Vi\`ete's fractal distributions and their momenta

TL;DR

This paper explores the solutions of Schr"oder-Poincaré polynomial equations, establishing a correspondence between zeros and discrete functions, and deriving explicit formulas for zeros and inverse branches using Weierstrass-Hadamard factorization.

Contribution

It introduces a novel approach to represent solutions of polynomial equations via infinite products and explicit formulas, advancing the understanding of their zeros and inverse functions.

Findings

Established a one-to-one correspondence between zeros of solutions and discrete functions.

Derived explicit momenta formulas for zeros.

Provided explicit branches of the multi-valued inverse function.

Abstract

Solutions of Schr\"oder-Poincar\'e's polynomial equations $f(az)=P(f(z))$ usually do not admit a simple closed-form representation in terms of known standard functions. We show that there is a one-to-one correspondence between zeros of $f$ and a set of discrete functions stable at infinity. The corresponding Vi\`ete-type infinite products for zeros of $f$ are also provided. This allows us to obtain a special kind of closed-form representation for $f$ based on the Weierstrass-Hadamard factorization. From this representation, it is possible to derive explicit momenta formulas for zeros. We discuss also the rate of convergence of WH-factorization and momenta formulas. Obtaining explicit closed-form expressions is the main motivation for this work. Finally, all the branches of the multi-valued function $f^{-1}$ are computed explicitly.

Equations

$$\frac{2}{\pi }=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2+\sqrt{2}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}...$$

$$\varepsilon_{0}\sqrt{2+\varepsilon_{1}\sqrt{2+\varepsilon_{2}\sqrt{2+...}}}=2\sin\biggl{(}\frac{\pi}{4}\sum_{n=0}^{+\infty}\frac{\varepsilon_{0}\varepsilon_{1}\varepsilon_{2}...\varepsilon_{n}}{2^{n}}\biggr{)},$$

$$f(z)=\underset{n\to \infty }{\lim }\underset{n}{\underbrace{P\circ ...\circ P}}(b+a^{-n}z),z\in \mathbb{C},$$

$$(P_0^{-1}{)}^{\circ n}(w):=\underset{n}{\underbrace{P_0^{-1}\circ ...\circ P_0^{-1}}}(w)\to b.$$

$$|(P_0^{-1}{)}^{\prime }(b)|=|a^{-1}|<1.$$

$$Q(z):=\frac{P(z)-P(b)}{z-b}=\frac{P(z)-b}{z-b}$$

$$\Sigma =\{\sigma :\mathbb{N}\to \{0,...,d-1\},\underset{n\to \infty }{\lim }{\sigma }_n=0\}.$$

$$z(\sigma )=-b\prod _{n=1}^{\infty }\frac{a}{Q(P_{{\sigma }_n}^{-1}\circ ...\circ P_{{\sigma }_1}^{-1}(0))}.$$

$$z=g_{\sigma }(w)=(w-b)\prod _{n=1}^{\infty }\frac{a}{Q(P_{{\sigma }_n}^{-1}\circ ...\circ P_{{\sigma }_1}^{-1}(w))},\sigma \in \Sigma .$$

$$z=g_{\sigma ,m}=a^m(P_{{\sigma }_1}^{-1}(b)-b)\prod _{n=2}^{\infty }\frac{a}{Q(P_{{\sigma }_n}^{-1}\circ ...\circ P_{{\sigma }_1}^{-1}(b))},m\in \mathbb{N}\cup \{0\},\sigma \in {\Sigma }^{\prime },$$

$$g_0(w)=\underset{n\to \infty }{\lim }{a}^n(\underset{n}{\underbrace{P_0^{-1}\circ ...\circ P_0^{-1}}}(w)-b).$$

$$g_0^{\prime }(w)=\prod _{n=1}^{\infty }\frac{a}{P^{\prime }(\underset{n}{\underbrace{P_0^{-1}\circ ...\circ P_0^{-1}}}(w))}.$$

$$f(z)=w+(b-w)\prod_{\sigma\in\Sigma}\biggl{(}1-\frac{z}{g_{\sigma}(w)}\biggr{)},\ \ z\in{\mathbb{C}}.$$

$$f(z)=b\prod_{\sigma\in\Sigma}\biggl{(}1+\frac{z}{b}\prod_{n=1}^{\infty}\frac{Q(P_{\sigma_{n}}^{-1}\circ...\circ P_{\sigma_{1}}^{-1}(0))}{a}\biggr{)},\ \ z\in{\mathbb{C}}.$$

$$f(z)=b+z\prod_{m\geqslant 0}\prod_{\sigma\in\Sigma^{\prime}}\biggl{(}1-\frac{z}{a^{m}g_{\sigma,0}}\biggr{)}=b+\biggl{(}\biggl{\{}\frac{z}{g_{\sigma,0}}\biggr{\}}_{\sigma\in\Sigma^{\prime}};\frac{1}{a}\biggr{)}_{\infty}z,\ \ z\in{\mathbb{C}},$$

$$\sum _{\sigma \in \Sigma }\prod _{n=1}^{\infty }\frac{Q(P_{{\sigma }_n}^{-1}\circ ...\circ P_{{\sigma }_1}^{-1}(w))}{a}=f^{\prime }(0)=1,\forall w\in \mathbb{C}\setminus \{b\}.$$

$$f^{\prime \prime }(0)$$

$$f^{(m)}(0)$$

$$B_{m,j}(x_{1},...,x_{m-j+1})=\sum\frac{m!}{k_{1}!...k_{m-j+1}!}\biggl{(}\frac{x_{1}}{1!}\biggr{)}^{k_{1}}...\biggl{(}\frac{x_{m-j+1}}{(m-j+1)!}\biggr{)}^{k_{m-j+1}},$$

$$\sum _{i=1}^{m-j+1}{k}_i=j,\sum _{i=1}^{m-j+1}i{k}_i=m,$$

$$\sum _{\sigma \in \Sigma }\prod _{n=1}^{\infty }\frac{Q(P_{{\sigma }_n}^{-1}\circ ...\circ P_{{\sigma }_1}^{-1}(w){)}^2}{a^2}$$

$$\sum _{\sigma \in \Sigma }\prod _{n=1}^{\infty }\frac{Q(P_{{\sigma }_n}^{-1}\circ ...\circ P_{{\sigma }_1}^{-1}(w){)}^m}{a^m}$$

$$\sum _{\sigma \in \Sigma }\frac{w-b}{g_{\sigma }(w)}=f^{\prime }(0)=1,\sum _{\sigma \ne \tau }\frac{(w-b{)}^2}{g_{\sigma }(w)g_{\tau }(w)}=\frac{f^{\prime \prime }(0)}{2!}=\frac{P^{\prime \prime }(b)}{2(a^2-a)}$$

$$g_{\sigma }(b)=\{\begin{array}{ll}0,& \sigma =0,\\ g_{{\sigma }^{\prime },m},& {\sigma }^{\prime }=\sigma {|}_{\{n:n⩾m\}},\mathrm{where}{\sigma }_m\ne 0\mathrm{and}{\sigma }_n=0\mathrm{for}n<m,\end{array}\sigma \in \Sigma .$$

$$|g_{\sigma }(w)|⩾C|a{|}^{|\mathrm{supp}\sigma |},\sigma \in \Sigma ,\sigma \ne 0.$$

$$\sum_{|\mathop{\mathrm{supp}}\nolimits\sigma|\geqslant N}|g_{\sigma}(w)|^{-m}=O((d|a|^{-m})^{N}),\ \ \prod_{|\mathop{\mathrm{supp}}\nolimits\sigma|\geqslant N}\biggl{|}1-\frac{z}{g_{\sigma}(w)}\biggr{|}=1+O(|z|(d|a|^{-1})^{N}),$$

$$Q(z)=\frac{2z^2-1-1}{z-1}=2z+2.$$

$$P_1^{-1}(w)=\sqrt{\frac{1+w}{2}},P_{-1}^{-1}(w)=-\sqrt{\frac{1+w}{2}}.$$

$$\sqrt{z}=r^{\frac{1}{2}}{e}^{\frac{i\vartheta }{2}}\mathrm{for}z=r{e}^{i\vartheta },r⩾0,\vartheta \in (-\pi ,\pi ].$$

$$|\sqrt{z_1}-\sqrt{z_2}|=\frac{|z_1-z_2|}{|\sqrt{z_1}+\sqrt{z_2}|}⩽\frac{|z_1-z_2|}{\sqrt{2}},z_1,z_2\in D,$$

$$\Sigma =\{\sigma :\mathbb{N}\to \{\pm 1\},\underset{n\to \infty }{\lim }{\sigma }_n=1\}.$$

Full text

1 Introduction and main results

The classical Viète’s formula

[TABLE]

uses nested square root radicals to represent the constant $\pi$. Wiki says ”By now many formulas similar to Viète’s involving either nested radicals or infinite products of trigonometric functions are known for $\pi$, as well as for other constants such as the golden ratio”, see, e.g., [1, 2, 4, 3]. In this note, we derive formulas for zeros of functions satisfying Schröder-Poincaré’s polynomial equations. In general, the formulas for zeros will involve various nested-radicals products similar to Viète’s. These formulas can be used in Weierstrass-Hadamard factorization to obtain various closed-form expressions.

Finally, looking through ”A chronology of continued square roots and other continued compositions” [11], I found paper [12], where a detailed analysis of real roots of $f$, satisfying $f(az)=f(z)^{2}+c$, is provided. Many interesting facts are presented in [11], e.g., an interesting story of the famous formula

[TABLE]

where $\varepsilon_{i}=-1,0,1$.

We assume facts about existence of entire solutions of SP-equation to be known, see, e.g., [6, 9]. Let $P$ be some polynomial of degree $d\geqslant 2$. Let $b$ be some its repelling point $P(b)=b$, with $|a|>1$ for $a:=P^{\prime}(b)$. Consider the entire solution $f$ of SP-equation $f(az)=P(f(z))$ satisfying $f(0)=b$, $f^{\prime}(0)=1$. This solution can be taken as

[TABLE]

see, e.g., [14]. Composition (1) converges uniformly in any compact subset of ${\mathbb{C}}$. For simplicity, let us assume $b\neq 0$. This is not a restriction, since $\widetilde{f}:=f+c$, $c\in{\mathbb{C}}$, also satisfies some polynomial SP-equation. Let $P_{0}^{-1}(w)$, $w\in{\mathbb{C}}$ be the principal branch of $P^{-1}$ analytic in some open domain containing $b$, where $P_{0}^{-1}(b)=b$. We assume also that

Hypothesis 1. For any $w\in{\mathbb{C}}$ the orbit

[TABLE]

This assumption means that point $b$ repelling for $P$ is attracting for $P_{0}^{-1}$. Note that once $(P_{0}^{-1})^{\circ k}(w)\in\{|w-b|<\delta\}$ for some small $\delta>0$ and some $k\in{\mathbb{N}}$, then $(P_{0}^{-1})^{\circ n}(w)$ stays in $\{|w-b|<\delta\}$ for $n>k$ and $(P_{0}^{-1})^{\circ n}(w)\to b$, since

[TABLE]

Let $P_{j}^{-1}$, $1\leqslant j\leqslant d-1$ be other branches of $P^{-1}$ so that $\{z_{j}(w)\}_{j=0}^{d-1}=\{P_{j}^{-1}(w)\}_{j=0}^{d-1}$ is a complete set of solutions of $P(z)=w$, defined for all $w\in{\mathbb{C}}$. For our research, it does not matter how the branches of $P^{-1}$ are numbered. There are only two things that we should pay close attention to: 1) analyticity of the principal branch $P_{0}^{-1}$ at an open neighbourhood of its attracting point $b$; 2) Hypothesis 1.

Introduce the polynomial

[TABLE]

and the set of discrete functions stable at infinity

[TABLE]

Theorem 1.1

The set of zeros of $f$ coincides with $\{z(\sigma)\}_{\sigma\in\Sigma}$, where

[TABLE]

Each zero is counted according to its multiplicity. In other words, the multiplicity of $z_{0}$ as zero of $f$ is $\#\{\sigma\in\Sigma:\ z(\sigma)=z_{0}\}$.

We may apply Theorem 1.1 to the function $F(z):=f(z)-w$ with some constant $w\in{\mathbb{C}}$, because $F$ also satisfies SP-equation $F(az)=P(F(z)+w)-w$ similar to that for $f$. We only should care about the assumption $f(0)-w=b-w\neq 0$, see after (1).

Corollary 1.2

All the solutions of $f(z)=w$, where $w\in{\mathbb{C}}\setminus\{b\}$, have the form

[TABLE]

Each solution is counted according to its multiplicity.

The case $w=b$ is special, because formal setting $w=b$ in (6) leads to $g_{\sigma}(b)=0$, $\forall\sigma\in\Sigma$ which seems doubt. The next theorem is devoted to the case $w=b$.

Theorem 1.3

All the solutions of $f(z)=b$ have the form $z=g_{0,0}=0$ or

[TABLE]

where $\Sigma^{\prime}=\{\sigma\in\Sigma:\ \sigma_{1}\neq 0\}$.

In fact, $\{g_{\sigma}\}_{\sigma\in\Sigma}$ are all the branches of super-multi-valued function $f^{-1}$. Depending on the choice of the branches $P^{-1}$, the functions $g_{\sigma}(w)$ may or may not be analytic. We can only state that the branch $g_{0}(w)$ ($\sigma=0$) is analytic in $\Omega\setminus\{b\}$, where $\Omega$ is some small neighborhood of $b$, e.g., considered in the remark after Hypothesis 1. Due to (1) and arguments presented before (31), we have

[TABLE]

Differentiating (8) and using $(P_{0}^{-1})^{\prime}(w)=1/P^{\prime}(P_{0}^{-1}(w))$, $w\in\Omega$ we obtain

[TABLE]

The convergence of the product (9), as well as (8), and (6), (5) is exponentially fast, as discussed in the beginning of Section 3.

The order of the entire function $f$ can be computed explicitly by substituting $\alpha e^{A|z|^{\rho}}$ into SP-equation $f(az)=P(f(z))$, see, e.g., [13]. Extracting leading terms after the substitution, we obtain $\rho=\frac{\ln d}{\ln|a|}$. If $\rho<1$ then the Weierstrass-Hadamard (WH) factorization for $f$ does not contain exponential factors, see [15].

Corollary 1.4

Suppose that $d<|a|$. If $w\neq b$ then WH-factorization for $f$ is

[TABLE]

In particular,

[TABLE]

If $w=b$ then

[TABLE]

where $(\{\alpha\}_{i\in R};\beta)_{\infty}:=\prod_{i\in R}\prod_{n=0}^{\infty}(1-\alpha_{i}\beta^{n})$ is the q-Pochhammer symbol.

Equation (10) allows us to compute explicitly momentum formulas for zeros of $f(z)-w$, for any fixed $w\neq b$. This new type of formulas will include both: infinite products and infinite sums. The first (negative) momentum formula for zeros follows from (10) immediately

[TABLE]

Let us note how to compute explicitly other momenta of zeros. First, differentiating $f(az)=P(f(z))$ at $z=0$ and using $f(0)=b$, $f^{\prime}(0)=1$, $P^{\prime}(b)=a$, we obtain recurrent formulas to determine all the derivatives:

[TABLE]

where $B_{m,j}$ are Bell polynomials. They are given by

[TABLE]

where the sum is taken over all sequences $k_{1}$, $k_{2}$, …, $k_{m-j+1}$ of non-negative integers such that the two conditions are satisfied:

[TABLE]

see more about Faà di Bruno’s formula for high order derivatives of compositions in, e.g., wiki. Now, differentiating $\ln(f(z)-w)$ at $z=0$ and using (11), we obtain the momenta formulas of high orders $m\geqslant 2$:

[TABLE]

Another type of Vieta formulas also follows from (10):

[TABLE]

and so on. There is a natural extension of $g_{\sigma}(w)$ to the case $w=b$:

[TABLE]

Then all the solutions $f(z)=w$, where $w\in{\mathbb{C}}$, have the form $z=g_{\sigma}(w)$, $\sigma\in\Sigma$. It is possible to estimate the remainder of series (19), (13), and products (10)-(12). They converge exponentially fast, regarding the length of the support of functions $\sigma\in\Sigma$.

Theorem 1.5

There is $C>0$, depending on the polynomial $P$ only, such that

[TABLE]

Moreover, if $d<|a|$ then

[TABLE]

for $N\to\infty$, $m\geqslant 1$, and any fixed $z\in{\mathbb{C}}$.

We will begin the next section with examples. The proof of the main result is placed in the final section.

2 Examples

1. Consider the case $P(z)=2z^{2}-1$. SP-equation is $f(az)=2f(z)^{2}-1$. We take $f(0)=b=1$, $f^{\prime}(0)=1$. Then $a=(2z^{2}-1)^{\prime}|_{z=b}=4$. Polynomial (3) is

[TABLE]

There are two branches of $P^{-1}$:

[TABLE]

We assume that

[TABLE]

The branch $P_{1}^{-1}$ is principal. It is analytically defined near the attracting (for $P_{1}^{-1}$) point $b$. Moreover, $(P_{1}^{-1})^{\circ n}(w)$ converges to its fixed point $b$ for any $w\in{\mathbb{C}}$, since $\sqrt{z}$ is a contraction mapping in the closed domain $D=\{z:\ \mathop{\mathrm{Re}}\nolimits z\geqslant 1/\sqrt{2}\}$:

[TABLE]

and $P_{1}^{-1}\circ P_{1}^{-1}({\mathbb{C}})\subset D$. Thus, Hypothesis 1 is satisfied and we can use Theorem 1.1 and its Corollaries. To parameterize zeros of $f$, we should use the set

[TABLE]

Then zeros of $f$ have form (5)

[TABLE]

Computations show

[TABLE]

and so on. This is in full agreement with expected values, since $f(z)=\cos\sqrt{-2z}$. In this case, the formulas for zeros are, in fact, modified Viète’s formulas, see also [1, 2]. The order of entire function $f$ is $1/2$. WH-factorization is

[TABLE]

2. Consider the case $P(z)=z^{2}-1$. SP-equation is $f(az)=f(z)^{2}-1$. We take $f(0)=b=\frac{\sqrt{5}+1}{2}$, $f^{\prime}(0)=1$. Then $a=(z^{2}-1)^{\prime}|_{z=b}=2b$. Polynomial (3) is

[TABLE]

There are two branches of $P^{-1}$:

[TABLE]

Again, using the arguments from Example 1, we can state that Hypothesis 1 is satisfied. To parametrize zeros of $f$, we should use the same set as in the previous example

[TABLE]

Then zeros of $f$ have the form

[TABLE]

The first negative zero $z(1,1,1,...)=-2C$ relates to the so-called Paris constant $C$ appearing in the approximation of the golden ratio by nested square root radicals, see [4, 5, 10]. Zeros of $f$ are also related to the polynomial dynamics generated by $P=z^{2}-1$ and, hence, approximate the corresponding Julia set growing up to infinity, see more in [7, 8, 9]. The zeros form impressive fractal structures, see Fig. 1. The order of entire function $f$ is $\ln 2/\ln a<1$. Hence, there is WH-factorization

[TABLE]

There are infinitely many complex zeros of multiplicities $2^{n}$ for any $n\geqslant 0$, see [10]. All the multiplicities are taken into account in WH-factorization mentioned above. The first, second and third momentum formulas for zeros, see (13), (18) and (19), are

[TABLE]

3. Let us consider the cubic SP-equation $f(az)=f(z)^{3}-6$, $f(0)=b=2$, $f^{\prime}(0)=1$. Then $a=3b^{2}=12$. The order of the entire function $f(z)$ is $\ln 3/\ln 12<1$. Let us skip the similar arguments as in the previous examples that show that the principal branch of $P^{-1}$ for $P(z)=z^{3}-6$ satisfies Hypothesis 1. So, we can use (13), (18) to obtain explicit momentum formulas

[TABLE]

[TABLE]

and so on. All of the momenta are rational numbers.

3 Proof of Theorems 1.1, 1.3, and 1.5

First of all let us show that infinite products (5) are well defined. Suppose that

[TABLE]

for some $w\in{\mathbb{C}}$ and $n\in{\mathbb{N}}$. If $n=1$ then (3) gives $w=b$. Consider the case $n>1$. We have that $P_{\sigma_{n}}^{-1}\circ...\circ P_{\sigma_{1}}^{-1}(w)\neq b$, since $Q(b)=P^{\prime}(b)=a\neq 0$. Next, if $P_{\sigma_{n}}^{-1}\circ...\circ P_{\sigma_{1}}^{-1}(w)\neq b$ then (3) and (24) give us

[TABLE]

which leads to $w=b$, since $P(b)=b$. Hence any denominator in (5) is non-zero, since $b\neq 0$ by the assumption from the beginning of the article. Due to analyticity of $P_{0}^{-1}$ in some open neighbourhood of its attracting point $b$, where $|(P_{0}^{-1})^{\prime}(b)|=|a^{-1}|<1$, we have that for any $\varepsilon>0$ there is $\delta>0$ such that for any $w\in\{|w-b|<\delta\}$:

[TABLE]

Identity and inequality (25) along with (2) and the stability condition $\lim\sigma_{n}=0$ in (4) lead to

[TABLE]

where small $\varepsilon>0$ is taken such that $c:=|a^{-1}|+\varepsilon<1$. Hence

[TABLE]

This guaranties the convergence of infinite products (5) for any $\sigma\in\Sigma$. The convergence of the products is exponentially fast, since $c<1$. Note that the same arguments give also the exponential rate of convergence of (9).

Let $\widetilde{z}$ be some zero of $f(z)$ of multiplicity $\widetilde{m}\in{\mathbb{N}}$, i.e.

[TABLE]

SP-equation gives

[TABLE]

Taking $n>\widetilde{m}$ such that $f^{\prime}(a^{-n}\widetilde{z})\neq 0$ (recall that $f^{\prime}(0)=1\neq 0$ and $|a|>1$), differentiating (28) at $z=\widetilde{z}$ and using (27), we obtain that

[TABLE]

Thus, $f(a^{-n}\widetilde{z})$ is a root of $P^{\circ n}$ of a multiplicity at least $\widetilde{m}$. This means that

[TABLE]

for at least $\widetilde{m}$ different $\sigma:\{1,...,n\}\to\{0,...,d\}$.

SP-equation can be written in the form $f(a^{-1}z)=P^{-1}(f(z))$. Since $f(0)=b$, the branch $P^{-1}$ should coincide with the principal branch $P_{0}^{-1}$ in a small neighbourhood of $b$, i.e. $f(a^{-1}z)=P_{0}^{-1}(f(z))$ for all sufficiently small $z$ (see also the remark before Hypothesis 1). Let $\widetilde{n}$ be such that $f(a^{-\widetilde{n}}\widetilde{z})$ belongs to this small neighbourhood of $b$. We assume also that $\widetilde{n}$ is large enough to satisfy (30) with at least $\widetilde{m}$ different $\sigma$. Then, by (30), we have

[TABLE]

Denote $\sigma_{n}=0$ for $n>\widetilde{n}$. Thus, using $f(0)=b$, $f^{\prime}(0)=1$, we get

[TABLE]

Like (30), identity

[TABLE]

holds for at least $\widetilde{m}$ different $\sigma\in\Sigma$.

Conversely, suppose that (33) holds for $\widetilde{m}$ different $\sigma\in\Sigma$. To finish the proof we need to show that $\widetilde{z}$ is a zero of $f$ of a multiplicity at least $\widetilde{m}$. Using (1) and the second identity in (32), we obtain

[TABLE]

since the convergence of (1) is uniform in any bounded domain. Hence $\widetilde{z}$ is a zero of $f$. Now, let $N$ be such that $P_{\sigma_{N}}^{-1}\circ...\circ P_{\sigma_{1}}^{-1}(0)$ is sufficiently close to $b$, where $f^{-1}$ is defined, so that

[TABLE]

We can do this because $P_{\sigma_{n}}^{-1}=P_{0}^{-1}$ for large $n$ by definition (4), and we are under Hypothesis 1. We also assume that $N$ is so large that (35) is valid for at least $\widetilde{m}$ different $\sigma:\{1,...,N\}\to\{0,...,d\}$ coinciding with the segments of those $\sigma\in\Sigma$ mentioned in (33), and, also, all $\sigma_{n}=0$ for $n>N$. For simplicity, in the previous sentence we use the same symbol for $\sigma\in\Sigma$ and for its segment $\sigma|_{\{1,...,N\}}$. Finally, it is assumed that $N$ is so large that

[TABLE]

see comments before (31). Using (36), (33), the assumption $\sigma_{n}=0$, $n>N$, and the arguments similar to (32), we obtain

[TABLE]

Hence, all $t_{\sigma}$ are equal to each other. Using (35), the remark after (35) about $\widetilde{m}$ different $\sigma$, and (37), we conclude that $f(a^{-N}\widetilde{z})$ is a zero of $P^{\circ N}$ of a multiplicity at least $\widetilde{m}$. Thus, differentiating $f(z)=P^{\circ N}(f(a^{-N}z))$ at $z=\widetilde{z}$, we obtain that $f^{(j)}(\widetilde{z})=0$, $j=0,...,\widetilde{m}-1$. Hence, $\widetilde{z}$ is a zero of $f$ of a multiplicity at least $\widetilde{m}$. The proof of Theorem 1.1 is finished.

To prove Theorem 1.3 let us note that

[TABLE]

where $P_{\sigma_{0}}^{-1}(w):=w$ and $g_{\sigma}(w)$ are defined in (6). Moreover, using the similar arguments as in the proof of Theorem 1.1, we can state that if $\sigma_{m}\neq 0$ then (38) is true for $w=b$. Now, choosing the first $m$ such that $\sigma_{m}\neq 0$, we finish the proof of Theorem 1.3.

Recall that $P_{0}^{-1}$ is the unique branch of $P^{-1}$ such that $P_{0}^{-1}(b)=b$ and $P_{0}^{-1}$ is analytic in the neighbourhood of $b$. Let $\delta>0$ be so small that

[TABLE]

Identities (6) and (38) give

[TABLE]

If $\sigma_{m}\neq 0$ then (39) leads to

[TABLE]

Since $f$ is continuous and $f(0)=b$, there is $C>0$ such that

[TABLE]

Thus, by (41) and (42), we have $|a^{-m}g_{\sigma}(w)|\geqslant C$, which gives (22). Using (22), we get

[TABLE]

By analogy, we can estimate the product in (23). The proof of Theorem 1.5 is finished.