EXPANSIONS IN NON-INTEGER BASES: LOWER, MIDDLE AND TOP ORDERS

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1 EXPANSIONS IN NON-INTEGER BASES: LOWER, MIDDLE AND TOP ORDERS NIKITA SIDOROV To the memory o Bill Parry ABSTRACT. Let q (1, 2); it is known that each x [0, 1/(q 1)] has an expansion o the orm x = n=1 a nq n with a n {0, 1}. It was shown in [4] that i q < ( 5 + 1)/2, then each x (0, 1/(q 1)) has a continuum o such expansions; however, i q > ( 5 + 1)/2, then there exist ininitely many x having a unique expansion [5]. In the present paper we begin the study o parameters q or which there exists x having a ixed inite number m > 1 o expansions in base q. In particular, we show that i q < q 2 = 1.71, then each x has either 1 or ininitely many expansions, i.e., there are no such q in (( 5+1)/2, q 2 ). On the other hand, or each m > 1 there exists γ m > 0 such that or any q (2 γ m, 2), there exists x which has exactly m expansions in base q. 1. INTRODUCTION AND SUMMARY Expansions o reals in non-integer bases have been studied since the late 1950s, namely, since the pioneering works by Rényi [14] and Parry [13]. The model is as ollows: ix q (1, 2) and call any 0-1 sequence (a n ) n 1 an expansion in base q or some x 0 i (1.1) x = a n q n. n=1 Note that x must belong to I q := [0, 1/(q 1)] and that or each x I q there is always at least one way o obtaining the a n, namely, via the greedy algorithm ( choose 1 whenever possible ) which until recently has been considered virtually the only option. In 1990 Erdős et al. [4] showed (among other things) that i q < G := ( 5 + 1)/ , then each x (0, 1/(q 1)) has in act 2 ℵ 0 expansions o the orm (1.1). I q = G, then each x I q has 2 ℵ 0 expansions, apart rom x = ng (mod 1) or n Z, each o which has ℵ 0 expansions in base q (see [17] or a detailed study o the space o expansions or this case). However, i q > G, then although a.e. x I q has 2 ℵ 0 expansions in base q [15], there always exist (at least countably many) reals having a unique expansion see [5]. Let U q denote the set o x I q which have a unique expansion in base q. The structure o the set U q is reasonably well understood; its main property is that U q is countable i q is not too ar rom the golden ratio, and uncountable o Hausdor dimension strictly between 0 and 1 Date: August 27, Mathematics Subject Classiication. 11A63. Key words and phrases. Beta-expansion, Cantor set, thickness, non-integer base. 1

2 2 NIKITA SIDOROV otherwise. More precisely, let q KL denote the Komornik-Loreti constant introduced in [7], which is deined as the unique solution o the equation m n x n = 1, 1 where m = (m n ) 0 is the Thue-Morse sequence m = , i.e., a ixed point o the morphism 0 01, The Komornik-Loreti constant is known to be the smallest q or which x = 1 has a unique expansions in base q (see [7]), and its numerical value is approximately It has been shown by Glendinning and the author in [5] that (1) U q is countable i q (G, q KL ), and each unique expansion is eventually periodic; (2) U q is a continuum o positive Hausdor dimension i q > q KL. Let now m N {ℵ 0 } and put B m = {q (G, 2) : x I q which has exactly m expansions in base q o the orm (1.1)}. It ollows rom the quoted theorem rom [5] that B 1 = (G, 2), but very little has been known about B m or m 2. The purpose o this paper is to begin a systematic study o these sets. Remark 1.1. It is worth noting that in [3] it has been shown that or each m N there exists an uncountable set E m o q such that the number x = 1 has m + 1 expansions in base q. The set E m (2 ε m, 2), where ε m is small. A similar result holds or m = ℵ 0. Note also that a rather general way to construct numbers q (1.9, 2) such that x = 1 has two expansions in base q, has been suggested in [8]. 2. LOWER ORDER: q CLOSE TO THE GOLDEN RATIO We will write x (a 1, a 2, ) q i (a n ) n 1 is an expansion o x in base q o the orm (1.1). Theorem 2.1. For any transcendental q (G, q KL ) we have the ollowing dichotomy: each x I q has either a unique expansion or a continuum o expansions in base q. Proo. We are going to exploit the idea o branching introduced in [16]. Let x I q have at least two expansions o the orm (1.1); then there exists the smallest n 0 such that x (a 1,, a n, a n+1, ) q and x (a 1,, a n, b n+1, ) q with a n+1 b n+1. We may depict this biurcation as shown in Fig. 1. I (a n+1, a n+2, ) q is not a unique expansion, then there exists n 2 > n with the same property, etc. As a result, we obtain a subtree o the binary tree which corresponds to the set o all expansions o x in base q, which we call the branching tree o x. It has been shown in [16, Theorem 3.6] that i q (G, q KL ) that or or all x, except, possibly, a countable set, the branching tree is in act the ull binary tree and hence x has 2 ℵ 0 expansions in base q; the issue is thus about these exceptional x s. 1 For the list o all constants used in the present paper, see Table 5.1 beore the bibliography.

3 EXPANSIONS IN NON-INTEGER BASES 3 a n+1 a n+2 a n2 a 1 a 2 a n 1 a n x b n+1 b n+2 b n 2 FIGURE 1. Branching and biurcations Note that or x to have at most countably many expansions in base q, its branching tree must have at least two branches which do not biurcate. In other words, there exist two expansions o x in base q, (a n ) n 1 and (b n ) n 1 such that (a k, a k+1, ) is a unique expansion and so is (b j, b j+1, ) or some k, j N. Without loss o generality, we may assume j = k, because the shit o a unique expansion is known to be a unique expansion [5]. Hence x = = k a i q i + q k r k (q) i=1 k b i q i + q k r k(q), i=1 where r k (q), r k (q) U q and r k (q) r k (q). (I they are equal, then q is obviously algebraic.) Since each unique expansion or q (G, q KL ) is eventually periodic ([5, Proposition 13]), we have U q Q(q), whence the equation (2.1) k (a i b i )q i = q k (r k (q) r k(q)) i=1 implies that q is algebraic, unless (2.1) is an identity. Assume it is an identity or some q; then it is an identity or all q > 1, because r k (q) = π(q) + ρ(q)/(1 q r ) and r k (q) = π (q) + ρ (q)/(1 q r ), where π, π, ρ, ρ are polynomials.

4 4 NIKITA SIDOROV Let j = min {i 1 : a i b i } < k. We multiply (2.1) by q j and get k a j b j + (a i b i )q j i q j k (r k (q) r k(q)), i=j+1 which is impossible, since q + implies a j b j = 0, a contradiction. The next question we are going to address in this section is inding the smallest element o B 2. Let q B m and denote by U q (m) the set o x I q which have m expansions in base q. Firstly, we give a simple characterization o the set B 2 : Lemma 2.2. A number q (G, 2) belongs to B 2 i and only i 1 U q U q. Proo. 1. Let q B 2 ; then there exists x having exactly two expansions in base q. Without loss o generality we may assume that there exist two expansions o x, with a 1 = 0 and with a 1 = 1. (Otherwise we shit the expansion o x until we obtain x having this property.) Note that x [ 1, 1 q q(q 1)] =: Jq the interval which is called the switch region in [2]. Conversely, i x J q, then it has a branching at n = 1. Since x has only two dierent expansions in base q, both shits o x, namely, qx (or a 1 = 0) and qx 1 (or a 1 = 1), must belong to U q, whence 1 U q U q. 2. Let y U q and y + 1 U q. We claim that x := (y + 1)/q belongs to U q (2). Note that y U q implies y J q, whence y < 1/q, because i y were greater than 1/(q(q 1)), we would have y + 1 > q2 q+1 > 1. q 1 q 1 Thus, y < 1/q, whence x J q, because y + 1 < 1/(q 1). Since x J q, it has at least two dierent expansions in base q, with a 1 = 0 and a 1 = 1, and shiting each o them yields qx = y + 1 and qx 1 = y, both having unique expansions. Hence there are only two possible expansions o x, i.e., x U q (2). This criterion, simple as it is, indicates the diiculties one aces when dealing with B 2 as opposed to the unique expansions; at irst glance, it may seem rather straightorward to veriy whether i a number x has a unique expansion, then so does x + 1 but this is not the case. The reason why this is actually hard is the act that typically adding 1 to a number alters the tail o its greedy expansion (which, o course, coincides with its unique expansion i x U q ) in a completely unpredictable manner so there is no way o telling whether x + 1 belongs to U q as well. Fortunately, i q is suiciently small, the set o unique expansions is very simple, and i q is close to 2, then U q is large enough to satisy U q U q = [ 1/(q 1), 1/(q 1)] see Section 4. Lemma 2.3. Let G < q q ; then any unique expansion belongs to the set {0 k (10), 1 k (01), 0, 1 } with k 0. Proo. I x q := ((2 q)/(q 1), 1), then, by [5, Section 4], each unique expansion or this range o q is either (10) or (01). I x I q \ q, then any unique expansion is o the orm 1 k ε or 0 k ε, where ε is a unique expansion o some y q ([5, Corollary 15]).

5 EXPANSIONS IN NON-INTEGER BASES 5 Proposition 2.4. The smallest element o B 2 is q 2, the appropriate root o (2.2) x 4 = 2x 2 + x + 1, with a numerical value Furthermore, B 2 (G, q ) = {q 2 }. Proo. Let q be the cubic unit which satisies (2.3) x 3 = 2x 2 x + 1, q We irst show that q B 2. By Lemma 2.2, it suices to produce y U q such that y + 1 U q as well. Note that q satisies x 4 = x 3 + x (together with 1); put y ( ) q. Then y + 1 ( ) q, both unique expansions by Lemma 2.3. Hence in B 2 q. This makes our search easier, because by Lemma 2.3, each unique expansion or q (G, q ) belongs to the set {0 k (10), 1 k (01), 0, 1 } with k 0. Let us show irst that the two latter cases are impossible or q (G, q ). Indeed, i x (10 ) q had exactly two expansions in base q, then the other expansion would be o the orm (01 k (01) ) q, which would imply 1 = 1/q + 1/q /q k + 1/q k+2 + 1/q k+4 + with k 1. I k 2, then 1 < 1/q + 1/q 2 + 1/q 4, i.e., q > q ; k = 1 implies q = G. The case o the tail 1 is completely analogous To simpliy our notation, put λ = q 1 (1/q, 1/G). So let x (0 l 1 (10) ) q and x + 1 (1 k 1 (01) ) q, both in U q, with l 1, k 1. Then we have 1 + λl 1 λ = λ λk λ + λk 1 1 λ. 2 Simpliying this equation yields (2.4) λ l + λ k = 2λ 2 + λ 1. In view o symmetry, we may assume k l. Case 1: l = 1. This implies λ k = 2λ 2 1, whence 2λ 2 1 > 0, i.e., λ > 1/ 2 > 1/G. Thus, there are no solutions o (2.4) lying in (1/q, 1/G) or this case. Case 2: l = 2. Here λ k = λ 2 + λ 1 > 0, whence λ > 1/G. Thus, there are no solutions here either. Case 3: l = 3. We have (2.5) λ k = λ 3 + 2λ 2 + λ 1. Note that the root o (2.5) as a unction o k is decreasing. For k = 3 the root is above 1/G, or k = 4 it is exactly 1/G. For k = 5 the root o (2.5) satisies x 5 = x 3 + 2x 2 + x 1, which can be actorized into x 4 + x 3 + 2x 2 = 1, i.e., the root is exactly 1/q 2. Finally, or k = 6 the root satisies x 6 = x 3 + 2x 2 + x 1, which actorizes into x 3 x 2 + 2x 1 = 0, i.e., λ = 1/q. For k > 6 the root o (2.5) lies outside the required range. Case 4: l = 4, k {4, 5}. For k = 4 the root o 2x 4 = 2x 2 + x 1 is < 1/q = I k = 5, then the root is 0.543, i.e., even smaller. Hence there are no appropriate solutions o (2.4) here.

6 6 NIKITA SIDOROV Case 5: I l 5 and k 5, then the LHS o (2.4) is less than 2G 5 < 0.2, whereas the RHS is greater than 2q 2 + q 1 1 > 0.21, whence there are no solutions o (2.4) in this case. I l = 4, k 6, then, similarly, λ k + λ l λ 4 + λ 6 < G 4 + G 6 < Thus, the only case which produces a root in the required range is Case 3, which yields 1/q 2. Hence (2.6) (G, q ) B 2 = {q 2 }. Remark 2.5. Let q = q 2 and let y (0000(10) ) q2 U q and y + 1 (11(01) ) q2 U q. We thus see that in this case the tail o the expansion does change, rom (10) to (01). (Not the period, though!) Also, the proo o Lemma 2.2 allows us to construct x U q (2) 2 explicitly, namely, x (011(01) ) q2 (10000(10) ) q2, i.e., x A slightly more detailed study o equation (2.4) shows that it has only a inite number o solutions λ (1/q KL, 1/G). In order to construct an ininite number o q B 2 (q, q KL ), one thus needs to consider unique expansions with tails dierent rom (01) : Proposition 2.6. The set B 2 (q, q KL ) is ininite countable. Proo. We are going to develop the idea we used to show that q B 2. Namely, let (q (n) ) n 1 be the sequence o algebraic numbers speciied by their greedy expansions o 1: q (1) : 1 (11 0 ) (1) q = G, q (2) : 1 ( ) (2) q = q q (3) : 1 ( ) q (3) q (n). : 1 (m 1,, m 2 n 0 ) (n) q, where (m n ) is the Thue-Morse sequence see Introduction. It is obvious that q (n) now deine the sequence z n as ollows: whence [5, Proposition 9] implies that z n U (n) q or all n 2. z n (0 2n (m 2 n 1 +1 m 2 n) ) (n) q, z n + 1 (m 1,, m 2 n 1(m 2 n 1 +1 m 2 n) ) (n) q. Lemma 2.7. We have B m B 2 or any natural m 3. q KL. We and z n + 1 U (n) q, whence by Lemma 2.2, q (n) B 2

7 EXPANSIONS IN NON-INTEGER BASES 7 Proo. I q B m or some natural m 3, then the branching argument immediately implies that there exists x U (m ) q, with 1 < m < m. Hence, by induction, there exists x U q (2). Thereore, B m B 2 or all m N \ {1}. Our next result shows that a weaker analogue o Theorem 2.1 holds without assuming q being transcendental, provided q < q. Theorem 2.8. For any q (G, q 2 ) (q 2, q ), each x I q has either a unique expansion or ininitely many expansions o the orm (1.1) in base q. Here G = 1+ 5 and q 2 2 and q are given by (2.2) and (2.3) respectively. Proo. It ollows immediately rom Proposition 2.4, Lemma 2.7 and relation (2.6) that B m (G, q ) {q 2 } or all m N \ {1}. Corollary 2.9. For q (G, q 2 ) (q 2, q ) each x J q has ininitely many expansions in base q. Proo. It suices to recall that each x J q has at least two expansions in base q and apply Theorem 2.8. It is natural to ask whether the claim o Theorem 2.8 can be strengthened in the direction o getting rid o q B ℵ0 so we could claim that a stronger version o Theorem 2.1 holds or q < q. It turns out that the answer to this question is negative. Notice irst that B ℵ0 B 2, since G B ℵ0 \ B 2. Our goal is to show that in act, B ℵ0 \ B 2 is ininite see Proposition 2.11 below. We begin with a useul deinition. Let x (a 1, a 2, ) q ; we say that a m is orced i there is no expansion o x in base q o the orm x (a 1,, a m 1, b m, ) q with b m a m. Lemma Let q > G and x (a 1,, a m, (01) ) q (b 1,, b k, a 1,, a m, (01) ) q, where a 1 b 1, and assume that a 2,, a m are orced in the irst expansion and b 2,, b k are orced in the second expansion. Then q B ℵ0. Proo. Since all the symbols in the irst expansion except a 1, are orced, the set o expansions or x in base q is as ollows: a 1,, a m, (01), b 1,, b k, a 1,, a m, (01), b 1,, b k, b 1,, b k, a 1,, a m, (01),. i.e., clearly ininite countable. The ladder branching pattern or x is depicted in Fig. 2. Proposition The set B ℵ0 (q 2, q ) is ininite countable. Proo. Deine q (n) as the unique positive solution o (10000(10) ) q (n) ( 0 11(01) n (10) ) q (n)

8 8 NIKITA SIDOROV x. FIGURE 2. A branching or countably many expansions (never mind the boxes or the moment) and put λ n = 1/q (n). A direct computation shows that λ 2n+1 n = 1 λ n 2λ 2 n + λ 3 n + λ 5 n, 1 λ n 2λ 2 n + λ 5 n whence λ n 1/q 2 (as 1/q 2 is a root o 1 x 2x 2 + x 3 + x 5 ), and consequently, q (n) q 2. By Lemma 2.10, i we show is that each symbol between the boxed 0 and the boxed 1 is orced, then q (n) B ℵ0. Let us prove it. Notice that i x (a 1, a 2, ) q, then a 1 similarly, a 1 = 1 is orced i 1 a kq k > 1/q(q 1). We need the ollowing = 0 is orced i and only i 1 a kq k < 1/q; Lemma (1) I q > G, m 0 and x (1(01) m 1 ) q, then the irst 1 is orced (where stands or an arbitrary tail). (2) I q > q 2, m 1 and x ((01) m 10000(10) ) q, then the irst 0 is orced. Proo. (1) By the above remark, we need to show that 1 q + 1 q q + 1 2m+1 q > 1 2m+2 q(q 1), which is equivalent to (with λ = 1/q < 1/G) 1 λ 2m+2 + λ 2m+1 > λ 1 λ 2 1 λ or 1 λ λ 2 > λ 2m+1 λ 2m+2 λ 2m+3, which is true, in view o 1 λ λ 2 > 0 and λ 2m+1 < 1.

9 (2) Putting λ = 1/q, we need to show that EXPANSIONS IN NON-INTEGER BASES 9 λ 2 + λ λ 2m + λ 2m+1 + λ2m+6 1 λ 2 < λ. This is equivalent to λ 2m < 1 λ λ2 1 λ λ 2 + λ. 5 The LHS in this inequality is a decreasing unction o m, and or m = 1 we have that it holds or λ < 0.59, whence q > q 2 suices. The proo o Proposition 2.11 now ollows rom the deinition o the sequence (q (n) ) n 1 and rom Lemma Remark The set B ℵ0 (G, q 2 ) is nonempty either: take q ω to be the appropriate root o x 5 = x 4 + x 3 + x 1, with the numerical value Then x (100(10) ) qω ( (10) ) qω, and similarly to the above, one can easily show that the three 1s between the boxed symbols are orced. Hence, by Lemma 2.10, q ω B ℵ0. The question whether in B ℵ0 = G, remains open. Remark The condition o q being transcendental in Theorem 2.1 is probably not necessary even or q > q. It would be interesting to construct an example o a amily o algebraic q (q, q KL ) or which the dichotomy in question holds. 3. MIDDLE ORDER: q JUST ABOVE q KL This case looks rather diicult or a hands-on approach, because, as we know, the set U q or q > q KL contains lots o transcendental numbers x, or which the tails o expansions in base q or x and x + 1 are completely dierent. However, a very simple argument allows us to link B 2 to the well-developed theory o unique expansions or x = 1. Following [7], we introduce U := {q (1, 2) : x = 1 has a unique expansion in base q}. Recall that in [7] it was shown that min U = q KL. Lemma 3.1. We have U B 2. Consequently, the set B 2 (q KL, q KL + δ) has the cardinality o the continuum or any δ > 0. Proo. Since x = 0 has a unique expansion in any base q, the irst claim is a straightorward corollary o Lemma 2.2. The second claim ollows rom the act that U (q KL, q KL + δ) has the cardinality o the continuum or any δ > 0, which in turn is a consequence o the act that the closure o U is a Cantor set see [9, Theorem 1.1].

10 10 NIKITA SIDOROV 4. TOP ORDER: q CLOSE TO m = 2. We are going to need the notion o thickness o a Cantor set. Our exposition will be adapted to our set-up; or a general case see, e.g., [1]. A Cantor set C R is usually constructed as ollows: irst we take a closed interval I and remove a inite number o gaps, i.e., open subintervals o I. As a result we obtain a inite union o closed intervals; then we continue the process or each o these intervals ad ininitum. Consider the nth level, L n ; we have a set o newly created gaps and a set o bridges, i.e., closed intervals connecting gaps. Each gap G at this level has two adjacent bridges, P and P. The thickness o C is deined as ollows: τ(c) = in n min min G L n { } P G, P, G where I denotes the length o an interval I. For example, i C is the standard middle-thirds Cantor set, then τ(c) = 1, because each gap is surrounded by two bridges o the same length. The reason why we need this notion is the theorem due to Newhouse [11] asserting that i C 1 and C 2 are Cantor sets, I 1 = conv(c 1 ), I 2 = conv(c 2 ), and τ(c 1 )τ(c 2 ) > 1 (where conv stands or convex hull), then C 1 + C 2 = I 1 + I 2, provided the length o I 1 is greater than the length o the maximal gap in C 2 and vice versa. In particular, i τ(c) > 1, then C + C = I + I. Notice that U q is symmetric about the centre o I q because whenever x (a 1, a 2, ) q, 1 one has x (1 a q 1 1, 1 a 2, ) q. Recall that Lemma 2.2 yields the criterion 1 U q U q or q B 2. Thus, we have U q = 1/(q 1) U q, whence U q U q = U q + U q 1/(q 1). Hence our criterion can be rewritten as ollows: (4.1) q B 2 q q 1 U q + U q. It has been shown in [5] that the Hausdor dimension o U q tends to 1 as q 2. Thus, one might speculate that or q large enough, the thickness o U q is greater than 1, whence by the Newhouse theorem, U q + U q = 2I q, which implies the RHS o (4.1). However, there are certain issues to be dealt with on this way. First o all, in [10] it has been shown that U q is not necessarily a Cantor set or q > q KL. In act, it may contain isolated points and/or be non-closed. This issue however is not really that serious because U q is known to dier rom a Cantor set by a countable or empty set [10], which is negligible in our set-up. A more serious issue is the act that even i the Hausdor dimension o a Cantor set is close to 1, its thickness can be very small. For example, i one splits one gap by adding a very small bridge, the thickness o a resulting Cantor set will become very small as well! In other words, τ is not at all an increasing unction with respect to inclusion. Nonetheless, the ollowing result holds: Lemma 4.1. Let T denote the real root o x 3 = x 2 + x + 1, T Then [ ] 2 (4.2) U q + U q = 0,, q T. q 1

11 EXPANSIONS IN NON-INTEGER BASES 11 Proo. Let Σ q denote the set o all sequences which provide unique expansions in base q. It has been proved in [5] that Σ q Σ q i q < q ; hence Σ T Σ q. Note that by [5, Lemma 4], Σ T can be described as ollows: it is the set o all 0-1 sequences which do not contain words 0111 and 1000 and also do not end with (110) or (001). Let Σ T Σ T denote the set o 0-1 sequences which do not contain words 0111 and Note that by the cited lemma, Σ T Σ q whenever q > T. Denote by π q the projection map rom {0, 1} N onto I q deined by the ormula π q (a 1, a 2, ) = a n q n, and put V q = π q ( Σ T ). Since Σ T is a perect set in the topology o coordinate-wise convergence, and since π 1 q Uq is a continuous bijection, π q : Σ T V q is a homeomorphism, whence V q is a Cantor set which is a subset o U q or q > T. I q = T, then π q (Σ T ) = π q ( Σ T ), hence the same conclusion about V T. In view o Newhouse s theorem, to establish (4.2), it suices to show that τ(v q ) > 1, because conv(v q ) = conv(u q ) = I q. To prove this, we need to look at the process o creation o gaps in I q. Note that any gap is the result o the words 000 and 111 in the symbolic space being orbidden. The irst gap thus arises between π q ([0110]) = [ λ 2 + λ 3, λ 2 + λ 3 + λ5 1 λ] and πq ([1001]) = [ λ + λ 4, λ + λ 4 + λ5 1 λ]. (Here, as above, λ = q 1 (1/2, 1/T ) and [i 1 i r ] denotes the corresponding cylinder in {0, 1} N.) The length o the gap is λ + λ 4 ( λ 2 + λ 3 + λ5 1 λ), which is signiicantly less than the length o either o its adjacent bridges. Furthermore, it is easy to see that any new gap on level n 5 always lies between π q ([a0110]) and π q ([a1001]), where a is an arbitrary 0-1 word o the length n 4 which contains neither 0111 nor The length o the gap is thus independent o a and equals λ n 3 + λ n λ n 2 λ n 1 λ n+1 1 λ. As or the bridges, to the right o this gap we have at least the union o the images o the cylinders [a1001], [a1010] and [a1011], which yields the length λ n 3 + λn 1 1 λ λn 3 λ n = λ n 1 1 λ λn. Hence n=1 gap bridge 1 1 λ λ2 + λ 3 λ4 λ 2 1 λ λ3 1 λ = 1 2λ + 2λ3 2λ 4 λ 2 λ 3 + λ 4. This raction is indeed less than 1, since this is equivalent to the inequality (4.3) 3λ 4 3λ 3 + λ 2 + 2λ 1 > 0, which holds or λ > 0.48.

12 12 NIKITA SIDOROV The bridge on the let o the gap is [π q (a010 ), π q (a01101 )], and its length is bridge 2 = λ n 2 + λ n 1 + λn+1 1 λ λn 2 = λ n 1 + λn+1 = λn 1 1 λ 1 λ λn = bridge 1, whence bridge 2 < gap, and we are done. As an immediate corollary o Lemma 4.1 and (4.1), we obtain Theorem 4.2. For any q [T, 2) there exists x I q which has exactly two expansions in base q. Remark 4.3. The constant T in the previous theorem is clearly not sharp inequality (4.3), which is the core o our proo, is essentially the argument or which we need a constant close to T. Considering U q directly (instead o V q ) should help decrease the lower bound in the theorem (although probably not by much) m 3. Theorem 4.4. For each m N there exists γ m > 0 such that Furthermore, or any ixed m N, (2 γ m, 2) B j, 2 j m. (4.4) lim q 2 dim H U (m) q = 1, where, as above, U (m) q denotes the set o x I q which have precisely m expansions in base q. Proo. Note irst that i q B m and 1 U q (m) U q, then q B m+1. Indeed, analogously to the proo o Lemma 2.2, i y U q and y + 1 U q (m), then (y + 1)/q lies in the interval J q, and the shit o its expansion beginning with 1, belongs to U q, and the shit o its expansion beginning with 0, has m expansions in base q. Similarly to Lemma 4.1, we want to show that or a ixed m 2, U (m) q U q = [ 1 q 1, 1 q 1 i q is suiciently close to 2. We need the ollowing result which is an immediate corollary o [6, Theorem 1]: Proposition. For each E > 0 there exists > 0 such that or any two Cantor sets C 1, C 2 R such that conv(c 1 ) = conv(c 2 ) and τ(c 1 ) >, τ(c 2 ) >, their intersection C 1 C 2 contains a Cantor set C with τ(c) > E. Let T k be the appropriate root o x k = x k 1 + x k x + 1. Then T k 2 as k +, and it ollows rom [5, Lemma 4] that Σ Tk is a Cantor set o 0-1 sequences which do not contain 10 k nor 01 k and do not end with (1 k 1 0) or (0 k 1 1). Similarly to the proo o Lemma 4.1, we introduce the sets Σ Tk and deine V (k) q ] = π q ( Σ Tk ) or q > T k. For the same reason as above, V q (k) is always a Cantor set or q T k. Using the same arguments as in the aorementioned proo, one can show that or any M > 1 there exists k N such that τ ( V q (k) ) > M or all q > Tk. More precisely, any gap which is created on the nth level is o the orm [π q (a01 k 1 0) + λ n+1 /(1 λ), π q (10 k 1 1)], while the

13 EXPANSIONS IN NON-INTEGER BASES 13 bridge on the right o this gap is at least [π q (10 k 1 1), π q (101 k 1 ) + λ n+1 /(1 λ)]; a simple computation yields (4.5) bridge gap λ2 + λ k λ k+1 2λ k λ k λ λ2 1 2λ +, k +, since λ T 1 k 1/2 as k +. The same argument works or the bridge on the let o the gap. We know that 1 (m) (U q q (U q 1)) is a subset o U q (m+1) J q ; we can also extend it to I q \ J q by adding any number 0s or any number o 1s as a preix to the expansion o any x 1 (m) (U q q (U q 1)). Thus, conv ( U q (m+1) ) = Iq, provided this set is nonempty. Let us show via an inductive method that U q (m+1) is nonempty or m 2. Consider U q (3) ; by the above, there exists k 3 such that or q > T k3, the intersection U q (U q 1) contains a Cantor set o thickness greater than 1. Extending it to the whole o I q, we obtain a Cantor set [ o thickness ] greater than 1 whose support is I q. This set is contained in U q (2), whence U q (2) U q = 1, 1, q 1 q 1 yielding that U q (3) or q > T k3. Finally, by increasing q, we make sure U q (3) contains a Cantor set o thickness greater than 1, which implies U q (4), etc. Thus, or any m 3 there exists k m such that U q (m) i q > T km. Putting γ m = 2 T km completes the proo o the irst claim o the theorem. To prove (4.4), notice that rom (4.5) it ollows that τ(v q (k) ) + as k +. Since or any Cantor set C, log 2 dim H (C) log(2 + 1/τ(C)) (see [12, p. 77]), we have dim H (V q (k) ) 1 as k +, whence dim H (U q (m) ) 1 as q 2, in view o T k 2 as k +. Remark 4.5. From the proo it is clear that the constructed sequence γ m 0 as m +. It would be interesting to obtain some bounds or γ m ; this could be possible, since we roughly know how depends on E in the proposition quoted in the proo. Namely, rom [6, Theorem 1] and the remark in p. 888 o the same paper, it ollows that or large E we have E. Finally, in view o γ m 0, one may ask whether actually B m. It turns out that m N ℵ 0 the answer to this question is airmative. Proposition 4.6. For q = T and any m N ℵ 0 there exists x m I q which has m expansions in base q. Proo. Let irst m N. We claim that x m (1(000) m (10) ) q U (m+1) q. Note irst that i some x (10001 ) q has an expansion (0, b 2, b 3, b 4, ) in base q, then b 2 = b 3 = b 4 = 1 because (01101 ) q < ( ) q, a straightorward check.

14 14 NIKITA SIDOROV Thereore, i x m (0, b 2, b 3, ) q or m = 1, then b 2 = b 3 = b 4 = 1, and since 1 = 1/q +1/q 2 +1/q 3, we have (b 5, b 6, ) q = ((10) ) q, the latter being a unique expansion. Hence x 1 has only two expansions in base q. For m 2, we still have b 2 = b 3 = 1, but b 4 can be equal to 0. This, however, prompts b 5 = b 6 = 1, and we can continue with b 3i 1 = b 3i = 1, b 3i+1 = 0 until b 3j 1 = b 3j = b 3j+1 = 1 or some j m, since (10 ) q ((011) j 10 ) q, whence (1(000) j (10) ) q > ((011) j 01 ) q or any j 1. Thus, any expansion o x m in base q is o the orm x m ((011) j 1(000) m j (10) ) q, 0 j m, i.e., x m U q (m+1). For m = ℵ 0, we have x = lim m x m (10 ) q. Notice that x ( ) q with the irst two 1s clearly orced so we can apply Lemma 2.10 to conclude that x U (ℵ 0) q, whence q = T B ℵ0 as well. Remark 4.7. The choice o the tail (10) in the proo is unimportant; we could take any other tail, as long as it is a unique expansion which begins with 1. Thus, or q = T, dim H U (m) q = dim H U q, m N. This seems to be a very special case, because typically one might expect a drop in dimension with m. Note that in [5] it has been shown that dim H U T = log G/ log T SUMMARY AND OPEN QUESTIONS Summing up, here is the list o basic properties o the set B 2 : The set B 2 (G, q KL ) is ininite countable and contains only algebraic numbers (the lower order 2 ). The latter claim is valid or B m with m 3, although it is not clear whether B m (G, q KL ) is nonempty. B 2 (q KL, q KL + δ) has the cardinality o the continuum or any δ > 0 (the middle order ). [T, 2) B 2 (the top order ), with a similar claim about B m with m 3. Here are a ew open questions: Is B 2 closed? Is B 2 (G, q KL ) a discrete set? Is it true that dim H (B 2 (q KL, q KL + δ)) > 0 or any δ > 0? Is it true that dim H (B 2 (q KL, q KL + δ)) < 1 or some δ > 0? What is the value o in B m or m 3? What is the smallest value q 0 such that U q + U q = 2I q or q q 0? Is in B ℵ0 = G? Does B ℵ0 contain an interval as well? 2 Our terminology is borrowed rom cricket.

15 EXPANSIONS IN NON-INTEGER BASES 15 Acknowledgement. The author is grateul to Boris Solomyak or stimulating questions and to Martijn de Vries and especially the anonymous reeree or indicating some errors in the preliminary version o this paper. q equation numerical value G x 2 = x q ω x 5 = x 4 + x 3 + x q 2 x 4 = 2x 2 + x q x 3 = 2x 2 x q KL 1 m nx n+1 = T x 3 = x 2 + x TABLE 5.1. The table o constants used in the text. REFERENCES [1] S. Astels, Thickness measures or Cantor sets, Electron. Res. Announc. Amer. Math. Soc. 5 (1999), [2] K. Dajani and C. Kraaikamp, Random β-expansions, Ergodic Theory Dynam. Systems 23 (2003), [3] P. Erdős, I. Joó, On the number o expansions 1 = q ni, Ann. Univ. Sci. Budapest 35 (1992), [4] P. Erdős, I. Joó and V. Komornik, Characterization o the unique expansions 1 = i=1 q ni and related problems, Bull. Soc. Math. Fr. 118 (1990), [5] P. Glendinning and N. Sidorov, Unique representations o real numbers in non-integer bases, Math. Res. Lett. 8 (2001), [6] B. Hunt, I. Kan and J. Yorke, When Cantor sets intersect thickly, Trans. Amer. Math. Soc. 339 (1993), [7] V. Komornik and P. Loreti, Unique developments in non-integer bases, Amer. Math. Monthly 105 (1998), [8] V. Komornik and P. Loreti, On the expansions in non-integer bases, Rend. Mat. Appl. 19 (1999), [9] V. Komornik and P. Loreti, On the topological structure o univoque sets, J. Number Theory 122 (2007), [10] V. Komornik and M. de Vries, Unique expansions o real numbers, preprint. [11] S. Newhouse, The abundance o wild hyperbolic sets and non-smooth stable sets or dieomorphisms, Inst. Hautes Études Sci. Publ. Math. 50 (1979), [12] J. Palis and F. Takens, Hyperbolicity and sensitive chaotic dynamics at homoclinic biurcations, Cambridge Univ. Press, Cambridge, [13] W. Parry, On the β-expansions o real numbers, Acta Math. Acad. Sci. Hung. 11 (1960), [14] A. Rényi, Representations or real numbers and their ergodic properties, Acta Math. Acad. Sci. Hung. 8 (1957) [15] N. Sidorov, Almost every number has a continuum o β-expansions, Amer. Math. Monthly 110 (2003), [16] N. Sidorov, Universal β-expansions, Period. Math. Hungar. 47 (2003), [17] N. Sidorov and A. Vershik, Ergodic properties o Erdős measure, the entropy o the goldenshit, and related problems, Monatsh. Math. 126 (1998), SCHOOL OF MATHEMATICS, THE UNIVERSITY OF MANCHESTER, OXFORD ROAD, MANCHESTER M13 9PL, UNITED KINGDOM. SIDOROV@MANCHESTER.AC.UK