# Periodic Points of Asymmetric Bernoulli Shifts

Authors: Yong-Guo Shi; Kai Chen; Wei Liao
Journal: Journal of Mathematics
 (2021)

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## Abstract

It is well-known that Sharkovskii’s theorem gives a complete structure of periodic order for a continuous self-map on a closed bounded interval. As a further study, a natural problem is how to determine the location and number of periodic points for a specific map. This paper considers the periodic points of asymmetric Bernoulli shift, which is a piecewise linear chaotic map.


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## Body

## 1. Introduction

In 1964, Sharkovskii [1] firstly introduced a special ordering on the set of positive integers. This ordering implies that if p⊲q and a continuous self-map of a closed bounded interval has a point of period p; then it has a point of period q. The least number with respect to this ordering is 3. Thus, if a map has a point of period 3, then it has points of any periods. In 1975, the latter result was rediscovered by Li and Yorke [2]. Then numerous papers are devoted to the study of interval maps (see e.g., [3–5] and references therein).Bifurcation points of some interval maps were studied in [6], and the limit behavior of orbits and probabilistic some problems were considered in [7, 8]. Recently, Ivanov in [9] considered an exact lower bound for the number of orbits of a given period for a self-map of a closed bounded interval.Consider the asymmetric Bernoulli shiftF:0,1⟶0,1 with a parameter 0<a<1, defined by(1)Fx=xa,0≤x≤a,x−a1−a,a<x≤1.Specially, whena=1/2, it is the Bernoulli Shift or the binary transformation, also known as doubling map or the binary transformation. Conjugacies between asymmetric Bernoulli shifts are constructed in [10].Given a positive integern, one interesting question is how to find all n-periodic points of F. The other is how many n-periodic points of F.In this paper, we study periodic orbits ofF. In the next section, we present dynamics of jumps of Fn. Section 3 recalls the real number representation, i.e, F-expansion. In Section 4, we use the F-expansion to give explicit formulas of Fnx for n∈ℕ, explicit formulas of jumps of Fnx, explicit formulae of fixed points of Fnx, and explicit formulas of all n-periodic points of Fx. The last section gives the number hn of periodic orbits of a given period n for F and the limit behavior of hn.


## 2. Dynamics of Jumps ofFn

Forn∈ℕ, let Fnx denote the n-th iterate of F, which is recursively defined by F0x=x and Fn=FFn−1x for x∈0,1.A pointc∈0,1 is called a jump of F if the one-sided limits, Fc− and Fc+, exist and are finite, but are not equal. The set of jumps of F is denoted by JF. One can see that(2)JF⊆JF2⊆⋯⊆JFn⊆JFn+1⊆⋯.Each element ofJFn+1\JFn must be a preimage under F of a point from JFn. More precisely,(3)JFn+1\JFn=F−1JFn\JFn.The mapF has the unique jump a. Put x1,0:=0, x1,1:=a, and x1,2:=1. Let I denote the unit interval 0,1, I1,1:=x1,0,x1,1, and I1,2:=x1,1,x1,2. One can see that Fn has 2n−1 jumps for n≥2 by induction. For i,j∈ℕ+, let xi,0:=0,xi,2i:=1, and xi,j denote the j th jumps of Fi in the following order:(4)0=xi,0<xi,1<xi,2<⋯<xi,j<⋯<xi,2i−1<xi,2i=1.PutIi,j:=xi,j−1,xi,j for every j∈1,2,3,…,2i. It is clear that Ii,j is the j-th monotonic interval of Fi.Lemma 1.
Forn≥1, the jumps of Fn and Fn−1 have the following relationship:(i)
Fxn,k=Fxn,2n−1+k=xn−1,kfor1≤k≤2n−1−1(ii)
xn,2n−1=xn−1,2n−2=a(iii)
FiIn,k=FiIn,2n−1+k=In−i,kfor1≤k≤2n−iand1≤i≤n−1Proof.
We first claim thata is a jump of Fn for every n≥1. In fact, since a is a jump of Fx, a is also a jump of Fnx for n≥2. Moreover, it is easy to check that Fna=1 for n≥2.
Next, we prove (i) and (ii) by induction. It is clear that these results holds forn=2.
Assume that these results hold forn=m≥2, i.e.,(i)
Fxm,k=Fxm,2m−1+k=xm−1,k for 1≤k≤2m−1−1(ii)
xm,2m−1=xm−1,2m−2=a
Now we shall prove these results hold forn=m+1. Denote 2m−1 jumps of Fm by(5)0<xm,1<xm,2<⋯<xm,k<⋯<xm,2m−1<1.
SinceF is strictly increasing on the subinterval I1,1 and FI1,1=0,1, for each k∈1,2,…,2m−1, there exists the unique point, denoted by xm+1,k, in I1,1 such that Fxm+1,k=xm,k. Since F is strictly increasing on I1,1, one can see that(6)0<xm+1,1<xm+1,2<⋯<xm+1,k<⋯<xm+1,2m−1<a.
Further, by the definition of jump,xm+1,k is a jump of Fm+1=Fm∘F for each k∈1,2,…,2m−1.
Similarly, sinceF is strictly increasing on the subinterval I1,2 and FI1,2=0,1, for each k∈1,2,…,2m−1, there exists the unique point, denoted by xm+1,2m+k, in I1,2 such that Fxm+1,2m+k=xm,k. Since F is strictly increasing on I1,2, one can see that(7)a<xm+1,2m+1<xm+1,2m+2<⋯<xm+1,2m+k<⋯<xm+1,2m+1−1<1.
Further, by the definition of jump,xm+1,2m+k is a jump of Fm+1=Fm∘F for each k∈1,2,…,2m−1. Let xm+1,2m denote a. Therefore,(i)
Fxm+1,k=Fxm+1,2m+k=xm,k for 1≤k≤2m−1(ii)
xm+1,2m=xm,2m−1=a
It follows from (i) that for1≤k≤2n−1,(8)FIn,k=FIn,2n−1+k=In−1,k.
Then for1≤k≤2n−i and 1≤i≤n−1,(9)FiIn,k=FiIn,2n−1+k=In−i,k.
This completes the proof.


## 3.F-Expansion

In this section, we will introduce a new real number representation.Definition 1.
A sequenceεkk∈ℕ+ of 0 and 1 is called the itinerary of x∈0,1 with respect to the asymmetric Bernoulli shift F:0,1⟶0,1 and a∈0,1, if, for k≥1,(10)εk=0,Fk−1x≤a,1,Fk−1x>a.In fact, the itinerary ofx∈0,1 with respect to F and a∈0,1 is just the F-expansion of a real x∈0,1. According to [10], or these two classic papers [11, 12], we have an expansion for x in powers of the numbers a and 1−a:(11)x=∑k=1∞εkak−sk−11−ask−1=∑k=1∞εkak1−aask−1,where s0=0 and sk:=∑j=1kεj for k≥1.Thus, everyx∈0,1 can be represented through its digit sequence εkk∈ℕ+. In this situation, write x=ε1,ε2,…,εk,… for short. One can see that every infinite F-expansion is unique, whereas each x∈0,1 with a finite F-expansion can be expanded in exactly two ways, namely, one immediately verifies that(12)x=ε1,…,εk−1,1=ε1,…,εk−1,0,1,1,1,….In the following, we employ a convention in which finite fractions such as(13)1,1,0,0,0,0,…=1,0,1,1,1,1,…,1,0,1,0,0,0,…=1,0,0,1,1,1,…,are represented as finite fractions with infinite zeros, as 1,1,0,0,0,… or 1,0,1,0,0,0,…, unless otherwise stated.Lemma 2.
Ifx=ε1,ε2,…,εk,…, then Fx=ε2,ε3,…,εk,….Proof.
It follows from a property of the asymmetric Bernoulli shiftFx that x≤a provided that ε1=0 in x=ε1,ε2,…,εk,…. One then finds(14)Fx=xa=∑k=1∞εkak−11−aask−1=∑k=2∞εkak−11−aask−1=∑k=1∞εk+1ak1−aask=ε2,ε3,…,εk,….
On the other hand, one hasa<x≤1 if ε1=1 and hence(15)Fx=x−a1−a=∑k=2∞εkak−sk−11−ask−1−1=∑k=1∞εk+1ak+1−sk1−ask−1=ε2,ε3,…,εk,….This shows that, from the perspective of symbolic dynamics,F corresponds to the shift map on the space 0,1ℕ+, at least for those points with an infinite F-expansion.One easily finds that the periodicity of the orbits is related to recurringF-expansions. For example,(16)1,0,1,1,1,1,0,1,1,1,…,is a recurring F-expansion with the recurring unit of the length 5, and hence, it is a 5-periodic point of F.


## 4. The Explicit Formula ofFn

SinceFn is a piecewise linear map, and Fn is strictly increasing on each subinterval In,k. One can obtain the explicit formula of Fn.Theorem 1.
Ifx=ε1,ε2,…,εk,…, then(17)Fnx=xan−sn1−asn−∑j=1nεjan−j−sn+sj−11−asn−sj+1,forn∈ℕ+.Proof.
We prove this result by mathematical induction.
We firstly consider the trivial casen=1. If x∈I1,1, then ε1=0 and s1=0. Thus,(18)Fx=xa1−s11−as1−∑j=11εja1−j−s1+sj−11−as1−sj+1=xa.
Ifx∈I1,2, then ε1=1 and s1=1. Thus,(19)Fx=xa1−s11−as1−∑j=11εja1−j−s1+sj−11−as1−sj+1=x1−a−a1−a.
Therefore, the result holds forn=1.
Assume that the result holds forn=m≥1, i.e.,(20)Fmx=xam−sm1−asm−∑j=1mεjam−j−sm+sj−11−asm−sj+1.
Now we shall prove that the result holds forn=m+1. If Fmx≤a; then εm+1=0 and sm+1=sm. Thus,(21)Fm+1x=1a⋅xam−sm1−asm−∑j=1mεjam−j−sm+sj−11−asm−sj+1=xam+1−sm+11−asm+1−∑j=1m+1εjam+1−j−sm+1+sj−11−asm+1−sj+1.
IfFmx>a; then εm+1=1 and sm+1=sm+1. Thus,(22)Fm+1x=11−a⋅xam−sm1−asm−∑j=1mεjam−j−sm+sj−11−asm−sj+1−a1−a=xam−sm1−asm+1−∑j=1mεjam−j−sm+sj−11−asm−sj+1+1−a1−a=xam+1−sm+11−asm+1−∑j=1mεjam+1−j−sm+1+sj−11−asm+1−sj+1−a1−a=xam+1−sm+11−asm+1−∑j=1m+1εjam+1−j−sm+1+sj−11−asm+1−sj+1.
Therefore, the result holds forn=m+1. The proof is completed.As a corollary, we present the exact formulas of these jumps ofFn.Corollary 1.
All jumps ofFn are given by(23)∑j=1nεjaj+1−sj1−asj−1,where εj=0 or 1, and not all are εj equal to 0.Proof.
If allεj are zero, then x1,0=0, and it is not a jump.
From Theorem2, solving Fnx=0, we can obtain all these jumps of Fn.Definition 2.
A pointx in X is called a periodic point of a self-mapping f:X⟶X if there exists an positive integer n such that(24)fnx=x.
The smallest positive integern satisfying the above is called the prime period or least period of the point x, the point x is called an n-periodic point of f, and the sequence x,fx,…,fn−1x is called an n-periodic orbit.
In particularly, an 1-periodic point is called a fixed point.The following corollary presents the exact formulas of all fixed points ofFn.Corollary 2.
All fixed points ofFn are given by(25)11−an−sn1−asn∑j=1nεjaj+1−sj1−asj−1,where εj=0 or 1.Proof.
Since the curve ofy=Fnx intersects the line of y=x at 2n points, Fnx has 2n fixed points. Solving Fnx=x, we have(26)x=11−an−sn1−asn∑j=1nεjaj+1−sj1−asj−1.


## 5. The Number ofn-Periodic Points

The fixed points ofF are the intersections of y=Fx and y=x, namely, two points x=0,1. The intersections of y=F2x and y=x have four points where there are two 2-periodic points, namely, two points(27)a21−a+a2,a1−a+a2.The other two intersectionsx=0 and 1 are the fixed points. The intersections of y=F3x and y=x have eight points where there are six 3-periodic points and two fixed points.In general, the intersections ofy=Fnx and y=x have 2n periodic points. If x is a p-periodic point of F, then p|n. Let hp denote the number of the p-periodic points. Then,(28)∑p|nhp=2n,for every integern≥1,where the sum extends over all positive divisors p of n.In order to obtain the exact numberhn of n-periodic points of F, we need to introduce the Möbius function and Möbius inversion formula (see, for example, [13, 14]).Define the Möblius functionμ:ℕ⟶1,0,1 by(29)μn=1,n=1,−1r,n=q1,q2,…,qr,q1<q2<⋯<qr,0,others.Thus, ifn1,n2=1, then μn1n2=μn1μn2, and for any n∈ℕ, there holds(30)∑k|nμk=1n.Lemma 3 (Möbius inversion formula).
Ifh and g are arithmetic functions, i.e., from ℕ to ℂ, satisfying(31)gn=∑d|nhd,for every integern≥1.Then,(32)hn=∑d|nμdgnd,for every integern≥1,where μ is the Möbius function and the sums extend over all positive divisors d of n.In effect, the originalhn can be determined given gn by using the inversion formula.Corollary 3.
The number ofn-periodic points of F is given by,(33)hn=∑d|nμnd2d,for every integern≥1.LetNn denote the number of n-periodic orbits of F. Then,(34)Nn=hnn=1n∑d|nμnd2d,for every integern≥1.The first severalNn are(35)N1=2,N2=1,N3=2,N4=3,N5=6,N6=9,N7=18,N8=30,while Nn for larger n are(36)N16=4080,N20=52377,N32=134215680,N64=288230376084602880.If the values just above are compared to(37)21616=4096,22020=52428.8,23232=134217728,26464=288230376151711744,one finds that the ratio of Nn and 2n/n approaches 1 as n⟶+∞.Now we shall prove that the ratio ofhn and 2n approaches 1 as n⟶+∞.Theorem 2.
Lethp be the number of the p-periodic points of F. Then,(38)limn⟶+∞hn2n=limn⟶+∞∑d|nμn/d2d2n=1,where μ is the Möbius function.Proof.
On one hand,(39)∑d|nμnd2d≤∑d|n2d=2n+∑d|n,d≠n2d≤2n+∑1≤d≤n/22d≤2n+2n/2+1−2.
On the other hand,(40)∑d|nμnd2d=2n+∑d|n,d≠nμnd2d≥2n−∑d|n,d≠n2d≥2n−∑1≤d≤n/22d≥2n−2n/2+1+2.
Consequently,(41)1=limn⟶+∞2n−2n/2+1+22n≤limn⟶+∞∑d|nμn/d2d2n≤limn⟶+∞2n+2n/2+1−22n=1.
By the squeeze theorem,(42)limn⟶+∞hn2n=limn⟶+∞∑d|nμn/d2d2n=1.



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