I understand the circularity ^_^, it was worded a little weird. I got excited when I thought of applying these methods to tetration and I didn't consider too much about the erradic behaviour of the tetration function in the imaginary plane. As mike pointed out it blows up as n grows in 1/{n-s}^e which is not what I would've expected :\.
However! using the slogarithm makes a lot more sense.
NOW I have something to say. Using fractional calculus, (and even carlson's theorem if you want), if a slogarithm satisfies the exponential bounds in a half plane it is UNIQUELY determined by the values it takes on at integers. This implies, that the tetration it provides is fully determined by the sequence of numbers \( a_n \) such that \( (^{a_n} e) = n \).
Now that's gotta be something interesting.
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.
However! using the slogarithm makes a lot more sense.
NOW I have something to say. Using fractional calculus, (and even carlson's theorem if you want), if a slogarithm satisfies the exponential bounds in a half plane it is UNIQUELY determined by the values it takes on at integers. This implies, that the tetration it provides is fully determined by the sequence of numbers \( a_n \) such that \( (^{a_n} e) = n \).
Now that's gotta be something interesting.
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.