Maybe the solution at z=0 to f(f(z))=-z+z^2

[Leo.W]

Im not sure how these relate but 3 ideas cross my mind.
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regards

tommy1729

2) about 1-periodic theta mapping, I don't think it'll help. If it would, then our mixed real-valued tetration for base \(0<b<e^{-e}\) would be easily to compute, but till today I can't compute those tetration. Maybe I'll give it a try. The key is to figure out how it behaves in an interval like [0,1], not at infinity, since the iterative method would bring us to different branches, we must develop a technique to iteratively (but not a limit from/at infinity) make an initial guess converge. This is our obstruction now.

3)Your idea is good, and it coincides with things in my last thread and my procedure, but not enough since we need a series containing terms like \(z^3\log(z)\). On the other hand, we do want it comes true as a good expression for our iteration, but the truth is, we can't even figure out the first term.
\(f^t(z)=(-1)^tz+O(z^2)\)
\(f^t(z)=(-1)^{-t}(z)+O(z^2)\)
\(f^t(z)=(-1)^{3t}(z)+O(z^2)\) and so on.
Which one is your answer?
In fact, the first term is enough to affect all other terms following and they all fit the basic semigroup's identity.
After that, my computation gives:
If we even assume a better series (asymp):
\(f^t(z)=f_1(t)z+(f_{2,0}(t)+f_{2,1}(t)\log(z))z^2+(f_{3,0}(t)+f_{3,1}(t)\log(z)+f_{3,2}(t)\log(z)^2)z^3+O(z^4\log(z)^3)\)
and we (A) force it satisfies the semigroup identities,
and if we assume (B1) \(f_1(t)=(-1)^{-t}\), then
\(f^t(z)=f_1(t)z+f_1(t)(f_1(t)-1)\frac{z^2}{2}+f_1(t)(f_1(t)-1)^2\frac{z^3}{4}+O(z^4)\)
which cancels all terms containing \(\log(z)\) and thus a Taylor, hence incorrect.
But we know that the series is correct anytime t is integer.
and if we assume (B2) \(f_1(t)=(-1)^t\), then the same thing happens,
\(f^t(z)=f_1(t)z+f_1(t)(f_1(t)-1)\frac{z^2}{2}+f_1(t)(f_1(t)-1)^2\frac{z^3}{4}+O(z^4)\)
This means our assumption goes wrong. There's contradiction. And by logics, the glitch of the system potentially happens at our condition (A). The semigroup identities. That means we must deal with tons of branches, multivalued multifiers and stuffs. It can be horrible.
The halfiterate can be computed in the same way. You assume coefficients and you earn one series.