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Let $\mathfrak{I}:=\big\{ \, f:=\sum_{k=0}^\infty f_k z^k \in\mathbb{C}[[z]]\; : \text{s.t. }\; f_0=0 \;\text{ and }\; f_1=1\big\}$. A most basic result about linearization states that, for any $f\in\mathfrak{I}$ and for any $\lambda\in\mathbb{C}$ not a root of unity, there exists a unique $h\in\mathfrak{I}$ that linearizes $f(\lambda z)$: $$h(\lambda z)=f(\lambda h(z)).$$ Existence and uniqueness of this conjugation in the setting of formal power series is indeed easily established, as the coefficients of $h$ can be determined inductively from the data $(f_1,f_2,\dots) $ expanding the composition: one finds $$h_1=1$$ $$h_{n+1}={1\over \lambda^{n}-1 } \sum_{k\ge2,\,j} f_k \lambda^{k-1} h_{j_1}\dots h_{j_k},$$ the latter sum being extended over all integers $k\ge2$ and multi-indices $j\in\mathbb{Z}_+^k$ of length $|j|:=j_1+\dots+j_k=n+1$, thus with all components $j_i\le n$. The above recursion implies that the coefficients $h_n$ should be rational functions of $\lambda$, with denominators in the form $\prod_{k=1}^{n-1}(\lambda^k-1)^{m_{n,k}}$, where the exponents $m_{k,n}$ may be larger than $1$, at least at a first glance. However, after experiments with various $f$ and up to $n=50$ I observed that the form of $h_n$ always shows the simpler denominator, with all $m_{k,n}=1$: $$h_n={P_n(\lambda)\over (\lambda-1)(\lambda^2-1)\dots(\lambda^{n-1}-1)},$$ for some polynomials $P_n(\lambda)$ of degree at most ${n\choose 2}$.

Is this really a general phenomenon? If so, why? Is there a simpler formula (maybe still recursive) for $h_n$ than the above recursion?

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    $\begingroup$ Late comment: I've seen this as a "general phenomenon" because I got used to look at this functional selfcomposition as a diagonalization problem of Carlemanmatrices. Carlemanmatrices are a special form, associated with some function $f(x)=c_0+c_1 x + c_2 x^2 + ... $ and in a first step rewritten as dot-product $f(x) = V(x) \cdot C $ where $V(x)$ is the vector $[1,x,x^2,x^3,...]$ and $C$ is the columnvector $[c_0,c_1,c_2,...]$. The Carlemanmatrix for $f(x)$, say $F$ is now constructed so that $V(x) \cdot F = V(f(x))$ (...cntd...) $\endgroup$
    – Gottfried Helms
    Commented Sep 30, 2022 at 6:17
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    $\begingroup$ (...) and obviously iterations of $f(x)$ are now expressible by powers of $F$. Fractional iterates of $f()$ by fractional powers of $F$, and this is, for $c_0=0$ and $c_1<>1$, easily computable by a recusive procedure. We get then $V(f°^h(x))= V(x) \cdot M \cdot D^h \cdot W$ where for easiness of notation I write $W=M^{-1}$. If I understand you correctly, $M$ is then the Carlemanmatrix for your function $h()$. Thus, this is a very general concept and the occurences of polynomials $P_n(\lambda)$ of degree at most $\binom{n}{2}$ is then the general phenomenon. (...cntd...) $\endgroup$
    – Gottfried Helms
    Commented Sep 30, 2022 at 6:25
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    $\begingroup$ (...cntd...) For the functions $t^x-1$ (with $t \ne \exp(1)$) I have detailed this in a workout go.helms-net.de/math/tetdocs/APT.htm with explicite display of coefficients for the polynomials (of small order). (for $t=\exp(1)$ resp $u=\log(t)$ having $u=1$ this cannot be used for fractional powers, one needs then application of the matrix-logarithm, another subject, another polynomials) $\endgroup$
    – Gottfried Helms
    Commented Sep 30, 2022 at 6:31

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You may find useful information in a recent article by D. Sauzin and al. "Explicit linearization of one-dimensional germs through tree-expansions" here, where they use "mould calculus" (introduced by J. Écalle 40 years ago) to write down the coefficients $h_n$ and explore the combinaotrial structure of their family.

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I think it is very unusual for the $h_n$s to have a simpler formulation. Even in the case of a quadratic polynomial, the conjugating function $h$ need not be a polynomial. I'm afraid I don't have a general explanation of that for you, other than noticing it by working through some examples.

There is however a simpler construction of $h$ as a limit, rather than as a power series that works when $|\lambda|\neq 1$, namely let $$h_n=\lambda^{-n}\circ f^n.$$ Then, $h_n\to h$. This formulation is fairly well known so, perhaps, you are already aware of it. It's quite easy to work with programmatically.

Note, however that the case $|\lambda|=1$ is more subtle than you've stated. The existence of the linearization is then equivalent to the the fixed point being in the Fatou set. Indeed, if $\lambda$ is a root of unity, the fixed point is in the Julia set but, even if $\lambda$ is not a root of unity the fixed point might still be in the Julia set.

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  • $\begingroup$ Thank you, yes, I am aware of what you are saying. Here I only look at the formal series expansion, with no issue of convergence, so the only condition on $\lambda$ is not being a root of unity. (However one can assume $|\lambda|\neq 1$ if useful). $\endgroup$
    – Pietro Majer
    Commented May 12, 2017 at 4:59
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    $\begingroup$ How does this address the question? As I understand it was not conjectured that $h$ is a polynomial. The question was about denominators of the coefficients of $h$. $\endgroup$
    – Alexandre Eremenko
    Commented May 12, 2017 at 5:59
  • $\begingroup$ @AlexandreEremenko You're right, of course. The reply started as a comment but ended up being longer than I anticipated. $\endgroup$
    – Mark McClure
    Commented May 12, 2017 at 12:20

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