Non-uniform Expansions of Real Numbers

Abstract

We introduce and study non-uniform expansions of real numbers, given by two non-integer bases.

1 Introduction

Expansions of reals numbers in non-integer bases are studied since the pioneering works of Rényi in the end of the 1950s and Parry in the 1960s, see [11,12,13]. In these works, especially the greedy algorithm that determines the digits of such expansions and the relationship of these expansions to symbolic dynamics is addressed. In the 1990s, a group of Hungarian mathematics led by Paul Erdős revived this file of research, see [3,4,5]. Beside other results, they proved that each $x\in (0,1/(1-q))$ has a continuum of expansions of the form $\sum _{i=1}^{\infty } q^{-n_{i}}$ if $1<q<G$, where G is the golden ratio. In the sequel, Sidorov [14] used ergodic theoretical methods to prove that for all $q\in (1,2)$ almost all $x\in (0,1/(1-q))$ have such an expansion. Moreover, Glendinning and Sidorov [7] proved that there always exist (at least countably many) reals having a unique expansion if $q>G$. Nowadays, especially dimensional theoretical aspects of expansions of reals numbers in non-integer bases are studied; see, for instance, [1, 9, 10]. In this paper, we introduce non-uniform expansions of real numbers, which may be viewed as expansions with respect to two non-integer bases. As far as we know, such expansions were not studied yet, although they constitute a natural generalisation. The rest of the paper is organized as follows: In the next section, we give two descriptions of non-uniform expansions of real numbers. In the following, we introduce a greedy, a lazy, and intermittent algorithms that give the digits of these expansions. In section four, we prove a theorem on the existence of a continuum of non-uniform expansions of real numbers, which is similar to the results in the uniform case we mentioned above. In the last section, we characterise real numbers which have a unique non-uniform expansion and prove a theorem on the cardinality of the set of such numbers.

2 The Expansions

Let $\Sigma =\{0,1\}^{{\mathbb {N}}}$ be the set of sequences of zeros and ones. Equipped with the metric:

$$\begin{aligned} d((s_{i}),(t_{i}))=\sum _{i=1}^{\infty }|s_{i}-t_{i}|2^{-i}, \end{aligned}$$

$\Sigma$ is a compact, perfect, and totally disconnected space. For $s=(s_{i})\in \Sigma$ and $n\in {\mathbb {N}}$, let $0_{n}(s)$ be the number of zeros and $1_{n}(s)$ be the number of ones in the sequence $(s_{1},\ldots ,s_{n})$.

Fix $\beta _{0},\beta _{1}\in (1/2,1)$ with $\beta _{0}\ge \beta _{1}$ and let $I=I_{\beta _{1}}=[0,\beta _{1}/(1-\beta _{1})]$.¹ We consider the map $\pi _{\beta _{0},\beta _{1}}:\Sigma \rightarrow I$ given by:

$$\begin{aligned} \pi _{\beta _{0},\beta _{1}}(s)=\pi _{\beta _{0},\beta _{1}}((s_{i}))=\sum _{i=1}^{\infty }s_{i}\beta _{0}^{0_{i}(s)}\beta _{1}^{1_{i}(s)}. \end{aligned}$$

Several times, we will use another description of this map, which we now describe. Let $T_{0},T_{1}:I\rightarrow I$ be the contractions given by:

$$\begin{aligned} T_{0}(x)=\beta _{0} x\qquad \text{ and } \qquad T_{1}(x)=\beta _{1}x+\beta _{1}.\end{aligned}$$

By induction, we have:

$$\begin{aligned} T_{s_{1}}\circ \cdots \circ T_{s_{n}}(x)=\beta _{0}^{0_{n}(s)}\beta _{1}^{1_{n}(s)}x+\sum _{i=1}^{n}s_{i}\beta _{0}^{0_{i}(s)}\beta _{1}^{1_{i}(s)}. \end{aligned}$$

Hence, for all $s\in \Sigma$ and all $x\in I$:

$$\begin{aligned} \pi _{\beta _{0},\beta _{1}}(s)=\pi _{\beta _{0},\beta _{1}}((s_{i}))=\lim _{n\rightarrow \infty }T_{s_{1}}\circ \cdots \circ T_{s_{n}}(x).\end{aligned}$$

Definition 2.1

We call a sequence $s\in \Sigma$ with $\pi _{\beta _{0},\beta _{1}}(s)=x$ a $(\beta _{0},\beta _{1})$-expansion of $x\in I$.

The following proposition guarantees the existence of $(\beta _{0},\beta _{1})$-expansions.

Proposition 2.1

The map $\pi _{\beta _{0},\beta _{1}}$ is continuous and surjective.

Proof

If $d((s_{i}),(t_{i}))<2^{-u}$, we have $s_{i}=t_{i}$ for $i=1,\ldots ,u$, which implies:

$$\begin{aligned} |\pi _{\beta _{0},\beta _{1}}(s_{i})-\pi _{\beta _{0},\beta _{1}}(t_{i})|<\beta _{0}^{u}\beta _{1}/(1-\beta _{1}). \end{aligned}$$

Hence, $\pi _{\beta _{0},\beta _{1}}$ is continuous. Note that:

$$\begin{aligned} T_{0}(I)\cup T_{1}(I) =[0,\beta _{0}\beta _{1}/(1-\beta _{1})]\cup [\beta _{1},\beta _{1}/(1-\beta _{1})]=I, \end{aligned}$$

since $\beta _{0}+\beta _{1}\ge 1$. This implies that:

$$\begin{aligned} \bigcup _{(s_{i})\in \{0,1\}^n} T_{s_{1}}\circ \cdots \circ T_{s_{n}}(I)=I. \end{aligned}$$

for every $n\ge 1$. Hence, for each $x\in I$, there is sequence $s\in \Sigma$, such that:

$$\begin{aligned} x\in T_{s_{1}}\circ \cdots \circ T_{s_{n}}(I), \end{aligned}$$

By the compactness of $\Sigma$, the map $\pi _{\beta _{0},\beta _{1}}$ is surjective $\square$

In the next section, we describe an algorithm which determines one $(\beta _{0},\beta _{1})$-expansion of $x\in I$.

3 The Greedy and the Lazy Algorithm

Using the notations of last section, we define a map $G:I\rightarrow I$ by:

$$\begin{aligned} G(x)&=\left\{ \begin{array}{ll} T_{0}^{-1}(x), &{} x\not \in T_{1}(I) \\ T_{1}^{-1}(x), &{} x \in T_{1}(I) \end{array}\right. \\&=\left\{ \begin{array}{ll} \beta _{0}^{-1}x, &{} x\in [0,\beta _{1}) \\ \beta _{1}^{-1}x-1, &{} x \in [\beta _{1},\beta _{1}/(1-\beta _{1})], \end{array}\right. \end{aligned}$$

see Fig. 1. For $x\in I$, we define the greedy expansion $g=g(x)=(g_{i})\in \Sigma$ with respect to $(\beta _{0},\beta _{1})$ by

$$\begin{aligned} g_{i}=\lfloor \beta _{1}^{-1}G^{i-1}(x)\rfloor ,\end{aligned}$$

where $\lfloor a\rfloor$ is the greatest integer not greater than a. We have the following.

Proposition 3.1

For all $x\in I$, the greedy expansion $g(x)\in \Sigma$ with respect to $(\beta _{0},\beta _{1})$ is an $(\beta _{0},\beta _{1})$-expansion of x; that is, $\pi _{\beta _{0},\beta _{1}}(g(x))=x$.

Proof

If $g_{i}=0$, we have $G^{i-1}(x)<\beta _{1}$, which implies $G^{i-1}(x)\not \in T_{1}(I)$ and $G^{i-1}(x)\in T_{g_{i}}(I)$. If $g_{i}=1$, we have $G^{i-1}(x)\ge \beta _{1}$, which implies $G^{i-1}(x)\in T_{1}(I)$; hence, again, $G^{i-1}(x)\in T_{g_{i}}(I)$. By the definition of G, we conclude:

$$\begin{aligned} x\in T_{g_{1}}\circ \cdots \circ T_{g_{i}}(I)\end{aligned}$$

for all $i\in {\mathbb {N}}$, but this implies $\pi _{\beta _{0},\beta _{1}}(g(x))=x$.$\square$

Fig. 1

The maps $T_{0}^{-1}$ and $T_{1}^{-1}$ in the case $\beta _{0}=3/4, \beta _{1}=2/3$

To define the lazy expansion, let $L:I\rightarrow I$ be given by:

$$\begin{aligned} L(x)&=\left\{ \begin{array}{ll} T_{0}^{-1}(x), &{} x\in T_{0}(I) \\ T_{1}^{-1}(x), &{} x \not \in T_{0}(I) \end{array}\right. \\&=\left\{ \begin{array}{ll} \beta _{0}^{-1}x, &{} x \in [0,\beta _{0}\beta _{1}/(1-\beta _{1})] \\ \beta _{1}^{-1}x-1, &{} x \in (\beta _{0}\beta _{1}/(1-\beta _{1}),\beta _{1}/(1-\beta _{1})], \end{array}\right. \end{aligned}$$

see Fig. 1. For $x\in I$, the lazy expansion $l=l(x)=(l_{i})\in \Sigma$ with respect to $(\beta _{1},\beta _{2})$ is given by:

$$\begin{aligned} l_{i}=\lceil (1-\beta _{1})(\beta _{0}\beta _{1})^{-1} L^{i-1}(x)\rceil -1,\end{aligned}$$

where $\lceil a\rceil$ is the smallest integer not smaller than a. We have the following.

Proposition 3.2

For all $x\in I$, the lazy expansion $l(x)\in \Sigma$ with respect to $(\beta _{1},\beta _{2})$ is an $(\beta _{0},\beta _{1})$-expansion of x; that is, $\pi _{\beta _{0},\beta _{1}}(l(x))=x$.

Proof

If $l_{i}=0$, we have $L^{i-1}(x)\le \beta _{0}\beta _{1}/(1-\beta _{1})$, which implies $L^{i-1}(x)\in T_{0}(I)$; hence, $L^{i-1}(x)\in T_{l_{i}}(I)$. If $l_{i}=1$, we have $L^{i-1}(x)>\beta _{0}\beta _{1}/(1-\beta _{1})$, which implies that $L^{i-1}(x)\not \in T_{0}(I)$; hence, again, $L^{i-1}(x)\in T_{l_{i}}(I)$. By the definition of L, we conclude:

$$\begin{aligned} x\in T_{l_{1}}\circ \cdots \circ T_{l_{i}}(I)\end{aligned}$$

for all $i\in {\mathbb {N}}$, which implies $\pi _{\beta _{0},\beta _{1}}(l(x))=x$.$\square$

For $\alpha \in (\beta _{1},\beta _{0}\beta _{1}/(1-\beta _{1})$, we may also consider intermediate expansions $m(x)=(m_{i})$ with respect to $(\beta _{1},\beta _{2})$ given by:

$$\begin{aligned} m_{i}=\lfloor \alpha ^{-1}M_{\alpha }^{i-1}(x)\rfloor ,\end{aligned}$$

where:

$$\begin{aligned}M_{\alpha }(x)=\left\{ \begin{array}{ll} \beta _{0}^{-1}x, &{} x\in [0,\alpha ) \\ \beta _{1}^{-1}x-1, &{} x \in [\alpha ,\beta _{1}/(1-\beta _{1})]\end{array}\right. .\end{aligned}$$

Again, these are $(\beta _{0},\beta _{1})$-expansions of $x\in I$.

4 A Continuum of Expansions

It is natural to ask how many $(\beta _{0},\beta _{1})$-expansions a real number in I has. It turns out that usually there is a continuum of such expansions:

Theorem 4.1

Let $\beta _{0},\beta _{1}\in (1/2,1)$ with $\beta _{0}\ge \beta _{1}$. We have:

(1) Almost all $x\in I$ have a continuum of $(\beta _{0},\beta _{1}$)-expansions.

(2) If $\beta _{1}^2+\beta _{0}> 1$, all $x\in I\backslash \{0,\beta _{1}/(1-\beta _{1})\}$ have a continuum of $(\beta _{0},\beta _{1})$-expansions.

Proof

We first prove (2). Let $J=(0,\beta _{1}/(1-\beta _{1}))$ and:

$$\begin{aligned} \Lambda _{0}=T_{0}(J)\cap T_{1}(J)=\left( \beta _{1},\beta _{0}\frac{\beta _{1}}{1-\beta _{1}}\right) . \end{aligned}$$

We recursively define $\Lambda _{n+1}=T_{0}(\Lambda _{n})\cup \Lambda \cup T_{1}(\Lambda _{n})$ and prove by induction:

$$\begin{aligned} \Lambda _{n}=\left( \beta _{0}^{n}\beta _{1},\left( \beta _{1}^{n}\left( \beta _{0}-1 \right) +1 \right) \frac{\beta _{1}}{1-\beta _{1}}\right) . \end{aligned}$$

We have:

$$\begin{aligned} T_{0}(\Lambda _{n})&\cup \Lambda _{n}\cup T_{0}(\Lambda _{n})=\left( \beta _{0}^{n+1}\beta _{1},\beta _{0}\left( \beta _{1}^{n}\left( \beta _{0}-1 \right) +1 \right) \frac{\beta _{1}}{1-\beta _{1}}\right) \\&\cup \left( \beta _{0}^{n}\beta _{1},\left( \beta _{1}^{n}\left( \beta _{0}-1 \right) +1 \right) \frac{\beta _{1}}{1-\beta _{1}}\right) \\&\cup \left( \beta _{0}^{n}\beta _{1}^2+\beta _{1},\left( \beta _{1}^{n+1}\left( \beta _{0}-1 \right) +1 \right) \frac{\beta _{1}}{1-\beta _{1}}\right) =\Lambda _{n+1}. \end{aligned}$$

In the last equation, we use $\beta _{1}^2+\beta _{0}>1$ and $\beta _{0}^2+\beta _{1}> 1$, which is true, since $\beta _{0}\ge \beta _{1}$. Note that $\bigcup _{n\ge 0} \Lambda _{n}=J$. Hence, for every $x\in J$, there is a $k\ge 0$ and a sequence $(s_{1},\ldots ,s_{k})\in \{0,1\}^{k}$, such that:

$$\begin{aligned} x=T_{s_{1}}\circ \cdots \circ T_{s_{k}}\circ T_{0}(x_{0}) \text{ and } x=T_{s_{1}}\circ \cdots \circ T_{s_{k}}\circ T_{1}(x_{1}), \end{aligned}$$

where $x_{0},x_{1}\in J$ and $x_{0}\not =x_{1}$. Hence, we obtain two expansions of x that differ in the $k+1$-digit. Applying the result to $x_{0}(x)$ and $x_{1}(x)$, we obtain four expansions of x. Here, we use that $x_{0}(x)$ and $x_{1}(x)$ are not at the boundary of J. Repeating this procedure $\aleph _{0}$ times, we see that there are $2^{\aleph _{0}}$ expansions of x. 

Now, we prove (1). Let $G:I\rightarrow I$ be the map associated with the greedy expansion from Sect. 3. G is a piecewise linear expanding interval map and such maps are known to have an ergodic measure, which is equivalent to the Lebesgue measure, see [2] and [8]. By Poincare recurrence theorem for almost all $x\in I$, there is a $k\ge 0$, such that $G^{k}(x)\in \Lambda _{0}$. Hence, for almost all $x\in J$, there is a $k\ge 0$ and a sequence $(s_{1},\ldots ,s_{k})\in \{0,1\}^{k}$, such that:

$$\begin{aligned} x=T_{s_{1}}\circ \cdots \circ T_{s_{k}}\circ T_{0}(x_{0}) \text{ and } x=T_{s_{1}}\circ \cdots \circ T_{s_{k}}\circ T_{1}(x_{1}), \end{aligned}$$

where $x_{0},x_{1}\in J$ and $x_{0}\not =x_{1}$. For almost all x, both numbers $x_{1}(x),x_{2}(x)$ have two different $(\beta _{0},\beta _{1})$-expansion, and hence, almost all x have four different expansions. We use here that the intersection of two sets of full measure has full measure. Repeating this procedure $\aleph _{0}$ times, we obtain $2^{\aleph _{0}}$ expansions for almost all $x\in I$, using the fact that a countable intersection of sets of full measure has full measure. $\square$

Obviously, the $(\beta _{0},\beta _{1})$-expansion of 0 and $\beta _{1}/(1-\beta _{1})$ is unique. Our theorem leaves the question open if there are numbers x in the interior of I that have a unique $(\beta _{0},\beta _{1})$-expansion. We will address this question in the following section.

5 Unique Expansions

We consider the shift map $\sigma :\{0,1\}^{{\mathbb {N}}}\rightarrow \{0,1\}^{{\mathbb {N}}}$ given by $\sigma ((s_{k}))=(s_{k+1})$. Using this map, we may characterise numbers which have a unique $(\beta _{0},\beta _{1})$-expansion as follows:

Proposition 5.1

The $(\beta _{0},\beta _{1})$-expansion $(s_{i})$ of x is unique if and only if:

$$\begin{aligned} \pi _{\beta _{0},\beta _{1}}(\sigma ^{k}(s_{i}))\in [0,\beta _{1})\cup (\beta _{0}\beta _{1}/(1-\beta _{1}),\beta _{1}/(1-\beta _{1})] \end{aligned}$$

for all $k\ge 0$.

Proof

$\pi _{\beta _{0},\beta _{1}}((s_{i}))=\pi _{\beta _{0},\beta _{1}}((t_{i}))$ with $(s_{i})\not =(t_{i})$ if and only if there exists a smallest $k\ge 0$, such that $s_{k+1}\not =t_{k+1}$ and $\pi _{\beta _{0},\beta _{1}}(\sigma ^{k}(s_{i}))=\pi _{\beta _{0},\beta _{1}}(\sigma ^{k}(t_{i}))$. However, this is equivalent to $\pi _{\beta _{0},\beta _{1}}(\sigma ^{k}(s_{i}))\in T_{0}(I)\cap T_{1}(I)=[\beta _{1},\beta _{0}\beta _{1}/(1-\beta _{1})]$. The proposition follows by contraposition. $\square$

Using this characterisation of points with unique expansion, we are able to prove:

Theorem 5.1

Let $\beta _{0},\beta _{1}\in (1/2,1)$ and $\beta _{0}\ge \beta _{1}$. (1) If $\beta _{0}(1+\beta _{1})<1$, there exist at least countable many $x\in I$, which have a unique $(\beta _{0},\beta _{1})$-expansion. (2) If $\beta _{0}(1+2\beta _{1}-\beta _{0}\beta _{1})<1$, there are uncountable many $x\in I$, which have a unique $(\beta _{0},\beta _{1})$-expansion. Moreover, the set of these x has positive Hausdorff dimension.

Proof

First, we prove (1). Consider the periodic sequence $p=(010101\ldots )$. Since $\beta _{0}(1+\beta _{1})<1$, we have:

$$\begin{aligned} \pi _{\beta _{0},\beta _{1}}(p)=\beta _{0}\beta _{1}/(1-\beta _{0}\beta _{1})<\beta _{1}. \end{aligned}$$

Note that $\beta _{0}(1+\beta _{1})<1$ implies $\beta _{1}(1+\beta _{0})<1$, since $\beta _{0}\ge \beta _{1}$. Hence, we have $\beta _{0}-\beta _{0}^2\beta _{1}<1-\beta _{1}$ and thus:

$$\begin{aligned} \pi _{\beta _{0},\beta _{1}}(\sigma (p))=\beta _{1}/(1-\beta _{0}\beta _{1})=\beta _{0}\beta _{1}/(\beta _{0}-\beta _{0}^2\beta _{1})>\beta _{0}\beta _{1}/(1-\beta _{1}). \end{aligned}$$

By proposition $x=\pi _{\beta _{0},\beta _{1}}(p)$ has a unique $(\beta _{0},\beta _{1})$-expansion. Obviously, the same is true for all x of the form $x=\pi _{\beta _{0},\beta _{1}}((0\ldots 0101010\ldots ))$, and there exist countable many of such x. Now, we prove (2). Let $V=\{01,10\}^{{\mathbb {N}}}$ and:

$$\begin{aligned} U=\bigcup _{k=0}^{\infty }\sigma ^{k}(V)=V\cup (\{0\}\times V)\cup (\{1\}\times V), \end{aligned}$$

where $\times$ is the cartesian product. We prove that $\pi _{\beta _{0},\beta _{1}}(U)\subseteq [0,\beta _{1})\cup (\beta _{0}\beta _{1}/(1-\beta _{1}),\beta _{1}/(1-\beta _{1})]$. The sequence $s\in U$ with $s_{1}=0$ that has the largest projection under $\pi _{\beta _{0},\beta _{1}}$ obviously is $s=(011010101\ldots )$. We have:

$$\begin{aligned} \pi _{\beta _{0},\beta _{1}}(s)=\beta _{1}\frac{\beta _{0}+\beta _{0}\beta _{1}-\beta _{0}^2\beta _{1}}{1-\beta _{0}\beta _{1}}<\beta _{1} \end{aligned}$$

by our assumption. The sequence $s\in U$ with $s_{1}=1$ that has the smallest projection under $\pi _{\beta _{0},\beta _{1}}$ obviously is $s=(1001010101\ldots )$. We have:

$$\begin{aligned} \pi _{\beta _{0},\beta _{1}}(s)=\beta _{1}+\frac{(\beta _{0}\beta _{1})^2}{1-\beta _{0}\beta _{1}}>\frac{\beta _{0}\beta _{1}}{1-\beta _{1}}.\end{aligned}$$

The inequality here is equivalent to $\beta _{1}(1+2\beta _{0}-\beta _{0}\beta _{1})<1$ which is true, since we assume $\beta _{0}\ge \beta _{1}$. It remains to show that the Hausdorff dimension of $A:=\pi _{\beta _{0},\beta _{1}}(V)$ is positive. Consider the maps:

$$\begin{aligned} F(x)=T_{0}\circ T_{1}(x)=\beta _{0}\beta _{1}x+\beta _{0}\beta _{1}\end{aligned}$$

and

$$\begin{aligned} H(x)=T_{1}\circ T_{0}(x)=\beta _{0}\beta _{1}x+\beta _{1},\end{aligned}$$

and let $\Lambda =(\pi _{\beta _{0},\beta _{1}}((01)^{\infty }),\pi _{\beta _{0},\beta _{1}}((10)^{\infty })$. We have $F(\Lambda )\subseteq \Lambda$ and $H(\Lambda )\subseteq \Lambda$ and $F(\Lambda )\cap H(\Lambda )=\emptyset$ by our assumptions on $\beta _{0}$ and $\beta _{1}$. In the language of fractal geometry, this means that (F, H) induce an iterated function system fulfilling the open set condition, see [6]. The attractor of this iterated function system is A, since $A=F(A)\cup H(A)$, and the classical formula for self-similar fractals gives:

$$\begin{aligned} \dim _{H}A=\frac{-\log (2)}{\log (\beta _{0}\beta _{1})}>0. \end{aligned}$$

$\square$