Kneser's Super Logarithm

[mike3]

So then it would seem to agree with the Kouznetsov function, then, wouldn't it, i.e. the \( \operatorname{tet}_e(z) \) function developed this way decays to approximately \( 0.318 \pm 1.337i \) as \( z \rightarrow \pm i \infty \)? Hmm. Does this Kneser method work for other bases, too?

Yes, and Yes. I don't think anyone's tried it though.

Can it be used at a complex base, e.g. \( 2 + 1.5i \)?

You could always generate the inverse Abel function from the fixed point for the complex base. But if f(z) is real valued, f(z+1) would have a complex value, so the Kneser mapping couldn't convert the Abel function into a real valued tetration. But it seems like it could generate an analytic complex base tetration where sexp(-1)=0, sexp(0)=1, sexp(1)=2+1.5i, sexp(2)=(2+1.5i)^(2+1.5i) and sexp(-2,-3,-4...)=singularity....

Yeah, that's what I'd be after, and I wouldn't expect it to be real-valued at the real axis (except at \( z = -1 \) and \( z = 0 \)). What would a graph of this function \( \operatorname{tet}_{2 + 1.5i}(x) \) look like along the real axis (two graphs, actually, one for real, one for imag)?

Can this also be done with the Cauchy integral formula method as well? Note that we need to know which two fixed points it should decay to at \( \pm i \infty \), and the presumed lack of conjugate symmetry (i.e. I expect that \( \operatorname{tet}_{2 + 1.5i}(\bar{z})\ \ne\ \bar{\operatorname{tet}_{2 + 1.5i}(z)} \) in general, due to it being not real-valued at the real axis at most points) would seem to complicate attempts to try to run this one numerically.