Iteration with two analytic fixed points

[tommy1729]

I think its not yet as mature as you wish, but keep going.
In the mean time I was constructing another counter example, this time without linear fractional mappings.
The idea is the following, we know that the Schröder iteration is given by:
\[ f^{\mathfrak{R} t}(x) = \chi^{-1}(c^t \chi(x)) \]
Instead for looking for functions \(f\) I just took \(\chi\) as my start of research, how does \(\chi\) need to look, such that \(f^{\mathfrak{R}t}\) is still analytic at a second fixed point.
For a second fixed point \(z_2\) it must be valid that:
\[ c^t \chi(z_2) = \chi(z_2) \]
which is basically satisfied for \(\chi(z_2)=0\) or \(\chi(z_2)=\pm\infty\).
So I took a function that satisfies all 3 values  : the tangent! And constructed the following function
\[ f(x) = \arctan(c\tan(x)) \] with the iteration \( f^{\mathfrak{R}t}(x) = \arctan(c^t\tan(x)) \). One has to be a bit careful with the choosen branch, so we choose:
\[ \arctan_0(c^t\tan(x)) + \pi\left\lfloor \frac{x+\frac{\pi}{2}}{\pi}\right\rfloor \]
This looks like:

So the main question is: Is this function analytic at \(\frac{\pi}{2} + k\pi \)?
Expressing tan with sin:
\[ \tan(x)=\frac{\sin(x)}{\sqrt{1-\sin(x)^2}}, \quad\arctan(x) = \arcsin\left(\frac{x}{\sqrt{1+x^2}}\right) \]
For brevity, write \(s\) for \(\sin(x)\)
\[\arctan(c\tan(x)) = \arcsin\left(\frac{\frac{cs}{\sqrt{1-s^2}}}{\sqrt{1+\frac{c^2s^2}{1-s^2}}}\right) = 
\arcsin\left(\frac{cs}{\sqrt{1-s^2+c^2s^2}}\right) = \arcsin\left(\frac{c\sin(x)}{\sqrt{1+\sin(x)^2(-1+c^2)}}\right)\]
So we see: nothing scary happens around \(x\approx \frac{\pi}{2}\), \(\sin(x)\approx 1\) just combination of analytic functions.
Hence \(f\) and \(f^{\mathfrak{R}t}\) is analytic on whole \(\mathbb{R}\) at infintely many fixed points.

Epic answer Bo !

This reminds me I posted a question about iterations of periodic functions + identity. 

But my main question is this :

How about polynomial cases ?

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&

Which bring me to the following " psychological " analysis ;

when we consider super( abel(x) + c ) 

we tend to jump to conclusions ...
(demonstration ; )

if x is a fixpoint then super(v) = x only at v around infinity.

in the limit to v we get super(v) = x.

reaching a limit at infinity means having derivative going to zero.

so the abel at x must have the derivative 1/0 , hence a log , pole or singularity.

so super( abel(x) + c) for nonzero c has the singularity of abel(x) , unless in special cases where the super cancels it 

( aka such as exp( ln(x) + c ) but not in say cube(cuberoot(x) + 1)  )

since the singularity or pole was not removed 

super ( abel(x) + c) ) for x a fixpoint must also have a singularity or pole.

we then believe 

super ( abel(fix) + 1/2 ) has a pole or singularity for fix.

and even if it is not at one fixpoints we believe it must be at another fix.

this psychological or rational failure might even happen unconsciously.

It is a very subtle mistake !!

Second " psychological " analysis :

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

it is pseudoperiodic near the fixpoint used for the regular iteration.

but away from the fixpoint it might become less and less speudoperiodic or even take ANOTHER pseudoperiodic when we get closer to the other fixpoint.

An easy proofs follows from using better asymtotics than linear for the regular itertion near a fixpoint ; then suddenly it is no longer periodic.

***

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.
with an analytic path from one fixpoint to the other ( repel and attract ).
and ofcourse the half-iterate not simply being a degree 2 polynomial.

Or a proof that it is not possible.

Or other degrees ofcourse.

I assume degree 4 is the smallest ( lower bound ) possible.



regards

tommy1729