Iteration and Iterative Roots of Fractional Polynomial Function

Abstract

Iteration is involved in the fields of dynamical systems and numerical computation and so forth. The computation of iteration is difficult for general functions (even for some simple functions such as linear fractional functions). In this paper, we discuss fractional polynomial function and use the method of conjugate similitude to obtain its expression of general iterate of order n under two different conditions. Furthermore, we also give iterative roots of order n for the function under two different conditions.

1. Introduction

Iteration is a repetition of the same operation. Given a nonempty set 𝑋 and a self-mapping 𝑓∶𝑋→𝑋, define 𝑓⁰(𝑥)=𝑥,𝑓¹(𝑥)=𝑓(𝑥),𝑓^𝑛(𝑥)=𝑓∘𝑓^𝑛−1(𝑥), where 𝑛∈ℤ⁺ and ∘ denotes the composition of mappings. 𝑓^𝑛 is called the 𝑛th iterate of 𝑓, and 𝑛 is the iterate index of 𝑓^𝑛 concerning 𝑓. Iteration is often observed in mathematics, science, engineering, and daily life, but the computation of iteration of some elementary functions is very complicated and sometimes rather difficult (see [1–7]), such as linear fractional functions 𝑓(𝑥)=(𝑎𝑥+𝑏)/(𝑥+𝑐), where 𝑎,𝑏,𝑐∈ℝ,𝑎𝑐−𝑏≠0. Using the numerical computation method, we only make some partitions on the defined interval of 𝑥 to obtain pointwise data and approximately curves of 𝑓^𝑛. Although computer algebra system such as Maple provided the symbol computational tool, we still need to calculate the 𝑛th iterate of 𝑓 for a given 𝑛, and the expression of iteration is complicated even for 𝑛=12 (see [8]). However, using the method of conjugate similar, we can effectively calculate its iteration of order 𝑛(see the following (*)). This example shows that computer is not universal, and we need to find good mathematical method.

Given mapping 𝑓 and 𝑔, if there exists invertible mapping ℎ such that 𝑓=ℎ⁻¹∘𝑔∘ℎ, then 𝑓 is conjugating to 𝑔. Obviously, if 𝑓=ℎ⁻¹∘𝑔∘ℎ, then 𝑓^𝑛=ℎ⁻¹∘𝑔^𝑛∘ℎ. We usually use this method to turn iteration of complicated function into iteration of simple function which is easy to get general iteration. We call it as the method of conjugation. For example, using the method of conjugation in reference [9], fractional linear function is conjugated to a linear function by conjugation function ℎ(𝑥)=1/(𝑥−𝑠), where 𝑠 is a root of the equation 𝑠²−(𝑎−𝑐)𝑠−𝑏=0. Thus, the 𝑛th iterate of the fractional linear function 𝑓 is 𝑓^𝑛(𝑥)=()𝑠+𝑎−𝑠^𝑛(𝑥−𝑠)((𝑎−𝑠)^𝑛−(𝑠+𝑐)^𝑛)𝑥₀(𝑥−𝑠)+(𝑠+𝑐)^𝑛,𝑎−𝑐2²()+𝑏≠0,𝑠+𝑎−𝑠)(𝑥−𝑠()𝑠,𝑛𝑥+𝑎−𝑛+1𝑎−𝑐2²+𝑏=0, (∗) where 𝑥₀=1/(𝑎−𝑐−2𝑠). By the same method, in reference [10], Jin et al. discuss that the 𝑛th iterate of polynomial function 𝑓(𝑥)=𝑎₀𝑥^𝑚+𝑎₁𝑥^𝑚−1+⋯+𝑎_𝑚, (1.1) where 𝑚∈ℤ,𝑚≥2,𝑎_𝑖∈ℂ,𝑖=0,1,…,𝑚, under the conditions that 𝑎_𝑖=𝑎₀𝐶^𝑖_𝑚(𝑎₁/𝑚𝑎₀)^𝑖,(𝑖=2,3,…,𝑚−1), 𝑎_𝑚=(𝑎₁/𝑚𝑎₀)[𝑎₀(𝑎₁/𝑚𝑎₀)^𝑚−1−1], and 𝑎₀≠0, is 𝑓^𝑛(𝑥)=𝑎^(𝑚𝑛₀^{−1)/(𝑚−1)}𝑎𝑥+¹𝑚𝑎₀^𝑚𝑛−𝑎₁𝑚𝑎₀, (1.2) where 𝐶^𝑖_𝑚 denotes the number of combination, that is, 𝐶^𝑖_𝑚=𝑚!/𝑖!(𝑚−𝑖)!.

Given a nonempty set 𝐼 and an integer 𝑛>0, an iterative root of order 𝑛 of a given self-mapping 𝑓∶𝐼→𝐼 is a self-mapping 𝜑∶𝐼→𝐼 such that 𝜑^𝑛(𝑥)=𝑓(𝑥),∀𝑥∈𝐼,𝑛∈ℤ⁺, (1.3) where 𝜑^𝑛 denotes the 𝑛th iterate of 𝜑, that is, 𝜑^𝑛=𝜑∘𝜑^𝑛−1. The problem of iterative roots of mapping is an important problem in the iteration theory (see [9, 11–13]). It was studied early from the 19th century, but great advances have been made since 1950s, most of which were given for monotone self-mappings on compact interval. For nonmonotonic cases, there are also some progress in references (see [14–19]).

In this paper, we study iteration and iterative roots of the fractional polynomial function 𝑓(𝑥)=𝑎_𝑘𝑥^𝑘+𝑎_𝑘−1𝑥^𝑘−1+𝑎_𝑘−2𝑥^𝑘−2+⋯+𝑎₁𝑥+𝑎₀𝑏_𝑘𝑥^𝑘+𝑏_𝑘−1𝑥^𝑘−1+𝑏_𝑘−2𝑥^𝑘−2+⋯+𝑏₁𝑥+𝑏₀, (1.4) where 𝑘∈ℤ,𝑘≥1,𝑎_𝑖,𝑏_𝑖∈ℝ,𝑖=0,1,…,𝑘. It can be treated as a nonmonotonic mapping on ℝ. Using the method of conjugation, we get the expression of 𝑓^𝑛 and iterative roots of order 𝑛 of 𝑓 under some conditions.

2. Iteration of Fractional Polynomial Functions

Theorem 2.1. Let either 𝑎₀≠0 when 𝑘 is odd or 𝑎₀>0 when 𝑘 is even. Suppose that the fractional polynomial function 𝑓 defined by (1.4) satisfies 𝑎_𝑖=𝐶^𝑖_𝑘(√^𝑘𝑎₀)^𝑘−𝑖,(𝑖=0,1,…,𝑘),𝑏_𝑗=−𝑎_𝑗/√^𝑘𝑎₀,(𝑗=0,1,…,𝑘−1), then 𝑓^𝑛(𝑥)=1−1/√^𝑘𝑎₀−𝑏−√^𝑘𝑎₀^{−𝑘(1−(−𝑘)𝑛)/(1+𝑘)}−1/𝑥−1/√^𝑘𝑎₀^{(−𝑘)𝑛}, (2.1) where 𝑏=−1/√^𝑘𝑎₀−𝑏_𝑘, when 𝑘=1, √^𝑘𝑎₀∶=𝑎₀; 𝑛∈ℤ⁺.

Proof. On the basis of what 𝑓 satisfies, the fractional polynomial function 𝑓 defined by (1.4) transforms into 𝑓(𝑥)=𝑥^𝑘+𝐶√_𝑘^𝑘−1𝑘𝑎₀𝑥^𝑘−1+𝐶√_𝑘^𝑘−2𝑘𝑎₀²𝑥^𝑘−2+⋯+𝐶√¹_𝑘^𝑘𝑎₀^𝑘−1𝑥+𝐶√⁰_𝑘^𝑘𝑎₀^𝑘−𝒜𝑥^𝑘−𝒜𝐶√_𝑘^𝑘−1𝑘𝑎₀𝑥^𝑘−1−𝒜𝐶√_𝑘^𝑘−2𝑘𝑎₀²𝑥^𝑘−2−⋯−𝒜𝐶√¹_𝑘^𝑘𝑎₀^𝑘−1𝑥−𝒜𝐶√⁰_𝑘^𝑘𝑎₀^𝑘−𝑏𝑥^𝑘=𝑥+√^𝑘𝑎₀^𝑘−𝒜𝑥+√^𝑘𝑎₀^𝑘−𝑏𝑥^𝑘, (2.2) where 𝒜denotes (1/√^𝑘𝑎₀). Set ℎ₁(𝑥)=1−1/𝑥, then the inverse of ℎ₁ is ℎ₁⁻¹(𝑥)=1/(1−𝑥), it follows that ℎ₁∘𝑓∘ℎ₁⁻¹(𝑥)=ℎ₁𝑓ℎ₁⁻¹(𝑥)=ℎ₁𝑓11−𝑥=ℎ₁1+√^𝑘𝑎₀−√^𝑘𝑎₀𝑥^𝑘−√^𝑘𝑎₀⁻¹1+√^𝑘𝑎₀−√^𝑘𝑎₀𝑥^𝑘=−𝑏_𝑘√𝑎₀+1_𝑘√𝑎₀+𝑏−√^𝑘𝑎₀^𝑘𝑥−1+√^𝑘𝑎_0𝑘√𝑎₀^−𝑘∶=𝑔(𝑥). (2.3) Set ℎ₂(𝑥)=𝑥−(1+√^𝑘𝑎₀)/√^𝑘𝑎₀, then the inverse of ℎ₂ is ℎ₂⁻¹(𝑥)=𝑥+(1+√^𝑘𝑎₀)/√^𝑘𝑎₀, it follows that ℎ₂∘𝑔∘ℎ₂⁻¹(𝑥)=ℎ₂𝑔ℎ₂⁻¹(𝑥)=ℎ₂𝑔𝑥+1+√^𝑘𝑎_0𝑘√𝑎₀=ℎ₂𝑏−√^𝑘𝑎₀^𝑘𝑥^−𝑘+^𝑘√𝑎₀+1_𝑘√𝑎₀=𝑏−√^𝑘𝑎₀^𝑘𝑥^−𝑘∶=𝐺(𝑥). (2.4) By induction, we obtain easily 𝐺^𝑛(𝑥)=𝑏(−√^𝑘𝑎₀)^{𝑘(1−(−𝑘)𝑛)/(1+𝑘)}𝑥^{(−𝑘)𝑛}. (2.5) By (2.4) and (2.5), we have 𝑔^𝑛(𝑥)=ℎ₂⁻¹∘𝐺^𝑛∘ℎ₂(𝑥)=𝑏(−√^𝑘𝑎₀)^{𝑘(1−(−𝑘)𝑛)/(1+𝑘)}𝑥−1+√^𝑘𝑎_0𝑘√𝑎₀^{(−𝑘)𝑛}+1+√^𝑘𝑎_0𝑘√𝑎₀. (2.6) By (2.3) and (2.6), we get 𝑓^𝑛(𝑥)=ℎ₁⁻¹∘𝑔^𝑛∘ℎ₁(𝑥)=1−1/√^𝑘𝑎₀−𝑏−√^𝑘𝑎₀^{−𝑘(1−(−𝑘)𝑛)/(1+𝑘)}−1/𝑥−1/√^𝑘𝑎₀^{(−𝑘)𝑛}. (2.7) This completes the proof.

Theorem 2.2. Let either 𝑏₀≠0 when 𝑘 is odd or 𝑏₀>0 when 𝑘 is even. Suppose that the fractional polynomial function 𝑓 defined by (1.4) satisfies 𝑏_𝑖=𝐶^𝑖_𝑘(^𝑘𝑏₀)^𝑘−𝑖(𝑖=0,1,2,…,𝑘), 𝑎_𝑗=(−^𝑘𝑏₀)𝑏_𝑗(𝑗=1,2,…,𝑘), then 𝑓^𝑛(𝑥)=𝑑^{(1−(−𝑘)𝑛)/(1+𝑘)}𝑥+^𝑘𝑏₀^{(−𝑘)𝑛}−^𝑘𝑏₀, (2.8) where 𝑑=𝑎₀+𝑏₀^𝑘𝑏₀; when 𝑘=1, ^𝑘𝑏₀∶=𝑏₀, 𝑛∈ℤ⁺.

Proof. On the basis of what 𝑓 satisfies, the fractional polynomial function 𝑓 defined by (1.4) transforms into 𝑓(𝑥)=−^𝑘𝑏₀𝑥^𝑘−^𝑘𝑏₀𝐶_𝑘^𝑘−1𝑘𝑏₀𝑥^𝑘−1−^𝑘𝑏₀𝐶_𝑘^𝑘−2𝑘𝑏₀²𝑥^𝑘−2−⋯−^𝑘𝑏₀𝐶¹_𝑘^𝑘𝑏₀^𝑘−1𝑥+𝑑−𝑏₀^𝑘𝑏₀𝑥^𝑘+𝐶_𝑘^𝑘−1𝑘𝑏₀𝑥^𝑘−1+𝐶_𝑘^𝑘−2𝑘𝑏₀²𝑥^𝑘−2+⋯+𝐶¹_𝑘^𝑘𝑏₀^𝑘−1𝑥+𝐶⁰_𝑘^𝑘𝑏₀^𝑘=−^𝑘𝑏₀𝑥^𝑘+𝐶_𝑘^𝑘−1𝑘𝑏₀𝑥^𝑘−1+𝐶_𝑘^𝑘−2𝑘𝑏₀²𝑥^𝑘−2+⋯+𝐶¹_𝑘^𝑘𝑏₀^𝑘−1𝑥+𝑏₀+𝑑𝑥+^𝑘𝑏₀^𝑘=−^𝑘𝑏₀𝑥+^𝑘𝑏₀^𝑘+𝑑𝑥+^𝑘𝑏₀^𝑘. (2.9) Set ℎ(𝑥)=𝑥+^𝑘𝑏₀, then the inverse of ℎ is ℎ⁻¹(𝑥)=𝑥−^𝑘𝑏₀, it follows that ℎ∘𝑓∘ℎ⁻¹(𝑥)=ℎ𝑓ℎ⁻¹(𝑥)=ℎ𝑓𝑥−^𝑘𝑏₀−=ℎ^𝑘𝑏₀𝑥^𝑘+𝑑𝑥^𝑘=ℎ−^𝑘𝑏₀+𝑑𝑥^−𝑘=𝑑𝑥^−𝑘∶=𝑔(𝑥). (2.10) By induction, we obtain easily 𝑔^𝑛(𝑥)=𝑑^{(1−(−𝑘)𝑛)/(1+𝑘)}𝑥^{(−𝑘)𝑛}. (2.11) By (2.10) and (2.11), we get 𝑓^𝑛(𝑥)=ℎ⁻¹∘𝑔^𝑛∘ℎ(𝑥)=𝑑^{(1−(−𝑘)𝑛)/(1+𝑘)}𝑥+^𝑘𝑏₀^{(−𝑘)𝑛}−^𝑘𝑏₀. (2.12) This completes the proof.

3. Iterative Roots of Fractional Polynomial Functions

We first give some useful lemmas.

Lemma 3.1. If 𝑓(𝑥)=𝑙(𝑥+𝑐)^𝑘−𝑐, where 𝑙,𝑘,𝑐∈ℝ,𝑙𝑘≠0, then 𝑓^𝑛(𝑥)=𝑙^𝑛𝑥+𝑙^𝑛𝑙𝑐−𝑐,𝑘=1,_{(1−𝑘𝑛)/(1−𝑘)}(𝑥+𝑐)^𝑘𝑛−𝑐,𝑘≠1, (3.1) where 𝑛∈ℤ⁺.

Proof. Set ℎ(𝑥)=𝑥+𝑐, then the inverse of ℎ is ℎ⁻¹(𝑥)=𝑥−𝑐, it follows that ℎ∘𝑓∘ℎ⁻¹(𝑥)=ℎ𝑓ℎ⁻¹(𝑥)=ℎ𝑓(𝑥−𝑐)=ℎ𝑙𝑥^𝑘−𝑐=𝑙𝑥^𝑘∶=𝑔(𝑥). (3.2) By induction, we obtain that the 𝑛th iterate of 𝑔 is 𝑔^𝑛(𝑥)=𝑙^𝑛𝑙𝑥,𝑘=1,_{(1−𝑘𝑛)/(1−𝑘)}𝑥^𝑘𝑛,𝑘≠1. (3.3) By (3.2) and (3.3), we get 𝑓^𝑛(𝑥)=ℎ⁻¹∘𝑔^𝑛(𝑥)=𝑙∘ℎ^𝑛𝑥+𝑙^𝑛𝑙𝑐−𝑐,𝑘=1,_{(1−𝑘𝑛)/(1−𝑘)}(𝑥+𝑐)^𝑘𝑛−𝑐,𝑘≠1. (3.4) This completes the proof.

Lemma 3.2. If 𝑓(𝑥)=𝑥^𝑘/(𝑎𝑥^𝑘−𝑏(𝑥−𝑐)^𝑘), where 𝑎,𝑏,𝑐,𝑘∈ℝ,𝑎𝑏𝑐𝑘≠0,𝑘≠1, satisfies 𝑎𝑐=1, then 𝑓^𝑛(𝑥)=11/𝑐−𝑏𝑐^{𝑘(1−𝑘𝑛)/(1−𝑘)}(−1/𝑥+1/𝑐)^𝑘𝑛, (3.5) where 𝑛∈ℤ⁺.

Proof. Set ℎ(𝑥)=1−1/𝑥, then the inverse of ℎ is ℎ⁻¹(𝑥)=1/(1−𝑥), it follows that ℎ∘𝑓∘ℎ⁻¹(𝑥)=ℎ𝑓ℎ⁻¹(𝑥)𝑓1=ℎ11−𝑥=ℎ𝑎−𝑏(1−𝑐+𝑐𝑥)^𝑘=1−𝑎+𝑏(1−𝑐+𝑐𝑥)^𝑘=𝑏𝑐^𝑘𝑥+1−𝑐𝑐^𝑘+1−𝑎∶=𝑔(𝑥). (3.6) By Lemma 3.1, when (1−𝑐)/𝑐=𝑎−1, that is, when 𝑎𝑐=1, we get that the 𝑛th iterate of 𝑔 is 𝑔^𝑛(𝑥)=𝑏𝑐^{𝑘(1−𝑘𝑛)/(1−𝑘)}𝑥+1−𝑐𝑐^𝑘𝑛−1−𝑐𝑐. (3.7) By (3.6) and (3.7), we get 𝑓^𝑛(𝑥)=ℎ⁻¹∘𝑔^𝑛(𝑥)=1∘ℎ1/𝑐−𝑏𝑐^{𝑘(1−𝑘𝑛)/(1−𝑘)}(−1/𝑥+1/𝑐)^𝑘𝑛. (3.8) This completes the proof.

In Theorems 2.1 and 2.2, we get the expression of 𝑓^𝑛 of the fractional polynomial function (1.4) under different conditions, they can be treated as a mapping which involves parameter 𝑛, then 𝑛 is extended from ℤ to ℚ, we can obtain the iterative roots. For example, we can get the iterative roots of the fractional polynomial function (1.4) by extending the results of Theorems 2.1 and 2.2.

Theorem 3.3. Suppose that the fractional polynomial function 𝑓 defined by (1.4) satisfies conditions in Theorem 2.1, then 𝑓 has iterative roots of any odd order𝑛: 𝜑(𝑥)=1−1/√^𝑘𝑎₀−𝑏−√^𝑘𝑎₀^{−𝑘(1−(−𝑘)1/𝑛)/(1+𝑘)}−1/𝑥−1/√^𝑘𝑎₀^{(−𝑘)1/𝑛}, (3.9) where 𝑛∈ℤ⁺.

Proof. In what follows, we only need to prove that 𝜑^𝑛(𝑥)=𝑓(𝑥) holds under the case that 𝑓 satisfies the conditions in Theorem 2.1 and 𝑛 is positive odd.
In fact, suppose that the fractional polynomial functions 𝑓 defined by (1.4) satisfy the conditions in Theorem 2.1, then we have 𝑓(𝑥)=𝑥^𝑘+𝐶√_𝑘^𝑘−1𝑘𝑎₀𝑥^𝑘−1+𝐶√_𝑘^𝑘−2𝑘𝑎₀²𝑥^𝑘−2+⋯+𝐶√¹_𝑘^𝑘𝑎₀^𝑘−1𝑥+𝐶√⁰_𝑘^𝑘𝑎₀^𝑘−𝒜𝑥^𝑘−𝒜𝐶√_𝑘^𝑘−1𝑘𝑎₀𝑥^𝑘−1−𝒜𝐶√_𝑘^𝑘−2𝑘𝑎₀²𝑥^𝑘−2−⋯−𝒜𝐶√¹_𝑘^𝑘𝑎₀^𝑘−1𝑥−𝒜𝐶√⁰_𝑘^𝑘𝑎₀^𝑘−𝑏𝑥^𝑘=𝑥+√^𝑘𝑎₀^𝑘−𝒜𝑥+√^𝑘𝑎₀^𝑘−𝑏𝑥^𝑘, (3.10) where 𝒜 denotes (1/√^𝑘𝑎₀).
Because 𝜑(𝑥)=1−1/√^𝑘𝑎₀−𝑏−√^𝑘𝑎₀^{−𝑘(1−(−𝑘)1/𝑛)/(1+𝑘)}−1/𝑥−1/√^𝑘𝑎₀^{(−𝑘)1/𝑛}=𝑥^{(−𝑘)1/𝑛}−1/√^𝑘𝑎₀𝑥^{(−𝑘)1/𝑛}−𝑏−√^𝑘𝑎₀^{−𝑘(1−(−𝑘)1/𝑛)/(1+𝑘)}−√^𝑘𝑎₀^{−(−𝑘)1/𝑛}𝑥+√^𝑘𝑎₀^{(−𝑘)1/𝑛}, (3.11) when 𝑛 is positive odd, by Lemma 3.2, we have 𝜑^𝑛(𝑥)=1−1/√^𝑘𝑎₀−𝑏−√^𝑘𝑎₀^{−𝑘(1+𝑘1/𝑛)/(1+𝑘)}−√^𝑘𝑎₀^𝑘1/𝑛−√^𝑘𝑎₀^{−𝑘1/𝑛(1−(−𝑘1/𝑛)𝑛)/(1+𝑘1/𝑛)}−1/𝑥−1/√^𝑘𝑎₀^{(−𝑘1/𝑛)𝑛}=1−1/√^𝑘𝑎₀−𝑏/−√^𝑘𝑎₀^𝑘−1/𝑥−1/√^𝑘𝑎₀^−𝑘=𝑥+√^𝑘𝑎₀^𝑘−1/√^𝑘𝑎₀𝑥+√^𝑘𝑎₀^𝑘−𝑏𝑥^𝑘=𝑓(𝑥). (3.12) Thus, 𝑓 has iterative roots of any odd order𝑛: 𝜑(𝑥)=1−1/√^𝑘𝑎₀−𝑏−√^𝑘𝑎₀^{−𝑘(1−(−𝑘)1/𝑛)/(1+𝑘)}−1/𝑥−1/√^𝑘𝑎₀^{(−𝑘)1/𝑛}, (3.13) This completes the proof.

Remark 3.4. When 𝑛 is positive even, the above 𝜑(𝑥) is not well defined, thus, under the condition in Theorem 3.3, 𝑓 has no iterative roots of any order 𝑛 in the form of 𝜑(𝑥).

By Theorems 2.1 and 3.3, we are easy to get the following corollary.

Corollary 3.5. Let 𝑓(𝑥)=(𝑥+𝑎₀)/(𝑏₁𝑥−1) and 𝑎₀≠0, then(1)the expression of 𝑓^𝑛 is 𝑓^𝑛(𝑥)=1−1/𝑎₀−1/𝑎²₀+𝑏₁/𝑎₀^{(1−(−1)𝑛)/2}−1/𝑥−1/𝑎₀^(−1)𝑛,𝑛∈ℤ⁺ (3.14)(2)𝑓 has iterative roots 𝑓 of any odd order 𝑛, that is, 𝑓^𝑛=𝑓, where 𝑛=2𝑚−1,𝑚∈ℤ⁺.

Theorem 3.6. Suppose that the fractional polynomial function 𝑓 defined by (1.4) satisfies conditions in Theorem 2.2, then 𝑓 has iterative roots of any odd order𝑛: 𝐹(𝑥)=𝑑^{(1−(−𝑘)1/𝑛)/(1+𝑘)}𝑥+^𝑘𝑏₀^{(−𝑘)1/𝑛}−^𝑘𝑏₀, (3.15) where 𝑛∈ℤ⁺.

Proof. In what follows, we only need to prove that 𝐹^𝑛(𝑥)=𝑓(𝑥) holds under the case that 𝑓 satisfies the conditions in Theorem 2.2 and 𝑛 is positive odd.
In fact, suppose that the fractional polynomial functions 𝑓 defined by (1.4) satisfy the conditions in Theorem 2.2, then we have 𝑓(𝑥)=−^𝑘𝑏₀𝑥^𝑘−^𝑘𝑏₀𝐶_𝑘^𝑘−1𝑘𝑏₀𝑥^𝑘−1−^𝑘𝑏₀𝐶_𝑘^𝑘−2𝑘𝑏₀²𝑥^𝑘−2−⋯−^𝑘𝑏₀𝐶¹_𝑘^𝑘𝑏₀^𝑘−1𝑥+𝑑−𝑏₀^𝑘𝑏₀𝑥^𝑘+𝐶_𝑘^𝑘−1𝑘𝑏₀𝑥^𝑘−1+𝐶_𝑘^𝑘−2𝑘𝑏₀²𝑥^𝑘−2+⋯+𝐶¹_𝑘^𝑘𝑏₀^𝑘−1𝑥+𝐶⁰_𝑘^𝑘𝑏₀^𝑘=−^𝑘𝑏₀𝑥^𝑘+𝐶_𝑘^𝑘−1𝑘𝑏₀𝑥^𝑘−1+𝐶_𝑘^𝑘−2𝑘𝑏₀²𝑥^𝑘−2+⋯+𝐶¹_𝑘^𝑘𝑏₀^𝑘−1𝑥+𝑏₀+𝑑𝑥+^𝑘𝑏₀^𝑘=−^𝑘𝑏₀𝑥+^𝑘𝑏₀^𝑘+𝑑𝑥+^𝑘𝑏₀^𝑘. (3.16) When 𝑛 is positive odd, by Lemma 3.1, we have 𝐹^𝑛(𝑥)=𝑑^{(1−(−𝑘)1/𝑛)/(1+𝑘)(1−[(−𝑘)1/𝑛]𝑛)/(1−(−𝑘)1/𝑛)}𝑥+^𝑘𝑏₀^{[(−𝑘)1/𝑛]𝑛}−^𝑘𝑏₀=𝑑𝑥+^𝑘𝑏₀^−𝑘−^𝑘𝑏₀=−^𝑘𝑏₀𝑥+^𝑘𝑏₀^𝑘+𝑑𝑥+^𝑘𝑏₀^𝑘=𝑓(𝑥). (3.17) Thus, 𝑓 has iterative roots of any odd order𝑛: 𝐹(𝑥)=𝑑^{(1−(−𝑘)1/𝑛)/(1+𝑘)}𝑥+^𝑘𝑏₀^{(−𝑘)1/𝑛}−^𝑘𝑏₀. (3.18) This completes the proof.

Remark 3.7. When 𝑛 is positive even, the above 𝐹(𝑥) is not well defined, thus, under the conditions in Theorem 3.6, 𝑓 has no iterative roots of any order 𝑛 in the form of 𝐹(𝑥).

By Theorems 2.2 and 3.6, we are easy to get the following corollary.

Corollary 3.8. Let 𝑓(𝑥)=(𝑎₁𝑥+𝑎₀)/(𝑥−𝑎₁) and 𝑎₀+𝑎²₁≠0, then(1)the expression of 𝑓^𝑛 is 𝑓^𝑛(𝑥)=𝑎₀+𝑎²₁^{(1−(−1)𝑛)/2}𝑥−𝑎₁^(−1)𝑛+𝑎₁,𝑛∈ℤ⁺ (3.19)(2)𝑓 has iterative roots 𝑓 of any odd order 𝑛, that is, 𝑓^𝑛=𝑓, where 𝑛=2𝑚−1,𝑚∈ℤ⁺.

4. Examples

We demonstrate our theorems with the following examples.

Example 4.1. Consider the fractional polynomial function 𝑓(𝑥)=(𝑥³+6𝑥²+12𝑥+8)/((9/2)𝑥³−3𝑥²−6𝑥−4), by Theorem 2.1, we know 𝑘=3,𝑎₀=8,𝑏=−5, thus iteration of order 𝑛 of the fractional polynomial functions𝑓: 𝑓^𝑛(𝑥)=1−1/2−(5/8)^{(1−(−3)𝑛)/4}(−1/2−1/𝑥)^(−3)𝑛. (4.1) By Theorem 3.3, we know 𝑓 has iterative roots of any odd order𝑛: 𝜑(𝑥)=1−1/2−(5/8)^{(1−(−3)1/𝑛)/4}(−1/2−1/𝑥)^(−3)1/𝑛, (4.2) where 𝑛∈ℤ⁺.

Example 4.2. Consider the fractional polynomial function 𝑓(𝑥)=(𝑥²+2𝑥+1)/(−2𝑥−1), by Theorem 2.1, we know 𝑘=2,𝑎₀=1,𝑏=−1, thus iteration of order 𝑛 of the fractional polynomial functions𝑓: 𝑓^𝑛(𝑥)=1−1−(−1)^{(1−(−2)𝑛)/3}(−1−1/𝑥)^(−2)𝑛. (4.3) By Theorem 3.3, we know 𝑓 has iterative roots of any odd order𝑛: 𝜑(𝑥)=1−1−(−1)^{(1−(−2)1/𝑛)/3}(−1−1/𝑥)^(−2)1/𝑛, (4.4) where 𝑛∈ℤ⁺.

Example 4.3. Consider the fractional polynomial function 𝑓(𝑥)=(2𝑥³−12𝑥²+24𝑥−13)/(𝑥³−6𝑥²+12𝑥−8), by Theorem 2.2, we know 𝑘=3,𝑎₀=−13,𝑎₃=2,𝑑=3, thus iteration of order 𝑛 of the fractional polynomial functions𝑓: 𝑓^𝑛(𝑥)=3^{(1−(−3)𝑛)/4}(𝑥−2)^(−3)𝑛+2. (4.5) By Theorem 3.6, we know 𝑓 has iterative roots of any odd order𝑛: 𝐹(𝑥)=3^{(1−(−3)1/𝑛)/4}(𝑥−2)^(−3)1/𝑛+2, (4.6) where 𝑛∈ℤ⁺.

Acknowledgment

This work is supported by Scientific Research Fund of Sichuan Provincial Education Department under Grant 11ZA160.