Slog(Exponential Factorial(x))

[JmsNxn]

what i wonder if how basechange(e,x) behaves as a function of slog(x).

basechange(e,x) = a0 + a1 slog(x) + a2 slog(x)^2 + ...

regards

tommy1729

I was always intrigued by the basechange function. But I'm confused how it's presented in most situations in this forum.

Let, \(b > \eta\). If we write:

\[
\text{tet}_b(z) = \text{tet}_e(B(z))\\
\]

Isn't it clear that locally a solution always exists to this equation?  The monodromy theorem guarantees this function always exist.

So That:

\[
\text{tet}_b(z) : \mathbb{C}/(-\infty,-2] \to \mathbb{C}\\
\]

Similarily with \(b=e\). Therefore \(B(z)\) always exists locally. In fact, we only need to worry about \(-\delta < \Re(z) < 1+\delta\) and \(|\Im(z)| < \delta\). Because the orbits of the exponential will cover everywhere else. This solution always exists. It is always holomorphic and \(1-1\) for small enough \(\delta\). Then, you have a function:

\[
\text{slog}_b(b^z) = \text{slog}_b(z) + 1 = B^{-1}(\text{slog}(e^z)) = B^{-1}(\text{slog}(z)) + 1\\
\]

Once you have a super logarithm on non trivial neighborhood of \([0,1]\)... I mean, just iterate the orbits and you have \(\text{slog}\) in the complex plane, with its poles. Now you have the tetration.

I'm confused why it's such a difficult problem on this forum. I'll admit. I have no fucking clue how to evaluate or compute this function \(B\). But it's pretty simple complex analytic theory to prove it exists. You just need the analytic implicit function theorem and a modest use of the monodromy theorem about the line \([0,1]\).

Maybe it's because I am assuming we are using Kneser in both situations. So it's not much more than an implicit solution to the equation \(b^y = e^u\). Is there something deep I am missing with the basechange theory.

I'm not talking about "the base change formula" also. That formula is non-analytic. And that should be settled.

Is there some reason this problem is still important that I'm not privy too, or too dumb to notice?

Regards, James