
find me a function that (a) satisfies f(e^n) = f(n)+1 (b) holomorphic in a half plane and satisfies our exponential bounds then DA DUM DA DUM f is a slogarithm.
1) HUH ?
\( \int_0 ^\infty t^{z-1} \sum_{n=0}^\infty a_n (-t)^n/n!\,dt = \G(z)slog(-z) \) for \( 0 < \Re(z) < \Re(L) \)
What is Re(L) ??
2) as for finding those a_n without a given slog or sexp :
Perhaps usefull is the approximation :
\( d slog(x) / dx \) ~ \( C (x ln(x) ln^{[2]}(x) ln^{[3]}(x) ...)^{-1} \)
from where you can estimate the a_n.
3)
* quote * :
find me a function that (a) satisfies f(e^n) = f(n)+1 (b) holomorphic in a half plane and ...
* end quote *
Probably NO and ! that is already a uniqueness criterion for sexp.
And I think it is also a uniqueness criterion for slog.
Isnt that f(x) 2pi i periodic ? Is that not a problem ?
Need to think about it more though ...
regards
tommy1729