On extension to "other" iteration roots

[JmsNxn]

Let's consider building up the real-to-real tetration base 1/2. We should notice that, if we denote  \( T:T(z+1)=f(T(z)),L:f(L)=L,s=f'(L)\to{T}(z)=\sum_{n=0}^\infty{a_ns^{nz}} \) and \( T_2:T_2(z+1)=f^2(z) \), then they should behave asymptotically like: \( T_2\(n)\sim{T(2n)} \) at all integers n congruent to 0 modulo 2, since s is a negetive number. How about n congruent to 1 modulo 2? Simple.
\( f(T_2\(n))\sim{T(2n+1)} \) will answer, due to the functional equation T satisfies.
Now we want a function that asymptotically behaves like T just at integers but also real-to-real, so that it preserves all integer values properly.
Consider a period 2 function \( P(z)=\sum_{m=0}^\infty{c_me^{m\pi{i}z}} \) which satisfies \( P(0)=1,P(1)=0 \), then set a new "superfunction" \( W(z)=P(z)T_2\(\frac{z}{2}\)+P(z+1)f(T_2(\frac{z-1}{2})) \), easily check that W preserves all integer values but different from T. 
Also we want W have the property real-to-real. Firstly, since s<0, \( s^2>0 \), we can force that the \( T_2 \) map is real-to-real, as written and computed in the way: \( T_2(z)=\sum_{n=0}^\infty{a_n(s^2)^{nz}} \).
Then we only need to determine P(z), make it a real-to-real function is a striaghtforward solution, in my practical computation I used \( P(z)=\frac{1+cos(\pi{z})}{2} \).

Actually this reminds me quite of the Fibonacci-Number extension described on Wikipedia (that I just read).
They also start with a function that equals on even numbers and a function that equals on odd numbers. And then they are combined with \(\cos(\pi x)\) and then you have already a real valued Fibonacci extension! (And I independently found that same formula in my Fibonacci iteration thread, with a similar reasoning)
And I wonder if you really need the iteration limit part, but you just do it as in the real valued Fibonacci extension:
Your complex valued superfunction T can be written as \(T(x)=\sigma^{-1}(s^x \sigma(z_0))\) with real valued Schröder function \(\sigma\).
and you simply make a real valued superfunction out of it via
\[ T_{\mathfrak{R}}(x) = \sigma^{-1}(\cos(\pi x)\;|s|^x \;\sigma(z_0)) \]
Because \(\cos(\pi 2 n)=1\) and \(\cos(\pi (2n+1))=-1\) it agrees on integer iterations with T but still satisfies the superfunction condition:
\begin{align}
T_{\mathfrak{R}}(x+1) &= \sigma^{-1}(-\cos(\pi x)\;|s|^x |s|\;\sigma(z_0))\\
                                   &= \sigma^{-1}(s\sigma(\sigma^{-1}(\cos(\pi x)\;|s|^x \;\sigma(z_0))))\\
                                   &= f(\sigma^{-1}(\cos(\pi x)\;|s|^x \;\sigma(z_0))\\
                                   & = f(T_{\mathfrak{R}}(x))
\end{align}
PS:
Can you correct one formula in your original post: You write \(T_2(z+1)=f^2(z)\) but I guess you mean \(T_2(z+1)=f^2(T(z))\) - quite puzzled me on first try to read ...

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 \(b = \eta^-\)).