Change of base formula using logarithmic semi operators

[sheldonison]

.....
Now to just solidify it a bit. Given by gottfried's result here:
http://math.eretrandre.org/tetrationforu...653&page=3

which he also got from mike3

the sequence \( a_n \)

\( a_1 = \ln(b)\\a_2 = \ln(\ln(b^b))\\a_3 = \ln(\ln(\ln(b^{b^b}))) \)

converges to a set value, and is reexpressable as:

\( a_2 = \ln(b) + \ln^{\circ 2}(b) \)
\( a_3 = (\ln(b) + \ln^{\circ 2}(b))\bigtriangleup_{-1}^e \ln^{\circ 3}(b) \)

which of course, looks a lot like our change of base formula when x is 1... So putting them together we get:

\( \text{sexp}_b(z) = \exp^{\circ z}(\ln^{\circ z}(\exp_b^{\circ z}(1)) \)

which is like coming around full circle so we can be sure this method works.

I'm not sure if this is the same thing or not, but earlier we looked at defining sexp_e(z) (or any other base) from cheta(z), where "k" is a constant chosen so the limit for sexp_e(0)=1. Here \( \text{sexp}_\eta \) and \( \text{slog}_\eta \) refer to the upper super exponential for base \( \eta \), cheta(z) and its inverse.
\( \text{sexpBaseChange}_e(z) = \lim_{n \to\infty}\log^{[n]}\text{sexp}_\eta(z+k+n) \)

\( k=\lim_{n \to \infty} {\text{slog}_\eta(\exp^{[n]}(1))-n \)

I'm not sure if this is the same as what you're getting at or not. We call this definition of sexp, the base change function, which is \( C_\infty \) at the real axis and, see this post, it is conjectured to be nowhere analytic. The base change function has many very interesting properties, in that at the real axis, there is a small 1-periodic wobble connecting it to the "correct" sexp function.
- Sheldon