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Orbits of N-expansions with a finite set of digits

Orbits of N-expansions with a finite set of digits For N ∈ N and α ∈ R such that 0 <α ≤ N − 1, we define I := [α, α + 1] and ≥2 α − − I := [α, α + 1) and investigate the continued fraction map T : I → I , which is α α α α N N defined as T (x ) := − d(x ), where d : I → N is defined by d(x ) := − α . α α x x For N ∈ N , for certain values of α, open intervals (a, b) ⊂ I exist such that for ≥7 α almost every x ∈ I there is an n ∈ N for which T (x)/ ∈ (a, b) for all n ≥ n . α 0 0 These gaps (a, b) are investigated using the square Υ := I × I , where the orbits α α T (x ), k = 0, 1, 2,... of numbers x ∈ I are represented as cobwebs. The squares Υ are the union of fundamental regions, which are related to the cylinder sets of the map T , according to the finitely many values of d in T . In this paper some clear α α conditions are found under which I is gapless. If I consists of at least five cylinder α α sets, it is always gapless. In the case of four cylinder sets there are usually no gaps, except for the rare cases that there is one, very wide gap. Gaplessness in the case of two or three cylinder sets depends on the position of the endpoints of I with regard to the fixed points of I under T . α α Keywords Continued fractions · Dynamical systems · Gaps Mathematics Subject Classification Primary 11J70; Secondary 37E05 Communicated by H. Bruin. B Jaap de Jonge c.j.dejonge@uva.nl Cor Kraaikamp C.Kraaikamp@tudelft.nl Hitoshi Nakada nakada@math.keio.ac.jp Korteweg-de Vries Institute for Mathematics, Science Park 105-107, 1098 XG Amsterdam, The Netherlands Department of Electrical Engineering, Mathematics and Computer Science, Delft University of Technology, Mekelweg 4, 2628 CD Delft, The Netherlands Department of Mathematics, Keio University, Yokohama 223-8522, Japan 123 80 J. de Jonge et al. 1 Introduction In 2008, Edward Burger and his co-authors introduced in [2] new continued frac- tion expansions, the so-called N -expansions, which are nice variations of the regular continued fraction (RCF) expansion. These N -expansions have been studied in vari- ous papers since; see [1,3,7]. In [8], a subclass of these N -expansions is introduced, for which the digit set is always finite. These particular N -expanions are defined as follows: For N ∈ N and α ∈ R such that 0 <α ≤ N − 1, let I := [α, α + 1] and ≥2 α I := [α, α + 1). Hereafter we denote by N the set of positive integers n ≥ k, while ≥k we write N if k = 1. We define the N -expansion map T : I → I (or I )as α α α T (x ) := − d(x ), (1) where d : I → N is defined by N N d(x ) := − α , if either x ∈ (α, α + 1] or both x = α and − α/ ∈ Z x α and N N d(α) = − α − 1, if − α ∈ Z. α α Note that if − α ∈ Z, we have that T (α) = α + 1. This is the only case in which the range of T is I and not I . α α For a fixed α ∈ (0, N − 1] and x ∈ I we define for n ∈ N n−1 d = d (x ) := d(T (x )). n n Note that for α ∈ (0, N − 1] fixed, there are only finitely many possibilities for each d . Applying (1), we obtain for every x ∈ I a continued fraction expansion of the form N N N x = T (x ) = = = ··· = , (2) d + T (x ) N N 1 α d + d + 1 1 d + T (x ) d + d + which we will throughout this paper write as x =[d , d , d ,...] (note that this 1 2 3 N ,α expansion is infinite for every x ∈ I , since 0 ∈ / I ); we will call the numbers d , i ∈ N, α α i the partial quotients or digits of this N-continued fraction expansion of x;see [3,8], 123 Orbits of N-expansions with a finite set of digits 81 Δ Δ 2 1 r r p α 2 f 1 2  f  α +1 2 1 1 2 Fig. 1 N = 51, α = 6 where these continued fractions (also with a finite set of digits) were introduced and elementary properties were studied (such as the convergence in reference [3]). In each cylinder set Δ := {x ∈ I ; d(x ) = i } of rank 1, with d ≤ i ≤ d , i α min max where d := d(α) is the largest partial quotient, and d := d(α + 1) the smallest max min one given N and α,the map T obviously has one fixed point f . As of now we will α i write simply ‘cylinder set’ for ‘cylinder set of rank 1’. It is easy to see that 4N + i − i f = f (N ) = , for d ≤ i ≤ d . (3) i i min max Note that − α ∈ Z if and only if for some d ∈ N we have that α = f , while ≥2 d+1 for any α ∈[ f , f ) we have that Δ =∅. 0 d+2 d+1 d+1 Given N ∈ N ,welet α = N − 1 be the largest value of α we consider. ≥2 max The reason for this is that for larger values of α we would have 0 as a partial quotient as well. Since T (x ) =−N /x and because 0 <α ≤ N − 1, we have |T (x )| > 1 α α on I . From this it follows that the fixed points act as repellers and that the maps T are expanding if 0 <α ≤ N − 1. This is equivalent to the convergence of the N -expansion of all x ∈ I . Each pair of consecutive cylinders sets (Δ ,Δ ) is divided by a discontinuity i i −1 N N point p (N,α) of T , satisfying − i = α,so p = . A cylinder set Δ is called i α i i p α+i full if T (Δ ) = I (or I ). If a cylinder set is not full, it contains either α (in which α i α case T (α) < α + 1) or α + 1 (in which case T (α + 1)>α), and is called incomplete. α α On account of our definition of T , cylinder sets will always be an interval, never consist of one single point. The main object of this paper is the sequence T (x ), n = 0, 1, 2,...,for x ∈ I , which is called the orbit of x under T . More specifically, we are interested in subsets of I that we will call gaps for such orbits. Before we will give a proper definition of ‘gap’, we will give an example of orbits of points in I for a pair {N,α}. Note that, since the fixed points are repellent, orbits cannot remain in one cylinder set indefinitely if x ∈ I is not a fixed point of T , so any orbit will show an infinite migration between α α cylinder sets. A naive approach is to compute the orbits of many points of I and obtain a plot of the asymptotic behaviour of these orbits by omitting the first, say hundred, iterations. Figure 1 shows such a plot for N = 51 and α = 6. It appears that there are parts of I (illustrated by dashed line segments) that are not visited by any orbit after many iterations of T . i i In fact, setting  = T (α) and r = T (α + 1), the orbit of any point—apart i i α α from the fixed points f and f —after once having left the interval (r , r ) ⊂ Δ or 1 2 2 1 2 ( , ) ⊂ Δ of Fig. 1, will never return to it. 1 2 1 For reasons of legibility we will usually omit suffices such as ‘(N )’, ‘(N,α)’or‘(N , d)’. 123 82 J. de Jonge et al. Fig. 2 N = 51, α = 6, x = 6.5 2 1 In order to get a better understanding of the orbits of N -expansions, it is useful to consider the graphs of T , which are drawn in the square Υ (= Υ ) := I × I .This α α N ,α α square is divided in rectangular sets of points  := {(x , y) ∈ Υ : d(x ) = i }, which i α are the two-dimensional fundamental regions associated with the one-dimensional cylinder sets we already use. We will call these regions in short cylinders. Now consider (x , T (x )) ∈ Υ . Then (x , T (x )) goes to (T (x ), T (x )) under T . Regarding this, T α α α α α α has one fixed point F := ( f , f ) in each  . We will denote the dividing line between i i i i and  by l , which is the set { p }×[α, α + 1), with p the discontinuity point i i −1 i i i between Δ and Δ . In case T (Δ ) = I , we will call the cylinder  full and i i −1 α i i the branch of the graph of T in  complete; if a cylinder is not full, we will call it α i and its associated branch of T incomplete. We will call the collection of Υ and its α α associated branches, fixed points and dividing lines an arrangement of Υ .If Υ is a α α union of full cylinders, we will call the associated arrangement also full. Figure 2 is an example of such an arrangement, in which a part of the cobweb is drawn associated with the orbit we investigated previously. The discontinuity point p = is now visible as a dividing line between Δ and Δ . 2 1 2 In [3,8] the arrangement for N = 4 and α = 1 is studied, consisting of two full cylinders  and  and not showing any gaps. On the other hand, the demonstration 1 2 5 13 of the interval ( , ) being a gap of the interval [2, 3] in the case (N,α) = (9, 2) 2 5 in [8] is done without referring to such an arrangement. In this paper, and even more so in the next paper, we will show that arrangements may considerably support the insight in the occurrence of gaps. We will now give a formal definition of gaps, which is slightly delicate, since − N T (I ) = I (or I if − α ∈ Z). α α α Definition 1 A maximal open interval (a, b) ⊂ I is called a gap of I if for almost α α every x ∈ I there is an n ∈ N for which T (x)/ ∈ (a, b) for all n ≥ n . α 0 0 123 Orbits of N-expansions with a finite set of digits 83 There are several papers where ‘gaps’ appear in number theoretic expansions. One of the first important examples is given by the so-called negative β-expansion, introduced and studied by Shunji Ito and Taizo Sadahiro [5]. For β> 1, let I := − , , and let the (−β)-transformation T : I → I be defined by β −β β β β+1 β+1 T (x ) =−βx − − βx . β + 1 Ito and Sadahiro characterise the admissible sequences of (−β)-expansions, give a necessary and sufficient condition for the (−β)-shift to be sofic, and explicitly deter- mine the invariant measure of the (−β)-transformations. In their Example 18, Ito and Sadahiro also noticed the existence of gaps, as intervals where the density of the invariant measure is zero. In [11], Lingmin Liao and Wolfgang Steiner studied nega- tive β-expansions defined on [0, 1], which are conjugate to the negative β-expansions defined by Ito and Sadahiro. In [11] a complete classification of the gaps and of the mixing properties of these negative β-expansions is given. For maps slightly more general than the negative β-maps the reader is referred to Franz Hofbauer’s paper [4]. In Theorem 2 we show that there is a unique absolutely continuous invariant prob- ability measure μ such that T is ergodic with respect to μ .Asin[11](fornegative α α α β-expansions), one also could define a gap as any interval in I where the density of μ is zero. Remark 1 In the example of Fig. 1 the intervals (r , r ) and ( , ) are gaps and 2 1 1 2 for x ∈ (r , r ) ∪ ( , )\{ f , f } there exists an n = n (x ) such that T (x)/ ∈ 2 1 1 2 1 2 0 0 (r , r ) ∪ ( , ) for n ∈ N . The ‘for almost’ formulation in the definition of ‘gap’ 2 1 1 2 ≥2 is necessary so as to exclude fixed points and pre-images of fixed points, i.e. points that are mapped under T to a fixed point, which may never leave an gap. In Sect. 5 we even find a class of gaps (a, b) such that for uncountably many x ∈ I and all n ∈ N ∪{0} we have T (x ) ∈ (a, b). Remark 2 Whenever we use the word ‘gap’ in relation to arrangements, we mean the gap of the associated interval I . In [8] computer simulations were used to get a more general impression of orbits of N -expansions. For a lot of values of α, plots such as Fig. 1 were stacked, for 0 <α ≤ α , so as to obtain graphs such as Fig. 3, with the values of α on the max vertical axis and at each height the corresponding interval I drawn. In the same paper, similar graphs are given for N = 9, 20, 36 and 100. In all cases it appears that ‘gaps’ such as in Fig. 1 appear for values of α equal to or not much smaller than α . max Since the plots in [8] are based on computer simulations, they do not actually show very small gaps (smaller than pixel size) nor clarify much the connection between the gaps for each N . Still, the suggestion is strong that for α sufficiently small there are no gaps. It also seems that for α large enough several disjoint gaps may occur. In Fig. 3 we see this for α near α = 50 − 1. max All ‘for all’ statements in this paper are with respect to Lebesgue measure. 123 84 J. de Jonge et al. Fig. 3 A simulation of intervals I with gaps if existent, for 0 <α ≤ 50 − 1and N = 50 In this paper we will not only investigate conditions for gaplessness, we will also show that simulations such as Fig. 3 fail to reveal the existence, for certain N and α, of one extremely large gap in plots such as Fig. 3 below the last visible gap. In a subsequent paper we will go into another very interesting property of orbits of N - expansions that is hardly revealed by simulations such as Fig. 3: the existence of large numbers of gaps for large N and α close to α . But now we will concentrate on max gaplessness. Remark 3 If no gaps exist with non-empty intersection with a cylinder set, we call the cylinder set gapless. In Sect. 2 we will consider two classes of arrangements that have no gaps: full arrangements and specific arrangements with more than three cylinders. The gapless- ness of the latter class, involving the proof of Theorem 4, for which some preliminary results will be presented shortly, is largely given in Sect. 2, but involve some intri- cacies for small values of N so as to finish it at the end of this paper. In Sect. 3 we will consider arrangements with two cylinders and in Sect. 4 we will concentrate on arrangements with three cylinders, but will prove a sufficient condition for gaplessness that is valid for arrangements with any number of cylinders larger than 2. Finally, in Sect. 5 we will prove a result on gaps in certain arrangements with four cylinders and we will finish the proof of Theorem 4. After that, it is merely a matter of checking that for N ∈{2, 3, 4, 5, 6} all arrangements are gapless. Although most results in this paper seem intuitively clear and obvious, our experi- ence with these continued fraction maps is they are surprisingly unruly. The simulations from Fig. 3 may serve as an example that things are not what they always seem to be. From these simulations one might draw the (erroneous) conclusion that the number of gaps is at most 3. In a forthcoming paper we show however that the number of gaps is a function of N and α, and is growing larger and larger as N tends to infinity. 123 Orbits of N-expansions with a finite set of digits 85 Although the proofs in this paper are often elementary, they require case distinctions and cumbersome calculations. 2 Full arrangements and arrangements with more than four cylinders If I consists of full cylinder sets only, we obviously have no gaps. In this situation the mutual relations between N , α and d(α) show a great coherence, as expressed in the following theorem: Theorem 1 The interval I consists of m full cylinder sets, with m ∈ N , if and only α ≥2 if there is a positive integer k such that ⎪ α = k, (4) N = mk(k + 1), d(α) = (m − 1)(k + 1). Proof of Theorem 1 Writing d := d(α), the interval I is the union of m full cylinder sets if and only if − d = α + 1, (5) − (d − m + 1) = α. α+1 Note that the first equation in (5) can be written as N = α + (d + 1)α, while the second equation in (5) equals N = α + (d + 2 − m)α + d + 1 − m. Subtracting the first of these equations from the last we find d + 1 − m α = . (6) m − 1 From (5), we have −(d + 1) + (d + 1) + 4N α = , (7) which yields that α is either a quadratic irrational or a rational number. Since (6) implies that α is a rational number we find that the integer (d + 1) + 4N must be a 2 2 square, i.e. there exists a positive integer s such that s = (d + 1) + 4N . Note that d is an even integer if and only if s is an odd integer if and only if s is an odd integer. Consequently we find that the numerator of α in (7)is always even, and (7) yields that α is a positive integer, say k. From the equations in (5) it follows that not only α = k but also α + 1 = k + 1 is a divisor of N . 123 86 J. de Jonge et al. From the definition of T in (1) (especially the case − α ∈ Z) we see that d = d(α) = − (k + 1). (8) On the other hand (6) yields that, since α = k, d = (m − 1)(k + 1), and from this and (8) we see that − (k + 1) = (m − 1)(k + 1), i.e.N = mk(k + 1). Conversely, let k be a positive integer such that the relations of (4) hold. Then both N /α and N /(α + 1) are positive integers, implying that all cylinder sets are full. Moreover, since d = d(α) = d is given by max N mk(k + 1) d = − α − 1 = − k − 1 = (m − 1)(k + 1), α k and d = d(α + 1) is given by min d(α + 1) = − α = mk − k = (m − 1)k, α + 1 it follows that there are d − (d − 1) = (m − 1)(k + 1) − (m − 1)k + 1 = (m − 1) + 1 = m max min full cylinder sets. Theorem 1 serves as a starting point for our investigation of orbits of N -expansions. The first thing we will do is give some preliminary results (in Sect. 2.1) that we need for proving (in Sect. 2.2) Theorem 3 and Theorem 4 on gaplessness of arrangements with at least five cylinders. 2.1 Preliminary results The first thing to pay attention to is the way N and α and d(α), the value of the largest partial quotient, are interdependent, which is illustrated by the following lemmas: Lemma 1 Given N and α,let d := d(α) be the largest possible digit. Then d ≥ N − 1 if and only if α< 1. 123 Orbits of N-expansions with a finite set of digits 87 The proof of this lemma is left to the reader. If α = α = N − 1, we have max ⎪ √ − ( 2 − 1) =4for N = 2; 2−1 ⎪ 3 ⎨ √ − ( 3 − 1) =3for N = 3; 3−1 d(α) = (9) ⎪ 4 ⎪ √ − ( 4 − 1) − 1 =2for N = 4; 4−1 ⎪ √ ⎩ N N +1 − ( N − 1) = 2 + =2for N ∈ N . ≥5 N −1 N −1 On the other hand we have, for N ∈ N, N ≥ 2 fixed: lim d(α) = lim − α =∞. α↓0 α↓0 α The following lemma provides for a lower bound for the rate of increase of d(α) compared with the rate of decrease of α. Lemma 2 Let N ∈ N be fixed and d := d(α). Then d is constant for α ∈[ f , f ), ≥2 d+1 d and d increases overall more than twice as fast as α decreases. Proof of Lemma 2 Starting from α , d increases by 1 each time α decreases beyond max a fixed point, i.e. if − α ∈ N. For the difference between two successive fixed points f and f we have d−1 d 2 2 4N + (d − 1) − (d − 1) 4N + d − d f − f = − d−1 d 2 2 2 2 4N + (d − 1) − 4N + d + 1 1 = < . 2 2 This finishes the proof. Closely related to the previous lemma is the following one, the proof of which is left to the reader. Lemma 3 Let d ∈ N and N ∈ N be fixed and let f (N ) be defined by the equation ≥2 ≥2 d N / f (N ) − d = f (N ) (so f (N ) is the fixed point of the map x → N /x − dfor d d d x ∈ (0, N /d)). Then f (N + 1) − f (N + 1)> f (N ) − f (N ). d−1 d d−1 d So, for d fixed, the distance between two consecutive fixed points increases if N increases. We have, in fact, for d fixed: lim ( f (N ) − f (N )) = ; d−1 d N →∞ 123 88 J. de Jonge et al. cf. the proof of Lemma 2.For N fixed, on the other hand, we have: lim f (N ) = 0. d→∞ While d(α) is a monotonously non-increasing function of α, the number of cylinder sets is not. The reason is obvious: starting from α = α , the number of cylinder max sets changes every time either α or α + 1 decreases beyond the value of a fixed point; in the first case, the number increases by 1, and in the second case, it decreases by 1. Since T (x ) =−N /x < 0 and T (x ) = 2N /x > 0on I , T (x ) is decreasing and α α α α convex on I , implying that a net increase of the number of cylinder sets. Still, for N and α large enough, it may take a long time of α decreasing from α before the max number of cylinder sets stops alternating between two successive numbers k ∈ N ≥2 and k + 1, and starts to alternate between the numbers k + 1 and k + 2. As an example, we take N = 100. If α decreases from α , the interval I consists of two cylinder max α sets until α decreases beyond f and cylinder set Δ emerges; then, if α + 1 decreases 3 3 beyond f , cylinder set Δ disappears and so on, until α decreases beyond f and Δ 1 1 8 9 emerges while Δ has not yet disappeared. In order to get a grip on counting the number of cylinder sets, the following arith- metic will be useful: a full cylinder set counts for 1, an incomplete left one counts for N N − d − α, and an incomplete right one for α + 1 − ( − d ), giving rise to max min α α+1 the following definition: Definition 2 Let N ∈ N and α ∈ R such that 0 <α ≤ N − 1 and T the ≥2 α N -continued fraction map. The branch number b(N,α) is defined as b(N,α) := d − d − 1 (the number of full cylinder sets save for the outermost ones) max min + − d − α (the length of the image of the leftmost cylinder set) max + α + 1 − − d (the length of the image of the rightmost cylinder set). min α + 1 From this the next lemma follows immediately: Lemma 4 For N ∈ N and 0 <α ≤ N − 1 we have ≥2 N N N b(N,α) = − = . α α + 1 α(α + 1) It follows that for fixed N the branch number b(N,α) is a strictly decreasing function of α. Remark 4 Applying Lemma 4, we find N 1 b(N,α ) = √ √ = 1 + √ . (10) max ( N − 1) N N − 1 The word ‘branch’ refers to the part of the graph of T on the concerning cylinder set. 123 Orbits of N-expansions with a finite set of digits 89 It follows that b(N,α) > 1 for all N ∈ N , so the number of cylinder sets is always ≥2 at least 2. On the other hand, from Lemma 4 it follows that the number of cylinder sets increases to infinity as α decreases from α to 0. Actually, we have infinitely max many digits only if α = 0. In this case the corresponding N -expansion is the greedy N -expansion, studied in [1,3]. The relation N /(α(α + 1)) = b yields 4N + 1 − 1 α = , (11) from which we derive that d(α) = d (or d(α) = d − 1 in case − α ∈ Z), where d is given by ⎢ ⎥ ⎢ ⎥ 4N (b − 1) + 1 + b + 1 ⎢ ⎥ ⎣ ⎦ d = . (12) 2.2 Gaplessness if the branch number is large enough So far, we merely discussed the way I is divided in cylinder sets, depending on the values of N,α, d(α) and the branch number b. In order to present some first results on sufficient conditions for gaplessness, we will zoom in on some ergodic properties of T . Lemma 5 If μ is an absolutely continuous invariant probability measure for T , then there exists a function h of bounded variation such that μ(A) = hdλ, λ − a.e., with λ the Lebesgue measure, i.e. any absolutely continuous invariant probability measure has a version of its density function of bounded variation. Proof of Lemma 5 Since inf |T | > 1, applying Theorem 1 from [9] immediately yields the assertion. Theorem 2 Let N ∈ N . Then there is a unique absolutely continuous invariant ≥2 probability measure μ such that T is ergodic with respect to μ . α α α Proof of Theorem 2 Let μ be a unique absolutely continuous invariant probability measure for T and choose its density function h of bounded variation. Then there exists an open interval J such that h(x)> 0 for any x ∈ J , since h has at most countably many discontinuity points. Consider {T J : n ≥ 0}. Since inf |T | > 1, there exists an n such that T (J ) includes a discontinuity point. (If necessary we may choose endpoints of J not in the preimages of discontinuity points of T .) We note that for see also page 185, Theorem 1 in [10]. 123 90 J. de Jonge et al. any measurable subset A ⊂ J with μ (A)> 0 equivalently λ(A)> 0, μ(T A)> 0 n +1 for any n ≥ 1. Now T (J ) includes two intervals J and J attached to α and α + 1 respectively. For any measurable subset B ⊂ J ∪ J of positive λ-measure, μ(B )> 0  r 0 −(n +1) 0, since otherwise we have a contradiction; μ(B ) = 0 and μ(T (B )) > 0 0 0 n +1 (since there is a B ⊂ J such that T (B ) = B ,μ(B )> 0). This shows that 1 1 0 1 any two absolutely continuous invariant probability measures μ and μ cannot have 1 2 disjoint supports (i.e. they cannot be singular to each other), which is equivalent to the uniqueness of the absolutely continuous invariant probability measure and hence its ergodicity. The next result follows directly from Theorem 2: Corollary 1 If iteration of T maps all open subintervals of I to the interval I , then α α I contains no gaps. Proof of Corollary 1 The assumption implies that the absolutely continuous invariant probability measure μ is equivalent to the Lebesgue measure, which implies that for any measurable subset A ⊂ I , μ(A) = 0 if and only if λ(A) = 0. Suppose that there is a gap J . Since J is an open interval, we have λ(J)> 0, thus μ(J)> 0. Since μ(I )< ∞ implies a.e. x ∈ J , there exists infinitely many positive integers n such that T (x ) ∈ J (by the Poincaré recurrence theorem), which contradicts the assumption that there is a gap. Before we present the first of two theorems on gaplessness, we note that in the case N = 2, the condition |T (x )| > 2 for all x ∈ I is not satisfied for any α ∈ (0, 2−1]. Theorem 3 Let N ∈ N , and let 0 <α ≤ N − 1. Let |T (x )| > 2 for all x ∈ I . ≥3 α Then I contains no gaps. Proof of Theorem 3 The condition implies N /(α + 1) > 2, yielding α< N /2 − 1. From Lemma 4 it follows that 2 2N b(N,α) > √ , 2N − 2 which is larger than 2 for all N ∈ N .So I consists of at least three cylinder sets. ≥3 α Since |T (x )| > 2 for all x ∈ I , there exists an ε> 0 such that for any open interval J that is contained in a cylinder set of T we have 0 α | | T (J ) ≥ (2 + ε)|J |, α 0 0 where |J | denotes the length (i.e. Lebesgue measure) of an interval J . If T (J ) contains two consecutive discontinuity points p , p of T , then α 0 i +1 i α (p , p ) ⊂ T (J ), i +1 i α 0 and we immediately have that O 2 I := (α, α + 1) = T (p , p ) ⊂ T (J ). α i +1 i 0 α α 123 Orbits of N-expansions with a finite set of digits 91 If T (J ) contains only one discontinuity point p of T , then T (J ) is the disjoint α 0 α α 0 union of two subintervals located in two adjacent cylinder sets: T (J ) = J ∪ J . α 0 1 2 Obviously, |T (J )|=|J |+|J |. α 0 1 2 Now select the larger of these two intervals J , J , and call this interval J . Then 1 2 |J |≥ (1 + )|J |. 1 0 In case T (J ) does not contain any discontinuity point of T ,weset J = T (J ). α 0 α 1 α 0 Induction yields that there exists an  ∈ N such that |J |≥ 1 + |J | , whenever T (J ) includes no more than one discontinuity point of T . But then there α −1 α must exist a k ∈ N such that T (J ) contains two (or more) consecutive discontinuity α k 2 O points of T , and we find that T (J ) = I . Applying Corollary 1, we conclude that α k α α there is no gap in I . Remark 5 As one of our referees remarked, the result from Theorem 3 is rather folklore. As to one of its first appearances we refer to Keith Wilkinson’s paper [13]; it might be interesting also to consult Gerhard Keller’s paper [6]. The next theorem, which is partly a corollary of the previous one, gives an even more explicit condition for gaplessness. Theorem 4 Let I consist of five cylinder sets or more. Then I has no gaps. α α Proof of Theorem 4, part I: Let I consist of five cylinder sets or more. Then b(N,α) > 3, implying 1 1 3 12N + 9 − 9 4N α< + 1 − (cf. (11)), in which case |T (α + 1)| > 3 − . 2 2 2N The second inequality yields that for N ∈ N we have |T (α + 1)| > 2 and, ≥18 applying Theorem 3, I is gapless. Now suppose N ∈{12,..., 17}. Then b(N,α) = 3 involves arrangements with four cylinders. In each of these cases, the smallest α such that I has not yet (i.e. decreasing from α ) consisted of five cylinder sets is f . α max 7 In all six cases (two of which are illustrated in Fig. 4)wehave |T ( f + 1)| > 2, yielding the gaplessness of I for arrangements with five or more cylinders in case N ∈{12,..., 17}. This finishes the proof of Theorem 4 for N ∈ N .For N ∈ ≥12 {2,..., 11} a similar approach does not work. We will use some ideas that we will 123 92 J. de Jonge et al. 6 5 4 3 6 5 4 3 N =17,α = f ; |T (α +1)| =2.0098 ··· N =12,α = f ; |T (α +1)| =2.0415 ··· 7 7 Fig. 4 Two arrangements illustrating the gaplessness of arrangements with five cylinders or more on account of Theorem 3 introduce and develop in the next sections and will finish the proof of Theorem 4 at the end of Sect. 5. In the following we will go into conditions for gaplessness of arrangements con- sisting of less than five cylinders. We will start with two cylinders and will use the results for arrangements with three and four cylinders. Remark 6 Since b(N,α) is a strictly decreasing function of α (cf. Lemma 4) and b(N,α ) = 1 + 1/( N − 1), the condition I = Δ ∪ Δ is never satisfied in max α d d−1 case N ∈{2, 3}. 3 Gaplessness if I consists of two cylinder sets In general, if the branch number is not much larger than 1 (which is if α is not much smaller than α ), the overall expanding power of T , determined by T (or |T |, max α α α which we will often use), is not enough to exclude the existence of gaps; we shall elaborate on this in a subsequent article. However, in the case of two cylinder sets I = Δ ∪ Δ , there is a very clear condition under which this power suffices: α d d−1 Theorem 5 Let I = Δ ∪ Δ .If T (α) ≥ f and T (α + 1) ≤ f , then I is α d d−1 α d−1 α d α gapless. Although the statement of 5 is intuitively clear, for the proof of Theorem 5 we need several results and lemmas that we will prove first. Then, immediately following Remark 11, we will prove Theorem 5 itself. Remark 7 In case I = Δ ∪ Δ and either T (α) < f or T (α + 1)> f ,itis α d d−1 α d−1 α d 2 2 easy to see that (T (α), T (α)) or (T (α + 1), T (α + 1)) is a gap, respectively. α α α α Since arrangements under the condition of Theorem 5 play an important role in this section, we introduce the following notations: 123 Orbits of N-expansions with a finite set of digits 93 2 1 2 1 3 2 3 2 3 2 F F F 2 2 F2 F F 2 3 F F 3 3 N =9,σ ≈ 1.01 N =17,σ ≈ 1.06 N =18,σ ≈ 1.27 N =24,σ ≈ 1.23 N =25,σ ≈ 1.22 3 2 4 3 4 3 5 4 5 4 F F F 3 4 F 4 F 5 F 4 3 F F 4 5 N =49,σ ≈ 1.20 N =50,σ ≈ 1.33 N =99,σ ≈ 1.26 N = 100,σ ≈ 1.36 N = 165,σ ≈ 1.29 Fig. 5 Arrangements in F (d), d ∈{2, 3, 4, 5},where α is maximal Definition 3 Let N ∈ N be fixed. For d ∈ N , we define F (d) as the family of all ≥4 ≥2 arrangements Υ such that I = Δ ∪ Δ , T (α) ≥ f and T (α + 1) ≤ f . N ,α α d d−1 α d−1 α d We will write F (d) in case α satisfies the equation T (α) = f , the root of which α d−1 we will henceforth denote by α(N , d). Remark 8 Note that for each N ∈ N and d ∈ N we have that α(N , d), if it exists, ≥4 ≥2 is the only value of α such that F (d) is not void. If the expanding power of T is large enough to exclude the existence of gaps for the largest α for which an arrangement in F (d) exists, there will not be gaps in any arrangement in F (d). We will now first show how to find these largest α, which takes some effort. If we have finished that, we will go into the expanding power of |T | in these arrangements with largest α. For 4 ≤ N ≤ 8, with d = 2, we have T (α )> f , while T (α + 1) = α max 1 α max max max α < f . Hence we see Υ ∈ F (2) and F (2) =∅.For N ∈ N we have max 2 N ,α ≥9 max T (α )< f .If d = 2 we can find α such that Υ in F (2) for each 9 ≤ N ≤ α max 1 N ,α max 17; see Fig. 5, where ten arrangements in various F (d) are drawn. Underneath each arrangement we have mentioned an approximation of σ(α) := |T (α + 1)|, which we will later return to. This σ is important, because it is the expanding power on the rightmost cylinder set that may be too weak to exclude gaps. If d = 2 and N ∈ N , the condition T (α) = f yields T (α + 1)> f , and d ≥18 α d−1 α d has to increase by 1 so as to find an arrangement in F (3).If d = 3, for 18 ≤ N ≤ 24 we find that the largest α is f − 1, in which case T (α + 1) = α< f and d−2 α d T (α) > f (so in this case the arrangement with the largest α is in F (3) but not α d−1 in F (3)); for 25 ≤ N ≤ 49, the largest α is such that T (α) = f .If N ∈ N ,the α 2 ≥50 family F (3) is empty and d has to increase further; see Fig. 5 once more. In the proof of Lemma 8 this approach (of exhausting F (d) for successive values of N and going to F (d + 1) for larger values of N ) will be formalised into a proof by induction. Due to (12) such an increase is always possible, no matter how large d and N become. Note that this inductive approach works since for each d only finitely many N exist such that there are α with Υ ∈ F (d). To see why this claim holds, note that for N ,α 123 94 J. de Jonge et al. fixed N and d, the smallest α for which d = d(α) = d is α , given by max d 4N + (d + 1) − (d + 1) α = f = ; d d+1 cf. (3). For this α it is not necessarily so that I = Δ ∪ Δ , i.e. that I consists α d d−1 α d d of two cylinder sets (e.g. if N = 2 and d = 5, there are five cylinder sets). However, if b(N,α ) ≤ 2, we know that I exists of two cylinder sets, the left one of which is d α full. According to Lemma 4, the branch number b(N,α ) satisfies N 4N b(N,α ) = = . 2 2 2 f ( f + 1) d+1 d+1 4N + (d + 1) − 2d 4N + (d + 1) + d − 1 Keeping d fixed and letting N →∞, we find lim b(N,α ) = 1. N →∞ In view of this and Lemma 3 (and the results mentioned directly thereafter), we choose N sufficiently large, such that for α ≥ α we have b(N,α) < 5/4 and f − f > d d−1 d 1/4. Now suppose that for such a sufficiently large value of N there exists an α ≥ α , such that α ∈ F (d). Then by Definition 2 of branch number and the assumption that α ∈ F (d), we have that b(N,α) ≥ 1 + f − f > , d−1 d which is impossible since for N sufficiently large, d fixed and α ≥ α we have b(N,α) ≤ b(N,α )< . It follows that for d fixed and N sufficiently large, F (d) is void. We will prove (in Lemma 8) that if N ∈ N there exists a minimal d ∈ N such ≥25 ≥3 that the arrangement in F (d) with α maximal lies in F (d). Before we will prove this, we will explain the relation between d and N for arrangements in F (d). In Fig. 5 we see that for N ∈{49, 99, 165} the arrangements in F are very similar, and that the arrangement for N = 100 is more similar to these than the arrangement for N = 50. Moreover, the last three arrangements look hardly curved. This is easy to understand, considering the following equations, where b(N,α) = b is fixed: √ √ 2 2 b( 4bN + b + 2N + b) b 4bN + b |T (α)|= and |T (α)|−|T (α + 1)|= . α α α 2N N Since (b 4bN + b )/N is a decreasing function of N , approaching 0 from above as N →∞, the second equation implies that for a fixed branch number b the branches become less curved as N increases; i.e., the curves approach linearity as N →∞ and b 123 Orbits of N-expansions with a finite set of digits 95 (a , 1) (a, a) (1, − + a +1) Fig. 6 The ‘limit graph’ of T , translated over (−α, −α), under the conditions I = Δ ∪ Δ and α α d d−1 N /α − d = f for N →∞ (and α, d →∞ accordingly) d−1 is fixed. Although in F (d) the branch number is not so much fixed as bounded between 1 and 2, we have a similar decrease of curviness as N increases. The arrangements for N ∈{49, 99, 165} in Fig. 5 suggest that (assuming T (α) = f , i.e. α = α(N , d))) α d−1 if N →∞ (and d →∞ and α →∞ accordingly), the difference f − T (α + 1) d α tends to 0, yielding a ‘limit graph’ of T that consists of two parallel line segments (the straightened branch curves of T ); see Fig. 6, obtained by translating the graph over (−α, −α). In this situation we have both a := T (α) (mod α) = f (mod α) α d−1 and T (α + 1) = f (also (mod α) in Fig. 6). Because in the limit both parts of the α d graph are linear with the same slope, we also have that (0, a + 1) lies on the prolonged right line segment, from which we derive that the line segments have slope −1/a.The line with equation y =− + a + 1 intersects the line y = 1at (a , 1) (so the dividing 2 1 line is x = a ) and intersects the line x = 1in 1, − + a + 1 , yielding the point 1 1 x − + a + 1, − + a + 1 on the line through (0, a) with equation y =− + a a a a (since T (α + 1) = f ). From this we derive 2a = 1, so a = 1/2. α d From Fig. 6 we almost immediately find that the branch number for the limit case is 2 and that the dividing line is at 1/2. We use this heuristic to find a formula describing the relation between N and d for arrangements in F (d) very precisely. Note that for arrangements similar to the limit graph we have 2 2 4N + (d − 1) − 4N + d + 1 1 + f − f = + 1 ≈ b(N,α) ≈ 2, d−1 d from which we derive √ √ N ≈ (4 + 3 2)(d − d) +2or d ≈ 1 + (6 2 − 8)(N − 2) + 1 . (13) 123 96 J. de Jonge et al. If, for d fixed, we determine arrangements in F (d) such that the difference f − T (α + 1) is positive and as small as possible according to our heuristic, the d α best function seems to be N = (4 + 3 2)(d − d), yielding the right N (after round- ing off to the nearest integer) for d ∈{3,..., 500}\{9, 50, 52, 68, 69, 80, 97, 129, 167, 185, 210, 231, 289, 330, 416, 440, 444, 479, 485}, in all of which cases the rounding off should have been up instead of down. For d = 2 we find N = √ √ 2(4 + 3 2)= 17, for d = 3wefind N =6(4 + 3 2)= 49, for d = 4 √ √ we find N =12(4 + 3 2)= 99 and or d = 5wefind N =20(4 + 3 2)= 165; see Fig. 5 once more. Although we do not know generally if rounding off to the nearest integer yields the right N , with (13) we can find a very good overall indication of the relation between d and N for arrangements in F (d) by looking at the difference between the image of α(N , d) + 1 and f ; see Definition 3. With some straightforward calculations we find that N 4N + (d − 1) − (d + 1) α(N , d) = . (14) 2(N − d) Applying (14), we write f (N ) − (N /(α(N , d) + 1) − (d − 1)) as 2 2 2 2 2 (N + dN + d) 4N + d − N 4N + (d − 1) − (N − d(d − 4)N − d(d − 2)) j (N ) := 2(N + dN + d) and, more generally, define 2 2 2 2 2 (x + dx + d) 4x + d − x 4x + (d − 1) − (x − d(d − 4)x − d(d − 2)) j (x ) := 2(x + dx + d) (15) for x ∈[25, ∞). We note that I = Δ ∪ Δ is equivalent to N /(α + 1) − (d − 1) ≥ α. In case α d d−1 α = α(N , d),wehave 2 2 N 4N + (d − 1) − (d − 1)N + 2d(d − 2)N + 2d(d − 1) − (d − 1) = . α(N , d) + 1 2(N + dN + d) (16) Applying (16), for the difference h (N ) := N /(α(N , d) + 1) − (d − 1) − α(N , d) we write 3 2 3 2 2 2N + 4dN + (2d − 5d + 3d)N + 2d (d − 1) − dN (2N + 1) 4N + (d − 1) h (N ) := , 2(N − d)(N + dN + d) (17) and, more generally, define 3 2 3 2 2 2x + 4dx + (2d − 5d + 3d)x + 2d (d − 1) − dx (2x + 1) 4x + (d − 1) h (x ) := , 2(x − d)(x + dx + d) (18) for x ∈[25, ∞). 123 Orbits of N-expansions with a finite set of digits 97 Now we can prove the lemma that is illustrated by the arrangements for N ∈ {17, 49, 99, 165} in Fig. 5. In order to so, we define for fixed d ∈ N ≥2 S(d) := {N ∈ N : I = Δ ∪ Δ , T (α) = f (N ) and T (α + 1) ≤ f (N )} ≥4 α d d−1 α d−1 α d and M := max S(d). √ √ 2 2 Lemma 6 Let d ∈ N . Then M ∈{(4 + 3 2)(d − d), (4 + 3 2)(d − d)}. ≥2 d Proof of Lemma 6 First we note that for d = 2, we have that (4 + 3 2)(d − d) equals 17, which corresponds with what we had already calculated and drawn in Fig. 5.Now let d ∈ N . First we have to show that h (M )> 0for M ∈{(4 + ≥3 d d d √ √ 2 2 3 2)(d − d), (4 + 3 2)(d − d)}, for this assures us that I = Δ ∪ Δ .We α d d−1 will leave this to the reader; it is merely very cumbersome to show, while technically straightforward. The only thing left to do is showing that j ((4 + 3 2)(d − d)) > 0; (19) j ((4 + 3 2)(d − d) + 1)< 0, since the first equation implies that j ((4 + 3 2)(d − d))> 0, while the second implies that j ((4 + 3 2)(d −d)+ 1)< 0. The work to be done is as cumbersome and straightforward as the previous work to be done for this proof and is left to the reader as well. Before we will show that for N ∈ N there are a d ∈ N and an α such that ≥25 ≥3 Υ ∈ F (d), we will prove the following lemma: N ,α Lemma 7 Let d ∈ N . Let N ∈ N be such that I = Δ ∪ Δ and ≥3 ≥25 α(M ,d) d d−1 T (α(M , d)) = f for M ∈{N , N + 1}. Then α(M ,d) d d−1 d T (α(N + 1, d) + 1) − α(N + 1, d)> T (α(N , d) + 1) − α(N , d), α(N +1,d) α(N ,d) i.e. h (N + 1)> h (N ). d d Proof of Lemma 7 We want to show that h (N ) from (17) is an increasing sequence, and do so by calculating the derivative of with x ∈[25, ∞), and then showing that h (x)> 0on [25, ∞). Although a little bit intricate, the work is straightforward and is left to the reader. Now we can prove the following lemma: We have throughout this paper frequently used (Wolfram) Mathematica for making intricate calculations, all of which are nonetheless algebraically basic. In relevant cases we think it will be evident if we did. 123 98 J. de Jonge et al. Lemma 8 Let N ∈{9,..., 17, 25, 26,...}. Then there are d ∈ N and α ∈ ≥2 (0, N − 1) such that I = Δ ∪ Δ ,T (α) = f and T (α + 1) ≤ f α d d−1 α d−1 α d (i.e. α = α(N , d)). Proof of Lemma 8 We will use induction on d.For N ∈{9,..., 17, 25, 26,... , 99} and d ∈{2, 3, 4} we refer to Fig. 5 and leave the calculations to the reader. Specifically, we have for 50 ≤ N ≤ 99 that Υ ∈ F (4). It is easily seen that Υ ∈ N ,α(N ,4) 99,α(99,5) ∗ ∗ F (5) as well. Due to Lemma 6, there is an N > 99 such that Υ ∈ F (5). 5 N ,α(N ,5) 5 5 Applying Lemma 7, we see that for all N ∈{99,..., N } we have Υ ∈ F (5). 5 N ,α(N ,5) For the induction step, let d ∈ N be such that there is an α for which Υ ∈ F (d), ≥5 N ,α where N is the largest such N possible, cf. Lemma 6. If we can show that for this N there is an α such that Υ ∈ F (d + 1), we are finished. This can be done by d N ,α showing that h ((4 + 3 2)(d − d) − 1)> 0, (20) d+1 forthisimplies h (N )> 0, in which case α is such that − (d + 1) = f , d+1 d i.e. α = α(N , d + 1). Although intricate, the calculations are straightforward and are left to the reader. Remark 9 Although Lemma 8 is about N in the first place, our approach is actually based on increasing d and then determining all N such that arrangements Υ ∈ N ,α F (d) exist. The proof of Lemma 8 yields the arrangements with the smallest d (and therefore the largest α)for which Υ ∈ F (d), as illustrated by the last five N ,α arrangements of Fig. 5. Example 1 For d = 4wehave M − 1 =(4 + 3 2)(d − d)= 98. Then ∗ ∗ Υ ∈ F (4) and Υ ∈ F (4), while Υ ∈ M −1,α(M −1,4) M ,α(M ,4) M +1,α(M +1,5) d d d d d d F (5); see Fig. 5. It follows immediately from our construction of α(M + 1, 5) that this is the largest α such that Υ ∈ F (5). M +1,α With manual calculations we can quickly calculate the expanding power of T in α + 1 for arrangements in F and F and N not too large, say N ∈ N , where the ≤49 smallest values are found where α is as large as possible. The next proposition gives a lower bound for |T (α + 1)| for such arrangements for most N . Proposition 1 Let N ∈{18}∪{50, 51,...}\{95,..., 99} and α ∈ (0, N − 1] such that I = Δ ∪ Δ for some d ∈ N,d ∈ N . Furthermore, suppose that T (α) ≥ α d d−1 ≥2 α f and T (α + 1) ≤ f . Then |T (α + 1)| > 2 = 1.259921 .... d−1 α d Proof of Proposition 1 Considering Lemma 8, we can confine ourselves to arrange- ments in F with α as large as possible. For α = α(N , d) (cf. (14)) we can write |T (α + 1)|= N /(α + 1) as k (N ) 4 2 3 2 2 3 2 2 2N + (d − 1) N + 2d(d − 1)N + 2d N + ((d − 1)N + 2dN ) 4N + (d − 1) = . 4 3 2 2 2 2(N + 2dN + d(d + 2)N + 2d N + d ) (21) 123 Orbits of N-expansions with a finite set of digits 99 It is not hard to find that, for d fixed, k is a decreasing sequence, with lim k (N ) = 1. However, from (13) it follows that if N →∞ we have N →∞ d that also d →∞ in a precise manner. Due to the previous lemmas, for each d we can confine ourselves to considering only N /(α + 1) for the largest N and α such that Υ ∈ F (d). Applying Lemma 6, an easy way to check if N ,α indeed |T (α(N , d) + 1)| > 2 is considering k (x ), with x ∈[100, ∞), α(N ,d) and then calculating k ((4 + 3 2)(d − d) + 1) for d ∈ N , which is amply d ≥5 larger than 2 = 1.2599 ··· . For the remaining cases d = 3 and N = 18 and for d = 4 and N ∈{50, 51,... , 94} it is easily checked manually that indeed |T (α(N , d) + 1)| > 2. α(N ,d) Remark 10 Considering our previous remarks concerning arrangements in F ,itmay be clear that lim N /(α(N , d) + 1) = 2. N →∞ Remark 11 The value 2 in the proof of Proposition 1 relates to the proof of Theorem 5 and also to the proofs of Proposition 9 and Theorem 3, where the numbers 2 and 2 have a similar importance. Considering the proof of Proposition 1, we could actually replace 2 by the smallest possible value, given by 94 20480015 + 320305 385 = = 1.2604 .... (α(94) + 1) 21233664 Finally we are ready to prove Theorem 5, stating that I = Δ ∪ Δ , with α d d−1 d := d(α), is gapless if T (α) ≥ f and T (α + 1) ≤ f . Considering Remark 11 α d−1 α d the value 2 in Proposition 1 can be replaced by 1.26, the third power of which is 2.000376. We will use this to stress that the gaplessness of Theorem 5 is actually relatively ample and does not require infinitesimal estimations. Proof of Theorem 5 First we note that the conditions imply N ∈ N .Now let Υ ∈ ≥4 N ,α F (d) and let K ⊂ I be any open interval. Since K expands under T , there is an α α n ∈ N∪{0} such that T (K ) contains for the first time a fixed point or the discontinuity point p , in the former case of which we are finished. So we assume that T (K )∩Δ = d d (b, p ]=: L, with f < b < p . Note that T (L) =[α, T (b)) ⊂[α, f ).For d d d α α d 2 2 2 T (L) = (T (b), T (α)], we similarly may assume that f < T (b)<α + 1 α d−1 α α α (since otherwise f ∈ T (L), and again we are done). d−1 3 3 3 Now suppose that T (L) contains p , excluding f ∈ T (L). Then T (L) = L ∪ d d 1 α α α 2 3 M , with L =[T (α), p ] and M = (p , T (b)). First we confine ourselves to N ∈ 1 1 d 1 d α α {18}∪{50,...}\{95,..., 99}. Since then |T (L)| > 2.000376|L| (cf. Remark 11), we have certainly |L | > 1.001|L| or |M | > 1.001|L|. If we consider the images of 1 1 2 3 L and M under T , T and T similarly as we did with the images of L, we find 1 1 α α α that due to expansiveness (see the proof of Theorem 3) there must be an m such that 3m 3m 3 f ∈ T (L ) or f ∈ T (M ) and we are finished. If T (L) does not contain d 1 d−1 1 α α α p , the expansion of L will only go on longer, yielding even larger L and M and 1 1 the reasoning would only be stronger that no gaps can exist. For N ∈{4,..., 17, 19, 20,... , 49, 95, 96,... , 99} a similar approach can be taken, but there is no useful general lower bound for |T (x )| on I . For these cases, 123 100 J. de Jonge et al. however, the moderate expanding power in Δ is easily made up for by a rela- d−1 tively strong expanding power in Δ , and the gaplessness is easily, although tediously, checked by hand (cf. Examples 2 and 3 below). This finishes the proof of Theorem 5. Example 2 In case N = 7, there exist α ∈ (0, 7 − 1] for which I = Δ ∪ Δ .The α 2 1 largest α for which Υ ∈ F (2) is α = 7 − 1, in which case |T (α + 1)|= 1. 7,α max However, |T ( f )|= 2.0938 ··· > 2, and the approach taken above even works due to the expanding power of T on [α, f ) alone. α 2 Example 3 In case N = 99, we have I = Δ ∪ Δ , and Υ ∈ F for α 4 3 99,α α = 99( 405 − 5)/190 = 7.8807 ··· . Then |T (α + 1)|= 1.2552 ··· , |T ( f )|= α α 1.3503 ··· and |T ( f )|= 1.4908 ··· . So for an interval (p , x ), with x ∈ (p , f ), 4 4 d 3 3 3 assuming that f ∈ / T (p , x ),wehave |T (p , x )| > 1.3503 ··· × 1.2552 ··· × 4 4 4 α α 1.4908 ··· × |(p , x )| 2|(p , x )|, implying enough expanding power for T to 4 4 exclude the existence of gaps. Remark 12 We can also prove that |T (x )| > 2on Δ for all arrangements under the assumptions of Theorem 5, but we cannot do without knowledge about the slope on Δ . d−1 Next we will make preparations for formulating a sufficient condition for gapless- ness in case I consists of three cylinder sets. Proving it involves more subtleties on the one hand, but will have a lot of similarities with the two-cylinder set case on the other hand. Once we have finished that, not much work remains to be done for gaplessness in case I consists of four or five cylinder sets. 4 A sufficient condition for gaplessness if I consists of three or four cylinder sets If I = Δ ∪...∪Δ , with m ∈{2, 3}, there is a sufficient condition for gaplessness α d d−m that resembles the condition for gaplessness in case I consists of two cylinder sets a lot: Theorem 6 Let I = Δ ∪ ... ∪ Δ , with m ∈{2, 3}. Then I is gapless if α d d−m α T (α) ≥ f or T (α + 1) ≤ f . α d−1 α d−m+1 We will prove this theorem in parts. In Sect. 4.1 we will prove Theorem 6 for m = 2; in Sect. 4.2 we will extend the result of Sect. 4.1 to m = 3; considering Theorem 4, extension to larger m is not useful. Remark 13 The difference between the ‘and’ of Theorem 5 and the ‘or’ of Theorem 6 has to do with the existence, in the latter case, of at least one full cylinder set. 123 Orbits of N-expansions with a finite set of digits 101 3 2 1 3 2 1 4 3 2 4 3 2 5 4 3 N =3,σ =1 N =4,σ ≈ 1.06 N =7,σ ≈ 1.46 N =9,σ ≈ 1.40 N =16,σ ≈ 1.64 |T (f )|≈ 1.77 |T (f )|≈ 1.64 |T (f )|≈ 2.09 |T (f )|≈ 1.92 |T (f )|≈ 2.08 α 1 α 1 α 2 α 2 α 3 |T (f )|≈ 3 |T (f )|≈ 2.62 |T (f )|≈ 2.95 |T (f )|≈ 2.62 |T (f )|≈ 2.62 2 2 3 3 4 α α α α α Fig. 7 Arrangements with largest α such that there is a d with I = Δ ∪ Δ ∪ Δ under the condition d d−1 d−2 T (α) ≥ f and T (α + 1) ≤ f α d−1 α d−1 4.1 Gaplessness if I consists of three cylinder sets Since we have m = 2, the condition T (α) ≥ f can be split in α d−1 ⎪ 1. T (α + 1) ≤ f ≤ T (α); α d−1 α 2. f ≤ T (α) ≤ T (α + 1); (22) d−1 α α 3. f ≤ T (α + 1) ≤ T (α); d−1 α α of course the condition T (α + 1) ≤ f can be split in a similar way. We will α d−1 prove Theorem 6 by proving gaplessness according to this distinction in three cases, associated with Lemmas 9, 10 and 11 respectively. The first of these is not very hard to prove: Lemma 9 Let I = Δ ∪ Δ ∪ Δ .If T (α) ≥ f and T (α + 1) ≤ f , α d d−1 d−2 α d−1 α d−1 then I is gapless. Proof of Lemma 9 The assumptions imply that b(N,α) > 2, yielding σ(α) =|T (α + 1)| > 2for N ∈ N .If N ∈ N ,welet K ⊂ I be any open interval. Since K ≥17 ≥17 α expands under T , there is an n ∈ N ∪{0}) such that T (K ) contains a fixed point or a discontinuity point p (with i ∈{0, 1}) , in the former case of which we are finished. d−i So we assume that T (K ) ⊃ L, where L = (b, p ], with f < b < p , d−i d−i d−i with i ∈{0, 1}.If T (L) contains a fixed point, we are finished. If T (L) does not α α contain a fixed point, then it cannot contain a discontinuity point, and we have that |T (L)| > 2|L|, implying enough expanding power of T to ensure gaplessness of at least one cylinder set (which might be non-full). Since both T (α) ≥ f and α d−1 T (α + 1) ≤ f ,itfollows that I is gapless. For 2 ≤ N ≤ 16 the slopes on I may α d−1 α α differ considerably: for some N , such as N = 7 and N = 16 we also have σ> 2, but if this is not he case, the steepness left of f is amply larger then 2; see Fig. 7 d−2 for some examples where α is as large as possible. This finishes the proof of Lemma 9 [cf. case 1 in (22)]. If I = Δ ∪Δ ∪Δ under the condition T (α) ≥ f and T (α+1)> f α d d−1 d−2 α d−1 α d−1 or under the condition T (α) < f and T (α + 1) ≤ f , I is gapless as well, α d−1 α d−1 α but this is much harder to prove. The following definition will be convenient: 123 102 J. de Jonge et al. 5 4 3 3 2 1 3 2 1 N =23,α =2.898 N =11, α =1.873 ··· N =7,α =1.54 |T (α +1)|≈ 1.51 |T (α +1)|≈ 1.33 |T (α +1)|≈ 1.09 α α α |T (α)|≈ 2.74 |T (α)|≈ 3.14 |T (α)|≈ 2.95 α α α Fig. 8 Arrangements with one very small cylinder Definition 4 Let I = Δ ∪ ... ∪ Δ , and 1 ≤ m ≤ d − 1. If T (α) ≤ f or α d d−m α d−1 T (α + 1) ≥ f , the cylinder set Δ respectively Δ is called small. α d−m+1 d d−m Taking a similar approach as in the proof of Theorem 5, one can show that the map T has enough expansive power to ensure that for any open interval K ⊂ I there exists α α a non-negative integer n such that T (K ) contains a fixed point. If this fixed point is in a non-small or even full cylinder set, we are done (as in the proofs of Theorem 5 and Lemma 9). However, if this fixed point is from the small cylinder set, then it only follows that every point of the small cylinder set is in the orbit under T of some point in K . Note this implies that the small cylinder set is gapless. So we may assume that the small cylinder set is gapless. Let us assume that the left cylinder set is small. We define L := T (Δ )\Δ . Since Δ is gapless, we have L = (p , T (α)]⊂ (p , f ), α d d d d d d−1 2 2 so T (L) =[T (α), α + 1).If T (α) ≤ f , we are finished, so we assume that α d−2 α α 2 2 3 3 T (α) > f . We then have T (L) = (T (α + 1), T (α)].If T (α) ≥ f we are d−2 α d−1 α α α α finished, since then f ∈ T (L). d−1 The question arises whether it is possible to keep avoiding fixed points if we go on with letting T work on L and its images (or similarly, if the right cylinder set is small, some interval R := T (Δ )\Δ ). We will argue that this is not possible in the α 1 1 two most plausible cases for gaps to exist, involving the least expansion. The first case is illustrated with two arrangements in Fig. 8, one where N = 23 and one where N = 11. In both arrangements one outer cylinder is very small while the other one is full or almost full. In the arrangement where N = 23, we see that L is a very narrow strip between p and T (α), T (L) is not so narrow anymore, and 5 α T (L) is definitely wide enough to make clear that avoiding fixed points f and f 4 3 is not possible. The middle arrangement, where N = 11, is an example of the case where Δ is small and Δ is actually full. Here we have that R := T (Δ )\Δ is d−2 d α 1 1 a very narrow strip between T (α + 1) and p and that T (R) is only slightly larger α 2 than T (Δ ), whence eventually there will be an n ∈ N such that f ∈ T (R) or α 1 3 f ∈ T (R). The rightmost arrangement in Fig. 8 is an illustration of the second plausible case for the existence of gaps: here Δ is small, while Δ is incomplete but not small. This 3 1 123 Orbits of N-expansions with a finite set of digits 103 4 3 2 N =30,α =3.6 |T (α +1)|≈ 1.42 |T (α)|≈ 2.31 T (R) V V T (R) α 1 2 R Fig. 9 Arrangement illustrating Lemma 10 3 2 1 4 3 2 N =17,α =2.6576 N =35,α =4.00167 |T (α +1)|≈ 1.27 |T (α +1)|≈ 1.40 α α |T (α)|≈ 2.41 |T (α)|≈ 2.19 α α 2 4 2 Fig. 10 Two arrangements in which almost T (α) < p < T (α + 1) (in fact, in both cases p = T (α)) α d α d α arrangement illustrates the role p might play in avoiding fixed points: in this case, 3 4 taking L := T (Δ )\Δ ,wehave T (L) = M ∪ M , with M =[T (α), p ] and α 3 3 1 2 1 2 α α 2 3 M = (p , T (α + 1)]. Since T (L) contains a discontinuity point, the expansion 2 2 α α under T is interrupted. If T (M ) would be a subset of T (Δ ) and T (M ) would α α 1 α 3 α 2 be a subset of T (L), the expansion would be finished and we would have three gaps: 3 4 2 2 (T (α), T (α + 1)), (T (α), T (α)) and (T (α + 1), T (α))—but this is not the case, α α α α α α as we will shortly prove. Of course arrangements exist such that one of the outer cylinders is small, fixed points are avoided (in the sense we used above) for a long time and it takes more of T working on L or R before one of the discontinuity points is captured. But in these cases the interruption of the expansion is even weaker than in the cases above. We will first show that arrangements such as the rightmost one of Fig. 8 exclude the existence of gaps (cf. Lemma 10) and will then consider cases such as the first two arrangements of Fig. 8 (cf. Lemma 11). Lemma 10 Let I = Δ ∪ Δ ∪ Δ .Then I is gapless if α d d−1 d−2 α f ≤ T (α) ≤ T (α + 1) or T (α) ≤ T (α + 1) ≤ f . d−1 α α α α d−1 123 104 J. de Jonge et al. ( , 1) a+1 (a, a) y = −(a +1)x + a (a, 1 − a ) y = −(a +1)x + a +1 Fig. 11 The ‘limit graph’ of T , translated over (−α, −α), under the conditions I = Δ ∪ Δ and α α d d−1 N /(N /α − d) − (d − 1) = p for N →∞ (and α, d →∞ accordingly). This ‘arrangement’ can be seen as one with three cylinders, where Δ (mod α), the one on the right, is infinitely small; see also the d−2 arrangements in Fig. 10 2 1 3 2 1 3 2 1 3 2 3 2 1 b(4,α)=2 b(5,α)=2 b(6,α)=2 b(11,α)=2 b(12,α)=2 2N +1−1 Fig. 12 Arrangements of Υ with b = 2; in each case α = N ,α Proof of Lemma 10 We will confine ourselves to the first case of this lemma, that is if f ≤ T (α) ≤ T (α + 1), since the second one is proved similarly (in d−1 α α fact, this case is slightly harder due to the smaller size of the absolute value of the derivatives). Regarding our observations above, we may assume that the small cylin- der set (which in this case is Δ ) is gapless (cf. the remarks after Definition 4). d−2 We will show that this implies the gaplessness of the other cylinder sets as well. We define R := T (Δ )\Δ and try to determine α such that p ∈ T (R) α d−2 d−2 d (see the remark immediately preceding this lemma). Necessary conditions for this 2 4 2 are T (α) < p < T (α + 1) (assuming that f ∈ / T (R) and f ∈ / T (R), d d α d−1 α α α since in either case we would be done). If these conditions are satisfied, we write 3 2 4 T (R) = V ∪ V , with V =[T (α), p ] and V = (p , T (α + 1)]. We will show 1 2 1 d 2 d α α α that we cannot have both T (V ) ⊂ T (R) and T (V ) ⊂ T (Δ ), which is neces- α 1 α α 2 α d−2 sary for limiting the expansion of R under T and so not eventually capturing f and α d f ; see Fig. 9. d−1 We take an approach that is similar to the proof of Theorem 5, for which several lemmas and a proposition where used, partially concerning a relation between N and d in the arrangements involved, partially concerning the slope in α + 1. In this proof we will not explicitly formulate similar statements as lemmas or propositions, nor do 123 Orbits of N-expansions with a finite set of digits 105 we prove them, since they require similar basic but very intricate calculations that we prefer to omit. In order to find the relationship between N and d for arrangements with the con- 2 4 ditions T (α) < p and T (α + 1)> p mentioned above, we refer to some more d d α α relevant arrangements, as shown in Fig. 10. In both cases in Fig. 10, α is such that T (α) = p , which is a value of α that is only a little larger than the values for which 2 4 T (α) < p and T (α + 1)> p . A ‘limit arrangement’ (where the third, rightmost d d α α cylinder is infinitely small), similar to the ‘limit arrangement’ used in the proof of 3 2 Theorem 5, is shown in Fig. 11. The assumptions yield a + a − 1 = 0, with real root a = 0.75487 ··· =: γ . Similar to the proof of Theorem 5 we then find that for arrangements as in Fig. 10 we have (d − 1)(d − 1 + γ)(1 + γ) N ≈ . Using this relationship, we can take a similar approach as in the proof of Proposi- tion 1. We leave out the tedious steps and confine ourselves to observing that the slope of the line segments in Fig. 11 is −(γ + 1) =−1.75487 ··· and that in 2 4 arrangements where T (α) < p and T (α + 1)> p ,wewill seethat theslope d d α α T (α + 1) approaches −(γ + 1) as N tends to infinity. However, for our proof the inequality |T (α + 1)| > ( 5 + 1) = 1.61803 ··· =: G suffices, which turns α 2 out to hold for N ∈ N . We will use this to show that for N ∈ N we have ≥273 ≥273 |T (R)| > |T (Δ )|+|T (R)|. From this it immediately follows that we cannot α d−2 α have that both T (V ) ⊂ T (R) and T (V ) ⊂ T (Δ ), and we are done with the α 1 α α 2 α d−2 proof of Lemma 11. Since |T (x )| is a decreasing function on I , and writing β := |Δ |,wehave α d−2 |T (Δ )| > |T (α + 1)|· β, so |R| >(|T (α + 1)|− 1)β. α d−2 α α It follows that |T (R)| >(|T (α + 1)|− 1) ·|T (p )|β, α d−1 α α that |T (R)| >(|T (α + 1)|− 1) ·|T (p )|·|T ( f )|β, d−1 d α α α α and finally that |T (R)| >(|T (α + 1)|− 1) ·|T (p )| ·|T ( f )|β. d−1 d α α α α We also have |T (Δ )| < |T (p )|β,so α d−2 d−1 |R| <(|T (p )|− 1)β and |T (R)| < |T ( f )|· (|T (p )|− 1)β. d−1 α d−1 d−1 α α α 123 106 J. de Jonge et al. It follows that |T (Δ )|+|T (R)| <(|T (p )|+|T ( f )|· (|T (p )|− 1))β α d−2 α d−1 d−1 d−1 α α α = (|T (p )|−|T ( f )|+|T ( f )|·|T (p )|)β d−1 d−1 d−1 d−1 α α α α < |T ( f )|·|T (p )|β. d−1 d−1 α α So, although crudely, we certainly have that |T (R)| > |T (Δ )|+|T (R)| if α d−2 α |T ( f )|·|T (p )| <(|T (α + 1)|− 1) ·|T (p )| ·|T ( f )|, d−1 d−1 d−1 d α α α α α that is, if |T ( f )| 1 <(|T (α + 1)|− 1) ·|T (p )|· . (23) d−1 α α |T ( f )| d−1 Since |T ( f )| (|T (α + 1)|− 1) ·|T (p )|· >(|T (α + 1)|− 1) ·|T (p )| d−1 d−1 α α α α |T ( f )| d−1 >(|T (α + 1)|− 1) ·|T (α + 1)|, α α we know that (23) holds for |T (α+1)| > G, which in turn holds for all N ∈ N .We ≥273 remark that this value is quite a wide upper bound, since we did a rough approximation. Still, checking that we cannot have both T (V ) ⊂ T (R) and T (V ) ⊂ T (Δ ) for α 1 α α 2 α d−2 smaller N is not that hard and is left to the reader. This finishes the proof of Lemma 10 [cf. case 2 in (22)]. Lemma 10 implies that in case I = Δ ∪ Δ ∪ Δ and f ≤ T (α) ≤ α d d−1 d−2 d−1 α T (α + 1) or T (α) ≤ T (α + 1) ≤ f the division of an interval containing p α α α d−1 d in two smaller ones cannot prevent an overall expansion that excludes any gaps. The other plausible case with three cylinder sets in which gaps might exist is if one outer cylinder set is very small, while the other one is full or nearly full, such that either 3 3 T (α + 1) ≥ T (α + 1) (if Δ is the small cylinder set) or T (α) ≤ T (α) (if Δ α d−2 α d α α is the small cylinder set). We will show that this is not possible either: Lemma 11 Let I = Δ ∪ Δ ∪ Δ . Then I is gapless if α d d−1 d−2 α f ≤ T (α + 1) ≤ T (α) or T (α + 1) ≤ T (α) ≤ f . d−1 α α α α d−1 Proof of Lemma 11 Taking into account our observations immediately following Defi- nition 4 and the arrangements of Fig. 8 for N = 23 and N = 11, we only have to prove that there are no α such that T (α) < T (α) is possible if T (α + 1) ≤ T (α) ≤ f α α α d−1 (in case Δ is small) or such that T (α + 1)> T (α + 1) is possible if f ≤ d α d−1 T (α + 1) ≤ T (α) (in case Δ is small). Note that the conditions T (α) < T (α) α α d−2 α and T (α +1)> T (α +1) imply that the branch number is slightly larger than 2. Now remember that I consists of m full cylinder sets if and only if α = k, N = mk(k + 1) and d = (m −1)(k +1) for some k ∈ N, cf. Theorem 1. Figure 12 shows for increasing 123 Orbits of N-expansions with a finite set of digits 107 values of N a sequence of arrangements where the branch number b is 2, from one full arrangement (here for N = 4) with two cylinders to the next one (here for N = 12). Since |T (α)| > |T (α + 1)|, the arrangements suggest that in case b is slightly larger α α than 2, the most favourable arrangement for T (α + 1) = T (α + 1) to have real roots 2 3 is if N = 2k + 2k − 1, where k ∈ N , while for T (α) = T (α) to have real roots is ≥2 α if N = 2k + 2k + 1, where k ∈ N. We will confine ourselves to investigating only the possibility of T (α + 1) = T (α + 1); the calculations for the other case are similar. So we will try and find out if for N = 2k + 2k − 1, d = k + 1, with k ∈ N ,the ≥2 positive root of T (α + 1) = T (α + 1) lies in I . To do this, we solve α α 2k + 2k − 1 2k + 2k − 1 − (k + 1) = − (k − 1), α + 1 2k + 2k − 1 − k 2k + 2k − 1 − (k − 1) α + 1 which is reducible to 3 2 2 4 2 5 4 3 2 (2k + 6k − k − 1)α + (2k + 5k + k − 2)α − (4k + 6k + 2k − 3k − k + 1) = 0, yielding 4 2 8 7 6 5 4 3 2 36k + 144k + 164k − 12k − 95k − 2k + 21k − 4k − (2k + 5k + k − 2) α = . 3 2 2(2k + 6k − k − 1) (24) A straightforward computation shows that this last expression is smaller than f , k+2 meaning that the root (24) lies outside I if I = Δ ∪ Δ ∪ Δ . Since N = α α k+1 k k−1 2k + 2k − 1 was the most favourable option for investigation, this finishes our proof [cf. case 3 in (22)]. Remark 14 The arrangement for N = 11 in Fig. 8 illustrates that the difference between T (α + 1) and T (α + 1) may be very small. 4.2 A sufficient condition for gaplessness in case I consists of four cylinder sets In the previous subsection we proved Theorem 6 for m = 2, by proving Lemmas 9, 10 and 11. In this subsection we will consider m = 3 and go into the analogons of Lemmas 9, 10 and 11. If I consists of four cylinder sets, the analogon of Lemma 9 is that arrangements I are gapless if I = Δ ∪ Δ ∪ Δ ∪ Δ while T (α) ≥ f and T (α + α α d d−1 d−2 d−3 α d−1 α 1) ≤ f . The analogon of Lemma 11 is that arrangements I are gapless if I = d−2 α α Δ ∪ Δ ∪ Δ ∪ Δ while f ≤ T (α + 1) ≤ T (α) or T (α + 1) ≤ d d−1 d−2 d−3 d−2 α α α T (α) ≤ f . In both cases branch numbers larger than 3 are involved, in which α d−1 case |T (α + 1)| > 2if N ∈ N (and Theorem 3 yields the desired result). The ≥18 cases 2 ≤ N ≤ 17 can be checked manually and are left to the reader; in Fig. 13 the arrangement for N = 11, associated with Lemma 11, illustrates that gaps are out of the question. 123 108 J. de Jonge et al. 5 4 3 2 5 4 3 6 5 4 7 6 5 N =11,α = f 6 N =15,α = f N =24,α = f − 1 N =35,α = f − 1 6 3 4 and T (α +1) = p and T (α)= p and T (α)= p α 5 α 5 α 6 Fig. 13 Four arrangements with two full cylinders The analogon of Lemma 10 is that arrangements I are gapless if I = Δ ∪Δ ∪ α α d d−1 Δ ∪ Δ while f ≤ T (α) ≤ T (α +1) or T (α) ≤ T (α +1) ≤ f .The d−2 d−3 d−1 α α α α d−2 arrangements for N = 15, N = 24 and N = 35 in Fig. 13 are interesting illustrations of the analogon of Lemma 10 in the case of two full cylinder sets instead of one. We will confine ourselves to the arrangement for N = 15; the other ones have similar properties. The arrangement for N = 15 is the boundary case for the situation where we have four cylinders, the left one of which (that would be Δ in this example) is extremely small and the right one is such that almost p ∈ T (T (Δ )\Δ ).The α 6 6 interesting thing is that this option would imply a quick interruption of the expansion of T (Δ )\Δ , involving two large gaps. But it is not really an option: the arrangement for α 6 6 N = 15 in Fig. 13 is exceptional among relatively small N (as well as the arrangements for N = 24 and N = 35 are), while for N > 36 we have |T (α + 1)| > 2if I = Δ ∪ Δ Δ and N /(α + 1) − (d − 2) = p or N /α − d = p . We derived α d d−1 d−2 d d this in a similar way as in the proof of Lemma 10 (see Fig. 11) or the preparations for Theorem 5 (see Fig. 6). Figure 14 shows the associated ‘limit graph’, from which it is easily found that a = (3 − 5), yielding branch number 2 + g, with g = 1/G the small golden section. With this, we conclude the proof of Theorem 6. In the next section we will prove that if I = Δ ∪ Δ ∪ Δ ∪ Δ gaps exist α d d−1 d−2 d−3 only in very rare cases and if they do, that they are very large. After that, we will finish the proof of Theorem 4, stating that all arrangements with five cylinders are gapless. 5 Gaplessness in case I contains two full cylinder sets If an arrangement contains three cylinders, and two of them are full, the arrangement is gapless according to Theorem 6. In this section we will proof that arrangements of four cylinders generally do not contain a gap either, save for special values of N.The core of this proof rests on values of α satisfying one of the equations 3 3 T (α) = T (α) (with root α ) and T (α + 1) = T (α + 1) (with root α ). α  α u α α We will show that for N such that α <α very large gaps exist for α ∈[α ,α ]. u  u 123 Orbits of N-expansions with a finite set of digits 109 Fig. 14 The ‘limit graph’ of T , (a, 1) translated over (−α, −α), under the conditions I = Δ ∪ Δ Δ and α d d−1 d−2 N /(α + 1) − (d − 2) = p for N →∞ (and α, d →∞ y = − x +3 accordingly) y = − x +1 (1,a) y = − x +2 The central theorem of this section is the following: Theorem 7 Let N ∈ N and I = Δ ∪ Δ ∪ Δ ∪ Δ . Then there is a gap ≥2 α d d−1 d−2 d−3 in I if and only if N = 2k + 2k − i, with k > 1 and i ∈{1, 2, 3}. Moreover, if there is a gap in I , the gap contains f and f , while Δ and Δ are gapless. α d−1 d−2 d d−3 Proof of Theorem 7 Suppose that there is a gap containing f and f in I and that d−1 d−2 α Δ and Δ are gapless. Then, as a sub-interval of a gap, the interval ( f , f ) is d d−3 d−1 d−2 a gap. Since f < f , N /( f + d − 1) = f and N /( f + d − 2) = f , d−1 d−2 d−1 d−1 d−2 d−2 we know that N N ( f , f )  , , d−1 d−2 f + d − 1 f + d − 2 d−2 d−1 where the larger open interval is a gap as well. What is more, the infinite sequence of intervals N N ( f , f )  , d−1 d−2 f + d − 1 f + d − 2 d−2 d−1 N N ,  ··· N N + d − 1 + d − 2 f +d−2 f +d−1 d−1 d−2 consists of the union of ( f , f ) with pre-images of ( f , f ) in Δ and d−1 d−2 d−1 d−2 d−1 Δ respectively and therefore of gaps containing f and f . It is contained in d−2 d−1 d−2 the closed interval [q, r ], with q =[d − 1, d − 2] ∈ Δ and r =[d − 2, d − 1] ∈ Δ , N ,α d−1 N ,α d−2 yielding 2 2 T (q) = q, T (q) = r , T (r ) = q and T (r ) = r . (25) α α α α 123 110 J. de Jonge et al. Since Δ and Δ are gapless, T (α) and T (α + 1) lie outside the interval (q, r ), d d−3 α α which is to say p < T (α) ≤ q and r ≤ T (α + 1)< p . d α α d−2 2 2 For the images of α under T this means that either T (α) ∈ Δ or T (α) ∈ α d−3 α α Δ , in the latter case of which we have, due to the expansiveness of T and the d−2 α equalities of (25), 2 3 |T (α) − q|≤|T (α) − r|≤|T (α) − q|, α α with equalities only in the case T (α) = q. From this we derive that 2 2 3 either T (α) ∈ Δ or T (α) ∈ Δ ∧ T (α) ≤ T (α) (26) d−3 d−2 α α α α and, similarly, that 2 2 3 either T (α + 1) ∈ Δ or T (α + 1) ∈ Δ ∧ T (α + 1) ≥ T (α + 1). (27) d d−1 α α α α In the following we will write α (N , m) (u for ‘upper’) for the positive root of the equation T (α + 1) = T (α + 1) (so T (α + 1) = r) and α (N , m) (l for ‘lower’) for α α the positive root of the equation T (α) = T (α) (so T (α) = q), with m the number α α of full cylinder sets; in the current case we have m = 2. Recall that I consists of m full cylinder sets if and only if α = k, N = mk(k + 1) and d = (m − 1)(k + 1) for some k ∈ N, cf. Theorem 1,soif m = 2, we have arrangements consisting of two full arrangements only for α = k, N = 2k(k + 1) and d = k + 1. If N is 2k(k + 1) − n, with n ∈ N, and α = k − x, with x ∈ R,wehave 2 2 2k + 2k − n 4xk − n + 2x − 2x 4xk − n b(N,α) = = 2 + > 2 + , 2 2 (k − x )(k + 1 − x ) (k − 1)(k + 1 − x ) k + x which is a little bit larger than 2 provided x and n are relatively small. For these arrangements we have 2 2 2k + 2k − n 2x + 2x − n d(α) = − (k − x ) = k + 2 + 3x + = k + 2 k − x k − x and 2 2 2k + 2k − n 2x + 2x − n d(α + 1) = − (k + 1 − x ) = k − 1 + 3x + k + 1 − x k + 1 − x = k − 1. So, for x and n relatively small, the arrangements consist of four cylinders, while the branch number is only a little bit larger than 2. We will now use this to finish the forward implication of Theorem 7. 123 Orbits of N-expansions with a finite set of digits 111 Fig. 15 Arrangement with N = 11,α = 1.8719 4 3 2 1 N =11,α =1.8719 ··· (k =2, i =1) T (α +1) = T (α +1) Since Δ decreases and Δ increases as α decreases, we see that the assumption d−3 d that there is a gap containing f and f in I implies α (N , 2) ≥ α (N , 2). d−1 d−2 α u We will shortly show that the only values of N for which α (N , 2) ≥ α (N , 2) are N = 2k + 2k − i, with k > 1 and i ∈{1, 2, 3}; in all cases d = k + 2. Although we could keep i as a variable in our calculations, we can limit ourselves to the case i = 3, since i = 3 is the least favourable value of i allowing for a gap, as is suggested in Figs. 12, 15, 16, 17 and 18. We will show that for i = 3 indeed α (N , 2) ≥ α (N , 2). Subsequently we will show that for 4 ≤ i ≤ 4k no gaps exist; the upper bound is 4k, 2 2 since 2k + 2k − 4k = 2(k − 1) + 2(k − 1), so as to confine the calculations to the group of arrangements where d = k + 2. So, let N = 2k + 2k − 3 and d = k + 2. Then α (N , 2) =[k + 2, k + 1, k] and α (N , 2) +1 =[k −1, k, k + 1]. Omitting straightforward calculations, we find that 2 4 2 (2k + 2k − 3) D − (2k + 3k + 3k − 6) α (2k + 2k − 3, 2) = 3 2 4k + 12k − 6k − 6 and 2 2 (2k + 2k − 3)( D − (k + 5k + 4)) α (2k + 2k − 3, 2) = , 4k − 18k − 8 Note that in Fig. 12 we have d = k + 1, while there is no small cylinder set Δ . k+2 We omit the suffix ‘N,α’ behind these expansions not only for eligibility but also because α has yet to 3 3 be determined as the root of T (α) = T (α) or T (α + 1) = T (α + 1). α α α α 123 112 J. de Jonge et al. Fig. 16 Arrangement with N = 11,α = 1.8687 4 3 2 1 N =11,α =1.8687 ··· (k =2,i =1) T (α)= T (α) with 4 3 2 2 2 D = 9k + 18k − 3k − 12k = (3k + 3k − 2) − 4. Since we assume that there is a gap containing f and f ,wehave α ≥ α . d−1 d−2 u Omitting the basic calculations, we find that this inequality holds for k ∈ N .Inthe ≥2 case k = 3 (and so N = 21), we have indeed 508032 − 192 α (2k + 2k − 3, 2) = = 2.7123 ··· > 2.7122 ··· 508032 − 588 = = α (2k + 2k − 3, 2); see Fig. 19. Some more basic calculations show that the cases N = 2k + 2k − 1 and N = 2k + 2k − 2 allow for larger intervals [α ,α ] where large gaps exist; see the next examples. √ √ 9075 − 26 99 − 9075 α (11, 2) = = 1.8719 ··· and α (11, 2) = = 1.8686 ··· 37 2 √ √ 1725 − 12 45 − 1725 α (10, 2) = = 1.7372 ··· and α (10, 2) = = 1.7334 ··· 17 2 √ √ 5103 − 22 27 − 567 α (9, 2) = = 1.5946 ··· and α (9, 2) = = 1.5941 ··· 31 2 123 Orbits of N-expansions with a finite set of digits 113 Fig. 17 Arrangement with N = 10,α = 1.7372 4 3 2 1 N =10,α =1.7372 ··· (k =2, i =2) T (α +1) = T (α +1) Fig. 18 Arrangement with N = 9,α = 1.5946 4 3 2 1 N =9,α =1.5946 ··· (k =2, i =3) T (α +1) = T (α +1) 228 − 5 α (8, 2) = = 1.4428 ··· and α (8, 2) = 9 − 57 = 1.4501 ··· We see that the intervals α − α decrease as N decreases, until (for N = 8) the ‘interval’ would have negative length, hence does not exist. 123 114 J. de Jonge et al. Fig. 19 Arrangement with N = 21,α = 2.7123 5 4 3 2 N =21,α =2.7123 α =2.7122 and α =2.7123 Now suppose N = 2k + 2k − 4 (note that voor k = 2we have N = 8). Then 3 2 (k + 2) D − (k + k + 2k + 4) α (2k + 2k − 4, 2) = 2(k + 4k + 2) and 3 2 (k − 1) D − (k + 4k − k − 4) α (2k + 2k − 4, 2) = , 2(k − 2k − 1) with 4 3 2 D = 9k + 18k − 7k − 16k. 2 2 There are no gaps provided α (2k + 2k − 4, 2) − α (2k + 2k − 4, 2)> 0. Once more we omit the calculations, finding that this inequality holds for k ∈ N ,sowe ≥2 conclude that there are no gaps in case N = 2k + 2k − 4. If we replace the number 4 in N = 2k + 2k − 4 by larger integers (if possible), there will not be any gaps either: the length of the ‘interval’ [α ,α ] would only become more negative. This concludes the proof that if I = Δ ∪ Δ ∪ Δ ∪ Δ and there is a gap containing f α d d−1 d−2 d−3 d−1 and f in I , then N = 2k + 2k − i, with k > 1 and i ∈{1, 2, 3}. d−2 α For the converse statement, we assume that N = 2k + 2k − i, with k ∈ N and i ∈{1, 2, 3}.Ifalso d = k + 2, then earlier in this proof we showed that only then α (N , 2) ≤ α (N , 2). We will show that for α such that α (N , 2) ≤ α ≤ α (N , 2) u  u there is a gap in I = Δ ∪ Δ ∪ Δ ∪ Δ containing both f and f . α d d−1 d−2 d−3 d−1 d−2 123 Orbits of N-expansions with a finite set of digits 115 Fig. 20 Arrangement with N = 20,α = 2.6124 5 4 3 2 N =20,α =2.6124 α =2.6132 ··· and α =2.6115 ··· As earlier in this proof, we set q =[d − 1, d − 2] and r =[d − 2, d − 1] . N ,α N ,α Set G = (q, r ), then clearly both f ∈ G and f ∈ G. Furthermore, by definition d−1 d−2 of α (N , 2) and α (N , 2) we have that for every α ∈[α (N , 2), α (N , 2)] that u  u T (α) ≤ q (and therefore T (α) ≥ r ) and that T (α + 1) ≥ r (and therefore T (α + 1) ≤ q). Note that T ((p , q)) = (r,α + 1) and that T ((r , p )) = (α, q). Then we have α d α d−2 c c c that T (G ) = G , where G is the complement of G in I . We are left to show that α α G = (q, r ) is a gap; i.e. that for almost all x ∈ G there exists an n = n(x ) such that n c T (x ) ∈ G . To show this, consider the map T : I → I , defined by α α −x (α+1) p −α ⎪ d ⎪ + , if x ∈ Δ ; p −α p −α d d − (d − 1), if x ∈ Δ ; d−1 T (x ) = (28) ⎪ − (d − 2), if x ∈ Δ ; d−2 ⎪ x (α+1) −α p ⎩ −x d−2 + , if x ∈ Δ . d−3 α+1−p α+1−p d−2 d−2 123 116 J. de Jonge et al. Table 1 The thin thread between α (N , 2)α (N , 2) having a gap or not N =91.594119 ··· 1.594686 ··· N = 21 2.712252 ··· 2.712310 ··· N = 37 3.776839 ··· 3.776851 ··· N = 57 4.817672 ··· 4.817675 ··· N =81.450165 ··· 1.442809 ··· N = 20 2.613247 ··· 2.611575 ··· N = 36 3.700989 ··· 3.700407 ··· N = 56 4.756087 ··· 4.755832 ··· So on Δ and on Δ we have that T is a straight line segment with negative slope, d d−3 through (α, α + 1) and (p ,α) on Δ , resp. through (p ,α + 1) and (α + 1,α) d d d−2 on Δ .For x ∈ Δ ∪ Δ we have that T (x ) = T (x ). To show that G is d−3 d−1 d−2 α a gap, it is enough to show the ergodicity of T . Then the maximality of G follows from the fact that the support of the absolutely continuous invariant measure is G , c c since T (G ) = G . The proof of the existence of the absolutely continuous invariant measure for T and its ergodicity is similar to the proof of Theorem 2. Here all branches are complete and the proof is rather simpler. Once we have the ergodicity of T,itis n c obvious that for a.e. x ∈ G there exists n = n (x ) such that z = T (x ) ∈ G . Then 0 0 z never returns in G under iterations of T . This finishes the proof of Theorem 7. We stress that the in case of N = 2k + 2k − 3 the intervals [α ,α ] on which gaps exist may be very small; see Fig. 19. On the other hand, in case N = 2k + 2k − 4, the gaplessness may be a very close call; see Fig. 20. Table 1 illustrates how fast these differences between α and α decrease as N increases: Remark 15 While a fixed point f is repellent for points within Δ , the fixed points i i in two adjacent cylinder sets behave mutually contracting for all other points in these cylinder sets. As a consequence, it may take quite some time before the orbit of points in the full cylinders of gap arrangements with four cylinders leave these full cylinders for the first time. As an example we take the gap arrangement for k = 50 (according to the notations used above). Then N = 2 · 50 + 2 · 50 − 3 = 5097, d = 52 and α ≈ α ≈ α ≈ 49.98019737. Table 2 shows for ten values of x between α and α + 1 the smallest n such that T (x)/ ∈ Δ ∪ Δ . What is more, there are uncountably 51 50 many x in the gap (a, b) that contains f and f such that T (x ) ∈ (a, b) for all d−1 d−2 n ∈ N∪{0}. Indeed, for any sequence (d , d ,..., d ,...) such that d ∈{d −1, d −2}, 1 2 n n with n ∈ N, we have that x =[d , d , d ,...] ∈ (a, b). 1 2 3 N ,α For the final, second part of the proof of Theorem 4, we will consider one by one all cases left, that is N ∈{2,..., 11}.If N = 11 and α ≥ f , I consists of five 7 α cylinder sets if and only if α ∈ ( f − 1, f ); see the left arrangement of Fig. 21, which 2 6 we already saw in Fig. 13. Since α (11, 3)>α (11, 3) (cf. page 5), we conclude on similar grounds as in the proof of Theorem 7, that the arrangement is gapless. If α ∈[ f , f − 1], the interval I consists of four cylinder sets, implying gaplessness 7 2 α 123 Orbits of N-expansions with a finite set of digits 117 Table 2 The difficulty of leaving the gap: with N = 5097, α = 49.98019737, for each of ten values of x ∈[α, α + 1] the smallest n is given such that T (x)/ ∈ Δ ∪ Δ 51 50 x 50 50.1 50.2 50.3 50.4 50.5 50.6 50.7 50.8 50.9 n 5417 2090 3568 1123 4776 185 5816 16231 5646 7604 ··· 5 4 3 2 5 4 3 2 7 6 5 4 3 2 13 2 1 N =7,α = 14 − 1 N =11,α = f N =7,α = f =1 N =2,α =0.15 6 6 Fig. 21 Borderline cases for part II of the proof of Theorem 4 because of Theorem 7. Since |T ( f + 1)|= 2.04 ··· , gaps are also excluded for all α ≤ f . A similar approach works for N = 10 (with |T ( f + 1)|= 2.03 ··· ), 7 7 N = 9 (with |T ( f + 1)|= 2.02 ··· ) and even N = 8, in which case f = 1, 7 7 |T ( f + 1)|= 2, and the arrangement with four cylinders is full. For N ∈{3,..., 7} we take a different approach, confining ourselves to the case N = 7; the cases N ∈{3,..., 6} are done similarly. We will omit most calculations, which are generally quite tedious and do hardly elucidate anything. So let N = 7. Then I consists of at least five cylinder sets if and only if α< f ; see the second α 6 arrangement of Fig. 21.Wehave |T (α + 1)|= 2for α = 14 − 1, in which case d = 2; see the third arrangement of Fig. 21. Now suppose 14−1 ≤ α< f = 1. min 6 We have |T ( f + 1)|= and |T (α)| > 7. Regarding these relatively large values, α 4 α it is not hard to understand that Δ is gapless. The part of the orbit of α + 1 under T 2 α in the third arrangement of Fig. 21 illustrates that even in the case of α = 14, the expansion of [T (α + 1), p ] under T clearly excludes the existence of gaps. α 3 α Finally, let N = 2. We have |T ( f )|= 2, indicating the rapid increase of |T | on α α I if α decreases. The large expansiveness of T left of f assures the gaplessness of α α 1 Δ . We will show that for any α ∈ (0, 2 − 1] the image of [T (α + 1), p ]) contains 1 α 2 most of the fixed points, implying the gaplessness of I ; see the last arrangement of Fig. 21 for an illustration of this. If T (α + 1) ≤ f this is quite obvious, so we assume α 2 T (α + 1)> f . Suppose that T (α + 1) = f ,for some s ∈ N . Then, omitting α 2 s ≥2 some basic calculations, we have α = (s + 1 − s + 8)/(2s − 7), whence 4s − 11s + (4s − 13) s + 8 − 15 d = d(α) = 2s − 7 2 2 4s − 13s + (4s − 13) s + 8 − 8 ≥ , 2s − 7 123 118 J. de Jonge et al. from which we derive that d ≥ 4s. This finishes the proof of Theorem 4. Remark 16 A paper worthwhile to read alongside our paper is [12]. As the title of this paper states, it deals with mixing properties of expanding Lorentz maps. There are many interesting results in [12] that embed the results from our paper in a more general context. Acknowledgements The first author of this paper is very thankful to the Dutch organisation for scientific research Nederlandse Organisatie voor Wetenschappelijk Onderzoek, which funded his research for this paper with a so-called Promotiebeurs voor Leraren, with grant number 023.003.036. The third author’s research was partially supported by the Japan Society for the Promotion of Science with Grant-in-Aid for Scientific Research (C) 20K03661. We also want to thank Julian Lyczak, then working at Leiden University, for his help with some programming in the preparatory phase of Sect. 5. We thank Niels Langeveld as well for providing us with Fig. 3. Finally, the authors would also like to thank the anonymous referees for very helpful suggestions which improved the presentation of this paper. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. References 1. Anselm, M., Weintraub, S.H.: A generalization of continued fractions. J. Number Theory 131(12), 2442–2460 (2011) 2. Burger, E.B., Gell-Redman, J., Kravitz, R., Walton, D., Yates, N.: Shrinking the period lengths of continued fractions while still capturing convergents. J. Number Theory 128(1), 144–153 (2008) 3. Dajani, K., Kraaikamp, C., Van der Wekken, N.: Ergodicity of N-continued fraction expansions. J. Number Theory 133(9), 3183–3204 (2013) 4. Hofbauer, F.A.: A two parameter family of piecewise linear transformations with negative slope. Acta Math. Univ. Comenian. (N.S.) 81(1), 15–30 (2012) 5. Ito, S., Sadahiro, T.: Beta-expansions with negative bases. Integers 9(A22), 239–259 (2009) 6. Keller, G.: Stochastic stability in some chaotic dynamical systems. Monatsh. Math. 94(4), 313–333 (1982) 7. Komatsu, T.: Shrinking the period length of quasi-periodic continued fractions. J. Number Theory 129(2), 358–366 (2009) 8. Kraaikamp, C., Langeveld, N.: Invariant measures for continued fraction algorithms with finitely many digits. JMAA 454(1), 106–126 (2017) 9. Lasota, A., Yorke, J.A.: On the existence of invariant measures for piecewise monotonic transforma- tions. TAMS 186, 481–488 (1973) 10. Li, T.-Y., Yorke, J.A.: Ergodic transformations from an interval into itself. TAMS 235, 183–192 (1978) 11. Liao, L., Steiner, W.: Dynamical properties of the negative beta-transformation. Ergodic Theory Dyn. Syst. 32(5), 1673–1690 (2012) 12. Oprocha, P., Potorski, P., Raith, P.: Mixing properties in expanding Lorenz maps. Adv. Math. 343, 712–755 (2019) 13. Wilkinson, K.M.: Ergodic properties of a class of piecewise linear transformations. Z. Wahrschein- lichkeitstheorie und Verw. Gebiete 31, 303–328 (1974/75) 123 Orbits of N-expansions with a finite set of digits 119 Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Monatshefte für Mathematik Springer Journals

Orbits of N-expansions with a finite set of digits

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Abstract

For N ∈ N and α ∈ R such that 0 <α ≤ N − 1, we define I := [α, α + 1] and ≥2 α − − I := [α, α + 1) and investigate the continued fraction map T : I → I , which is α α α α N N defined as T (x ) := − d(x ), where d : I → N is defined by d(x ) := − α . α α x x For N ∈ N , for certain values of α, open intervals (a, b) ⊂ I exist such that for ≥7 α almost every x ∈ I there is an n ∈ N for which T (x)/ ∈ (a, b) for all n ≥ n . α 0 0 These gaps (a, b) are investigated using the square Υ := I × I , where the orbits α α T (x ), k = 0, 1, 2,... of numbers x ∈ I are represented as cobwebs. The squares Υ are the union of fundamental regions, which are related to the cylinder sets of the map T , according to the finitely many values of d in T . In this paper some clear α α conditions are found under which I is gapless. If I consists of at least five cylinder α α sets, it is always gapless. In the case of four cylinder sets there are usually no gaps, except for the rare cases that there is one, very wide gap. Gaplessness in the case of two or three cylinder sets depends on the position of the endpoints of I with regard to the fixed points of I under T . α α Keywords Continued fractions · Dynamical systems · Gaps Mathematics Subject Classification Primary 11J70; Secondary 37E05 Communicated by H. Bruin. B Jaap de Jonge c.j.dejonge@uva.nl Cor Kraaikamp C.Kraaikamp@tudelft.nl Hitoshi Nakada nakada@math.keio.ac.jp Korteweg-de Vries Institute for Mathematics, Science Park 105-107, 1098 XG Amsterdam, The Netherlands Department of Electrical Engineering, Mathematics and Computer Science, Delft University of Technology, Mekelweg 4, 2628 CD Delft, The Netherlands Department of Mathematics, Keio University, Yokohama 223-8522, Japan 123 80 J. de Jonge et al. 1 Introduction In 2008, Edward Burger and his co-authors introduced in [2] new continued frac- tion expansions, the so-called N -expansions, which are nice variations of the regular continued fraction (RCF) expansion. These N -expansions have been studied in vari- ous papers since; see [1,3,7]. In [8], a subclass of these N -expansions is introduced, for which the digit set is always finite. These particular N -expanions are defined as follows: For N ∈ N and α ∈ R such that 0 <α ≤ N − 1, let I := [α, α + 1] and ≥2 α I := [α, α + 1). Hereafter we denote by N the set of positive integers n ≥ k, while ≥k we write N if k = 1. We define the N -expansion map T : I → I (or I )as α α α T (x ) := − d(x ), (1) where d : I → N is defined by N N d(x ) := − α , if either x ∈ (α, α + 1] or both x = α and − α/ ∈ Z x α and N N d(α) = − α − 1, if − α ∈ Z. α α Note that if − α ∈ Z, we have that T (α) = α + 1. This is the only case in which the range of T is I and not I . α α For a fixed α ∈ (0, N − 1] and x ∈ I we define for n ∈ N n−1 d = d (x ) := d(T (x )). n n Note that for α ∈ (0, N − 1] fixed, there are only finitely many possibilities for each d . Applying (1), we obtain for every x ∈ I a continued fraction expansion of the form N N N x = T (x ) = = = ··· = , (2) d + T (x ) N N 1 α d + d + 1 1 d + T (x ) d + d + which we will throughout this paper write as x =[d , d , d ,...] (note that this 1 2 3 N ,α expansion is infinite for every x ∈ I , since 0 ∈ / I ); we will call the numbers d , i ∈ N, α α i the partial quotients or digits of this N-continued fraction expansion of x;see [3,8], 123 Orbits of N-expansions with a finite set of digits 81 Δ Δ 2 1 r r p α 2 f 1 2  f  α +1 2 1 1 2 Fig. 1 N = 51, α = 6 where these continued fractions (also with a finite set of digits) were introduced and elementary properties were studied (such as the convergence in reference [3]). In each cylinder set Δ := {x ∈ I ; d(x ) = i } of rank 1, with d ≤ i ≤ d , i α min max where d := d(α) is the largest partial quotient, and d := d(α + 1) the smallest max min one given N and α,the map T obviously has one fixed point f . As of now we will α i write simply ‘cylinder set’ for ‘cylinder set of rank 1’. It is easy to see that 4N + i − i f = f (N ) = , for d ≤ i ≤ d . (3) i i min max Note that − α ∈ Z if and only if for some d ∈ N we have that α = f , while ≥2 d+1 for any α ∈[ f , f ) we have that Δ =∅. 0 d+2 d+1 d+1 Given N ∈ N ,welet α = N − 1 be the largest value of α we consider. ≥2 max The reason for this is that for larger values of α we would have 0 as a partial quotient as well. Since T (x ) =−N /x and because 0 <α ≤ N − 1, we have |T (x )| > 1 α α on I . From this it follows that the fixed points act as repellers and that the maps T are expanding if 0 <α ≤ N − 1. This is equivalent to the convergence of the N -expansion of all x ∈ I . Each pair of consecutive cylinders sets (Δ ,Δ ) is divided by a discontinuity i i −1 N N point p (N,α) of T , satisfying − i = α,so p = . A cylinder set Δ is called i α i i p α+i full if T (Δ ) = I (or I ). If a cylinder set is not full, it contains either α (in which α i α case T (α) < α + 1) or α + 1 (in which case T (α + 1)>α), and is called incomplete. α α On account of our definition of T , cylinder sets will always be an interval, never consist of one single point. The main object of this paper is the sequence T (x ), n = 0, 1, 2,...,for x ∈ I , which is called the orbit of x under T . More specifically, we are interested in subsets of I that we will call gaps for such orbits. Before we will give a proper definition of ‘gap’, we will give an example of orbits of points in I for a pair {N,α}. Note that, since the fixed points are repellent, orbits cannot remain in one cylinder set indefinitely if x ∈ I is not a fixed point of T , so any orbit will show an infinite migration between α α cylinder sets. A naive approach is to compute the orbits of many points of I and obtain a plot of the asymptotic behaviour of these orbits by omitting the first, say hundred, iterations. Figure 1 shows such a plot for N = 51 and α = 6. It appears that there are parts of I (illustrated by dashed line segments) that are not visited by any orbit after many iterations of T . i i In fact, setting  = T (α) and r = T (α + 1), the orbit of any point—apart i i α α from the fixed points f and f —after once having left the interval (r , r ) ⊂ Δ or 1 2 2 1 2 ( , ) ⊂ Δ of Fig. 1, will never return to it. 1 2 1 For reasons of legibility we will usually omit suffices such as ‘(N )’, ‘(N,α)’or‘(N , d)’. 123 82 J. de Jonge et al. Fig. 2 N = 51, α = 6, x = 6.5 2 1 In order to get a better understanding of the orbits of N -expansions, it is useful to consider the graphs of T , which are drawn in the square Υ (= Υ ) := I × I .This α α N ,α α square is divided in rectangular sets of points  := {(x , y) ∈ Υ : d(x ) = i }, which i α are the two-dimensional fundamental regions associated with the one-dimensional cylinder sets we already use. We will call these regions in short cylinders. Now consider (x , T (x )) ∈ Υ . Then (x , T (x )) goes to (T (x ), T (x )) under T . Regarding this, T α α α α α α has one fixed point F := ( f , f ) in each  . We will denote the dividing line between i i i i and  by l , which is the set { p }×[α, α + 1), with p the discontinuity point i i −1 i i i between Δ and Δ . In case T (Δ ) = I , we will call the cylinder  full and i i −1 α i i the branch of the graph of T in  complete; if a cylinder is not full, we will call it α i and its associated branch of T incomplete. We will call the collection of Υ and its α α associated branches, fixed points and dividing lines an arrangement of Υ .If Υ is a α α union of full cylinders, we will call the associated arrangement also full. Figure 2 is an example of such an arrangement, in which a part of the cobweb is drawn associated with the orbit we investigated previously. The discontinuity point p = is now visible as a dividing line between Δ and Δ . 2 1 2 In [3,8] the arrangement for N = 4 and α = 1 is studied, consisting of two full cylinders  and  and not showing any gaps. On the other hand, the demonstration 1 2 5 13 of the interval ( , ) being a gap of the interval [2, 3] in the case (N,α) = (9, 2) 2 5 in [8] is done without referring to such an arrangement. In this paper, and even more so in the next paper, we will show that arrangements may considerably support the insight in the occurrence of gaps. We will now give a formal definition of gaps, which is slightly delicate, since − N T (I ) = I (or I if − α ∈ Z). α α α Definition 1 A maximal open interval (a, b) ⊂ I is called a gap of I if for almost α α every x ∈ I there is an n ∈ N for which T (x)/ ∈ (a, b) for all n ≥ n . α 0 0 123 Orbits of N-expansions with a finite set of digits 83 There are several papers where ‘gaps’ appear in number theoretic expansions. One of the first important examples is given by the so-called negative β-expansion, introduced and studied by Shunji Ito and Taizo Sadahiro [5]. For β> 1, let I := − , , and let the (−β)-transformation T : I → I be defined by β −β β β β+1 β+1 T (x ) =−βx − − βx . β + 1 Ito and Sadahiro characterise the admissible sequences of (−β)-expansions, give a necessary and sufficient condition for the (−β)-shift to be sofic, and explicitly deter- mine the invariant measure of the (−β)-transformations. In their Example 18, Ito and Sadahiro also noticed the existence of gaps, as intervals where the density of the invariant measure is zero. In [11], Lingmin Liao and Wolfgang Steiner studied nega- tive β-expansions defined on [0, 1], which are conjugate to the negative β-expansions defined by Ito and Sadahiro. In [11] a complete classification of the gaps and of the mixing properties of these negative β-expansions is given. For maps slightly more general than the negative β-maps the reader is referred to Franz Hofbauer’s paper [4]. In Theorem 2 we show that there is a unique absolutely continuous invariant prob- ability measure μ such that T is ergodic with respect to μ .Asin[11](fornegative α α α β-expansions), one also could define a gap as any interval in I where the density of μ is zero. Remark 1 In the example of Fig. 1 the intervals (r , r ) and ( , ) are gaps and 2 1 1 2 for x ∈ (r , r ) ∪ ( , )\{ f , f } there exists an n = n (x ) such that T (x)/ ∈ 2 1 1 2 1 2 0 0 (r , r ) ∪ ( , ) for n ∈ N . The ‘for almost’ formulation in the definition of ‘gap’ 2 1 1 2 ≥2 is necessary so as to exclude fixed points and pre-images of fixed points, i.e. points that are mapped under T to a fixed point, which may never leave an gap. In Sect. 5 we even find a class of gaps (a, b) such that for uncountably many x ∈ I and all n ∈ N ∪{0} we have T (x ) ∈ (a, b). Remark 2 Whenever we use the word ‘gap’ in relation to arrangements, we mean the gap of the associated interval I . In [8] computer simulations were used to get a more general impression of orbits of N -expansions. For a lot of values of α, plots such as Fig. 1 were stacked, for 0 <α ≤ α , so as to obtain graphs such as Fig. 3, with the values of α on the max vertical axis and at each height the corresponding interval I drawn. In the same paper, similar graphs are given for N = 9, 20, 36 and 100. In all cases it appears that ‘gaps’ such as in Fig. 1 appear for values of α equal to or not much smaller than α . max Since the plots in [8] are based on computer simulations, they do not actually show very small gaps (smaller than pixel size) nor clarify much the connection between the gaps for each N . Still, the suggestion is strong that for α sufficiently small there are no gaps. It also seems that for α large enough several disjoint gaps may occur. In Fig. 3 we see this for α near α = 50 − 1. max All ‘for all’ statements in this paper are with respect to Lebesgue measure. 123 84 J. de Jonge et al. Fig. 3 A simulation of intervals I with gaps if existent, for 0 <α ≤ 50 − 1and N = 50 In this paper we will not only investigate conditions for gaplessness, we will also show that simulations such as Fig. 3 fail to reveal the existence, for certain N and α, of one extremely large gap in plots such as Fig. 3 below the last visible gap. In a subsequent paper we will go into another very interesting property of orbits of N - expansions that is hardly revealed by simulations such as Fig. 3: the existence of large numbers of gaps for large N and α close to α . But now we will concentrate on max gaplessness. Remark 3 If no gaps exist with non-empty intersection with a cylinder set, we call the cylinder set gapless. In Sect. 2 we will consider two classes of arrangements that have no gaps: full arrangements and specific arrangements with more than three cylinders. The gapless- ness of the latter class, involving the proof of Theorem 4, for which some preliminary results will be presented shortly, is largely given in Sect. 2, but involve some intri- cacies for small values of N so as to finish it at the end of this paper. In Sect. 3 we will consider arrangements with two cylinders and in Sect. 4 we will concentrate on arrangements with three cylinders, but will prove a sufficient condition for gaplessness that is valid for arrangements with any number of cylinders larger than 2. Finally, in Sect. 5 we will prove a result on gaps in certain arrangements with four cylinders and we will finish the proof of Theorem 4. After that, it is merely a matter of checking that for N ∈{2, 3, 4, 5, 6} all arrangements are gapless. Although most results in this paper seem intuitively clear and obvious, our experi- ence with these continued fraction maps is they are surprisingly unruly. The simulations from Fig. 3 may serve as an example that things are not what they always seem to be. From these simulations one might draw the (erroneous) conclusion that the number of gaps is at most 3. In a forthcoming paper we show however that the number of gaps is a function of N and α, and is growing larger and larger as N tends to infinity. 123 Orbits of N-expansions with a finite set of digits 85 Although the proofs in this paper are often elementary, they require case distinctions and cumbersome calculations. 2 Full arrangements and arrangements with more than four cylinders If I consists of full cylinder sets only, we obviously have no gaps. In this situation the mutual relations between N , α and d(α) show a great coherence, as expressed in the following theorem: Theorem 1 The interval I consists of m full cylinder sets, with m ∈ N , if and only α ≥2 if there is a positive integer k such that ⎪ α = k, (4) N = mk(k + 1), d(α) = (m − 1)(k + 1). Proof of Theorem 1 Writing d := d(α), the interval I is the union of m full cylinder sets if and only if − d = α + 1, (5) − (d − m + 1) = α. α+1 Note that the first equation in (5) can be written as N = α + (d + 1)α, while the second equation in (5) equals N = α + (d + 2 − m)α + d + 1 − m. Subtracting the first of these equations from the last we find d + 1 − m α = . (6) m − 1 From (5), we have −(d + 1) + (d + 1) + 4N α = , (7) which yields that α is either a quadratic irrational or a rational number. Since (6) implies that α is a rational number we find that the integer (d + 1) + 4N must be a 2 2 square, i.e. there exists a positive integer s such that s = (d + 1) + 4N . Note that d is an even integer if and only if s is an odd integer if and only if s is an odd integer. Consequently we find that the numerator of α in (7)is always even, and (7) yields that α is a positive integer, say k. From the equations in (5) it follows that not only α = k but also α + 1 = k + 1 is a divisor of N . 123 86 J. de Jonge et al. From the definition of T in (1) (especially the case − α ∈ Z) we see that d = d(α) = − (k + 1). (8) On the other hand (6) yields that, since α = k, d = (m − 1)(k + 1), and from this and (8) we see that − (k + 1) = (m − 1)(k + 1), i.e.N = mk(k + 1). Conversely, let k be a positive integer such that the relations of (4) hold. Then both N /α and N /(α + 1) are positive integers, implying that all cylinder sets are full. Moreover, since d = d(α) = d is given by max N mk(k + 1) d = − α − 1 = − k − 1 = (m − 1)(k + 1), α k and d = d(α + 1) is given by min d(α + 1) = − α = mk − k = (m − 1)k, α + 1 it follows that there are d − (d − 1) = (m − 1)(k + 1) − (m − 1)k + 1 = (m − 1) + 1 = m max min full cylinder sets. Theorem 1 serves as a starting point for our investigation of orbits of N -expansions. The first thing we will do is give some preliminary results (in Sect. 2.1) that we need for proving (in Sect. 2.2) Theorem 3 and Theorem 4 on gaplessness of arrangements with at least five cylinders. 2.1 Preliminary results The first thing to pay attention to is the way N and α and d(α), the value of the largest partial quotient, are interdependent, which is illustrated by the following lemmas: Lemma 1 Given N and α,let d := d(α) be the largest possible digit. Then d ≥ N − 1 if and only if α< 1. 123 Orbits of N-expansions with a finite set of digits 87 The proof of this lemma is left to the reader. If α = α = N − 1, we have max ⎪ √ − ( 2 − 1) =4for N = 2; 2−1 ⎪ 3 ⎨ √ − ( 3 − 1) =3for N = 3; 3−1 d(α) = (9) ⎪ 4 ⎪ √ − ( 4 − 1) − 1 =2for N = 4; 4−1 ⎪ √ ⎩ N N +1 − ( N − 1) = 2 + =2for N ∈ N . ≥5 N −1 N −1 On the other hand we have, for N ∈ N, N ≥ 2 fixed: lim d(α) = lim − α =∞. α↓0 α↓0 α The following lemma provides for a lower bound for the rate of increase of d(α) compared with the rate of decrease of α. Lemma 2 Let N ∈ N be fixed and d := d(α). Then d is constant for α ∈[ f , f ), ≥2 d+1 d and d increases overall more than twice as fast as α decreases. Proof of Lemma 2 Starting from α , d increases by 1 each time α decreases beyond max a fixed point, i.e. if − α ∈ N. For the difference between two successive fixed points f and f we have d−1 d 2 2 4N + (d − 1) − (d − 1) 4N + d − d f − f = − d−1 d 2 2 2 2 4N + (d − 1) − 4N + d + 1 1 = < . 2 2 This finishes the proof. Closely related to the previous lemma is the following one, the proof of which is left to the reader. Lemma 3 Let d ∈ N and N ∈ N be fixed and let f (N ) be defined by the equation ≥2 ≥2 d N / f (N ) − d = f (N ) (so f (N ) is the fixed point of the map x → N /x − dfor d d d x ∈ (0, N /d)). Then f (N + 1) − f (N + 1)> f (N ) − f (N ). d−1 d d−1 d So, for d fixed, the distance between two consecutive fixed points increases if N increases. We have, in fact, for d fixed: lim ( f (N ) − f (N )) = ; d−1 d N →∞ 123 88 J. de Jonge et al. cf. the proof of Lemma 2.For N fixed, on the other hand, we have: lim f (N ) = 0. d→∞ While d(α) is a monotonously non-increasing function of α, the number of cylinder sets is not. The reason is obvious: starting from α = α , the number of cylinder max sets changes every time either α or α + 1 decreases beyond the value of a fixed point; in the first case, the number increases by 1, and in the second case, it decreases by 1. Since T (x ) =−N /x < 0 and T (x ) = 2N /x > 0on I , T (x ) is decreasing and α α α α convex on I , implying that a net increase of the number of cylinder sets. Still, for N and α large enough, it may take a long time of α decreasing from α before the max number of cylinder sets stops alternating between two successive numbers k ∈ N ≥2 and k + 1, and starts to alternate between the numbers k + 1 and k + 2. As an example, we take N = 100. If α decreases from α , the interval I consists of two cylinder max α sets until α decreases beyond f and cylinder set Δ emerges; then, if α + 1 decreases 3 3 beyond f , cylinder set Δ disappears and so on, until α decreases beyond f and Δ 1 1 8 9 emerges while Δ has not yet disappeared. In order to get a grip on counting the number of cylinder sets, the following arith- metic will be useful: a full cylinder set counts for 1, an incomplete left one counts for N N − d − α, and an incomplete right one for α + 1 − ( − d ), giving rise to max min α α+1 the following definition: Definition 2 Let N ∈ N and α ∈ R such that 0 <α ≤ N − 1 and T the ≥2 α N -continued fraction map. The branch number b(N,α) is defined as b(N,α) := d − d − 1 (the number of full cylinder sets save for the outermost ones) max min + − d − α (the length of the image of the leftmost cylinder set) max + α + 1 − − d (the length of the image of the rightmost cylinder set). min α + 1 From this the next lemma follows immediately: Lemma 4 For N ∈ N and 0 <α ≤ N − 1 we have ≥2 N N N b(N,α) = − = . α α + 1 α(α + 1) It follows that for fixed N the branch number b(N,α) is a strictly decreasing function of α. Remark 4 Applying Lemma 4, we find N 1 b(N,α ) = √ √ = 1 + √ . (10) max ( N − 1) N N − 1 The word ‘branch’ refers to the part of the graph of T on the concerning cylinder set. 123 Orbits of N-expansions with a finite set of digits 89 It follows that b(N,α) > 1 for all N ∈ N , so the number of cylinder sets is always ≥2 at least 2. On the other hand, from Lemma 4 it follows that the number of cylinder sets increases to infinity as α decreases from α to 0. Actually, we have infinitely max many digits only if α = 0. In this case the corresponding N -expansion is the greedy N -expansion, studied in [1,3]. The relation N /(α(α + 1)) = b yields 4N + 1 − 1 α = , (11) from which we derive that d(α) = d (or d(α) = d − 1 in case − α ∈ Z), where d is given by ⎢ ⎥ ⎢ ⎥ 4N (b − 1) + 1 + b + 1 ⎢ ⎥ ⎣ ⎦ d = . (12) 2.2 Gaplessness if the branch number is large enough So far, we merely discussed the way I is divided in cylinder sets, depending on the values of N,α, d(α) and the branch number b. In order to present some first results on sufficient conditions for gaplessness, we will zoom in on some ergodic properties of T . Lemma 5 If μ is an absolutely continuous invariant probability measure for T , then there exists a function h of bounded variation such that μ(A) = hdλ, λ − a.e., with λ the Lebesgue measure, i.e. any absolutely continuous invariant probability measure has a version of its density function of bounded variation. Proof of Lemma 5 Since inf |T | > 1, applying Theorem 1 from [9] immediately yields the assertion. Theorem 2 Let N ∈ N . Then there is a unique absolutely continuous invariant ≥2 probability measure μ such that T is ergodic with respect to μ . α α α Proof of Theorem 2 Let μ be a unique absolutely continuous invariant probability measure for T and choose its density function h of bounded variation. Then there exists an open interval J such that h(x)> 0 for any x ∈ J , since h has at most countably many discontinuity points. Consider {T J : n ≥ 0}. Since inf |T | > 1, there exists an n such that T (J ) includes a discontinuity point. (If necessary we may choose endpoints of J not in the preimages of discontinuity points of T .) We note that for see also page 185, Theorem 1 in [10]. 123 90 J. de Jonge et al. any measurable subset A ⊂ J with μ (A)> 0 equivalently λ(A)> 0, μ(T A)> 0 n +1 for any n ≥ 1. Now T (J ) includes two intervals J and J attached to α and α + 1 respectively. For any measurable subset B ⊂ J ∪ J of positive λ-measure, μ(B )> 0  r 0 −(n +1) 0, since otherwise we have a contradiction; μ(B ) = 0 and μ(T (B )) > 0 0 0 n +1 (since there is a B ⊂ J such that T (B ) = B ,μ(B )> 0). This shows that 1 1 0 1 any two absolutely continuous invariant probability measures μ and μ cannot have 1 2 disjoint supports (i.e. they cannot be singular to each other), which is equivalent to the uniqueness of the absolutely continuous invariant probability measure and hence its ergodicity. The next result follows directly from Theorem 2: Corollary 1 If iteration of T maps all open subintervals of I to the interval I , then α α I contains no gaps. Proof of Corollary 1 The assumption implies that the absolutely continuous invariant probability measure μ is equivalent to the Lebesgue measure, which implies that for any measurable subset A ⊂ I , μ(A) = 0 if and only if λ(A) = 0. Suppose that there is a gap J . Since J is an open interval, we have λ(J)> 0, thus μ(J)> 0. Since μ(I )< ∞ implies a.e. x ∈ J , there exists infinitely many positive integers n such that T (x ) ∈ J (by the Poincaré recurrence theorem), which contradicts the assumption that there is a gap. Before we present the first of two theorems on gaplessness, we note that in the case N = 2, the condition |T (x )| > 2 for all x ∈ I is not satisfied for any α ∈ (0, 2−1]. Theorem 3 Let N ∈ N , and let 0 <α ≤ N − 1. Let |T (x )| > 2 for all x ∈ I . ≥3 α Then I contains no gaps. Proof of Theorem 3 The condition implies N /(α + 1) > 2, yielding α< N /2 − 1. From Lemma 4 it follows that 2 2N b(N,α) > √ , 2N − 2 which is larger than 2 for all N ∈ N .So I consists of at least three cylinder sets. ≥3 α Since |T (x )| > 2 for all x ∈ I , there exists an ε> 0 such that for any open interval J that is contained in a cylinder set of T we have 0 α | | T (J ) ≥ (2 + ε)|J |, α 0 0 where |J | denotes the length (i.e. Lebesgue measure) of an interval J . If T (J ) contains two consecutive discontinuity points p , p of T , then α 0 i +1 i α (p , p ) ⊂ T (J ), i +1 i α 0 and we immediately have that O 2 I := (α, α + 1) = T (p , p ) ⊂ T (J ). α i +1 i 0 α α 123 Orbits of N-expansions with a finite set of digits 91 If T (J ) contains only one discontinuity point p of T , then T (J ) is the disjoint α 0 α α 0 union of two subintervals located in two adjacent cylinder sets: T (J ) = J ∪ J . α 0 1 2 Obviously, |T (J )|=|J |+|J |. α 0 1 2 Now select the larger of these two intervals J , J , and call this interval J . Then 1 2 |J |≥ (1 + )|J |. 1 0 In case T (J ) does not contain any discontinuity point of T ,weset J = T (J ). α 0 α 1 α 0 Induction yields that there exists an  ∈ N such that |J |≥ 1 + |J | , whenever T (J ) includes no more than one discontinuity point of T . But then there α −1 α must exist a k ∈ N such that T (J ) contains two (or more) consecutive discontinuity α k 2 O points of T , and we find that T (J ) = I . Applying Corollary 1, we conclude that α k α α there is no gap in I . Remark 5 As one of our referees remarked, the result from Theorem 3 is rather folklore. As to one of its first appearances we refer to Keith Wilkinson’s paper [13]; it might be interesting also to consult Gerhard Keller’s paper [6]. The next theorem, which is partly a corollary of the previous one, gives an even more explicit condition for gaplessness. Theorem 4 Let I consist of five cylinder sets or more. Then I has no gaps. α α Proof of Theorem 4, part I: Let I consist of five cylinder sets or more. Then b(N,α) > 3, implying 1 1 3 12N + 9 − 9 4N α< + 1 − (cf. (11)), in which case |T (α + 1)| > 3 − . 2 2 2N The second inequality yields that for N ∈ N we have |T (α + 1)| > 2 and, ≥18 applying Theorem 3, I is gapless. Now suppose N ∈{12,..., 17}. Then b(N,α) = 3 involves arrangements with four cylinders. In each of these cases, the smallest α such that I has not yet (i.e. decreasing from α ) consisted of five cylinder sets is f . α max 7 In all six cases (two of which are illustrated in Fig. 4)wehave |T ( f + 1)| > 2, yielding the gaplessness of I for arrangements with five or more cylinders in case N ∈{12,..., 17}. This finishes the proof of Theorem 4 for N ∈ N .For N ∈ ≥12 {2,..., 11} a similar approach does not work. We will use some ideas that we will 123 92 J. de Jonge et al. 6 5 4 3 6 5 4 3 N =17,α = f ; |T (α +1)| =2.0098 ··· N =12,α = f ; |T (α +1)| =2.0415 ··· 7 7 Fig. 4 Two arrangements illustrating the gaplessness of arrangements with five cylinders or more on account of Theorem 3 introduce and develop in the next sections and will finish the proof of Theorem 4 at the end of Sect. 5. In the following we will go into conditions for gaplessness of arrangements con- sisting of less than five cylinders. We will start with two cylinders and will use the results for arrangements with three and four cylinders. Remark 6 Since b(N,α) is a strictly decreasing function of α (cf. Lemma 4) and b(N,α ) = 1 + 1/( N − 1), the condition I = Δ ∪ Δ is never satisfied in max α d d−1 case N ∈{2, 3}. 3 Gaplessness if I consists of two cylinder sets In general, if the branch number is not much larger than 1 (which is if α is not much smaller than α ), the overall expanding power of T , determined by T (or |T |, max α α α which we will often use), is not enough to exclude the existence of gaps; we shall elaborate on this in a subsequent article. However, in the case of two cylinder sets I = Δ ∪ Δ , there is a very clear condition under which this power suffices: α d d−1 Theorem 5 Let I = Δ ∪ Δ .If T (α) ≥ f and T (α + 1) ≤ f , then I is α d d−1 α d−1 α d α gapless. Although the statement of 5 is intuitively clear, for the proof of Theorem 5 we need several results and lemmas that we will prove first. Then, immediately following Remark 11, we will prove Theorem 5 itself. Remark 7 In case I = Δ ∪ Δ and either T (α) < f or T (α + 1)> f ,itis α d d−1 α d−1 α d 2 2 easy to see that (T (α), T (α)) or (T (α + 1), T (α + 1)) is a gap, respectively. α α α α Since arrangements under the condition of Theorem 5 play an important role in this section, we introduce the following notations: 123 Orbits of N-expansions with a finite set of digits 93 2 1 2 1 3 2 3 2 3 2 F F F 2 2 F2 F F 2 3 F F 3 3 N =9,σ ≈ 1.01 N =17,σ ≈ 1.06 N =18,σ ≈ 1.27 N =24,σ ≈ 1.23 N =25,σ ≈ 1.22 3 2 4 3 4 3 5 4 5 4 F F F 3 4 F 4 F 5 F 4 3 F F 4 5 N =49,σ ≈ 1.20 N =50,σ ≈ 1.33 N =99,σ ≈ 1.26 N = 100,σ ≈ 1.36 N = 165,σ ≈ 1.29 Fig. 5 Arrangements in F (d), d ∈{2, 3, 4, 5},where α is maximal Definition 3 Let N ∈ N be fixed. For d ∈ N , we define F (d) as the family of all ≥4 ≥2 arrangements Υ such that I = Δ ∪ Δ , T (α) ≥ f and T (α + 1) ≤ f . N ,α α d d−1 α d−1 α d We will write F (d) in case α satisfies the equation T (α) = f , the root of which α d−1 we will henceforth denote by α(N , d). Remark 8 Note that for each N ∈ N and d ∈ N we have that α(N , d), if it exists, ≥4 ≥2 is the only value of α such that F (d) is not void. If the expanding power of T is large enough to exclude the existence of gaps for the largest α for which an arrangement in F (d) exists, there will not be gaps in any arrangement in F (d). We will now first show how to find these largest α, which takes some effort. If we have finished that, we will go into the expanding power of |T | in these arrangements with largest α. For 4 ≤ N ≤ 8, with d = 2, we have T (α )> f , while T (α + 1) = α max 1 α max max max α < f . Hence we see Υ ∈ F (2) and F (2) =∅.For N ∈ N we have max 2 N ,α ≥9 max T (α )< f .If d = 2 we can find α such that Υ in F (2) for each 9 ≤ N ≤ α max 1 N ,α max 17; see Fig. 5, where ten arrangements in various F (d) are drawn. Underneath each arrangement we have mentioned an approximation of σ(α) := |T (α + 1)|, which we will later return to. This σ is important, because it is the expanding power on the rightmost cylinder set that may be too weak to exclude gaps. If d = 2 and N ∈ N , the condition T (α) = f yields T (α + 1)> f , and d ≥18 α d−1 α d has to increase by 1 so as to find an arrangement in F (3).If d = 3, for 18 ≤ N ≤ 24 we find that the largest α is f − 1, in which case T (α + 1) = α< f and d−2 α d T (α) > f (so in this case the arrangement with the largest α is in F (3) but not α d−1 in F (3)); for 25 ≤ N ≤ 49, the largest α is such that T (α) = f .If N ∈ N ,the α 2 ≥50 family F (3) is empty and d has to increase further; see Fig. 5 once more. In the proof of Lemma 8 this approach (of exhausting F (d) for successive values of N and going to F (d + 1) for larger values of N ) will be formalised into a proof by induction. Due to (12) such an increase is always possible, no matter how large d and N become. Note that this inductive approach works since for each d only finitely many N exist such that there are α with Υ ∈ F (d). To see why this claim holds, note that for N ,α 123 94 J. de Jonge et al. fixed N and d, the smallest α for which d = d(α) = d is α , given by max d 4N + (d + 1) − (d + 1) α = f = ; d d+1 cf. (3). For this α it is not necessarily so that I = Δ ∪ Δ , i.e. that I consists α d d−1 α d d of two cylinder sets (e.g. if N = 2 and d = 5, there are five cylinder sets). However, if b(N,α ) ≤ 2, we know that I exists of two cylinder sets, the left one of which is d α full. According to Lemma 4, the branch number b(N,α ) satisfies N 4N b(N,α ) = = . 2 2 2 f ( f + 1) d+1 d+1 4N + (d + 1) − 2d 4N + (d + 1) + d − 1 Keeping d fixed and letting N →∞, we find lim b(N,α ) = 1. N →∞ In view of this and Lemma 3 (and the results mentioned directly thereafter), we choose N sufficiently large, such that for α ≥ α we have b(N,α) < 5/4 and f − f > d d−1 d 1/4. Now suppose that for such a sufficiently large value of N there exists an α ≥ α , such that α ∈ F (d). Then by Definition 2 of branch number and the assumption that α ∈ F (d), we have that b(N,α) ≥ 1 + f − f > , d−1 d which is impossible since for N sufficiently large, d fixed and α ≥ α we have b(N,α) ≤ b(N,α )< . It follows that for d fixed and N sufficiently large, F (d) is void. We will prove (in Lemma 8) that if N ∈ N there exists a minimal d ∈ N such ≥25 ≥3 that the arrangement in F (d) with α maximal lies in F (d). Before we will prove this, we will explain the relation between d and N for arrangements in F (d). In Fig. 5 we see that for N ∈{49, 99, 165} the arrangements in F are very similar, and that the arrangement for N = 100 is more similar to these than the arrangement for N = 50. Moreover, the last three arrangements look hardly curved. This is easy to understand, considering the following equations, where b(N,α) = b is fixed: √ √ 2 2 b( 4bN + b + 2N + b) b 4bN + b |T (α)|= and |T (α)|−|T (α + 1)|= . α α α 2N N Since (b 4bN + b )/N is a decreasing function of N , approaching 0 from above as N →∞, the second equation implies that for a fixed branch number b the branches become less curved as N increases; i.e., the curves approach linearity as N →∞ and b 123 Orbits of N-expansions with a finite set of digits 95 (a , 1) (a, a) (1, − + a +1) Fig. 6 The ‘limit graph’ of T , translated over (−α, −α), under the conditions I = Δ ∪ Δ and α α d d−1 N /α − d = f for N →∞ (and α, d →∞ accordingly) d−1 is fixed. Although in F (d) the branch number is not so much fixed as bounded between 1 and 2, we have a similar decrease of curviness as N increases. The arrangements for N ∈{49, 99, 165} in Fig. 5 suggest that (assuming T (α) = f , i.e. α = α(N , d))) α d−1 if N →∞ (and d →∞ and α →∞ accordingly), the difference f − T (α + 1) d α tends to 0, yielding a ‘limit graph’ of T that consists of two parallel line segments (the straightened branch curves of T ); see Fig. 6, obtained by translating the graph over (−α, −α). In this situation we have both a := T (α) (mod α) = f (mod α) α d−1 and T (α + 1) = f (also (mod α) in Fig. 6). Because in the limit both parts of the α d graph are linear with the same slope, we also have that (0, a + 1) lies on the prolonged right line segment, from which we derive that the line segments have slope −1/a.The line with equation y =− + a + 1 intersects the line y = 1at (a , 1) (so the dividing 2 1 line is x = a ) and intersects the line x = 1in 1, − + a + 1 , yielding the point 1 1 x − + a + 1, − + a + 1 on the line through (0, a) with equation y =− + a a a a (since T (α + 1) = f ). From this we derive 2a = 1, so a = 1/2. α d From Fig. 6 we almost immediately find that the branch number for the limit case is 2 and that the dividing line is at 1/2. We use this heuristic to find a formula describing the relation between N and d for arrangements in F (d) very precisely. Note that for arrangements similar to the limit graph we have 2 2 4N + (d − 1) − 4N + d + 1 1 + f − f = + 1 ≈ b(N,α) ≈ 2, d−1 d from which we derive √ √ N ≈ (4 + 3 2)(d − d) +2or d ≈ 1 + (6 2 − 8)(N − 2) + 1 . (13) 123 96 J. de Jonge et al. If, for d fixed, we determine arrangements in F (d) such that the difference f − T (α + 1) is positive and as small as possible according to our heuristic, the d α best function seems to be N = (4 + 3 2)(d − d), yielding the right N (after round- ing off to the nearest integer) for d ∈{3,..., 500}\{9, 50, 52, 68, 69, 80, 97, 129, 167, 185, 210, 231, 289, 330, 416, 440, 444, 479, 485}, in all of which cases the rounding off should have been up instead of down. For d = 2 we find N = √ √ 2(4 + 3 2)= 17, for d = 3wefind N =6(4 + 3 2)= 49, for d = 4 √ √ we find N =12(4 + 3 2)= 99 and or d = 5wefind N =20(4 + 3 2)= 165; see Fig. 5 once more. Although we do not know generally if rounding off to the nearest integer yields the right N , with (13) we can find a very good overall indication of the relation between d and N for arrangements in F (d) by looking at the difference between the image of α(N , d) + 1 and f ; see Definition 3. With some straightforward calculations we find that N 4N + (d − 1) − (d + 1) α(N , d) = . (14) 2(N − d) Applying (14), we write f (N ) − (N /(α(N , d) + 1) − (d − 1)) as 2 2 2 2 2 (N + dN + d) 4N + d − N 4N + (d − 1) − (N − d(d − 4)N − d(d − 2)) j (N ) := 2(N + dN + d) and, more generally, define 2 2 2 2 2 (x + dx + d) 4x + d − x 4x + (d − 1) − (x − d(d − 4)x − d(d − 2)) j (x ) := 2(x + dx + d) (15) for x ∈[25, ∞). We note that I = Δ ∪ Δ is equivalent to N /(α + 1) − (d − 1) ≥ α. In case α d d−1 α = α(N , d),wehave 2 2 N 4N + (d − 1) − (d − 1)N + 2d(d − 2)N + 2d(d − 1) − (d − 1) = . α(N , d) + 1 2(N + dN + d) (16) Applying (16), for the difference h (N ) := N /(α(N , d) + 1) − (d − 1) − α(N , d) we write 3 2 3 2 2 2N + 4dN + (2d − 5d + 3d)N + 2d (d − 1) − dN (2N + 1) 4N + (d − 1) h (N ) := , 2(N − d)(N + dN + d) (17) and, more generally, define 3 2 3 2 2 2x + 4dx + (2d − 5d + 3d)x + 2d (d − 1) − dx (2x + 1) 4x + (d − 1) h (x ) := , 2(x − d)(x + dx + d) (18) for x ∈[25, ∞). 123 Orbits of N-expansions with a finite set of digits 97 Now we can prove the lemma that is illustrated by the arrangements for N ∈ {17, 49, 99, 165} in Fig. 5. In order to so, we define for fixed d ∈ N ≥2 S(d) := {N ∈ N : I = Δ ∪ Δ , T (α) = f (N ) and T (α + 1) ≤ f (N )} ≥4 α d d−1 α d−1 α d and M := max S(d). √ √ 2 2 Lemma 6 Let d ∈ N . Then M ∈{(4 + 3 2)(d − d), (4 + 3 2)(d − d)}. ≥2 d Proof of Lemma 6 First we note that for d = 2, we have that (4 + 3 2)(d − d) equals 17, which corresponds with what we had already calculated and drawn in Fig. 5.Now let d ∈ N . First we have to show that h (M )> 0for M ∈{(4 + ≥3 d d d √ √ 2 2 3 2)(d − d), (4 + 3 2)(d − d)}, for this assures us that I = Δ ∪ Δ .We α d d−1 will leave this to the reader; it is merely very cumbersome to show, while technically straightforward. The only thing left to do is showing that j ((4 + 3 2)(d − d)) > 0; (19) j ((4 + 3 2)(d − d) + 1)< 0, since the first equation implies that j ((4 + 3 2)(d − d))> 0, while the second implies that j ((4 + 3 2)(d −d)+ 1)< 0. The work to be done is as cumbersome and straightforward as the previous work to be done for this proof and is left to the reader as well. Before we will show that for N ∈ N there are a d ∈ N and an α such that ≥25 ≥3 Υ ∈ F (d), we will prove the following lemma: N ,α Lemma 7 Let d ∈ N . Let N ∈ N be such that I = Δ ∪ Δ and ≥3 ≥25 α(M ,d) d d−1 T (α(M , d)) = f for M ∈{N , N + 1}. Then α(M ,d) d d−1 d T (α(N + 1, d) + 1) − α(N + 1, d)> T (α(N , d) + 1) − α(N , d), α(N +1,d) α(N ,d) i.e. h (N + 1)> h (N ). d d Proof of Lemma 7 We want to show that h (N ) from (17) is an increasing sequence, and do so by calculating the derivative of with x ∈[25, ∞), and then showing that h (x)> 0on [25, ∞). Although a little bit intricate, the work is straightforward and is left to the reader. Now we can prove the following lemma: We have throughout this paper frequently used (Wolfram) Mathematica for making intricate calculations, all of which are nonetheless algebraically basic. In relevant cases we think it will be evident if we did. 123 98 J. de Jonge et al. Lemma 8 Let N ∈{9,..., 17, 25, 26,...}. Then there are d ∈ N and α ∈ ≥2 (0, N − 1) such that I = Δ ∪ Δ ,T (α) = f and T (α + 1) ≤ f α d d−1 α d−1 α d (i.e. α = α(N , d)). Proof of Lemma 8 We will use induction on d.For N ∈{9,..., 17, 25, 26,... , 99} and d ∈{2, 3, 4} we refer to Fig. 5 and leave the calculations to the reader. Specifically, we have for 50 ≤ N ≤ 99 that Υ ∈ F (4). It is easily seen that Υ ∈ N ,α(N ,4) 99,α(99,5) ∗ ∗ F (5) as well. Due to Lemma 6, there is an N > 99 such that Υ ∈ F (5). 5 N ,α(N ,5) 5 5 Applying Lemma 7, we see that for all N ∈{99,..., N } we have Υ ∈ F (5). 5 N ,α(N ,5) For the induction step, let d ∈ N be such that there is an α for which Υ ∈ F (d), ≥5 N ,α where N is the largest such N possible, cf. Lemma 6. If we can show that for this N there is an α such that Υ ∈ F (d + 1), we are finished. This can be done by d N ,α showing that h ((4 + 3 2)(d − d) − 1)> 0, (20) d+1 forthisimplies h (N )> 0, in which case α is such that − (d + 1) = f , d+1 d i.e. α = α(N , d + 1). Although intricate, the calculations are straightforward and are left to the reader. Remark 9 Although Lemma 8 is about N in the first place, our approach is actually based on increasing d and then determining all N such that arrangements Υ ∈ N ,α F (d) exist. The proof of Lemma 8 yields the arrangements with the smallest d (and therefore the largest α)for which Υ ∈ F (d), as illustrated by the last five N ,α arrangements of Fig. 5. Example 1 For d = 4wehave M − 1 =(4 + 3 2)(d − d)= 98. Then ∗ ∗ Υ ∈ F (4) and Υ ∈ F (4), while Υ ∈ M −1,α(M −1,4) M ,α(M ,4) M +1,α(M +1,5) d d d d d d F (5); see Fig. 5. It follows immediately from our construction of α(M + 1, 5) that this is the largest α such that Υ ∈ F (5). M +1,α With manual calculations we can quickly calculate the expanding power of T in α + 1 for arrangements in F and F and N not too large, say N ∈ N , where the ≤49 smallest values are found where α is as large as possible. The next proposition gives a lower bound for |T (α + 1)| for such arrangements for most N . Proposition 1 Let N ∈{18}∪{50, 51,...}\{95,..., 99} and α ∈ (0, N − 1] such that I = Δ ∪ Δ for some d ∈ N,d ∈ N . Furthermore, suppose that T (α) ≥ α d d−1 ≥2 α f and T (α + 1) ≤ f . Then |T (α + 1)| > 2 = 1.259921 .... d−1 α d Proof of Proposition 1 Considering Lemma 8, we can confine ourselves to arrange- ments in F with α as large as possible. For α = α(N , d) (cf. (14)) we can write |T (α + 1)|= N /(α + 1) as k (N ) 4 2 3 2 2 3 2 2 2N + (d − 1) N + 2d(d − 1)N + 2d N + ((d − 1)N + 2dN ) 4N + (d − 1) = . 4 3 2 2 2 2(N + 2dN + d(d + 2)N + 2d N + d ) (21) 123 Orbits of N-expansions with a finite set of digits 99 It is not hard to find that, for d fixed, k is a decreasing sequence, with lim k (N ) = 1. However, from (13) it follows that if N →∞ we have N →∞ d that also d →∞ in a precise manner. Due to the previous lemmas, for each d we can confine ourselves to considering only N /(α + 1) for the largest N and α such that Υ ∈ F (d). Applying Lemma 6, an easy way to check if N ,α indeed |T (α(N , d) + 1)| > 2 is considering k (x ), with x ∈[100, ∞), α(N ,d) and then calculating k ((4 + 3 2)(d − d) + 1) for d ∈ N , which is amply d ≥5 larger than 2 = 1.2599 ··· . For the remaining cases d = 3 and N = 18 and for d = 4 and N ∈{50, 51,... , 94} it is easily checked manually that indeed |T (α(N , d) + 1)| > 2. α(N ,d) Remark 10 Considering our previous remarks concerning arrangements in F ,itmay be clear that lim N /(α(N , d) + 1) = 2. N →∞ Remark 11 The value 2 in the proof of Proposition 1 relates to the proof of Theorem 5 and also to the proofs of Proposition 9 and Theorem 3, where the numbers 2 and 2 have a similar importance. Considering the proof of Proposition 1, we could actually replace 2 by the smallest possible value, given by 94 20480015 + 320305 385 = = 1.2604 .... (α(94) + 1) 21233664 Finally we are ready to prove Theorem 5, stating that I = Δ ∪ Δ , with α d d−1 d := d(α), is gapless if T (α) ≥ f and T (α + 1) ≤ f . Considering Remark 11 α d−1 α d the value 2 in Proposition 1 can be replaced by 1.26, the third power of which is 2.000376. We will use this to stress that the gaplessness of Theorem 5 is actually relatively ample and does not require infinitesimal estimations. Proof of Theorem 5 First we note that the conditions imply N ∈ N .Now let Υ ∈ ≥4 N ,α F (d) and let K ⊂ I be any open interval. Since K expands under T , there is an α α n ∈ N∪{0} such that T (K ) contains for the first time a fixed point or the discontinuity point p , in the former case of which we are finished. So we assume that T (K )∩Δ = d d (b, p ]=: L, with f < b < p . Note that T (L) =[α, T (b)) ⊂[α, f ).For d d d α α d 2 2 2 T (L) = (T (b), T (α)], we similarly may assume that f < T (b)<α + 1 α d−1 α α α (since otherwise f ∈ T (L), and again we are done). d−1 3 3 3 Now suppose that T (L) contains p , excluding f ∈ T (L). Then T (L) = L ∪ d d 1 α α α 2 3 M , with L =[T (α), p ] and M = (p , T (b)). First we confine ourselves to N ∈ 1 1 d 1 d α α {18}∪{50,...}\{95,..., 99}. Since then |T (L)| > 2.000376|L| (cf. Remark 11), we have certainly |L | > 1.001|L| or |M | > 1.001|L|. If we consider the images of 1 1 2 3 L and M under T , T and T similarly as we did with the images of L, we find 1 1 α α α that due to expansiveness (see the proof of Theorem 3) there must be an m such that 3m 3m 3 f ∈ T (L ) or f ∈ T (M ) and we are finished. If T (L) does not contain d 1 d−1 1 α α α p , the expansion of L will only go on longer, yielding even larger L and M and 1 1 the reasoning would only be stronger that no gaps can exist. For N ∈{4,..., 17, 19, 20,... , 49, 95, 96,... , 99} a similar approach can be taken, but there is no useful general lower bound for |T (x )| on I . For these cases, 123 100 J. de Jonge et al. however, the moderate expanding power in Δ is easily made up for by a rela- d−1 tively strong expanding power in Δ , and the gaplessness is easily, although tediously, checked by hand (cf. Examples 2 and 3 below). This finishes the proof of Theorem 5. Example 2 In case N = 7, there exist α ∈ (0, 7 − 1] for which I = Δ ∪ Δ .The α 2 1 largest α for which Υ ∈ F (2) is α = 7 − 1, in which case |T (α + 1)|= 1. 7,α max However, |T ( f )|= 2.0938 ··· > 2, and the approach taken above even works due to the expanding power of T on [α, f ) alone. α 2 Example 3 In case N = 99, we have I = Δ ∪ Δ , and Υ ∈ F for α 4 3 99,α α = 99( 405 − 5)/190 = 7.8807 ··· . Then |T (α + 1)|= 1.2552 ··· , |T ( f )|= α α 1.3503 ··· and |T ( f )|= 1.4908 ··· . So for an interval (p , x ), with x ∈ (p , f ), 4 4 d 3 3 3 assuming that f ∈ / T (p , x ),wehave |T (p , x )| > 1.3503 ··· × 1.2552 ··· × 4 4 4 α α 1.4908 ··· × |(p , x )| 2|(p , x )|, implying enough expanding power for T to 4 4 exclude the existence of gaps. Remark 12 We can also prove that |T (x )| > 2on Δ for all arrangements under the assumptions of Theorem 5, but we cannot do without knowledge about the slope on Δ . d−1 Next we will make preparations for formulating a sufficient condition for gapless- ness in case I consists of three cylinder sets. Proving it involves more subtleties on the one hand, but will have a lot of similarities with the two-cylinder set case on the other hand. Once we have finished that, not much work remains to be done for gaplessness in case I consists of four or five cylinder sets. 4 A sufficient condition for gaplessness if I consists of three or four cylinder sets If I = Δ ∪...∪Δ , with m ∈{2, 3}, there is a sufficient condition for gaplessness α d d−m that resembles the condition for gaplessness in case I consists of two cylinder sets a lot: Theorem 6 Let I = Δ ∪ ... ∪ Δ , with m ∈{2, 3}. Then I is gapless if α d d−m α T (α) ≥ f or T (α + 1) ≤ f . α d−1 α d−m+1 We will prove this theorem in parts. In Sect. 4.1 we will prove Theorem 6 for m = 2; in Sect. 4.2 we will extend the result of Sect. 4.1 to m = 3; considering Theorem 4, extension to larger m is not useful. Remark 13 The difference between the ‘and’ of Theorem 5 and the ‘or’ of Theorem 6 has to do with the existence, in the latter case, of at least one full cylinder set. 123 Orbits of N-expansions with a finite set of digits 101 3 2 1 3 2 1 4 3 2 4 3 2 5 4 3 N =3,σ =1 N =4,σ ≈ 1.06 N =7,σ ≈ 1.46 N =9,σ ≈ 1.40 N =16,σ ≈ 1.64 |T (f )|≈ 1.77 |T (f )|≈ 1.64 |T (f )|≈ 2.09 |T (f )|≈ 1.92 |T (f )|≈ 2.08 α 1 α 1 α 2 α 2 α 3 |T (f )|≈ 3 |T (f )|≈ 2.62 |T (f )|≈ 2.95 |T (f )|≈ 2.62 |T (f )|≈ 2.62 2 2 3 3 4 α α α α α Fig. 7 Arrangements with largest α such that there is a d with I = Δ ∪ Δ ∪ Δ under the condition d d−1 d−2 T (α) ≥ f and T (α + 1) ≤ f α d−1 α d−1 4.1 Gaplessness if I consists of three cylinder sets Since we have m = 2, the condition T (α) ≥ f can be split in α d−1 ⎪ 1. T (α + 1) ≤ f ≤ T (α); α d−1 α 2. f ≤ T (α) ≤ T (α + 1); (22) d−1 α α 3. f ≤ T (α + 1) ≤ T (α); d−1 α α of course the condition T (α + 1) ≤ f can be split in a similar way. We will α d−1 prove Theorem 6 by proving gaplessness according to this distinction in three cases, associated with Lemmas 9, 10 and 11 respectively. The first of these is not very hard to prove: Lemma 9 Let I = Δ ∪ Δ ∪ Δ .If T (α) ≥ f and T (α + 1) ≤ f , α d d−1 d−2 α d−1 α d−1 then I is gapless. Proof of Lemma 9 The assumptions imply that b(N,α) > 2, yielding σ(α) =|T (α + 1)| > 2for N ∈ N .If N ∈ N ,welet K ⊂ I be any open interval. Since K ≥17 ≥17 α expands under T , there is an n ∈ N ∪{0}) such that T (K ) contains a fixed point or a discontinuity point p (with i ∈{0, 1}) , in the former case of which we are finished. d−i So we assume that T (K ) ⊃ L, where L = (b, p ], with f < b < p , d−i d−i d−i with i ∈{0, 1}.If T (L) contains a fixed point, we are finished. If T (L) does not α α contain a fixed point, then it cannot contain a discontinuity point, and we have that |T (L)| > 2|L|, implying enough expanding power of T to ensure gaplessness of at least one cylinder set (which might be non-full). Since both T (α) ≥ f and α d−1 T (α + 1) ≤ f ,itfollows that I is gapless. For 2 ≤ N ≤ 16 the slopes on I may α d−1 α α differ considerably: for some N , such as N = 7 and N = 16 we also have σ> 2, but if this is not he case, the steepness left of f is amply larger then 2; see Fig. 7 d−2 for some examples where α is as large as possible. This finishes the proof of Lemma 9 [cf. case 1 in (22)]. If I = Δ ∪Δ ∪Δ under the condition T (α) ≥ f and T (α+1)> f α d d−1 d−2 α d−1 α d−1 or under the condition T (α) < f and T (α + 1) ≤ f , I is gapless as well, α d−1 α d−1 α but this is much harder to prove. The following definition will be convenient: 123 102 J. de Jonge et al. 5 4 3 3 2 1 3 2 1 N =23,α =2.898 N =11, α =1.873 ··· N =7,α =1.54 |T (α +1)|≈ 1.51 |T (α +1)|≈ 1.33 |T (α +1)|≈ 1.09 α α α |T (α)|≈ 2.74 |T (α)|≈ 3.14 |T (α)|≈ 2.95 α α α Fig. 8 Arrangements with one very small cylinder Definition 4 Let I = Δ ∪ ... ∪ Δ , and 1 ≤ m ≤ d − 1. If T (α) ≤ f or α d d−m α d−1 T (α + 1) ≥ f , the cylinder set Δ respectively Δ is called small. α d−m+1 d d−m Taking a similar approach as in the proof of Theorem 5, one can show that the map T has enough expansive power to ensure that for any open interval K ⊂ I there exists α α a non-negative integer n such that T (K ) contains a fixed point. If this fixed point is in a non-small or even full cylinder set, we are done (as in the proofs of Theorem 5 and Lemma 9). However, if this fixed point is from the small cylinder set, then it only follows that every point of the small cylinder set is in the orbit under T of some point in K . Note this implies that the small cylinder set is gapless. So we may assume that the small cylinder set is gapless. Let us assume that the left cylinder set is small. We define L := T (Δ )\Δ . Since Δ is gapless, we have L = (p , T (α)]⊂ (p , f ), α d d d d d d−1 2 2 so T (L) =[T (α), α + 1).If T (α) ≤ f , we are finished, so we assume that α d−2 α α 2 2 3 3 T (α) > f . We then have T (L) = (T (α + 1), T (α)].If T (α) ≥ f we are d−2 α d−1 α α α α finished, since then f ∈ T (L). d−1 The question arises whether it is possible to keep avoiding fixed points if we go on with letting T work on L and its images (or similarly, if the right cylinder set is small, some interval R := T (Δ )\Δ ). We will argue that this is not possible in the α 1 1 two most plausible cases for gaps to exist, involving the least expansion. The first case is illustrated with two arrangements in Fig. 8, one where N = 23 and one where N = 11. In both arrangements one outer cylinder is very small while the other one is full or almost full. In the arrangement where N = 23, we see that L is a very narrow strip between p and T (α), T (L) is not so narrow anymore, and 5 α T (L) is definitely wide enough to make clear that avoiding fixed points f and f 4 3 is not possible. The middle arrangement, where N = 11, is an example of the case where Δ is small and Δ is actually full. Here we have that R := T (Δ )\Δ is d−2 d α 1 1 a very narrow strip between T (α + 1) and p and that T (R) is only slightly larger α 2 than T (Δ ), whence eventually there will be an n ∈ N such that f ∈ T (R) or α 1 3 f ∈ T (R). The rightmost arrangement in Fig. 8 is an illustration of the second plausible case for the existence of gaps: here Δ is small, while Δ is incomplete but not small. This 3 1 123 Orbits of N-expansions with a finite set of digits 103 4 3 2 N =30,α =3.6 |T (α +1)|≈ 1.42 |T (α)|≈ 2.31 T (R) V V T (R) α 1 2 R Fig. 9 Arrangement illustrating Lemma 10 3 2 1 4 3 2 N =17,α =2.6576 N =35,α =4.00167 |T (α +1)|≈ 1.27 |T (α +1)|≈ 1.40 α α |T (α)|≈ 2.41 |T (α)|≈ 2.19 α α 2 4 2 Fig. 10 Two arrangements in which almost T (α) < p < T (α + 1) (in fact, in both cases p = T (α)) α d α d α arrangement illustrates the role p might play in avoiding fixed points: in this case, 3 4 taking L := T (Δ )\Δ ,wehave T (L) = M ∪ M , with M =[T (α), p ] and α 3 3 1 2 1 2 α α 2 3 M = (p , T (α + 1)]. Since T (L) contains a discontinuity point, the expansion 2 2 α α under T is interrupted. If T (M ) would be a subset of T (Δ ) and T (M ) would α α 1 α 3 α 2 be a subset of T (L), the expansion would be finished and we would have three gaps: 3 4 2 2 (T (α), T (α + 1)), (T (α), T (α)) and (T (α + 1), T (α))—but this is not the case, α α α α α α as we will shortly prove. Of course arrangements exist such that one of the outer cylinders is small, fixed points are avoided (in the sense we used above) for a long time and it takes more of T working on L or R before one of the discontinuity points is captured. But in these cases the interruption of the expansion is even weaker than in the cases above. We will first show that arrangements such as the rightmost one of Fig. 8 exclude the existence of gaps (cf. Lemma 10) and will then consider cases such as the first two arrangements of Fig. 8 (cf. Lemma 11). Lemma 10 Let I = Δ ∪ Δ ∪ Δ .Then I is gapless if α d d−1 d−2 α f ≤ T (α) ≤ T (α + 1) or T (α) ≤ T (α + 1) ≤ f . d−1 α α α α d−1 123 104 J. de Jonge et al. ( , 1) a+1 (a, a) y = −(a +1)x + a (a, 1 − a ) y = −(a +1)x + a +1 Fig. 11 The ‘limit graph’ of T , translated over (−α, −α), under the conditions I = Δ ∪ Δ and α α d d−1 N /(N /α − d) − (d − 1) = p for N →∞ (and α, d →∞ accordingly). This ‘arrangement’ can be seen as one with three cylinders, where Δ (mod α), the one on the right, is infinitely small; see also the d−2 arrangements in Fig. 10 2 1 3 2 1 3 2 1 3 2 3 2 1 b(4,α)=2 b(5,α)=2 b(6,α)=2 b(11,α)=2 b(12,α)=2 2N +1−1 Fig. 12 Arrangements of Υ with b = 2; in each case α = N ,α Proof of Lemma 10 We will confine ourselves to the first case of this lemma, that is if f ≤ T (α) ≤ T (α + 1), since the second one is proved similarly (in d−1 α α fact, this case is slightly harder due to the smaller size of the absolute value of the derivatives). Regarding our observations above, we may assume that the small cylin- der set (which in this case is Δ ) is gapless (cf. the remarks after Definition 4). d−2 We will show that this implies the gaplessness of the other cylinder sets as well. We define R := T (Δ )\Δ and try to determine α such that p ∈ T (R) α d−2 d−2 d (see the remark immediately preceding this lemma). Necessary conditions for this 2 4 2 are T (α) < p < T (α + 1) (assuming that f ∈ / T (R) and f ∈ / T (R), d d α d−1 α α α since in either case we would be done). If these conditions are satisfied, we write 3 2 4 T (R) = V ∪ V , with V =[T (α), p ] and V = (p , T (α + 1)]. We will show 1 2 1 d 2 d α α α that we cannot have both T (V ) ⊂ T (R) and T (V ) ⊂ T (Δ ), which is neces- α 1 α α 2 α d−2 sary for limiting the expansion of R under T and so not eventually capturing f and α d f ; see Fig. 9. d−1 We take an approach that is similar to the proof of Theorem 5, for which several lemmas and a proposition where used, partially concerning a relation between N and d in the arrangements involved, partially concerning the slope in α + 1. In this proof we will not explicitly formulate similar statements as lemmas or propositions, nor do 123 Orbits of N-expansions with a finite set of digits 105 we prove them, since they require similar basic but very intricate calculations that we prefer to omit. In order to find the relationship between N and d for arrangements with the con- 2 4 ditions T (α) < p and T (α + 1)> p mentioned above, we refer to some more d d α α relevant arrangements, as shown in Fig. 10. In both cases in Fig. 10, α is such that T (α) = p , which is a value of α that is only a little larger than the values for which 2 4 T (α) < p and T (α + 1)> p . A ‘limit arrangement’ (where the third, rightmost d d α α cylinder is infinitely small), similar to the ‘limit arrangement’ used in the proof of 3 2 Theorem 5, is shown in Fig. 11. The assumptions yield a + a − 1 = 0, with real root a = 0.75487 ··· =: γ . Similar to the proof of Theorem 5 we then find that for arrangements as in Fig. 10 we have (d − 1)(d − 1 + γ)(1 + γ) N ≈ . Using this relationship, we can take a similar approach as in the proof of Proposi- tion 1. We leave out the tedious steps and confine ourselves to observing that the slope of the line segments in Fig. 11 is −(γ + 1) =−1.75487 ··· and that in 2 4 arrangements where T (α) < p and T (α + 1)> p ,wewill seethat theslope d d α α T (α + 1) approaches −(γ + 1) as N tends to infinity. However, for our proof the inequality |T (α + 1)| > ( 5 + 1) = 1.61803 ··· =: G suffices, which turns α 2 out to hold for N ∈ N . We will use this to show that for N ∈ N we have ≥273 ≥273 |T (R)| > |T (Δ )|+|T (R)|. From this it immediately follows that we cannot α d−2 α have that both T (V ) ⊂ T (R) and T (V ) ⊂ T (Δ ), and we are done with the α 1 α α 2 α d−2 proof of Lemma 11. Since |T (x )| is a decreasing function on I , and writing β := |Δ |,wehave α d−2 |T (Δ )| > |T (α + 1)|· β, so |R| >(|T (α + 1)|− 1)β. α d−2 α α It follows that |T (R)| >(|T (α + 1)|− 1) ·|T (p )|β, α d−1 α α that |T (R)| >(|T (α + 1)|− 1) ·|T (p )|·|T ( f )|β, d−1 d α α α α and finally that |T (R)| >(|T (α + 1)|− 1) ·|T (p )| ·|T ( f )|β. d−1 d α α α α We also have |T (Δ )| < |T (p )|β,so α d−2 d−1 |R| <(|T (p )|− 1)β and |T (R)| < |T ( f )|· (|T (p )|− 1)β. d−1 α d−1 d−1 α α α 123 106 J. de Jonge et al. It follows that |T (Δ )|+|T (R)| <(|T (p )|+|T ( f )|· (|T (p )|− 1))β α d−2 α d−1 d−1 d−1 α α α = (|T (p )|−|T ( f )|+|T ( f )|·|T (p )|)β d−1 d−1 d−1 d−1 α α α α < |T ( f )|·|T (p )|β. d−1 d−1 α α So, although crudely, we certainly have that |T (R)| > |T (Δ )|+|T (R)| if α d−2 α |T ( f )|·|T (p )| <(|T (α + 1)|− 1) ·|T (p )| ·|T ( f )|, d−1 d−1 d−1 d α α α α α that is, if |T ( f )| 1 <(|T (α + 1)|− 1) ·|T (p )|· . (23) d−1 α α |T ( f )| d−1 Since |T ( f )| (|T (α + 1)|− 1) ·|T (p )|· >(|T (α + 1)|− 1) ·|T (p )| d−1 d−1 α α α α |T ( f )| d−1 >(|T (α + 1)|− 1) ·|T (α + 1)|, α α we know that (23) holds for |T (α+1)| > G, which in turn holds for all N ∈ N .We ≥273 remark that this value is quite a wide upper bound, since we did a rough approximation. Still, checking that we cannot have both T (V ) ⊂ T (R) and T (V ) ⊂ T (Δ ) for α 1 α α 2 α d−2 smaller N is not that hard and is left to the reader. This finishes the proof of Lemma 10 [cf. case 2 in (22)]. Lemma 10 implies that in case I = Δ ∪ Δ ∪ Δ and f ≤ T (α) ≤ α d d−1 d−2 d−1 α T (α + 1) or T (α) ≤ T (α + 1) ≤ f the division of an interval containing p α α α d−1 d in two smaller ones cannot prevent an overall expansion that excludes any gaps. The other plausible case with three cylinder sets in which gaps might exist is if one outer cylinder set is very small, while the other one is full or nearly full, such that either 3 3 T (α + 1) ≥ T (α + 1) (if Δ is the small cylinder set) or T (α) ≤ T (α) (if Δ α d−2 α d α α is the small cylinder set). We will show that this is not possible either: Lemma 11 Let I = Δ ∪ Δ ∪ Δ . Then I is gapless if α d d−1 d−2 α f ≤ T (α + 1) ≤ T (α) or T (α + 1) ≤ T (α) ≤ f . d−1 α α α α d−1 Proof of Lemma 11 Taking into account our observations immediately following Defi- nition 4 and the arrangements of Fig. 8 for N = 23 and N = 11, we only have to prove that there are no α such that T (α) < T (α) is possible if T (α + 1) ≤ T (α) ≤ f α α α d−1 (in case Δ is small) or such that T (α + 1)> T (α + 1) is possible if f ≤ d α d−1 T (α + 1) ≤ T (α) (in case Δ is small). Note that the conditions T (α) < T (α) α α d−2 α and T (α +1)> T (α +1) imply that the branch number is slightly larger than 2. Now remember that I consists of m full cylinder sets if and only if α = k, N = mk(k + 1) and d = (m −1)(k +1) for some k ∈ N, cf. Theorem 1. Figure 12 shows for increasing 123 Orbits of N-expansions with a finite set of digits 107 values of N a sequence of arrangements where the branch number b is 2, from one full arrangement (here for N = 4) with two cylinders to the next one (here for N = 12). Since |T (α)| > |T (α + 1)|, the arrangements suggest that in case b is slightly larger α α than 2, the most favourable arrangement for T (α + 1) = T (α + 1) to have real roots 2 3 is if N = 2k + 2k − 1, where k ∈ N , while for T (α) = T (α) to have real roots is ≥2 α if N = 2k + 2k + 1, where k ∈ N. We will confine ourselves to investigating only the possibility of T (α + 1) = T (α + 1); the calculations for the other case are similar. So we will try and find out if for N = 2k + 2k − 1, d = k + 1, with k ∈ N ,the ≥2 positive root of T (α + 1) = T (α + 1) lies in I . To do this, we solve α α 2k + 2k − 1 2k + 2k − 1 − (k + 1) = − (k − 1), α + 1 2k + 2k − 1 − k 2k + 2k − 1 − (k − 1) α + 1 which is reducible to 3 2 2 4 2 5 4 3 2 (2k + 6k − k − 1)α + (2k + 5k + k − 2)α − (4k + 6k + 2k − 3k − k + 1) = 0, yielding 4 2 8 7 6 5 4 3 2 36k + 144k + 164k − 12k − 95k − 2k + 21k − 4k − (2k + 5k + k − 2) α = . 3 2 2(2k + 6k − k − 1) (24) A straightforward computation shows that this last expression is smaller than f , k+2 meaning that the root (24) lies outside I if I = Δ ∪ Δ ∪ Δ . Since N = α α k+1 k k−1 2k + 2k − 1 was the most favourable option for investigation, this finishes our proof [cf. case 3 in (22)]. Remark 14 The arrangement for N = 11 in Fig. 8 illustrates that the difference between T (α + 1) and T (α + 1) may be very small. 4.2 A sufficient condition for gaplessness in case I consists of four cylinder sets In the previous subsection we proved Theorem 6 for m = 2, by proving Lemmas 9, 10 and 11. In this subsection we will consider m = 3 and go into the analogons of Lemmas 9, 10 and 11. If I consists of four cylinder sets, the analogon of Lemma 9 is that arrangements I are gapless if I = Δ ∪ Δ ∪ Δ ∪ Δ while T (α) ≥ f and T (α + α α d d−1 d−2 d−3 α d−1 α 1) ≤ f . The analogon of Lemma 11 is that arrangements I are gapless if I = d−2 α α Δ ∪ Δ ∪ Δ ∪ Δ while f ≤ T (α + 1) ≤ T (α) or T (α + 1) ≤ d d−1 d−2 d−3 d−2 α α α T (α) ≤ f . In both cases branch numbers larger than 3 are involved, in which α d−1 case |T (α + 1)| > 2if N ∈ N (and Theorem 3 yields the desired result). The ≥18 cases 2 ≤ N ≤ 17 can be checked manually and are left to the reader; in Fig. 13 the arrangement for N = 11, associated with Lemma 11, illustrates that gaps are out of the question. 123 108 J. de Jonge et al. 5 4 3 2 5 4 3 6 5 4 7 6 5 N =11,α = f 6 N =15,α = f N =24,α = f − 1 N =35,α = f − 1 6 3 4 and T (α +1) = p and T (α)= p and T (α)= p α 5 α 5 α 6 Fig. 13 Four arrangements with two full cylinders The analogon of Lemma 10 is that arrangements I are gapless if I = Δ ∪Δ ∪ α α d d−1 Δ ∪ Δ while f ≤ T (α) ≤ T (α +1) or T (α) ≤ T (α +1) ≤ f .The d−2 d−3 d−1 α α α α d−2 arrangements for N = 15, N = 24 and N = 35 in Fig. 13 are interesting illustrations of the analogon of Lemma 10 in the case of two full cylinder sets instead of one. We will confine ourselves to the arrangement for N = 15; the other ones have similar properties. The arrangement for N = 15 is the boundary case for the situation where we have four cylinders, the left one of which (that would be Δ in this example) is extremely small and the right one is such that almost p ∈ T (T (Δ )\Δ ).The α 6 6 interesting thing is that this option would imply a quick interruption of the expansion of T (Δ )\Δ , involving two large gaps. But it is not really an option: the arrangement for α 6 6 N = 15 in Fig. 13 is exceptional among relatively small N (as well as the arrangements for N = 24 and N = 35 are), while for N > 36 we have |T (α + 1)| > 2if I = Δ ∪ Δ Δ and N /(α + 1) − (d − 2) = p or N /α − d = p . We derived α d d−1 d−2 d d this in a similar way as in the proof of Lemma 10 (see Fig. 11) or the preparations for Theorem 5 (see Fig. 6). Figure 14 shows the associated ‘limit graph’, from which it is easily found that a = (3 − 5), yielding branch number 2 + g, with g = 1/G the small golden section. With this, we conclude the proof of Theorem 6. In the next section we will prove that if I = Δ ∪ Δ ∪ Δ ∪ Δ gaps exist α d d−1 d−2 d−3 only in very rare cases and if they do, that they are very large. After that, we will finish the proof of Theorem 4, stating that all arrangements with five cylinders are gapless. 5 Gaplessness in case I contains two full cylinder sets If an arrangement contains three cylinders, and two of them are full, the arrangement is gapless according to Theorem 6. In this section we will proof that arrangements of four cylinders generally do not contain a gap either, save for special values of N.The core of this proof rests on values of α satisfying one of the equations 3 3 T (α) = T (α) (with root α ) and T (α + 1) = T (α + 1) (with root α ). α  α u α α We will show that for N such that α <α very large gaps exist for α ∈[α ,α ]. u  u 123 Orbits of N-expansions with a finite set of digits 109 Fig. 14 The ‘limit graph’ of T , (a, 1) translated over (−α, −α), under the conditions I = Δ ∪ Δ Δ and α d d−1 d−2 N /(α + 1) − (d − 2) = p for N →∞ (and α, d →∞ y = − x +3 accordingly) y = − x +1 (1,a) y = − x +2 The central theorem of this section is the following: Theorem 7 Let N ∈ N and I = Δ ∪ Δ ∪ Δ ∪ Δ . Then there is a gap ≥2 α d d−1 d−2 d−3 in I if and only if N = 2k + 2k − i, with k > 1 and i ∈{1, 2, 3}. Moreover, if there is a gap in I , the gap contains f and f , while Δ and Δ are gapless. α d−1 d−2 d d−3 Proof of Theorem 7 Suppose that there is a gap containing f and f in I and that d−1 d−2 α Δ and Δ are gapless. Then, as a sub-interval of a gap, the interval ( f , f ) is d d−3 d−1 d−2 a gap. Since f < f , N /( f + d − 1) = f and N /( f + d − 2) = f , d−1 d−2 d−1 d−1 d−2 d−2 we know that N N ( f , f )  , , d−1 d−2 f + d − 1 f + d − 2 d−2 d−1 where the larger open interval is a gap as well. What is more, the infinite sequence of intervals N N ( f , f )  , d−1 d−2 f + d − 1 f + d − 2 d−2 d−1 N N ,  ··· N N + d − 1 + d − 2 f +d−2 f +d−1 d−1 d−2 consists of the union of ( f , f ) with pre-images of ( f , f ) in Δ and d−1 d−2 d−1 d−2 d−1 Δ respectively and therefore of gaps containing f and f . It is contained in d−2 d−1 d−2 the closed interval [q, r ], with q =[d − 1, d − 2] ∈ Δ and r =[d − 2, d − 1] ∈ Δ , N ,α d−1 N ,α d−2 yielding 2 2 T (q) = q, T (q) = r , T (r ) = q and T (r ) = r . (25) α α α α 123 110 J. de Jonge et al. Since Δ and Δ are gapless, T (α) and T (α + 1) lie outside the interval (q, r ), d d−3 α α which is to say p < T (α) ≤ q and r ≤ T (α + 1)< p . d α α d−2 2 2 For the images of α under T this means that either T (α) ∈ Δ or T (α) ∈ α d−3 α α Δ , in the latter case of which we have, due to the expansiveness of T and the d−2 α equalities of (25), 2 3 |T (α) − q|≤|T (α) − r|≤|T (α) − q|, α α with equalities only in the case T (α) = q. From this we derive that 2 2 3 either T (α) ∈ Δ or T (α) ∈ Δ ∧ T (α) ≤ T (α) (26) d−3 d−2 α α α α and, similarly, that 2 2 3 either T (α + 1) ∈ Δ or T (α + 1) ∈ Δ ∧ T (α + 1) ≥ T (α + 1). (27) d d−1 α α α α In the following we will write α (N , m) (u for ‘upper’) for the positive root of the equation T (α + 1) = T (α + 1) (so T (α + 1) = r) and α (N , m) (l for ‘lower’) for α α the positive root of the equation T (α) = T (α) (so T (α) = q), with m the number α α of full cylinder sets; in the current case we have m = 2. Recall that I consists of m full cylinder sets if and only if α = k, N = mk(k + 1) and d = (m − 1)(k + 1) for some k ∈ N, cf. Theorem 1,soif m = 2, we have arrangements consisting of two full arrangements only for α = k, N = 2k(k + 1) and d = k + 1. If N is 2k(k + 1) − n, with n ∈ N, and α = k − x, with x ∈ R,wehave 2 2 2k + 2k − n 4xk − n + 2x − 2x 4xk − n b(N,α) = = 2 + > 2 + , 2 2 (k − x )(k + 1 − x ) (k − 1)(k + 1 − x ) k + x which is a little bit larger than 2 provided x and n are relatively small. For these arrangements we have 2 2 2k + 2k − n 2x + 2x − n d(α) = − (k − x ) = k + 2 + 3x + = k + 2 k − x k − x and 2 2 2k + 2k − n 2x + 2x − n d(α + 1) = − (k + 1 − x ) = k − 1 + 3x + k + 1 − x k + 1 − x = k − 1. So, for x and n relatively small, the arrangements consist of four cylinders, while the branch number is only a little bit larger than 2. We will now use this to finish the forward implication of Theorem 7. 123 Orbits of N-expansions with a finite set of digits 111 Fig. 15 Arrangement with N = 11,α = 1.8719 4 3 2 1 N =11,α =1.8719 ··· (k =2, i =1) T (α +1) = T (α +1) Since Δ decreases and Δ increases as α decreases, we see that the assumption d−3 d that there is a gap containing f and f in I implies α (N , 2) ≥ α (N , 2). d−1 d−2 α u We will shortly show that the only values of N for which α (N , 2) ≥ α (N , 2) are N = 2k + 2k − i, with k > 1 and i ∈{1, 2, 3}; in all cases d = k + 2. Although we could keep i as a variable in our calculations, we can limit ourselves to the case i = 3, since i = 3 is the least favourable value of i allowing for a gap, as is suggested in Figs. 12, 15, 16, 17 and 18. We will show that for i = 3 indeed α (N , 2) ≥ α (N , 2). Subsequently we will show that for 4 ≤ i ≤ 4k no gaps exist; the upper bound is 4k, 2 2 since 2k + 2k − 4k = 2(k − 1) + 2(k − 1), so as to confine the calculations to the group of arrangements where d = k + 2. So, let N = 2k + 2k − 3 and d = k + 2. Then α (N , 2) =[k + 2, k + 1, k] and α (N , 2) +1 =[k −1, k, k + 1]. Omitting straightforward calculations, we find that 2 4 2 (2k + 2k − 3) D − (2k + 3k + 3k − 6) α (2k + 2k − 3, 2) = 3 2 4k + 12k − 6k − 6 and 2 2 (2k + 2k − 3)( D − (k + 5k + 4)) α (2k + 2k − 3, 2) = , 4k − 18k − 8 Note that in Fig. 12 we have d = k + 1, while there is no small cylinder set Δ . k+2 We omit the suffix ‘N,α’ behind these expansions not only for eligibility but also because α has yet to 3 3 be determined as the root of T (α) = T (α) or T (α + 1) = T (α + 1). α α α α 123 112 J. de Jonge et al. Fig. 16 Arrangement with N = 11,α = 1.8687 4 3 2 1 N =11,α =1.8687 ··· (k =2,i =1) T (α)= T (α) with 4 3 2 2 2 D = 9k + 18k − 3k − 12k = (3k + 3k − 2) − 4. Since we assume that there is a gap containing f and f ,wehave α ≥ α . d−1 d−2 u Omitting the basic calculations, we find that this inequality holds for k ∈ N .Inthe ≥2 case k = 3 (and so N = 21), we have indeed 508032 − 192 α (2k + 2k − 3, 2) = = 2.7123 ··· > 2.7122 ··· 508032 − 588 = = α (2k + 2k − 3, 2); see Fig. 19. Some more basic calculations show that the cases N = 2k + 2k − 1 and N = 2k + 2k − 2 allow for larger intervals [α ,α ] where large gaps exist; see the next examples. √ √ 9075 − 26 99 − 9075 α (11, 2) = = 1.8719 ··· and α (11, 2) = = 1.8686 ··· 37 2 √ √ 1725 − 12 45 − 1725 α (10, 2) = = 1.7372 ··· and α (10, 2) = = 1.7334 ··· 17 2 √ √ 5103 − 22 27 − 567 α (9, 2) = = 1.5946 ··· and α (9, 2) = = 1.5941 ··· 31 2 123 Orbits of N-expansions with a finite set of digits 113 Fig. 17 Arrangement with N = 10,α = 1.7372 4 3 2 1 N =10,α =1.7372 ··· (k =2, i =2) T (α +1) = T (α +1) Fig. 18 Arrangement with N = 9,α = 1.5946 4 3 2 1 N =9,α =1.5946 ··· (k =2, i =3) T (α +1) = T (α +1) 228 − 5 α (8, 2) = = 1.4428 ··· and α (8, 2) = 9 − 57 = 1.4501 ··· We see that the intervals α − α decrease as N decreases, until (for N = 8) the ‘interval’ would have negative length, hence does not exist. 123 114 J. de Jonge et al. Fig. 19 Arrangement with N = 21,α = 2.7123 5 4 3 2 N =21,α =2.7123 α =2.7122 and α =2.7123 Now suppose N = 2k + 2k − 4 (note that voor k = 2we have N = 8). Then 3 2 (k + 2) D − (k + k + 2k + 4) α (2k + 2k − 4, 2) = 2(k + 4k + 2) and 3 2 (k − 1) D − (k + 4k − k − 4) α (2k + 2k − 4, 2) = , 2(k − 2k − 1) with 4 3 2 D = 9k + 18k − 7k − 16k. 2 2 There are no gaps provided α (2k + 2k − 4, 2) − α (2k + 2k − 4, 2)> 0. Once more we omit the calculations, finding that this inequality holds for k ∈ N ,sowe ≥2 conclude that there are no gaps in case N = 2k + 2k − 4. If we replace the number 4 in N = 2k + 2k − 4 by larger integers (if possible), there will not be any gaps either: the length of the ‘interval’ [α ,α ] would only become more negative. This concludes the proof that if I = Δ ∪ Δ ∪ Δ ∪ Δ and there is a gap containing f α d d−1 d−2 d−3 d−1 and f in I , then N = 2k + 2k − i, with k > 1 and i ∈{1, 2, 3}. d−2 α For the converse statement, we assume that N = 2k + 2k − i, with k ∈ N and i ∈{1, 2, 3}.Ifalso d = k + 2, then earlier in this proof we showed that only then α (N , 2) ≤ α (N , 2). We will show that for α such that α (N , 2) ≤ α ≤ α (N , 2) u  u there is a gap in I = Δ ∪ Δ ∪ Δ ∪ Δ containing both f and f . α d d−1 d−2 d−3 d−1 d−2 123 Orbits of N-expansions with a finite set of digits 115 Fig. 20 Arrangement with N = 20,α = 2.6124 5 4 3 2 N =20,α =2.6124 α =2.6132 ··· and α =2.6115 ··· As earlier in this proof, we set q =[d − 1, d − 2] and r =[d − 2, d − 1] . N ,α N ,α Set G = (q, r ), then clearly both f ∈ G and f ∈ G. Furthermore, by definition d−1 d−2 of α (N , 2) and α (N , 2) we have that for every α ∈[α (N , 2), α (N , 2)] that u  u T (α) ≤ q (and therefore T (α) ≥ r ) and that T (α + 1) ≥ r (and therefore T (α + 1) ≤ q). Note that T ((p , q)) = (r,α + 1) and that T ((r , p )) = (α, q). Then we have α d α d−2 c c c that T (G ) = G , where G is the complement of G in I . We are left to show that α α G = (q, r ) is a gap; i.e. that for almost all x ∈ G there exists an n = n(x ) such that n c T (x ) ∈ G . To show this, consider the map T : I → I , defined by α α −x (α+1) p −α ⎪ d ⎪ + , if x ∈ Δ ; p −α p −α d d − (d − 1), if x ∈ Δ ; d−1 T (x ) = (28) ⎪ − (d − 2), if x ∈ Δ ; d−2 ⎪ x (α+1) −α p ⎩ −x d−2 + , if x ∈ Δ . d−3 α+1−p α+1−p d−2 d−2 123 116 J. de Jonge et al. Table 1 The thin thread between α (N , 2)α (N , 2) having a gap or not N =91.594119 ··· 1.594686 ··· N = 21 2.712252 ··· 2.712310 ··· N = 37 3.776839 ··· 3.776851 ··· N = 57 4.817672 ··· 4.817675 ··· N =81.450165 ··· 1.442809 ··· N = 20 2.613247 ··· 2.611575 ··· N = 36 3.700989 ··· 3.700407 ··· N = 56 4.756087 ··· 4.755832 ··· So on Δ and on Δ we have that T is a straight line segment with negative slope, d d−3 through (α, α + 1) and (p ,α) on Δ , resp. through (p ,α + 1) and (α + 1,α) d d d−2 on Δ .For x ∈ Δ ∪ Δ we have that T (x ) = T (x ). To show that G is d−3 d−1 d−2 α a gap, it is enough to show the ergodicity of T . Then the maximality of G follows from the fact that the support of the absolutely continuous invariant measure is G , c c since T (G ) = G . The proof of the existence of the absolutely continuous invariant measure for T and its ergodicity is similar to the proof of Theorem 2. Here all branches are complete and the proof is rather simpler. Once we have the ergodicity of T,itis n c obvious that for a.e. x ∈ G there exists n = n (x ) such that z = T (x ) ∈ G . Then 0 0 z never returns in G under iterations of T . This finishes the proof of Theorem 7. We stress that the in case of N = 2k + 2k − 3 the intervals [α ,α ] on which gaps exist may be very small; see Fig. 19. On the other hand, in case N = 2k + 2k − 4, the gaplessness may be a very close call; see Fig. 20. Table 1 illustrates how fast these differences between α and α decrease as N increases: Remark 15 While a fixed point f is repellent for points within Δ , the fixed points i i in two adjacent cylinder sets behave mutually contracting for all other points in these cylinder sets. As a consequence, it may take quite some time before the orbit of points in the full cylinders of gap arrangements with four cylinders leave these full cylinders for the first time. As an example we take the gap arrangement for k = 50 (according to the notations used above). Then N = 2 · 50 + 2 · 50 − 3 = 5097, d = 52 and α ≈ α ≈ α ≈ 49.98019737. Table 2 shows for ten values of x between α and α + 1 the smallest n such that T (x)/ ∈ Δ ∪ Δ . What is more, there are uncountably 51 50 many x in the gap (a, b) that contains f and f such that T (x ) ∈ (a, b) for all d−1 d−2 n ∈ N∪{0}. Indeed, for any sequence (d , d ,..., d ,...) such that d ∈{d −1, d −2}, 1 2 n n with n ∈ N, we have that x =[d , d , d ,...] ∈ (a, b). 1 2 3 N ,α For the final, second part of the proof of Theorem 4, we will consider one by one all cases left, that is N ∈{2,..., 11}.If N = 11 and α ≥ f , I consists of five 7 α cylinder sets if and only if α ∈ ( f − 1, f ); see the left arrangement of Fig. 21, which 2 6 we already saw in Fig. 13. Since α (11, 3)>α (11, 3) (cf. page 5), we conclude on similar grounds as in the proof of Theorem 7, that the arrangement is gapless. If α ∈[ f , f − 1], the interval I consists of four cylinder sets, implying gaplessness 7 2 α 123 Orbits of N-expansions with a finite set of digits 117 Table 2 The difficulty of leaving the gap: with N = 5097, α = 49.98019737, for each of ten values of x ∈[α, α + 1] the smallest n is given such that T (x)/ ∈ Δ ∪ Δ 51 50 x 50 50.1 50.2 50.3 50.4 50.5 50.6 50.7 50.8 50.9 n 5417 2090 3568 1123 4776 185 5816 16231 5646 7604 ··· 5 4 3 2 5 4 3 2 7 6 5 4 3 2 13 2 1 N =7,α = 14 − 1 N =11,α = f N =7,α = f =1 N =2,α =0.15 6 6 Fig. 21 Borderline cases for part II of the proof of Theorem 4 because of Theorem 7. Since |T ( f + 1)|= 2.04 ··· , gaps are also excluded for all α ≤ f . A similar approach works for N = 10 (with |T ( f + 1)|= 2.03 ··· ), 7 7 N = 9 (with |T ( f + 1)|= 2.02 ··· ) and even N = 8, in which case f = 1, 7 7 |T ( f + 1)|= 2, and the arrangement with four cylinders is full. For N ∈{3,..., 7} we take a different approach, confining ourselves to the case N = 7; the cases N ∈{3,..., 6} are done similarly. We will omit most calculations, which are generally quite tedious and do hardly elucidate anything. So let N = 7. Then I consists of at least five cylinder sets if and only if α< f ; see the second α 6 arrangement of Fig. 21.Wehave |T (α + 1)|= 2for α = 14 − 1, in which case d = 2; see the third arrangement of Fig. 21. Now suppose 14−1 ≤ α< f = 1. min 6 We have |T ( f + 1)|= and |T (α)| > 7. Regarding these relatively large values, α 4 α it is not hard to understand that Δ is gapless. The part of the orbit of α + 1 under T 2 α in the third arrangement of Fig. 21 illustrates that even in the case of α = 14, the expansion of [T (α + 1), p ] under T clearly excludes the existence of gaps. α 3 α Finally, let N = 2. We have |T ( f )|= 2, indicating the rapid increase of |T | on α α I if α decreases. The large expansiveness of T left of f assures the gaplessness of α α 1 Δ . We will show that for any α ∈ (0, 2 − 1] the image of [T (α + 1), p ]) contains 1 α 2 most of the fixed points, implying the gaplessness of I ; see the last arrangement of Fig. 21 for an illustration of this. If T (α + 1) ≤ f this is quite obvious, so we assume α 2 T (α + 1)> f . Suppose that T (α + 1) = f ,for some s ∈ N . Then, omitting α 2 s ≥2 some basic calculations, we have α = (s + 1 − s + 8)/(2s − 7), whence 4s − 11s + (4s − 13) s + 8 − 15 d = d(α) = 2s − 7 2 2 4s − 13s + (4s − 13) s + 8 − 8 ≥ , 2s − 7 123 118 J. de Jonge et al. from which we derive that d ≥ 4s. This finishes the proof of Theorem 4. Remark 16 A paper worthwhile to read alongside our paper is [12]. As the title of this paper states, it deals with mixing properties of expanding Lorentz maps. There are many interesting results in [12] that embed the results from our paper in a more general context. Acknowledgements The first author of this paper is very thankful to the Dutch organisation for scientific research Nederlandse Organisatie voor Wetenschappelijk Onderzoek, which funded his research for this paper with a so-called Promotiebeurs voor Leraren, with grant number 023.003.036. The third author’s research was partially supported by the Japan Society for the Promotion of Science with Grant-in-Aid for Scientific Research (C) 20K03661. We also want to thank Julian Lyczak, then working at Leiden University, for his help with some programming in the preparatory phase of Sect. 5. We thank Niels Langeveld as well for providing us with Fig. 3. Finally, the authors would also like to thank the anonymous referees for very helpful suggestions which improved the presentation of this paper. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. References 1. Anselm, M., Weintraub, S.H.: A generalization of continued fractions. J. Number Theory 131(12), 2442–2460 (2011) 2. 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Journal

Monatshefte für MathematikSpringer Journals

Published: May 1, 2022

Keywords: Continued fractions; Dynamical systems; Gaps; Primary 11J70; Secondary 37E05

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