UNIQUE REPRESENTATIONS OF REAL NUMBERS IN NON-INTEGER BASES

PAUL GLENDINNING AND NIKITA SIDOROV

1. INTRODUCTION. If q ∈ (1, 2), then it is natural to consider those x which have unique expansions in base q of the form

(1)  x = ∑_{n=1}^{∞} ε_{n}q^{−n}

with ε_{n} ∈ {0, 1}, i.e. ε = (ε_{n})_{1}^{∞} ∈ Σ where Σ = ∏_{1}^{∞}{0, 1}. The special case of x = 1 has received a lot of attention (see [5, 6, 7]), and recently Komornik and Loreti [10] have shown that there is a smallest q ∈ (1, 2), denoted by q_{c}, for which there is only one representation of x = 1 in this form. Moreover, they show that q_{c} is the unique solution of the equation ∑_{1}^{∞} 𝔪_{n}x^{−n+1} = 1, where 𝔪 = (𝔪_{n})_{1}^{∞} is the famous Thue-Morse sequence

(2)  𝔪 = 0110 1001 1001 0110 1001 0110 0110 1001 …

(see, e.g., [3]). From this they obtain the numerical value

q_{c} = 1.787231650 …,

Recently, Allouche and Cosnard [1] have shown that q_{c} is transcendental (see also [2]).

In this paper we return to the case of general x in (1). Given q ∈ (1, 2) we aim to describe some properties of the set of all x ≥ 0 for which the representation (1) with ε ∈ Σ exists and is unique.

Our first remark is that since ε ∈ Σ, x must belong to [0, 1/(q − 1)] and we exclude the trivial cases x = 0 (where ε necessarily equals 0^{∞}) and x = 1/(q − 1) (where ε = 1^{∞}). Secondly, if q ≤ G, where G = \frac{1+\sqrt{5}}{2} = 1.618 …, then the set of such x’s is known to be empty – see [6, Theorem 3]. So, we will consider only q ∈ (G, 2) and x ∈ (0, 1/(q − 1)).

If a sequence ε is a unique representation, it must be both the greedy (“choose 1 whenever you have a choice”) and lazy (“choose 0 whenever you have a choice”) expansion of x (for more details see [6]).

The characterization of both expansions is well known. Let the sequence (d_{n})_{1}^{∞} be defined as follows: let 1 = ∑_{1}^{∞} d′_{k}q^{−k} be the greedy expansion of 1, i.e, d′_{1} = [q] = 1, d′_{2} = [q{q}], etc. If the tail of the sequence (d′_{n}) differs from 0^{∞}, then we put d_{n} ≡ d′_{n}. Otherwise let k = max {j : d′_{j} > 0}, and (d_{1}, d_{2}, …) := (d′_{1}, …, d′_{k−1}, d′_{k} − 1)^{∞}. In the seminal paper [11] it is shown that if x < 1 and (1) is the greedy expansion, then (ε_{n}, ε_{n+1}, …) is lexicographically less (notation: ≺) than (d_{1}, d_{2}, …). Similarly, if x > \frac{2−q}{q−1} = \frac{1}{q−1} − 1, then (\overline{ε}_{n}, \overline{ε}_{n+1}, …) ≺ (d_{1}, d_{2}, …), where the bar denotes the reflection, i.e., \overline{1} = 0, \overline{0} = 1.

Date: January 15, 2001.

It is natural to restrict x to the interval Δ_{q} = (\frac{2−q}{q−1}, 1); however, in the end we return to the interval (0, 1/(q − 1)) – see Corollary 14. Let

A_{q} := {x ∈ Δ_{q} : the representation (1) is unique}.

Proposition 1. [5] The set A_{q} has Lebesgue measure zero for any q ∈ (G, 2).

One might think that the Hausdorff dimension of A_{q} is also always zero but this is not the case for q close to 2. Let us formulate the main theorem of the present paper. Theorem 2. The cardinality of the set A_{q} is

(i) 2, if q ∈ (G, q_{f}], where q_{f} is the root of the equation

x^{3} = 2x − x + 1, q_{f} = 1.7548776 …;

(ii) ℵ_{0} if q ∈ (q_{f}, q_{c}) and (iii) 2^{ℵ_{0}} if q ∈ [q_{c}, 2). Moreover, if q ∈ (q_{c}, 2), then A_{q} is a Cantor set (i.e., it is perfect and nowhere

dense), and dim_{H}(A_{q}) > 0.

Thus, for this problem the “Komornik-Loreti constant” q_{c} is critical as well – it separates a Cantor set of positive Hausdorff dimension from a countable set! Remark 3. Theorem 2 (i) is implicitly contained in [4], where some partial results on the structure of U_{q} were also obtained; however, for the reader’s convenience we are going to give the proof of this item as well (especially as it is very simple).

Note that formula (1) may be regarded as a map π_{q} : Σ → ℝ and by definition, it is one-to-one on the preimage of A_{q} which we will denote by U_{q}. Our tool will be the following lemma which describes the set U_{q} (see above): Lemma 4. (see [11, 6])

U_{q} = \{ε ∈ Σ : (ε_{n}, ε_{n+1}, …) ≺ (d_{1}, d_{2}, …) and (\overline{ε}_{n}, \overline{ε}_{n+1}, …) ≺ (d_{1}, d_{2}, …), n ∈ ℕ\}.

2. ABOVE THE CRITICAL VALUE. We will begin with the proof of Theorem 2 (iii). From Lemma 4 it follows that the set U_{q} is a non-decreasing function of q, i.e., if q′ ≤ q, then U_{q′} ⊂ U_{q} (this is because the sequence (d_{n}) is an increasing function of q – see [11]) . Therefore, to prove Theorem 2 (iii), it suffices to construct a sequence q_{n} ↘ q_{c} such that U_{q_{n}} is uncountable and dim_{H}(π_{q}(U_{q_{n}})) > 0 for any q ∈ (q_{n}, 2). We will call the sequences from U_{q} q-admissible. Let the notation q : 1 ∼ ε_{1}ε_{2} … indicate the fact that (d_{n}) ≡ (ε_{n}), where (d_{n}) = (d_{n})(q) is as above.

We define q_{n} as the number for which q_{n} : 1 ∼ 𝔪_{2}𝔪_{3} … 𝔪_{2^{n}}(\overline{𝔪}_{1} … \overline{𝔪}_{2^{n}})^{∞}, i.e., q_{1} : 1 ∼ 1(10)^{∞}, q_{2} : 1 ∼ 110(1001)^{∞} , etc. It is obvious from (2) that q_{n} ↘ q_{c}, because q_{c} : 1 ∼ 𝔪_{2}𝔪_{3} . . .. Let σ denote the shift on sequences; following [10], we denote σ(𝔪) by δ, i.e.,

(3)  δ = 1101 0011 0010 1101 0010 …

We are going to prove a number of technical lemmas first. We will use the following relations: δ_{2^{n}} = 1, δ_{2^{n}+j} = \overline{δ}_{j} for 1 ≤ j < 2^{n} (see [10]). Lemma 5. We have δ_{k} … δ_{k+2^{n}−1} ≺ δ_{1} … δ_{2^{n}} for any n ≥ 1 and any k ∈ {2, …, 2^{n}}.

Proof. By [10, Lemma 1], δ_{k}δ_{k+1} … ≺ δ_{1}δ_{2} …, whence δ_{k} … δ_{k+2^{n}−1} ⪯ δ_{1} … δ_{2^{n}}. Assume that n ≥ 2 and δ_{k} … δ_{k+2^{n}−1} = δ_{1} … δ_{2^{n}}. Then δ_{k} = δ_{k+1} = 1, whence δ_{k−1} = 0, and we can use the same argument as in the proof of [10, Lemma 1]. Namely, we have k−1 = 2^{s}+2^{m}+j, where s ≤ n−1, m < s and 0 ≤ j < 2^{m}. It was shown in the proof in question that if δ_{k−1} = 0, then δ_{k} … δ_{k+2^{m+1}−j−1} ≺ δ_{j+1} … δ_{2^{m}} ⪯ δ_{1} … δ_{2^{m}−j}, whence δ_{k} … δ_{k+2^{n}−1} ≺ δ_{1} … δ_{2^{n}}, because m + 1 < n. □

Remark 6. Actually, from the above proof it follows that δ_{k} … δ_{k+2^{n}−2} ≺ δ_{1} … δ_{2^{n}−1}. In a similar way one can show that \overline{δ}_{k} … \overline{δ}_{k+2^{n}−2} ≺ δ_{1} … δ_{2^{n}−1} for k < 2^{n}. Lemma 7. If δ_{k} … δ_{2^{n}−1} = δ_{1} … δ_{2^{n}−k} or \overline{δ}_{k} … \overline{δ}_{2^{n}−1} = δ_{1} … δ_{2^{n}−k} for k ≤ 2^{n} − 1, then δ_{2^{n}−k+1} = 1.

Proof. We are going to prove both statements simultaneously using induction on n. For n = 1, 2 it is a direct inspection; let n ≥ 3 and assume the claim to be true for any i ≤ n and δ_{k} … δ_{2^{n+1}−1} = δ_{1} … δ_{2^{n+1}−k}. Note first that if k < 2^{n}, then δ_{k} … δ_{2^{n}+k−1} = δ_{1} … δ_{2^{n}}, which contradicts Lemma 5. Hence k > 2^{n} (it is obvious that k ≠ 2^{n}); we have \overline{δ}_{k−2^{n}} … \overline{δ}_{2^{n}−1} = δ_{1} … δ_{2^{n+1}−k}, and we can apply the inductive hypothesis for l = k−2^{n}. Hence δ_{2^{n+1}−k+1} = 1. The case \overline{δ}_{k} … \overline{δ}_{2^{n+1}−1} = δ_{1} … δ_{2^{n+1}−k} is considered in the same way. □

Lemma 8. We have \overline{δ}_{k} … \overline{δ}_{2^{n}−1}δ_{2^{n}} … δ_{2^{n+1}−1} ≺ δ_{1} … δ_{2^{n+1}−k} for any n ≥ 1 and any k ∈ {1, …, 2^{n} − 1}.

Proof. By Remark 6, \overline{δ}_{k} … \overline{δ}_{2^{n}−1} ⪯ δ_{1} … δ_{2^{n}−k}. If this inequality is strict, we are done. Assume we have an equality here. Note first that by (3), k ≥ 5; next, by Lemma 7, δ_{2^{n}−k+1} = 1, and it suffices to show that δ_{2^{n}+1} … δ_{2^{n+1}−1} ≺ δ_{2^{n}−k+2} … δ_{2^{n+1}−k}, which is equivalent to the inequality \overline{δ}_{1} … \overline{δ}_{2^{n}−1} ≺ δ_{2^{n}−k+2} … δ_{2^{n+1}−k}, which in turn is equivalent to \overline{δ}_{2^{n}−k+2} … \overline{δ}_{2^{n+1}−k} ≺ δ_{1} … δ_{2^{n}−1}. Since 2^{n} − k + 2 < 2^{n}, the latter inequality follows from Remark 6. □

Now we are ready to prove the main statement of this item. Let v_{n} := 𝔪_{1} … 𝔪_{2^{n}}. Proposition 9. If q > q_{n} for some n ≥ 2, then U_{q} ⊃ V_{n} := ∏_{1}^{∞}\{v_{n}, \overline{v}_{n}\}.

Proof. We need to show that any 0-1 sequence which is constructed of blocks v_{n} and \overline{v}_{n} is q-admissible for any q > q_{n}. Let V′_{n} := \{σ^{k}ε : ε ∈ V_{n}, k ≥ 0\}. We claim that the maximal sequence in V′_{n} is σ(v_{n})(\overline{v}_{n})^{∞} – which is exactly the expansion of 1 for q_{n}. Let us prove this.

As v_{n} starts with 0 and \overline{v}_{n} starts with 1, it is obvious that we must look for the largest element in V′_{n} among the sequences of the form σ^{k}(v_{n})(\overline{v}_{n})^{∞} or σ^{k}(\overline{v}_{n})(\overline{v}_{n})^{∞} for some k < 2^{n}. The cases k = 0, 1 are straightforward; let k ≥ 2.

Let us first show that σ^{k}(v_{n})(\overline{v}_{n})^{∞} ≺ σ(v_{n})(\overline{v}_{n})^{∞}. It suffices to prove the strict inequality for the segments of length 2^{n}, i.e. δ_{k} … δ_{k+2^{n}−1} ≺ δ_{1} … δ_{2^{n}}, which is exactly the claim of Lemma 5. Similarly, taking the segments of length 2^{n+1} − k, we see that the inequality σ^{k}(\overline{v}_{n})(\overline{v}_{n})^{∞} ≺ σ(v_{n})(\overline{v}_{n})^{∞} follows from Lemma 8.

By the symmetry of V_{n}, the lexicographically smallest sequence in V′_{n} will be σ(\overline{v}_{n})v_{n}^{∞}. Both conditions of Lemma 4 are satisfied, whence V_{n} ⊂ U_{q}. □

Thus, U_{q} (and therefore, A_{q}) has the cardinality of the continuum for q > q_{c}. To show that the Hausdorff dimension of A_{q} is positive, it suffices to prove that dim_{H}(π_{q}(V_{n})) > 0 for any q > q_{n} (where V_{n} is as in Proposition 9). In fact, there is an explicit formula for the latter dimension, namely,

(4)  dim_{H}(π_{q}(V_{n})) = \frac{log 2}{2^{n} log q} < 1.

This follows from the following simple lemma: Lemma 10. Let Q > 2 and π_{Q} be as above, i.e., π_{Q}(ε) = ∑_{k=1}^{∞} ε_{k}Q^{−k}. Then

dim_{H} (π_{Q}(∏_{1}^{∞}\{a, b\})) = \frac{log 2}{log Q}

for any a ≠ b.

Proof. Let a < b; it is a direct inspection that π_{Q}(∏_{1}^{∞}{a, b}) is the middle-third Cantor set in [\frac{a}{Q−1}, \frac{b}{Q−1}] with the middle part of length \frac{(b−a)(Q−2)}{Q(Q−1)} and therefore it is homothetic to the middle-third Cantor set in [0, 1] with the middle part of length 1−2/Q. The Hausdorff dimension of the latter set is known to be equal to log 2/ log Q (the proof is essentially the same as for the classical middle-third Cantor set – see, e.g., [8, Chapter 1]). □

To obtain (4), it suffices to apply this lemma to our framework with q > q_{n}, Q := q^{2^{n}} > 2 (as n ≥ 1), a := ∑_{1}^{2^{n}} 𝔪_{i}q^{−i}, b := ∑_{1}^{2^{n}} \overline{𝔪}_{i}q^{−i}. This proves that dim_{H}(A_{q}) > 0.

To conclude the proof of Theorem 2 for q > q_{c}, we need to show that A_{q} is a Cantor set. Note first that from Proposition 1 it follows that A_{q} is nowhere dense. Furthermore, if we endow Σ with its natural weak topology (the topology of coordinate-wise convergence), then the map π_{q} : Σ → ℝ is continuous. From Lemma 4 it is easy to deduce that U_{q} is closed, whence A_{q} is closed as well. Hence π_{q}|_{U_{q}} is a homeomorphism, as U_{q} is compact.

Since A_{q} and U_{q} are homeomorphic, it suffices to show that U_{q} does not contain isolated points. Let q > q_{n} and assume first that ε ∈ U_{q} has a tail different from (01)^{∞}. Then there exists an infinite number of indices k such that ε_{k−2}ε_{k−1}ε_{k} = 110 or 001. In view of the symmetry of U_{q} both cases are similar; let it be 110, for instance. Then it is easy to see that (ε_{1}, …, ε_{k}, v_{n}^{∞}) and (ε_{1}, …, ε_{k}, v_{n}(\overline{v}_{n})^{∞}) are both q-admissible. This follows from the q-admissibility of ε and of (1, 1, 0, v_{n}^{∞}). Therefore, ε cannot be an isolated point.

Assume now that ε = (ε_{1}, …, ε_{k}, (01)^{∞}) is q-admissible. Then it is a direct inspection that (ε_{1}, …, ε_{k}, (01)^{k}, v_{n}^{∞}) and (ε_{1}, …, ε_{k}, (01)^{k}(\overline{v}_{n})^{∞}) are both q-admissible as well for any k ≥ 2. Hence A_{q} is a Cantor set if q > q_{c}. Remark 11. Note that actually there exists an explicit formula for the Hausdorff dimension of A_{q}. Namely, let u_{n}(q) denote the number of 0-1 words (ε_{1} … ε_{n}), each of which can be extended to some sequence in U_{q}. Then

(5)  dim_{H}(A_{q}) = lim_{n→+∞} \frac{log u_{n}(q)}{n log q}.

This can be shown by using methods of dynamical systems theory and will appear in [9]. One of the consequences of this formula is that the function q ↦ dim_{H}(A_{q}) is continuous. However, it is not Hölder continuous, as dim_{H}(A_{q}) ≥ C/| log(q − q_{c})| when q > q_{c} for some C > 0, because, as is easy to see, q_{n} − q_{c} = O(q_{c}^{−2^{n}}) and by Proposition 9 and formula (4), dim_{H}(A_{q_{n}}) ≥ 2^{−n}.

3. BELOW THE CRITICAL VALUE. Let us first prove Theorem 2 (i). Let q ∈ (G, q_{f}] and suppose ε ∈ U_{q} and ε_{k} = ε_{k+1} = 1 for some k ≥ 1. Since q_{f} : 1 ∼ (1100)^{∞}, we have by Lemma 4 and the monotonicity of U_{q}, (ε_{k+2}, ε_{k+3}, …) = ((0011)^{∞}), which is not q-admissible, as q ≤ q_{f}. In view of the symmetry of U_{q}, the same argument works in the case ε_{k} = ε_{k+1} = 0. Therefore, there cannot be two consecutive equal symbols in U_{q}, whence (01)^{∞} and (10)^{∞} are the only possible elements of U_{q}. However, as q > G, both these sequences are indeed q-admissible, because G : 1 ∼ (10)^{∞}.

It remains to show that U_{q} is countable for q ∈ (q_{f}, q_{c}) and has the cardinality of the continuum for q = q_{c}. Lemma 12. If q ≤ q_{c}, then for any q-admissible sequence ε such that ε_{1} … ε_{2^{n}} = δ_{1} … δ_{2^{n}} for some n ≥ 1, necessarily ε_{2^{n}+1} … ε_{2^{n+1}−1} = δ_{2^{n}+1} … δ_{2^{n+1}−1}.

Proof. By monotonicity, it suffices to consider the case q = q_{c}. Since ε is q_{c}- admissible and q_{c} : 1 ∼ δ, we have by Lemma 4, ε_{2^{n}+1} … ε_{2^{n+1}−1} ⪯ δ_{2^{n}+1} … δ_{2^{n+1}−1} = \overline{δ}_{1} … \overline{δ}_{2^{n}−1}. Hence δ_{1} … δ_{2^{n}−1} ⪯ \overline{ε}_{2^{n}+1} … \overline{ε}_{2^{n+1}−1}. At the same time, since ε is q_{c}- admissible, \overline{ε}_{2^{n}+1} … \overline{ε}_{2^{n+1}−1} ⪯ δ_{1} … δ_{2^{n}−1}, whence \overline{ε}_{2^{n}+1} … \overline{ε}_{2^{n+1}−1} = δ_{1} … δ_{2^{n}−1}, and ε_{2^{n}+1} … ε_{2^{n+1}−1} = δ_{2^{n}+1} … δ_{2^{n+1}−1}. □

Let w_{n} := σ(v_{n})1 = δ_{1} … δ_{2^{n}}, n ≥ 0.

Proposition 13. If q_{f} < q < q_{c}, then there exists n = n(q) such that U_{q} contains only eventually periodic sequences with the period w_{k}\overline{w}_{k} for some k ≤ n.

Proof. Suppose ε is q-admissible and ε_{m} … ε_{m+2^{k}−1} = w_{k} or \overline{w}_{k} for some m ≥ 1, k ≥ 1 (this always happens unless ε = (10)^{∞} or (01)^{∞}). Let it be w_{k}, for instance. Then the next 2^{k} − 1 symbols are determined by Lemma 12. For ε_{m+2^{k+1}−1} there is a choice. If it equals 1, then ε_{m} … ε_{m+2^{k+1}−1} = w_{k+1}; otherwise ε_{m} … ε_{m+2^{k+1}−1} = w_{k}\overline{w}_{k}. Therefore, either we obtain the period (w_{k}\overline{w}_{k}) or we will meet w_{k+1}. As q is strictly less than q_{c}, the latter event cannot occur for arbitrarily large k, because w_{n} cannot be the prefix of a q-admissible sequence starting with some n. Therefore, the periodic tail of the form described above, is inevitable for a q-admissible sequence if q < q_{c}. □

Hence A_{q} is at most countable if q < q_{c}. However, as q > q_{f}, it is easy to construct a countable subset of U_{q}; for example, take all the sequences of the form (10)^{k}(1100)^{∞} with k ∈ ℕ. Thus, card A_{q} = ℵ_{0}, which proves Theorem 2 (ii). From the proof of Proposition 13 and by monotonicity, U_{q_{c}} contains all words of the form

(w_{1}\overline{w}_{1})^{k_{1}} … (w_{n}\overline{w}_{n})^{k_{n}} …, k_{j} ∈ ℤ_{+}

and therefore has the cardinality of the continuum. The main theorem is proven.

Corollary 14. The set A′_{q} of all x ∈ (0, 1/(q − 1)) that have a unique representation of the form (1), is countable for q ∈ (G, q_{c}) and uncountable with positive Hausdorff dimension for q ∈ (q_{c}, 2).

Proof. It suffices to note that every sequence ε′ from U′_{q} = π_{q}^{−1}(A′_{q}) is of the form 0^{k}ε or 1^{k}ε, where ε ∈ U_{q} and k ≥ 0. This follows, for instance, from the characterization of the greedy expansions for all x ∈ (0, 1/(q−1)) analogous to Parry’s theorem. Namely, ε′ is the greedy expansion of some x ∈ (0, 1/(q − 1)) if and only if (ε′_{n+1}, ε′_{n+2}, …) ≺ (d_{1}, d_{2}, …) whenever ε′_{n} = 0 – see [7]. □

Remark 15. Note that if q < q_{c}, then A_{q} is still closed but not perfect. Indeed, any sequence whose tail is (w_{n}\overline{w}_{n})^{∞} for n = n(q), is an isolated point. However, A_{q_{c}} is a Cantor set of zero Hausdorff dimension. We leave the proof of this claim to the reader as a simple exercise.

Example 16. Let T = 1.839286 … be the “tribonacci number”, i.e., the positive root of the equation x^{3} = x^{2} + x + 1. It lies above q_{c}, hence by Theorem 2, the set U_{T} must have the cardinality of the continuum. Let, as above, u_{n} denote the number of 0-1 words that can be extended to sequences in U_{T} – which is exactly the set of 0-1 sequences without three consecutive equal symbols. Hence it is easy to see that u_{1} = 2, u_{2} = 4, u_{3} = 6, …, and by induction, u_{n} = 2F_{n+1}, where (F_{n})_{1}^{∞} is the Fibonacci sequence! Therefore, from the formula (5) it follows that

dim_{H}(A_{T}) = lim_{n→∞} \frac{log(2F_{n+1})}{n log T} = \frac{log G}{log T} = 0.789677 …

A similar pattern occurs for the “multinacci” number T_{n}, i.e., the appropriate root of the equation x^{n} = x^{n−1} + x^{n−2} + ⋯ + x + 1, n ≥ 3. Namely, dim_{H}(A_{T_{n}}) = \frac{log T_{n−1}}{log T_{n}}, whence it follows that 1 − dim_{H}(A_{q}) ≍ 2 − q as q ↗ 2. For more details see [9].

ACKNOWLEDGEMENT. The second author’s research was supported by the EPSRC grant number GR/L98923. We would like to express our gratitude to Jean-Paul Allouche for his generous help with the history of the problem, moral support and numerous remarks and suggestions. We are also indebted to Vilmos Komornik for suggesting Corollary 14.

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Department of Mathematics, UMIST, P.O. Box 88, Manchester M60 1QD, United Kingdom. E-mail: P.A.Glendinning@umist.ac.uk

Department of Mathematics, UMIST, P.O. Box 88, Manchester M60 1QD, United Kingdom. E-mail: Nikita.A.Sidorov@umist.ac.uk