On the Eventually Periodic Continued β-Fractions and Their Lévy Constants

Abstract

In this paper, we consider continued $\beta$-fractions with golden ratio base $\beta$. We show that if the continued $\beta$-fraction expansion of a non-negative real number is eventually periodic, then it is the root of a quadratic irreducible polynomial with the coefficients in $\mathbb{Z}\left[\beta \right]$ and we conjecture the converse is false, which is different from Lagrange’s theorem for the regular continued fractions. We prove that the set of Lévy constants of the points with eventually periodic continued $\beta$-fraction expansion is dense in [$\mathbf{c}$, +∞), where $\mathbf{c}=\frac{1}{2}log\frac{\beta +2-\sqrt{5\beta +1}}{2}$.

1. Introduction

As early as the seventeenth century, many scientists have studied the representation of numbers, such as continued fractions and decimal expansions. Among them, continued fractions have a long standing tradition of “noble service” to number theory. Consider the transformation G defined over $[0,1)\to [0,1)$ by

$$G\left(x\right)=\frac{1}{x}-\lfloor \frac{1}{x}\rfloor ,x\in (0,1)\mathrm{and}G\left(0\right)=0,$$

(1)

where $\lfloor t\rfloor$ denotes the greatest integer less than or equal to t. The continued fraction of a real number x has the following recursive form:

$$x=a_0\left(x\right)+\frac{1}{a_1\left(x\right)+\frac{1}{a_2\left(x\right)+\frac{1}{\ddots }}}:=[a_0\left(x\right);a_1\left(x\right),a_2\left(x\right),\dots ],$$

(2)

where $a_0\left(x\right)=\lfloor x\rfloor$ and $a_i\left(x\right)=\lfloor \frac{1}{G^{i-1}\left(x\right)}\rfloor$ for $i\ge 1$. For any $n\ge 1$, we denote by $\frac{p_n\left(x\right)}{q_n\left(x\right)}$ the canonical representation of the segment $[a_0\left(x\right);a_1\left(x\right),a_2\left(x\right),\dots ,a_n\left(x\right)]$ of the continued fraction, where $p_n\left(x\right)$ and $q_n\left(x\right)$ are relatively coprime integers. We say $\frac{p_n\left(x\right)}{q_n\left(x\right)}$ is the nth convergent of the continued fraction expansion of x.

For a real number x, considering the pointwise asymptotic behavior of the sequence $\{q_n\left(x\right),n\ge 1\}$, if

$$L\left(x\right)=\underset{n\to \infty }{lim}\frac{log{q}_n\left(x\right)}{n}$$

exists, such limit is called the Lévy constant of x, denoted by $L\left(x\right)$. Notice that if x has a Lévy constant, then $L\left(x\right)\ge log\frac{\sqrt{5}+1}{2}$. Lévy [1] has obtained the following result:

$$\underset{n\to \infty }{lim}\frac{log{q}_n\left(x\right)}{n}=\frac{{\pi }^2}{12log2}$$

for almost all $x\in \mathbb{R}$ in the sense of Lebesgue measure.

Lévy constant is tied to the quadratic irrational. It is known that, for a quadratic irrationality, the Lévy constant exists (see [2]). A classical result due to Lagrange asserts that: the continued fraction expansion of a positive real number x is a quadratic irrational if and only if it is eventually periodic (see [3,4,5]). That is,

$$x=[a_0;a_1,\dots ,a_m,\overline{a_{m+1},\dots ,a_{n+m}}],$$

(3)

where $\overline{a_{m+1},\dots ,a_{m+n}}$ denotes the periodic sequence. Lévy constant characterizes the mean value of the elements $a_{m+1},\dots ,a_{m+n}$ in the periodic part of expansion (3).

Many results on the Lévy constant have been studied by some authors, see [2,6,7,8,9]. Some other limit theorems of $\{q_n\left(x\right),n\ge 1\}$ in the continued fraction expansion of real numbers have been extensively studied. For example, Misevičius [10] got the corresponding rate of convergence and the central limit theorem of $\{q_n\left(x\right),n\ge 1\}$ was provided by [11]. We can also see some other results in [12,13].

Naturally, one introduced a similar algorithm of continued fractions by changing the decimal base by a real number $\beta$ to obtain a new expansion of real numbers, the continued $\beta$-fractions, where the sequence of partial quotients consists of $\beta$-integers (see [14]). We are interested in the connection among the numbers with eventually periodic continued $\beta$-fraction expansion, quadratic irrational, and the corresponding Lévy constant.

We recall the notion about $\beta$-expansion necessary for defining $\beta$-integer. $\beta$-expansion was introduced by Rényi [15] in 1957 and is a new approach to the theory of expansions. Let $\beta >1$ be a real number. Define the $\beta$-transformation $T_{\beta }:[0,1)⟶[0,1)$ as follows

$$T_{\beta }\left(x\right)=\beta x-\lfloor \beta x\rfloor =\left\{\beta x\right\}, x\in [0,1)$$

where $\left\{\beta x\right\}$ is the fraction part of $\beta x$. Let $x_i$ = $\lfloor \beta T_{\beta }^{i-1}\left(x\right)\rfloor$, this leads to $\beta$-expansion of x

$$x=\sum _{i\ge 1}\frac{x_i}{{\beta }^i}.$$

If $\beta \in \mathbb{N}$, the $\beta$-expansion of x is eventually periodic if and only if $x\in \mathbb{Q}$. For a Pisot number $\beta >1$, the $\beta$-expansion of x is eventually periodic if and only if $x\in \mathbb{Q}\left(\beta \right)\cap [0,1)$, where $\mathbb{Q}\left(\beta \right)$ is the smallest field containing $\mathbb{Q}$ and $\beta$, see [16,17].

Any $x>1$ can be uniquely expanded as $x={\sum }_{k=0}^n{v}_{-k}{\beta }^k+{\sum }_{k=1}^{\infty }{v}_k{\beta }^{-k}$. The sum of which consists of non-negative powers of $\beta$ is called $\beta$-integer part of x and is denoted by ${\lfloor x\rfloor }_{\beta }$. The sum of negative powers of $\beta$ is called $\beta$-fraction part of x and is denoted by ${\left\{x\right\}}_{\beta }$.

The real number $x\ge 0$, such that $x={\lfloor x\rfloor }_{\beta }$, is called non-negative $\beta$-integer, which is a generalization of the notion of non-negative integers and plays a fundamental role in the theory of continued $\beta$- fraction. The set of non-negative $\beta$-integers is denoted by ${\mathbb{Z}}_{\beta }^{+}$. That is,

$${\mathbb{Z}}_{\beta }^{+}=\{\mathrm{all} \beta -\mathrm{integers}\}=\{\xi :\exists x\ge 0, \mathrm{s.t.} \xi ={\lfloor x\rfloor }_{\beta }\}.$$

When $\beta =\frac{\sqrt{5}+1}{2}$, which is the root of the equation $x^2-x-1=0$, we know

$${\mathbb{Z}}_{\beta }^{+}=\{m+n\beta |m,n\in \mathbb{Z},m,n\ge 0,-1<m-\frac{n}{\beta }<\beta \}$$

(see [18]), and then obviously ${\mathbb{Z}}_{\beta }^{+}$ is a subset of $\mathbb{Z}\left[\beta \right]:=\{m+n\beta |m,n\in \mathbb{Z}\}$. Easily, we can obtain that ($\mathbb{Z}\left[\beta \right],+,\times )$ is a ring, where $+,\times$ are usual addition and multiplication in $\mathbb{R}$.

Definition 1 (Continued $\beta$-fraction transformation).

Define $T:[0,1)⟶[0,1)$, as

$$T\left(0\right):=0\mathit{and}T\left(x\right):=1/x-{\lfloor 1/x\rfloor }_{\beta }\mathit{if}x\in (0,1),$$

(4)

where ${\lfloor x\rfloor }_{\beta }$ is the largest β-integer not exceeding x.

Then the continued $\beta$-fraction of a non-negative real number x has the following form:

$$\begin{array}{cc}\hfill x& ={\lfloor x\rfloor }_{\beta }+\frac{1}{{\lfloor \frac{1}{x}\rfloor }_{\beta }+\frac{1}{\lfloor \frac{1}{T\left(x\right)}{\rfloor }_{\beta }+\ddots +\frac{1}{\lfloor \frac{1}{T^{k-1}\left(x\right)}{\rfloor }_{\beta }+T^k\left(x\right)}}}:=[a_0;a_1,a_2,\dots ,a_k+T^k\left(x\right)],\hfill \end{array}$$

or

$$\begin{array}{cc}\hfill x& ={\lfloor x\rfloor }_{\beta }+\frac{1}{{\lfloor \frac{1}{x}\rfloor }_{\beta }+\frac{1}{\lfloor \frac{1}{T\left(x\right)}{\rfloor }_{\beta }+\ddots }}:=[a_0;a_1,a_2,\dots ],\hfill \end{array}$$

where $a_0={\lfloor x\rfloor }_{\beta }$, $a_i=\lfloor \frac{1}{T^{i-1}\left(x\right)}{\rfloor }_{\beta }$ for any $i\ge 1$. For the case $\beta =\frac{\sqrt{5}+1}{2}$, Bernat [14] studied the real numbers with finite expansions, more precisely, he has proven that the continued $\beta$-fraction of x is finite if and only if $x\in \mathbb{Q}\left(\beta \right)$. Feng et al. [19] proved that the metric properties of continued $\beta$-fraction expansion. For more results on the continued $\beta$-fraction with Laurent series in finite fields, we refer the reader to [20,21].

Our first result shows that any number with eventually periodic continued $\beta$-fraction expansion is a root of an irreducible quadratic polynomial with coefficients in $\mathbb{Z}\left[\beta \right]$.

Theorem 1.

Let $\beta =\frac{\sqrt{5}+1}{2}$. If the continued β-fraction expansion of a positive number u is eventually periodic, then there exists an irreducible quadratic polynomial $a{x}^2+bx+c=0$, where $a,b,c\in \mathbb{Z}\left[\beta \right]$ with solution u.

The proof of Theorem 1 can be found in Section 3.

Remark 1.

When $\beta =10$, $\mathbb{Z}\left[\beta \right]=\mathbb{Z}$, then the continued β-fraction expansion becomes the regular continued fraction expansion. Analog to Lagrange’s Theorem in a regular continued fraction, a natural question is whether the converse of Theorem 1 holds. We conjecture that it is not true for $\beta =\frac{\sqrt{5}+1}{2}$; that is, there exists a positive number, which is a solution of irreducible quadratic polynomial, but its continued β-fraction expansion is not eventually periodic. We still cannot construct such number. The conjecture is based on the fact that $\mathbb{Z}\left[\beta \right]$ is dense in $\mathbb{R}$, which is different from the case $\beta =10$, where $\mathbb{Z}$ is isolated from $\mathbb{R}$.

Similar to the regular continued fraction expansion, we can define the denominator $q_n^{\beta }\left(x\right)$ of the convergent of x with respect to the continued $\beta$-fraction expansion (the detailed definition will be given in Section 2). Define the Lévy constant of x by $L\left(x\right)$ := $\underset{n\to \infty }{lim}\frac{1}{n}log{q}_n^{\beta }\left(x\right)$ if the limit exists. Let $\beta =\frac{\sqrt{5}+1}{2}$ and

$$\mathbf{B}=\left\{L\right(x):\mathrm{the\; continued} \beta -\mathrm{fraction\; expansion\; of} x \mathrm{is\; eventually\; periodic}\}.$$

Theorem 2.

The set $\mathbf{B}$ is dense in $[\mathbf{c},+\infty )$, where $\mathbf{c}=\frac{1}{2}log\frac{\beta +2-\sqrt{5\beta +1}}{2}$.

We give the proof of Theorem 2 in Section 4.

Remark 2.

In the classical case ($\beta =10$), similarly with Theorem 2, Wu [8] has proven that the set of corresponding Lévy constants of quadratic irrationals is dense in $[log\frac{\sqrt{5}+1}{2},\infty )$.

2. Convergents and Admissibility

Definition 2 (the rule for the formation of the convergents).

For arbitrary $k\ge 1$, $a_i\in {\mathbb{Z}}_{\beta }^{+}$ for any $i\ge 1$ and $a_0\in {\mathbb{Z}}_{\beta }^{+}\bigcup \left\{0\right\}$, let $p_{-1}=1,q_{-1}=0,p_0=a_0$, $q_0=1$ and define

$$\begin{array}{cc}\hfill & p_k=a_k{p}_{k-1}+p_{k-2},\hfill \\ \hfill & q_k=a_k{q}_{k-1}+q_{k-2}.\hfill \end{array}$$

(5)

Remark 3.

Note that $p_k$ and $q_k$ depend on β and $a_0,a_1,\dots ,a_k$, it is more rigorous to write $p_k^{\beta }(a_0,a_1,\dots ,a_k)$ and $q_k^{\beta }(a_0,a_1,\dots ,a_k)$ instead of $p_k$ and $q_k$. However hereafter, in order to avoid burdening of notation, we prefer to use the notation $p_k$ and $q_k$.

Definition 3 (Admissibility).

A finite sequence $(a_1,a_2,\dots ,a_n)\in {({\mathbb{Z}}_{\beta }^{+})}^n$ is called admissible, if there exists $x\ge 0$, such that $a_i\left(x\right)=a_i$ for any $i=1,2,\dots ,n$. Moreover, a sequence $(a_1,a_2,\dots )\in {({\mathbb{Z}}_{\beta }^{+})}^{\mathbb{N}}$ is called admissible, if there exists $x\ge 0$, such that $a_i\left(x\right)=a_i$ for all $i\in \mathbb{N}$.

Propositions 1 and 2 are well known, their proofs are contained in [22] (Theorem 2.1).

Proposition 1.

For any sequence $(a_0,a_1,a_2,\dots )$ with $a_i\in$${\mathbb{Z}}_{\beta }^{+}$ or any $i\ge 1$ and $a_0\in {\mathbb{Z}}_{\beta }^{+}\bigcup \left\{0\right\}$. Then for any $n\in \mathbb{N}$,

$$\frac{p_n}{q_n}=[a_0;a_1,a_2\dots ,a_n]$$

and $p_n,q_n\in \mathbb{Z}\left[\beta \right]$.

Proposition 2.

For all $k\ge 0$,

$$p_k{q}_{k-1}-q_k{p}_{k-1}={(-1)}^{k-1}.$$

Let ${\mathbb{Z}}_{\beta }^{+}=\{k_0<k_1<k_2<\dots <k_i<\dots \}$. Then $k_{i+1}-k_i=\beta -1$ or 1. Moreover, ${T|}_{(\frac{1}{k_{i+1}},\frac{1}{k_i})}=[0,1)$, for $k_{i+1}-k_i=1$; ${T|}_{(\frac{1}{k_{i+1}},\frac{1}{k_i})}=[0,\beta -1)$, for $k_{i+1}-k_i=\beta -1$. Therefore, we can sieve the points $k_i$ with $k_{i+1}-k_i=\beta -1$ from ${\mathbb{Z}}_{\beta }^{+}$, denoted by ${\mathsf{\Lambda }}_{\beta }$. That is to say,

$${\mathsf{\Lambda }}_{\beta }=\{k_i\in {\mathbb{Z}}_{\beta }^{+}:k_{i+1}-k_i=\beta -1\}.$$

Thus, ${\mathsf{\Lambda }}_{\beta }\subset {\mathbb{Z}}_{\beta }^{+}$.

Remark 4.

In fact, ${\mathsf{\Lambda }}_{\beta }=1+{\beta }^2{\mathbb{Z}}_{\beta }^{+}$, since any β-integer whose expansion ends with the digit 1 is the image of an element of ${\mathbb{Z}}_{\beta }^{+}$ under the map $x\mapsto \beta x^2+1$.

Proposition 3

([19]). The finite sequence $(a_1,a_2,\dots ,a_n)\in {\left({\mathbb{Z}}_{\beta }^{+}\right)}^n$ is admissible if and only if either $a_k\in {\mathsf{\Lambda }}_{\beta }$, but $a_{k+1}\ne 1$ or $a_k\notin {\mathsf{\Lambda }}_{\beta }$, $1\le k\le n-1$.

Proposition 4.

The sequence $(a_1,a_2,\dots )\in {\left({\mathbb{Z}}_{\beta }^{+}\right)}^{\mathbb{N}}$ is admissible if and only if either $a_k\in {\mathsf{\Lambda }}_{\beta }$ but $a_{k+1}\ne 1$ or $a_k\notin {\mathsf{\Lambda }}_{\beta }$, for $k\ge 1$.

Proof. 

The proof of the ‘only if’ part is similar to Proposition 3; thus, we omit its proof.

For the ‘if’ part, let $(a_1,a_2,\dots )\in {({\mathbb{Z}}_{\beta }^{+})}^{\mathbb{N}}$, which satisfies either $a_k\in {\mathsf{\Lambda }}_{\beta }$ but $a_{k+1}\ne 1$ or $a_k\notin {\mathsf{\Lambda }}_{\beta }$, $k\ge 1$. Denote $x={\displaystyle \frac{1}{a_1+{\displaystyle \frac{1}{a_2+\ddots }}}}$. By the induction and Proposition 3, we conclude that $(a_1,a_2,\dots )$ is admissible. □

Remark 5.

Proposition 4 tells us that not any sequence in ${({\mathbb{Z}}_{\beta }^{+})}^{\mathbb{N}}$ is admissible for continued β-fraction expansion and the sequence $\left(k_{i}1\right)$$(k_i\in {\mathsf{\Lambda }}_{\beta })$ is not admissible, which is different from the regular continued fraction ($\beta =10$).

3. Eventually Continued β-Fractions

Denote by $\mathbf{A}$ the set of real numbers, which are the non-negative roots of quadratic polynomials with coefficients in $\mathbb{Z}\left[\beta \right]$. That is to say, $\mathbf{A}=\{x\in {\mathbb{R}}^{+}:f\left(t\right)=a{t}^2+bt+c,f\left(t\right)$ is an irreducible polynomial, $f\left(x\right)=0,a,b,c\in \mathbb{Z}\left[\beta \right]\}$.

Proposition 5

([19]). For arbitrary $u\in {\mathbb{R}}^{+},k\ge 0$,

$$u=\frac{p_k\left(u\right)+p_{k-1}\left(u\right)T^k\left(u\right)}{q_k\left(u\right)+q_{k-1}\left(u\right)T^k\left(u\right)}.$$

The following is the proof of Theorem 1.

Proof of Theorem 1. 

Before the proof, we will give some notations: $u=[a_0;a_1,a_2,\dots ]$, $u_n=[a_n;a_{n+1},\dots ]$. Clearly, we notice that, if $u=[a_1,a_2,\dots ]$, then $u_{n+1}=\frac{1}{T^n\left(u\right)}$.

We divide the proof into two cases according to the purely periodic or eventually periodic of the continued $\beta$-fraction expansion of u.

Assume first that $u=\left[\overline{a_0;a_1,\dots ,a_k}\right]$ has a purely periodic continued $\beta$-fraction, so that $u_{k+1}=u_0=u$.

Then,

$$\begin{array}{c}\hfill u=\frac{p_k+p_{k-1}{T}^k\left(u\right)}{q_k+q_{k-1}{T}^k\left(u\right)}=\frac{p_k+p_{k-1}\frac{1}{u_{k+1}}}{q_k+q_{k-1}\frac{1}{u_{k+1}}}=\frac{u_{k+1}{p}_k+p_{k-1}}{u_{k+1}{q}_k+q_{k-1}}=\frac{u{p}_k+p_{k-1}}{u{q}_k+q_{k-1}},\end{array}$$

where the first equality can be obtained from Proposition 5, and the second equality is due to $u_{n+1}=\frac{1}{T^n\left(u\right)}$. So

$$u^2{q}_k+u(q_{k-1}-p_k)-p_{k-1}=0.$$

(6)

Now we show that the quadratic polynomial (6) is irreducible by contradiction. Suppose there exist $a,b\in \mathbb{Z}\left[\beta \right]$ such that $au+b=0$ then $u\in \mathbb{Q}\left(\beta \right)$. Therefore, the continued $\beta$-fraction expansion of u is finite. It is contradicted. Then $u\in \mathbf{A}$.

Now assume that:

$u=[a_0;a_1,\dots ,a_{N-1},\overline{a_N,\dots ,a_{N+k}}]$, then $u_N=\left[\overline{a_N;a_{N+1},\dots ,a_{N+k}}\right]$.

By Proposition 5, $u=\frac{u_N{p}_{N-1}+p_{N-2}}{u_N{q}_{N-1}+q_{N-2}}$, then we know $u_N=\frac{p_{N-2}-u{q}_{N-2}}{u{q}_{N-1}-p_{N-1}}$ and $\mathbb{Q}(u,\beta )=\mathbb{Q}(u_N,\beta )$ (where $\mathbb{Q}(u,\beta )$ is the smallest field containing $u,\mathbb{Q}$ and $\beta$). Noting that $u_N$ is purely periodic, from the case (1) that we proved $u_N\in \mathbf{A}$. Thus, there exists $a,b,c\in \mathbb{Z}\left[\beta \right]$, such that

$$a{u}_N^2+b{u}_N+c=0,$$

(7)

then we use $\frac{p_{N-2}-u{q}_{N-2}}{u{q}_{N-1}-p_{N-1}}$ to replace $u_N$ in (7). So, we get

$$R{u}^2+Su+T=0$$

(8)

where $R=a{q}_{N-2}^2-b{q}_{N-1}{q}_{N-2}+c{q}_{N-1}^2,S=-2a{p}_{N-2}{q}_{N-2}+b(p_{N-2}{q}_{N-1}+p_{N-1}{q}_{N-2})-2c{p}_{N-1}{q}_{N-1}$ and $T=a{p}_{N-2}^2-b{p}_{N-2}{p}_{N-1}+c{p}_{N-1}^2$. Now, we prove the quadratic polynomial (8) is irreducible by contradiction. Assume there exists $r,s\in \mathbb{Z}\left[\beta \right]$ such that $ru+s=0$ then $u\in \mathbb{Q}\left(\beta \right)$. Therefore, the continued $\beta$-fraction expansion of u is finite, which contradicts with the infiniteness of the continued $\beta$-fraction expansion of u. Then we have $u\in \mathbf{A}$. □

4. Lévy Constant

In this section, we write $q_n\left(x\right)$ as the denominator of the convergent of the continued $\beta$-fraction expansion of x. We call $\underset{n\to +\infty }{lim\; inf}\frac{log{q}_n\left(x\right)}{n}$ the lower-Lévy constant of x, denoted by $L_{*}\left(x\right)$ and we call $\underset{n\to +\infty }{lim\; sup}\frac{log{q}_n\left(x\right)}{n}$ the upper-Lévy constant of x, denoted by $L^{*}\left(x\right)$. If $L_{*}\left(x\right)=L^{*}\left(x\right)$, we call it the Lévy constant of x, denoted by $L\left(x\right)$.

Proposition 6.

(1) 

For any $n\ge 0$, ${\displaystyle \prod _{i=0}^n}{T}^i\left(x\right)={(-1)}^n(x{q}_n-p_n)$.

(2) 

For arbitrary $n\ge 0$, $q_n|x{q}_{n-1}-p_{n-1}|=\frac{q_n}{q_n+q_{n-1}{T}^n\left(x\right)}$.

(3) 

$\frac{1}{2}\le q_n|x{q}_{n-1}-p_{n-1}|\le 1$.

Proof. 

(1)

For any $x\in [0,1]\backslash \mathbb{Q}\left(\beta \right)$, $x=\frac{p_n+p_{n-1}{T}^n\left(x\right)}{q_n+q_{n-1}{T}^n\left(x\right)}$, then $T^n\left(x\right)=\frac{p_n-x{q}_n}{x{q}_{n-1}-p_{n-1}}$.

Therefore,

$$\begin{array}{c}x=\frac{p_0-x{q}_0}{x{q}_{-1}-p_{-1}}\\ T\left(x\right)=\frac{p_1-x{q}_1}{x{q}_0-p_0}\\ T^2\left(x\right)=\frac{p_2-x{q}_2}{x{q}_1-p_1}\\ ⋮\\ T^n\left(x\right)=\frac{p_n-x{q}_n}{x{q}_{n-1}-p_{n-1}}\end{array}$$

then ${\displaystyle \prod _{i=0}^n}{T}^i\left(x\right)={(-1)}^{n+1}(p_n-x{q}_n)={(-1)}^n(x{q}_n-p_n)$.

(2)

Since $x=\frac{p_n+p_{n-1}{T}^n\left(x\right)}{q_n+q_{n-1}{T}^n\left(x\right)}$, we can get

$$\begin{array}{cc}\hfill \left|x-\frac{p_{n-1}}{q_{n-1}}\right|& = \left|\frac{p_n+p_{n-1}{T}^n\left(x\right)}{q_n+q_{n-1}{T}^n\left(x\right)}-\frac{p_{n-1}}{q_{n-1}}\right|\hfill \\ & =\frac{\left|p_n{q}_{n-1}+p_{n-1}{q}_{n-1}{T}^n\left(x\right)-p_{n-1}{q}_n-p_{n-1}{q}_{n-1}{T}^n\left(x\right)\right|}{q_{n-1}(q_n+q_{n-1}{T}^n\left(x\right))}\hfill \\ & =\frac{\left|p_n{q}_{n-1}-p_{n-1}{q}_n\right|}{q_{n-1}(q_n+q_{n-1}{T}^n\left(x\right))}=\frac{1}{q_{n-1}(q_n+q_{n-1}{T}^n\left(x\right))}.\hfill \end{array}$$

Multiplying by $q_n{q}_{n-1}$ in the two sides of the above equality gives

$$q_n|x{q}_{n-1}-p_{n-1}|=\frac{q_n}{q_n+q_{n-1}{T}^n\left(x\right)}.$$

(3)

Since

$$q_n\left| x{q}_{n-1}-p_{n-1}\right|=\frac{q_n}{q_n+q_{n-1}{T}^n\left(x\right)}\ge \frac{q_n}{q_n+q_n{T}^n\left(x\right)}=\frac{1}{1+T^n\left(x\right)}\ge \frac{1}{2}$$

and

$$q_n\left| x{q}_{n-1}-p_{n-1}\right|=\frac{q_n}{q_n+q_{n-1}{T}^n\left(x\right)}\le \frac{q_n}{q_n}=1,$$

we have $\frac{1}{2}\le q_n|x{q}_{n-1}-p_{n-1}|\le 1$. □

Proposition 7.

If the continued β-fraction expansion of x is eventually periodic, so x has a Lévy constant.

Proof. 

By Proposition 6 (3), we can get

$$1\le \frac{1}{q_n|x{q}_{n-1}-p_{n-1}|}\le 2.$$

That is to say

$$|x{q}_{n-1}-p_{n-1}|\le \frac{1}{q_n}\le 2|x{q}_{n-1}-p_{n-1}|.$$

Taking the logarithm, we obtain

$$log|x{q}_{n-1}-p_{n-1}|\le -log{q}_n\le log2|x{q}_{n-1}-p_{n-1}|.$$

Evidently, the following inequality

$$\frac{log|x{q}_{n-1}-p_{n-1}|}{n}\le -\frac{log{q}_n}{n}\le \frac{log2+log|x{q}_{n-1}-p_{n-1}|}{n},$$

is true. So we only need to show $\underset{n\to \infty }{lim}\frac{log|x{q}_{n-1}-p_{n-1}|}{n}$ exists, for the x whose continued $\beta$-fraction is eventually periodic.

Now assume that: $x=[a_1,\dots ,a_k,\overline{a_{k+1},\dots ,a_{k+N}}]$. For any $n\ge k$, there exists $p\in \mathbb{N}$, such that $k+pN\le n<k+(p+1)N$,

$$\begin{array}{cc}& \frac{log|x{q}_{n-1}-p_{n-1}|}{n}=\frac{{\displaystyle \sum _{i=0}^{k-1}}log{T}^i\left(x\right)+p{\displaystyle \sum _{i=k}^{k+N-1}}log{T}^i\left(x\right)+{\displaystyle \sum _{i=k+Np}^{n-1}}log{T}^i\left(x\right)}{n}\hfill \\ & \le \frac{{\displaystyle \sum _{i=0}^{k-1}}log{T}^i\left(x\right)}{k+Np}+\frac{p{\displaystyle \sum _{i=k}^{k+N-1}}log{T}^i\left(x\right)}{k+Np}+\frac{{\displaystyle \sum _{i=k+Np}^{k+N(p+1)-1}}log{T}^i\left(x\right)}{k+Np},\hfill \end{array}$$

where the first equality can be obtained from Proposition 6 (1). Notice that,

$${\displaystyle \sum _{i=0}^{k-1}}log{T}^i\left(x\right)+{\displaystyle \sum _{i=k+Np}^{k+N(p+1)-1}}log{T}^i\left(x\right)={\displaystyle \sum _{i=0}^{k+N-1}}log{T}^i\left(x\right)=log\left|x{q}_{k+N-1}-p_{k+N-1}\right|.$$

Moreover,

$$\begin{array}{cc}\hfill {\displaystyle \prod _{i=0}^{k+N-1}}{T}^i\left(x\right)=\left|x{q}_{k+N-1}-p_{k+N-1}\right|& =\frac{1}{q_{k+N}+q_{k+N-1}{T}^{k+N}\left(x\right)}>\frac{1}{q_{k+N}+q_{k+N-1}}\hfill \\ & >\frac{1}{2q_{k+N-1}},\hfill \end{array}$$

and $T^i\left(x\right)\in (0,1)$, ${\displaystyle \prod _{i=0}^{k+N-1}}{T}^i\left(x\right)<1$ then we have $-log2q_{k+N-1}<{\displaystyle \sum _{i=0}^{k+N-1}}log{T}^i\left(x\right)<0$. So $\frac{{\displaystyle \sum _{i=0}^{k+N-1}}log{T}^i\left(x\right)}{k+pN}\to 0(asp\to \infty )$. Therefore,

$$\begin{array}{cc}\hfill \frac{log|x{q}_{n-1}-p_{n-1}|}{n}& \le \frac{{\sum }_{i=0}^{k-1}log{T}^i\left(x\right)}{k+Np}+\frac{p{\sum }_{i=k}^{k+N-1}log{T}^i\left(x\right)}{k+Np}+\frac{{\sum }_{i=k+Np}^{k+N(p+1)-1}log{T}^i\left(x\right)}{k+Np}\hfill \\ & ⟶\frac{{\sum }_{i=k}^{k+N-1}log{T}^i\left(x\right)}{N}(asp\to \infty ).\hfill \end{array}$$

On the other hand,

$$\begin{array}{cc}\hfill \frac{log|x{q}_{n-1}-p_{n-1}|}{n}& =\frac{{\displaystyle \sum _{i=0}^{k-1}}log{T}^i\left(x\right)+p{\displaystyle \sum _{i=k}^{k+N-1}}log{T}^i\left(x\right)+{\displaystyle \sum _{i=k+Np}^{n-1}}log{T}^i\left(x\right)}{n}\hfill \\ & \ge \frac{{\displaystyle \sum _{i=0}^{k-1}}log{T}^i\left(x\right)}{k+N(p+1)}+\frac{p{\displaystyle \sum _{i=k}^{k+N-1}}log{T}^i\left(x\right)}{k+N(p+1)}\hfill \\ & ⟶\frac{{\displaystyle \sum _{i=k}^{k+N-1}}log{T}^i\left(x\right)}{N}(asp\to \infty ),\hfill \end{array}$$

(similarly, $\frac{{\displaystyle \sum _{i=0}^{k-1}}log{T}^i\left(x\right)}{k+N(p+1)}\to 0$ as $p\to \infty$). So $\underset{n\to \infty }{lim}\frac{log|x{q}_{n-1}-p_{n-1}|}{n}=\frac{{\displaystyle \sum _{i=k}^{k+N-1}}log{T}^i\left(x\right)}{N}$.

Therefore, we know the proposition is true. □

Set $\beta =\frac{\sqrt{5}+1}{2}$, $x_1=\left[\overline{1,\beta }\right]$ and $x_2=\left[\overline{\beta ,1}\right]$, then $T\left(x_1\right)=x_2=\frac{-\beta +\sqrt{5\beta +1}}{2\beta }$, $T\left(x_2\right)=x_1=\frac{-\beta +\sqrt{5\beta +1}}{2}$. Then by the proof of Proposition 7, we obtain

$$L\left(x_1\right)=\frac{{\displaystyle \sum _{i=0}^1}log{T}^i\left(x_1\right)}{2}=\frac{1}{2}(log{x}_1+log{x}_2)=\frac{1}{2}log\left(x_1{x}_2\right)=\frac{1}{2}log\frac{\beta +2-\sqrt{5\beta +1}}{2}.$$

Using the same method, we can calculate $L\left(x_2\right)=\frac{1}{2}log\frac{\beta +2-\sqrt{5\beta +1}}{2}=L\left(x_1\right)$. Denote $\mathbf{c}=\frac{1}{2}log\frac{\beta +2-\sqrt{5\beta +1}}{2}$.

Proposition 8.

For arbitrary $x\ge 0$, $L_{*}\left(x\right)\ge \mathbf{c}$.

Proof. 

We want to prove that, for any $x>0$, $L_{*}\left(x\right)\ge \mathbf{c}$ holds, (i.e., $L_{*}\left(x\right)\ge L\left(x_1\right)$). That is to say, $\underset{n\to \infty }{lim\; inf}\frac{log{q}_n\left(x\right)}{n}\ge \underset{n\to \infty }{lim}\frac{log{q}_n\left(x_1\right)}{n}$ holds. In fact, we can show the more strict result: for any $x>0$, $q_n\left(x\right)\ge q_n\left(x_1\right)$ holds, which we will prove by contradiction.

Suppose there exist $x>0$ and $n\in \mathbb{N}$, such that $q_n\left(x\right)<q_n\left(x_1\right)$, let $n_0$ be the first position such that $q_n\left(x\right)<q_n\left(x_1\right)$. That is,

$$q_k\left(x\right)\ge q_k\left(x_1\right)\mathrm{for}\mathrm{all}1\le k\le n_0-1.$$

Thus by $q_n=a_n{q}_{n-1}+q_{n-2}$, we can get $a_{n_0}\left(x\right)<a_{n_0}\left(x_1\right)$. If $n_0$ is odd, then $a_{n_0}\left(x_1\right)=1$ due to $x_1=\left[\overline{1,\beta }\right]$. Therefore $a_{n_0}\left(x\right)<1$, which contradicts with $a_{n_0}\left(x\right)\in {\mathbb{Z}}_{\beta }^{+}\backslash \left\{0\right\}$. If $n_0$ is even, then $a_{n_0}\left(x_1\right)=\beta$ and $a_{n_0}\left(x\right)=1$. Thus

$$q_{n_0-1}\left(x\right)+q_{n_0-2}\left(x\right)<\beta q_{n_0-1}\left(x_1\right)+q_{n_0-2}\left(x_1\right),$$

that is to say,

$$(a_{n_0-1}+1)q_{n_0-2}\left(x\right)+q_{n_0-3}\left(x\right)<(\beta +1)q_{n_0-2}\left(x_1\right)+q_{n_0-3}\left(x_1\right),$$

but $q_{n_0-2}\left(x\right)\ge q_{n_0-2}\left(x_1\right),q_{n_0-3}\left(x\right)\ge q_{n_0-3}\left(x_1\right)$. Therefore, $a_{n_0-1}=1$ and $a_{n_0-1}\left(x\right)a_{n_0}\left(x\right)=11$, which means that the concatenation of $a_{n_0-1}\left(x\right)$ and $a_{n_0}\left(x\right)$. This contradicts admissibility of the sequence of continued $\beta$-fraction. Thus, for arbitrary $x\ge 0$, $L_{*}\left(x\right)\ge \mathbf{c}$. □

The following lemma is similar to [23], we present it here for completeness.

Lemma 1.

Let $(a_1,a_2\dots ,a_n)$ be admissible, $a_i\in {\mathbb{Z}}_{\beta }^{+}\backslash \left\{0\right\}$. For any $n\ge 1$, and $1\le k\le n$, we have

$$\frac{a_k+1}{2}\le \frac{q_n(a_1,a_2,\dots ,a_n)}{q_{n-1}(a_1,a_2,\dots ,a_{k-1},a_{k+1},\dots ,a_n)}\le a_k+1.$$

Proof. 

By Definition 2,

$$\frac{q_k(a_1,\dots ,a_k)}{q_{k-1}(a_1,\dots ,a_{k-1})}=\frac{a_k{q}_{k-1}(a_1,\dots ,a_{k-1})+q_{k-2}(a_1,\dots ,a_{k-2})}{q_{k-1}(a_1,\dots ,a_{k-1})}\le a_k+1,$$

$$\frac{q_k(a_1,\dots ,a_k)}{q_{k-1}(a_1,\dots ,a_{k-1})}\ge a_k\ge \frac{a_k+1}{2},$$

$$\begin{array}{cc}& \frac{q_{k+1}(a_1,\dots ,a_{k+1})}{q_k(a_1,\dots ,a_{k-1},a_{k+1})}=\frac{a_{k+1}{q}_k(a_1,\dots ,a_k)+q_{k-1}(a_1,\dots ,a_{k-1})}{a_{k+1}{q}_{k-1}(a_1,\dots ,a_{k-1})+q_{k-2}(a_1,\dots ,a_{k-2})}\hfill \\ & =\frac{(a_{k+1}{a}_k+1)q_{k-1}(a_1,\dots ,a_{k-1})+a_{k+1}{q}_{k-2}(a_1,\dots ,a_{k-2})}{a_{k+1}{q}_{k-1}(a_1,\dots ,a_{k-1})+q_{k-2}(a_1,\dots ,a_{k-2})}\hfill \\ & \le \frac{a_{k+1}{a}_k{q}_{k-1}(a_1,\dots ,a_{k-1})}{a_{k+1}{q}_{k-1}(a_1,\dots ,a_{k-1})}+\frac{q_{k-1}(a_1,\dots ,a_{k-1})+a_{k+1}{q}_{k-2}(a_1,\dots ,a_{k-2})}{a_{k+1}{q}_{k-1}(a_1,\dots ,a_{k-1})+q_{k-2}(a_1,\dots ,a_{k-2})}\hfill \\ & \le a_k+1,\hfill \end{array}$$

$$\begin{array}{cc}\hfill \frac{q_{k+1}(a_1,\dots ,a_{k+1})}{q_k(a_1,\dots ,a_{k-1},a_{k+1})}& =\frac{(a_{k+1}{a}_k+1)q_{k-1}(a_1,\dots ,a_{k-1})+a_{k+1}{q}_{k-2}(a_1,\dots ,a_{k-2})}{a_{k+1}{q}_{k-1}(a_1,\dots ,a_{k-1})+q_{k-2}(a_1,\dots ,a_{k-2})}\hfill \\ & \ge \frac{(a_{k+1}{a}_k+1)q_{k-1}(a_1,\dots ,a_{k-1})}{(a_{k+1}+1)q_{k-1}(a_1,\dots ,a_{k-1})}\hfill \\ & =\frac{a_{k+1}{a}_k+1}{a_{k+1}+1}\ge \frac{a_k+1}{2}.\hfill \end{array}$$

Using Definition 2 and induction, we get the desired result. □

Now we begin to prove Theorem 2.

Proof of Theorem 2. 

For any $\mathbf{c}<\lambda <\infty$ and any $0<ϵ<\lambda -\mathbf{c}$, choose $N\in \mathbb{N}$ large enough, such that

$$2e^{(\lambda -ϵ-\mathbf{c})(2N-1)+\mathbf{c}}<e^{(\lambda +ϵ-\mathbf{c})(2N-1)+\mathbf{c}}-1,$$

and choosing $1\ne b\in {\mathbb{Z}}_{\beta }^{+}$, such that

$$2e^{(\lambda -ϵ-\mathbf{c})(2N-1)+\mathbf{c}}-1\le b\le e^{(\lambda +ϵ-\mathbf{c})(2N-1)+\mathbf{c}}-1.$$

Let $x\in (0,1)$, such that $x=[0;\overline{\underset{2(N-1)}{\underbrace{\beta ,1,\dots ,\beta ,1}},b}]$.

For any $n\ge 2N-1$, there exists $k\in \mathbb{N}$, such that $n=k(2N-1)+2p$ and $p=0,1,\dots ,N-1$. By Lemma 1, we have

$$q_n\left(x\right)\le q_{n-k}(\beta ,1,\dots ,\beta ,1){(b+1)}^k\le c_1{e}^{\mathbf{c}(n-k)}{(b+1)}^k$$

(9)

and

$$q_n\left(x\right)\ge q_{n-k}(\beta ,1,\dots ,\beta ,1){\left(\frac{b+1}{2}\right)}^k\ge c_2{e}^{\mathbf{c}(n-k)}{\left(\frac{b+1}{2}\right)}^k,$$

(10)

where $c_1,c_2$ in (9) and (10) are positive constants, which do not depend on n.

Thus, by (9) and (10), we have

$$\begin{array}{cc}\hfill L\left(x\right)\le & \underset{n\to \infty }{lim\; sup}\frac{log\left(c_1{e}^{\mathbf{c}(n-k)}{(b+1)}^k\right)}{n}\hfill \\ & =\underset{n\to \infty }{lim\; sup}\frac{k(log(b+1)-\mathbf{c})}{n}+\mathbf{c}\hfill \\ & \le \underset{n\to \infty }{lim\; sup}\frac{k(log(b+1)-\mathbf{c})}{k(2N-1)}+\mathbf{c}\hfill \\ & =\frac{(log(b+1)-\mathbf{c})}{(2N-1)}+\mathbf{c}\le \lambda +ϵ\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill L\left(x\right)& \ge \underset{n\to \infty }{lim\; inf}\frac{log\left(c_2{e}^{\mathbf{c}(n-k)}{\left(\frac{b+1}{2}\right)}^k\right)}{n}\hfill \\ & =\underset{n\to \infty }{lim\; inf}\frac{k(log\left(\frac{b+1}{2}\right)-\mathbf{c})}{n}+\mathbf{c}\hfill \\ & \ge \underset{n\to \infty }{lim\; inf}\frac{k(log\left(\frac{b+1}{2}\right)-\mathbf{c})}{k(2N-1)+2(N-1)}+\mathbf{c}\hfill \\ & =\frac{log\frac{b+1}{2}-\mathbf{c}}{2N-1}+\mathbf{c}\ge \lambda -ϵ.\hfill \end{array}$$

Therefore $L\left(x\right)\in [\lambda -ϵ,\lambda +ϵ]$. Then $\mathbf{B}$ is dense in $[\mathbf{c},+\infty )$. □

5. Conclusions

Continued fractions have a number of remarkable properties related to real numbers. In view of these nice properties, it is natural to ask for the existence of a similar process in other contexts. The continued $\beta$-fraction was then introduced. In this article, we explored the relations among the numbers with eventually periodic continued $\beta$-fraction expansion, quadratic irrational, and Lévy constant. Two theorems are obtained and shed some light on the continued $\beta$-fraction. We also conjectured that there is no analogue of Lagrange’s theorem in the continued $\beta$-fraction, and this conjecture will be studied in our future work.