Iteration with two analytic fixed points

[bo198214]

How about polynomial cases ?

Depends on what you mean by polynomial case. If the Schröder/Kœnigs function is a polynomial we can do the same as with tangent, but not quite:
What one needs to create two fixed points is two values that are either 0,\(\pm\infty\). However not all combinations do work.
Best combination is \(\{0,\infty\}\) or \(\{0,-\infty\}\) and the function being analytic at \(x_0\), and (I call) quasi-analytic at \(x_\infty\) i.e. its real valued left and right of \(x_\infty\) and its holomorphic on a vicinity \(V\setminus\{x_\infty\}\) (or can be continued from left to right avoiding \(x_\infty\) and the function is strictly increasing/decreasing.
I think if you have these ingredients then the resulting function iteration is analytic at both fixed points for all t.
Similar consideration for \(\{0,-\infty\}\), the case \(\{\pm\infty,\mp\infty\}\) is a combination of the previous cases as the function needs to pass through 0.

The remaining cases (0,0) and \((\pm\infty,\pm\infty)\) don't work well typically, because the function is not injective (has a min/max).
Maybe its possible but I doubt.

So in case of a polynomial we have a lot of zeros but only one \(\infty\) at \(\infty\). So you don't have the case \(\{0,\pm\infty\}\).
If you however would allow rational functions, one can quite do a similar construction as with the tangent.

       
And this linear fractional mapping stuff works like this (these functions always have a pole and zero and hence the case \(\{0,\pm\infty\}\) is paramount).

But if you mean: iterating a polynomial at one of its fixed points - well this method of taking \(\chi^{-1}(c^t\chi(x))\) does mostly not yield polynomials. So this approach would not much help here.

I somewhat considered a degree 2 case with 2 fixpoints and its entire superfunction. 
Not quite the same as here but in the same spirit , the fermat superfunction.

https://math.eretrandre.org/tetrationfor...hp?tid=809&

Yes, I also stumbled over a similar case as f(x) = (x-1)^2 + 1 recently.
The regular iteration is \(f^{\mathfrak{R}t}(x) = (x-1)^{2^t} + 1\), though at its left fixed point (at 1) it is not analytic
which has to do with its Schröder/Koenigs function \(\log(x-1)\) is not real valued left of 1 (see my previous preconditions).

Second " psychological " analysis :

just because the regalar iterations ( koenigs function ) suggest a periodic superfunction , this is not true !!

Is it not??? The super function is \(z\mapsto\chi^{-1}(c^z \cdot \chi(z_0))\) and \(c^z\) is periodic with period \(\frac{2\pi i}{\log( c)}\) ... or do you mean that the choice of the branch of \(\chi^{-1}\) can make it non-periodic (like the tangent construction, but there the choice of the branch depended on \(z_0\)).

it would be nice to see a degree 4 polynomial where they agree on the half-iterate at 2 fixpoints and are analytic at the fixpoints too.

Well as I said, the applied method would not work for this case ... but let the hunt be open!