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

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)/2 1.61803, 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, 2008. 2000 Mathematics Subject Classiication. 11A63. Key words and phrases. Beta-expansion, Cantor set, thickness, non-integer base. 1

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 = 0110 1001 1001 0110, i.e., a ixed point o the morphism 0 01, 1 10. 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 1.78723. 1 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.

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 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]).

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 1.71064. 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 1.75488 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 2 + 1 (together with 1); put y (0000010101 ) q. Then y + 1 (11010101 ) 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 2 + + 1/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 1 2 1 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 0.565 < 1/q = 0.569. I k = 5, then the root is 0.543, i.e., even smaller. Hence there are no appropriate solutions o (2.4) here.

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 < 0.202. 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 0.64520. 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 (1101 0 ) (2) q = q q (3) : 1 (1101 0011 0 ) 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

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 2.10. 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 2.11. 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 1 1 0000(10) ) q (n)

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 2.12. (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 + + 1 3 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.

(2) Putting λ = 1/q, we need to show that EXPANSIONS IN NON-INTEGER BASES 9 λ 2 + λ 4 + + 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 2.12. Remark 2.13. 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 1.68042. Then x (100(10) ) qω ( 0 111 1 00(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 2.14. 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 NIKITA SIDOROV 4. TOP ORDER: q CLOSE TO 2 4.1. 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 1.83929. Then [ ] 2 (4.2) U q + U q = 0,, q T. q 1

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 1000. 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 1000. 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 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). 4.2. 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 2 + + 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

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+1 + 1 2λ 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 < (100010 ) q, a straightorward check.

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 (011 10 ) 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 0.78968. 5. 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.

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