On extension to "other" iteration roots

[bo198214]

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}
Or with your formula \({T_\mathfrak{R}}(z)=\sum_{n=0}^\infty{a_n(\cos(\pi x)\;|s|^z)^n}\)
And also as you hinted already the pseudo iteration \(f^{\circ x}(z)=\sigma^{-1}(\cos(\pi x)\;|s|^x \;\sigma(z))\) does not satisfy \(f^{\circ s+t}=f^{\circ s}\circ f^{\circ t}\).

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 ...