On extension to "other" iteration roots

[bo198214]

Ooo that is nice! Very straight to the point.  I was going to say it wouldn't analytic at first, but then I realized \(s\) is constant. I've never seen that before.

I would say the only disparaging fact about this expansion is that if you are going to expand in \(s\) (if we want the multiplier to be locally holomorphic) we'd have to be more clever. I think it would make sense to generalize this to \(s^x \theta_s(x)\) to try and get holomorphy in \(s\), which shouldn't be too hard if you fiddle around.

This would also make the real valued superfunction for base \(b = \eta^- - \delta\) for \(\delta > 0\)! And you could probably get holomorphy in \(b\) if you played your cards right! (with some kind of problem at \(b=1\) and \(b = \eta^-\)).

hey james, no intention to disappoint u but, if you're trying to build a real valued superfunction for base less than \(e^{-e}\) as tetration, you'd be careful about which method to use.
1. the "P method" here doesn't work anymore. The "P method" is only a merging tech, and it requires the merged function has only one limiting value as a fixed point, if you're taking the lower bases, there'll be 2 fixed point, thus the asymptotic expansion will not work
2. any real tetration for such bases generated by asymptotic expansions (at infty) cannot be continuous, this is very easy to prove, it follows ideas below:
  I. If you're using an asymptotic (at infty), there must be 2 limiting value, which is contradictionary if you're not considering oscillative function
  II. If you're considering oscillative function, there can only be 2 limiting value with discontinu(ous)ity, because it always follows \(T(z+1)=b^{T(z)}\), if any continuity was assumed, it'll contradict with (T(z+1)=b^{T(z)}\) because there can only be 2 limiting value, not a interval like sin(z)
So you must have a discontinuity by generating from an asymp
3. even if kneser's double dagger method or other contour-integral-like methods can work, it'll be difficult to choose the initial guess and also make the function to fit \(tet_b(0)=1\)
This is because such realvalued tet function must be oscillative, and kneser-like methods won't discern about where is the startpoint, the initial conditions \(T(0)=1, T(0)=b^b, T(0)=b^{b^b},\cdot\) cant be discerned easily, furthermore there must be tons of branch cuts for the 2 fixed points makes the Tet semi-periodic about a real period, so, Very Hard, besides it's not proven or manifested about its availability, because this time the method also walk across the "2 limiting value" case
maybe your beta method works, looking forward

It looks like my improved/simplified version of your "P method" yields the same result as your "P method" and can also handle the case \(b<\eta_-\), except for the condition \(\text{tet}_b(0)=1\) because it is a repelling fixed point and if you start to iterate you can not use 1.