Continued fractions with low complexity: Transcendence measures and quadratic approximation. Yann BUGEAUD

Transcription

1 Continued fractions with ow compexity: Transcendence measures and quadratic approximation Yann BUGEAUD Abstract. We estabish measures of non-quadraticity and transcendence measures for rea numbers whose sequence of partia quotients has subinear bock compexity. The main new ingredient is an improvement of Liouvie s inequaity giving a ower bound for the distance between two distinct quadratic rea numbers. Furthermore, we discuss the gap between Maher s exponent w 2 and Koksma s exponent w2. 1. Introduction A we-known open question in Diophantine approximation asks whether the continued fraction expansion of an irrationa agebraic number ξ either is utimatey periodic (this is the case if, and ony if, ξ is a quadratic irrationa), or it contains arbitrariy arge partia quotients. As a very sma step towards its resoution, we have recenty estabished in [20] two new combinatoria transcendence criteria for continued fraction expansions (we refer the reader to [6] for references to earier works). One of these criteria impies Theorem Bu beow, which states that the sequence of partia quotients a 1, a 2,... of an agebraic number [0; a 1, a 2,...] of degree at east three cannot be too simpe, in the foowing sense. The compexity function of an infinite sequence a = (a ) 1 of positive integers, which we wi often view as the infinite word a = a 1 a 2..., is the function n p(n, a) defined by p(n, a) = Card{(a, a +1,..., a +n 1 ) : 1}, for n 1, which counts the number of distinct bocks of ength n in the word a. Observe that the sequence (p(n, a)) n 1 is non-decreasing and that p(n, a) is infinite for every n 1 if the sequence (a ) 1 is unbounded. Furthermore, Morse and Hedund [30, 31] proved that (p(n, a)) n 1 is bounded if a is utimatey periodic and that, otherwise, it satisfies p(n, a) n + 1 for n 1. The atter inequaity is sharp since there exist uncountaby many words a for which p(n, a) = n + 1 for n Mathematics Subject Cassification : 11J70, 11J82, 11J87. Keywords: Continued fraction, transcendence, Schmidt s Subspace Theorem. 1

2 Theorem Bu. Let a = (a ) 1 be a sequence of positive integers which is not utimatey periodic and satisfies p(n, a) im inf < +. (1.1) n + n Then, the rea number is transcendenta. [0; a 1, a 2,..., a,...] One of the purposes of the present work is to study the accuracy with which rea numbers whose sequence of partia quotients satisfies a sighty stronger assumption than (1.1) are approximated by agebraic numbers of bounded degree. The quaity of approximation is measured by means of the functions w d introduced in 1932 by Maher [29]. For every integer d 1 and every rea number ξ, we denote by w d (ξ) the supremum of the rea numbers w for which 0 < P (ξ) < H(P ) w has infinitey many soutions in integer poynomias P (X) of degree at most d. Here, H(P ) stands for the naïve height of the poynomia P (X), that is, the maximum of the absoute vaues of its coefficients. Theorem 3.2 beow gives a necessary and sufficient condition on the infinite word a = (a ) 1 with im sup n + p(n, a) n < +, (1.2) which ensures that the rea number ξ := [0; a 1, a 2,...] satisfies w d (ξ) < + for every d 1. Its proof spits into two parts. To bound w d (ξ) for d 3, we use a genera method described in [5], based on a quantitative version of the Schmidt Subspace Theorem, and which has been aready appied successfuy in [6] to a certain cass of continued fractions. The main novety in the present paper is the method deveoped to contro w 2 (ξ), based on a refinement of Liouvie s inequaity giving a ower bound for the distance between two distinct quadratic rea numbers. Theorem 2.2 beow shows that, if (1.2) hods, then w 2 (ξ) is finite if, and ony if, the Diophantine exponent of a (a purey combinatoria quantity associated with a; see Section 2) is finite. Theorems 3.2 and 3.3 give transcendence measures for a cass of transcendenta numbers defined by their continued fraction expansion. The first resuts of this type were proved by A. Baker [13] in Since (1.2) is satisfied when a is an automatic sequence (see Sections 5 and 8), we get straightforwardy new resuts on agebraic approximation to rea numbers whose sequence of partia quotients can be generated by a finite automaton. We show that these numbers are either quadratic, or S- or T -numbers in Maher s cassification, which is recaed in Section 3. Shorty after Maher, Koksma [28] introduced in 1939 the exponents of approximation wd. For every integer d 1 and every rea number ξ, we denote by w d (ξ) the supremum of the rea numbers w for which 0 < ξ α < H(α) w 1 2

3 has infinitey many soutions in agebraic numbers α of degree at most d. Here, H(α) stands for the naïve height of the minima defining poynomia of α over Z. The exponents w 1 and w1 coincide and, for every rea number ξ, we have w d(ξ) w d (ξ) w d(ξ) + d 1, for d 2. (1.3) For more resuts on w d and wd, the reader is directed to Chapter 3 of [16]. R. C. Baker [14] was the first to estabish the existence of rea numbers ξ for which w d (ξ) differs from wd (ξ) for every integer d 2. To this end, he used a variant of Schmidt s construction of T -numbers [37]. By means of a sight modification of Baker s proof, Bugeaud [17] estabished that w 2 w2 takes any vaue in [0, 1). In view of (1.3), this is neary best possibe. Another purpose of the present work is to provide a new, fairy simpe and constructive proof of the atter resut. For any δ in (0, 1], we give expicit exampes of rea numbers ξ defined by their continued fraction expansion satisfying w 2 (ξ) = w2(ξ) + δ. The present paper is organized as foows. Our new resuts are stated in Sections 2 to 5. Measures of non-quadraticity and transcendence measures for continued fractions with ow compexity are given in Sections 2 and 3, respectivey. Section 4 is devoted to the study of the gap between the functions w 2 and w2. The resuts of Sections 2 and 3 are appied in Section 5 to automatic continued fractions and to a cass of morphic continued fractions, these two notions being defined in Section 8. Various resuts on continued fractions are gathered in Section 6. Section 7 is devoted to our improvement on Liouvie s inequaity and to two of its appications to bound, under various assumptions, the vaues of the functions w 2 and w2. A combinatoria auxiiary emma is the object of Section 9. Our main resuts are proved in Sections 10 and 11. Section 12 is devoted to an extension of Theorems 2.2 and 3.2 to a famiy of continued fractions with unbounded partia quotients. If nothing ese is specified, we use the notation A B (resp. A a B) when A is ess than some absoute constant (resp. some constant depending at most on a) times B. We write A B when A B and B A hod simutaneousy. 2. Quadratic approximation to continued fractions with ow compexity The foowing notation wi be used throughout this text. Let A be a finite or infinite set. The ength of a word W on the aphabet A, that is, the number of etters composing W, is denoted by W. For any positive integer k, we write W k for the word W... W (k times repeated concatenation of the word W ). More generay, for any positive rea number x, we denote by W x the word W x W, where W is the prefix of W of ength (x x ) W. Here, and in a what foows, y and y denote, respectivey, the integer part and the upper integer part of the rea number y. Let a = (a ) 1 be a sequence of eements from A that we identify with the infinite word a 1 a 2... a... Let ρ 1 be a rea number. We say that a satisfies Condition ( ) ρ if there exist two sequences of finite words (U n ) n 1, (V n ) n 1, and a sequence (w n ) n 1 of rea numbers such that: (i) For n 1, the word U n V w n n is a prefix of a; (ii) For n 1, we have U n V w n n / U n V n ρ; 3

4 (iii) The sequence ( V w n n ) n 1 is increasing. The Diophantine exponent of a, introduced in [3] and denoted by Dio(a), is the supremum of the rea numbers ρ for which a satisfies Condition ( ) ρ. It is cear from the definition that 1 Dio(a) + and that the Diophantine exponent of an utimatey periodic sequence is infinite. The converse is not true: it is easy to construct sequences whose Diophantine exponent is infinite but which are not utimatey periodic. The Diophantine exponent of a can be viewed as a measure of periodicity of a. We stress that it is independent of the aphabet on which a is written. We define the Diophantine exponent of an irrationa rea number to be the Diophantine exponent of its sequence of partia quotients. Definition 2.1. Let ξ := [0; a 1, a 2,..., a,...] be an irrationa rea number. The Diophantine exponent of ξ, denoted by Dio(ξ), is the Diophantine exponent of the infinite word a 1 a 2... By truncating the continued fraction expansion of an irrationa rea number ξ and competing then by periodicity, one can construct good quadratic approximations to ξ which aow us to bound w 2(ξ) from beow. An easy cacuation (see Section 11) shows that w 2(ξ) Dio(ξ) 1, (2.1) if (q 1/ ) 1 converges, where (p /q ) 1 denotes the sequence of convergents to ξ. This simpe argument does not yied any upper bound for w2(ξ). However, Theorem 2.2 asserts that, when the continued fraction expansion of ξ has subinear compexity, then it is possibe to bound Dio(ξ) from beow in terms of w2(ξ). Theorem 2.2. Let κ 2 and A 3 be integers. Let a = (a ) 1 be a sequence of positive integers bounded by A for which there exists an integer n 0 such that p(n, a) κn, for n n 0. (2.2) If the Diophantine exponent of a is finite, then the rea number satisfies ξ := [0; a 1, a 2,..., a,...] max{2, Dio(ξ) 1} w 2(ξ) w 2 (ξ) κ 3 Dio(ξ) (og(a + 1)) 4. (2.3) Let (p /q ) 1 denote the sequence of convergents to ξ. If (q 1/ ) 1 converges, then we have w 2 (ξ) κ 3 Dio(ξ). (2.4) 4

5 Let b 2 be an integer and a = (a ) 1 an infinite word on {0, 1,..., b 1} satisfying (1.2) and which is not utimatey periodic. Théorème 2.1 from [7] asserts that the rea number ζ := a b 1 is a Liouvie number (that is, it satisfies w 1 (ζ) = + ) if, and ony if, the Diophantine exponent of a is infinite. Theorem 2.2 above provides the anaogue of this resut for continued fraction expansions. The fact that w2(ξ) aways exceeds 2 when ξ is irrationa and not quadratic was proved by Davenport and Schmidt [25]. It is expained at the end of the proofs of Lemma 9.1 and of Theorem 2.2 that one can repace og(a + 1) in (2.3) by the quantity ( im sup + ) ( og q / im inf + ) og q. Consequenty, one gets the upper bound (2.4) when the sequence (q 1/ ) 1 converges. We stress that this bound does not depend on the aphabet on which a is written, a fact which was not pointed out previousy. At the end of [9], the authors noted that the cassica argument based on triange inequaities and Liouvie s inequaity α β 0.03 H(α) 2 H(β) 2, (2.5) vaid for distinct quadratic numbers α and β, is not powerfu enough to yied Theorem 2.2. Fortunatey, in the present situation, we are abe to consideraby improve (2.5), since one of the quadratic numbers invoved is very cose to its Gaois conjugate; see Lemma 7.1. The assumption (2.2) can be sighty reaxed and Theorem 2.2 can be extended to a cass of continued fractions with unbounded partia quotients and having repetitive patterns, provided that, however, the sequence (q 1/ ) 1 remains bounded; see Section 12. By combining (1.3) and (2.1), every irrationa number ξ satisfies w 2 (ξ) Dio(ξ) 1. We concude this section with a sharpening of this inequaity. Theorem 2.3. Let ξ be an irrationa number with bounded partia quotients. (p /q ) 1 denote the sequence of convergents to ξ. Set Then we have m = im inf + q1/ and M = im sup q 1/. + w 2 (ξ) og m (Dio(ξ) + 1) 1. og M In particuar, when the sequence (q 1/ ) 1 converges, then Let w 2 (ξ) Dio(ξ). (2.6) 5

6 The proof of Theorem 2.3 depends on Lemma 6.1. We show in Section 4 that inequaities (2.1) and (2.6) are sharp. 3. Transcendence measures for continued fractions with ow compexity Maher s cassification of rea numbers is based on the functions w d defined in the Introduction. For a rea number ξ, we set w(ξ) = im sup d (w d (ξ)/d) and, according to Maher [29], we say that ξ is an A-number, if w(ξ) = 0; S-number, if 0 < w(ξ) < ; T -number, if w(ξ) = and w d (ξ) < for any integer d 1; U-number, if w(ξ) = and w d (ξ) = for some integer d 1. Two transcendenta rea numbers beonging to different casses are agebraicay independent. The A-numbers are precisey the agebraic numbers and, in the sense of the Lebesgue measure, amost a numbers are S-numbers. The existence of T -numbers remained an open probem during neary forty years, unti it was confirmed by Schmidt, see Chapter 3 of [16] for references and further resuts. The set of U-numbers can be further divided in countaby many subcasses according to the vaue of the smaest integer d for which w d (ξ) is infinite. Definition 3.1. Let m 1 be an integer. A rea number ξ is a U m -number if, and ony if, w m (ξ) is infinite and w d (ξ) is finite for d = 1,..., m 1. We estabish the foowing resut, which can be viewed as the anaogue for continued fraction expansions of Théorème 1.1 of [7]. Theorem 3.2. Let a = (a ) 1 be a sequence of positive integers such that im sup n + p(n, a) n < +, (3.1) and set ξ := [0; a 1,..., a,...]. If Dio(ξ) is finite, then ξ is either an S-number or a T -number; otherwise, ξ is either quadratic or a U 2 -number. Moreover, if Dio(ξ) is finite, then there exists a constant c, depending ony on ξ, such that w d (ξ) exp(c (og 3d) 5 (og og 3d) 4 ), for d 1. The proof of Theorem 3.2 spits into two parts. Since the approximation by quadratic numbers has been det with in Theorem 2.2, it ony remains for us to contro the quaity of approximation by agebraic numbers of fixed degree at east equa to three. To do this, we use the Quantitative Subspace Theorem, foowing the genera method introduced in 6

7 [5] and aready appied to a restricted casses of continued fractions in [6]. There are, however, some additiona technica difficuties. Section 10 is devoted to this part of the proof of Theorem 3.2. The dependence on ξ of the constant c can be made more precise and expressed ony in terms of Dio(ξ) and im sup n + p(n, a)/n, provided that the sequence (q 1/ ) 1 converges, where q is the denominator of the -th convergent to ξ. A. Baker [13] was the first to estabish transcendence measures for a cass of continued fractions. He proved the foowing resut. Theorem Ba. Consider the continued fraction ξ = [0; a 1,... a n0 1, a n0,..., a n0 +r 0 1, a n1,..., a n1 +r }{{} 1 1,...], }{{} λ 0 times λ 1 times where the notation impies that n k = n k 1 + λ k 1 r k 1 and the λ k s indicate the number of times a bock of partia quotients is repeated (it is understood that two bocks which correspond to consecutive k are not identica). Suppose that the sequences (a n ) n 1 and (r n ) n 0 are respectivey bounded by A and K. Set L = im sup λ k /λ k 1, k + = im inf k + λ k/λ k 1. If L is infinite and > 1, then ξ is a U 2 -number. Furthermore, if L is finite and > exp(4a K ), then ξ is either an S-number or a T -number. Theorem Ba was extended in [6], where it is proved that the assumption > exp(4a K ) can be repaced by the weaker condition > 1. The techniques of the present work aow us to make a further generaization, that is, to remove the assumption that the sequence (r n ) n 0 has to be bounded. Theorem 3.3. Consider the continued fraction ξ = [0; a 1,... a n0 1, a n0,..., a n0 +r 0 1, a n1,..., a n1 +r }{{} 1 1,...], }{{} λ 0 times λ 1 times where the notation impies that n k = n k 1 + λ k 1 r k 1 and the λ k s indicate the number of times a bock of partia quotients is repeated (it is understood that two bocks which correspond to consecutive k are not identica). Suppose that the sequence (a n ) n 1 is bounded and that λ k tends to infinity with k. Set L = im sup k + λ k r k λ 1 r λ k 1 r k 1, = im inf k + λ k r k λ 1 r λ k 1 r k 1. If L is infinite, then ξ is a U 2 -number. If L is finite and is positive, then ξ is either an S-number or a T -number. Theorem 3.3 extends Theorem 3.2 of [6]. It is not a particuar case of Theorem 3.2 above since the assumption (3.1) may not be satisfied. The important point is that, however, Lemma 9.1 can be appied, since the foowing condition: Baker s paper quoted on page 884 of [6] is not the paper containing Theorem Ba. 7

8 There is a positive integer κ such that, for every sufficienty arge integer n, there is a word of ength n having two occurrences in the prefix of ength (κ + 1)n of the infinite word composed of the partia quotients of ξ is satisfied. The assumption (3.1) can be sighty reaxed and Theorem 3.2 can be extended to a cass of continued fractions with unbounded partia quotients and having repetitive patterns, provided that, however, the sequence (q 1/ ) 1 remains bounded; see Section On the gap between the exponents w 2 and w 2 The gap between the exponents w 2 and w 2 defined in the Introduction was investigated in [14, 15, 17]. It foows from (1.3) that the set of vaues taken by the function w 2 w 2 is contained in the cosed interva [0, 1], and we proved in [17] that this set incudes the haf open interva [0, 1). The constructions in [14, 15, 17] are variants of Schmidt s compicated construction of T -numbers [37] and do not give expicit rea numbers ξ for which w 2 (ξ) differs from w 2(ξ). Lemma 6.1 beow, on the continued fraction expansions of conjugate quadratic numbers, enabes us to construct quite easiy expicit exampes of rea numbers ξ with prescribed vaues for w 2 (ξ), w 2(ξ) and their difference w 2 (ξ) w 2(ξ). Theorem 4.1. Let w 3 be a rea number. Let b, c be distinct positive integers. Define the sequence (a n,w ) n 1 by setting a n,w = c if there exists an integer j such that n = w j and a n,w = b otherwise. Set ξ w := [0; a 1,w, a 2,w, a 3,w,...]. Then, Dio(ξ w ) = w and ξ w is either an S- or a T -number. Furthermore, if w (5+ 17)/2, then we have w 2(ξ w ) = w 1 and w 2 (ξ w ) = w. (4.1) It is very ikey that (4.1) remains true for 3 w < (5 + 17)/2, but this seems to be difficut to prove. Theorem 4.1 shows that inequaities (2.1) and (2.6) are both sharp. We can modify our construction to give, for every δ in (0, 1), expicit rea numbers ξ satisfying w 2 (ξ) w 2(ξ) = δ. Theorem 4.2. Let w 3 be a rea number. Let b, c, d be distinct positive integers. Let η be a positive rea number with η < w/4. For j 1, set m j = ( w j+1 w j 1 )/ ηw j. Define the sequence (a n,w,η ) n 1 by setting a n,w,η = c if there exists an integer j such that n = w j, by setting a n,w,η = d if there exist positive integers j and m = 1,..., m j such that n = w j + m ηw j, and by setting a n,w,η = b otherwise. Set ξ w,η := [0; a 1,w,η, a 2,w,η, a 3,w,η,...]. 8

9 Then, Dio(ξ w,η ) = w/(1 + η) and ξ w,η is either an S- or a T -number. Furthermore, if w 16, then we have w 2(ξ w,η ) = 2w 2 η 2 + η and w 2 (ξ w,η ) = 2w η 2 + η, hence w 2 (ξ w,η ) w 2(ξ w,η ) = η. We state an immediate consequence of the combination of (1.3) with Theorems 4.1 and 4.2 (or with [17]). Theorem 4.3. The set of vaues taken by the function w 2 w 2 is equa to the cosed interva [0, 1]. The rea numbers defined in Theorems 4.1 and 4.2 are the first expicit exampes of rea numbers ξ for which w 2(ξ) and w 2 (ξ) differ. They are aso the first exampes of bady approximabe numbers with this property. 5. Agebraic approximation to automatic continued fractions An infinite sequence a = (a ) 1 is an automatic sequence if it can be generated by a finite automaton, that is, if there exists an integer k 2 such that a is a finite-state function of the representation of in base k, for every 1. Let b 2 be an integer. In 1968, Cobham [23] asked whether a rea number whose b-ary expansion can be generated by a finite automaton (in the seque, such a rea number is caed a b-ary automatic number) is aways either rationa or transcendenta. A positive answer to Cobham s question was recenty given in [4], by means of a combinatoria transcendence criterion estabished in [8]. We addressed in [1] the anaogous question for continued fraction expansions and gave a positive answer to it in [20]. Namey, we proved that a rea number whose continued fraction expansion can be generated by a finite automaton (in the seque, such a rea number is caed an automatic number) is aways either quadratic or transcendenta. More Diophantine properties of b-ary irrationa automatic rea numbers are known. Adamczewski and Cassaigne [9] estabished that the Diophantine exponent of any nonutimatey periodic automatic sequence is finite (Lemma 8.3 beow) and deduced that no b-ary automatic number is a Liouvie number. By Theorem 2.2, we obtain ikewise a measure of non-quadraticity for automatic rea numbers which are not quadratic. Precise definitions of automatic and morphic sequences and of various quantities and notions associated with them are postponed to Section 8. Theorem 5.1. Let k 2 be an integer and a = (a ) 1 an infinite sequence of positive integers generated by a k-automaton. Let A 3 be an upper bound for the sequence a = (a ) 1. Let m be the cardinaity of the k-kerne of the sequence a and et I be the interna aphabet associated with a. Then, the rea number ξ := [0; a 1, a 2,..., a,...] 9

10 satisfies w 2 (ξ) (CardI) 6 k m+3 (og(a + 1)) 4. (5.1) Theorem 5.1 extends Theorem 7.3 of [9], vaid ony for a restricted cass of automatic numbers. Theorem 5.1 is an immediate consequence of Theorem 2.2, inequaity (8.1) and Lemma 8.3. We can sighty improve Theorem 5.1 by arguing directy (as was done in [9] for showing that b-ary automatic numbers are not Liouvie numbers) rather than appying the genera Lemma 9.1. The bound (5.1) depends on three parameters CardI, k and m which appear naturay in the study of automatic sequences. It aso depends, and this is rather ennoying, on the aphabet on which the sequence is written. However, one can remove the atter dependence when the automatic sequence is generated by a primitive morphism. Theorem 5.2. Let k 2 be an integer and et a = (a ) 1 be a non-utimatey periodic infinite sequence of positive integers generated by a primitive morphism σ on an aphabet of cardinaity b 2. Let v denote the width of σ. Then the rea number satisfies ξ := [0; a 1, a 2,..., a,...] w 2 (ξ) v 12b 6 b 12 Dio(ξ). We stress that the upper bound in Theorem 5.2 does not depend on the aphabet on which the morphic sequence is written. Theorem 5.2 is an immediate consequence of Theorem 2.2 combined with Lemmas 8.4 to 8.6. The Thue Morse sequence abbabaabbaababbabaab... defined on the aphabet {a, b} is a cassica exampe of an automatic sequence satisfying the assumption of Theorem 5.2. Further exampes are given in [12]. Adamczewski and Bugeaud [7] showed that irrationa b-ary automatic numbers are either S- or T -numbers in Maher s cassification of numbers. Since the compexity function of every automatic sequence a grows at most ineary with n (see inequaity (8.1) beow), Theorem 3.2 impies the anaogous resut for automatic numbers which are not quadratic. Theorem 5.3. Let a = (a ) 1 be an automatic sequence (or a sequence generated by a primitive morphism) of positive integers which is not utimatey periodic and set ξ := [0; a 1, a 2,..., a,...]. Then, ξ is a transcendenta number and there exists a constant c, depending ony on ξ, such that w d (ξ) exp(c (og 3d) 5 (og og 3d) 4 ), for d 1. In particuar, ξ is either an S-number or a T -number. The part of Theorem 5.3 deaing with sequences generated by a primitive morphism foows from Theorem 3.2 combined with Lemmas 8.4 and 8.6. There have been recenty severa new resuts on the rationa approximation to rea numbers whose expansion in some integer base is an automatic sequence [10, 18, 21]. Motivated by these works, we address and briefy discuss the foowing probem. 10

11 Probem 5.4. To determine the set of vaues taken by the exponents w 2 and w 2 at automatic continued fractions. With a suitabe modification of the construction given in [18], one can construct expicity, for every sufficienty arge rationa number p/q, an automatic continued fraction ξ satisfying w2(ξ) = p/q. It seems to be a difficut and chaenging probem to show that the sets of vaues taken by the functions w2 and w 2 at automatic continued fractions incude every rationa number greater than or equa to 2. Bugeaud and Laurent [22] have computed the vaues of w 2 (ξ) and w2(ξ) for certain morphic continued fractions ξ and found that these vaues are quadratic numbers. It was proved in [19] that the Thue Morse Maher number ξ t, whose sequence of binary digits is the Thue Morse sequence on {0, 1} starting with 0, satisfies w 1 (ξ t ) = 1. We address the anaoguous question for continued fraction expansions. Probem 5.5. Let a and b be distinct positive integers. Let ξ t,a,b be the rea number whose sequence of partia quotients is the Thue Morse sequence on {a, b} starting with a. To compute w 2(ξ t,a,b ) and w 2 (ξ t,a,b ). It is known [35] that w2(ξ t,a,b ) 7/3. In view of Schmidt s theorem [36], this is sufficient to concude that ξ t,a,b is transcendenta, a resut first proved by M. Quefféec [34] (see aso [11]). Since this ower bound is sma, one cannot prove that this is the exact vaue by arguing as in Lemma 7.3. We suspect, however, that 7/3 is not the correct vaue and pan to return to this question in a further work. 6. Continued fractions We gather in this section various resuts on continued fraction expansions, but we assume that the reader is aready quite famiiar with the subject (otherwise, he is directed e.g. to [33] or to Chapter 1 of [16]). In this and the next sections, we use the notation [0; a 1,..., a r, a r+1,..., a r+s ] := [0; U, V ], where U = a 1... a r and V = a r+1... a r+s, to indicate that the bock of partia quotients a r+1,..., a r+s is repeated infinitey many times. We aso denote by ζ the Gaois conjugate of a quadratic rea number ζ. The key observation for our main resuts is given in the foowing emma. Lemma 6.1. Let ξ be a quadratic rea number with utimatey periodic continued fraction expansion ξ = [0; a 1,..., a r, a r+1,..., a r+s ], with r 3 and s 1, and denote by ξ its Gaois conjugate. Let (p /q ) 1 denote the sequence of convergents to ξ. If a r a r+s, then we have ξ ξ a 2 r max{a r 2, a r 1 } q 2 r. 11

12 Proof. By a theorem of Gaois (see [33], page 83), the Gaois conjugate of τ = [a r+1 ; a r+2,..., a r+s, a r+1 ] is the quadratic number τ = [0; a r+s,..., a r+2, a r+1 ]. Since we get ξ ξ = ξ = p rτ + p r 1 q r τ + q r 1 and ξ = p rτ + p r 1 q r τ + q r 1, Assume that a r a r+s. Using the mirror formua we see that τ τ (q r τ + q r 1 ) q r τ + q r 1 1 q r q r τ + q r 1. (6.1) q r 1 /q r = [0; a r, a r 1,..., a 1 ], q r τ + q r 1 = [0; a r, a r 1,..., a 1 ] [0; a r+s,..., a r+2, a r+1 ] q r. If max{a r, a r+s } 2 min{a r, a r+s }, then one gets q r τ + q r 1 q r / min{a r, a r+s }. Otherwise, if a r < a r+s, then an easy computation shows that q r τ + q r 1 1 a r + 1/(1 + 1/(a r 2 + 1)) 1 a r+s + 1/(a r+s 1 + 1) q r q r a 2 r q r a 2 r max{a 1 r 2, a 1 r+s 1 } 1 a r 2, whie, if a r > a r+s, then the simiar estimate q r τ + q r 1 q r a 2 r max{a 1 r 1, a 1 r+s 2 } q r a 2 r 1 a r 1 hods. By (6.1), this competes the proof of the emma. We dispay an eementary resut on utimatey periodic continued fraction expansions, whose proof can be found in [33]. 12

13 Lemma 6.2. Let ξ be a quadratic rea number with utimatey periodic continued fraction expansion ξ = [0; a 1,..., a r, a r+1,..., a r+s ], (6.2) and denote by (p /q ) 1 the sequence of its convergents. Then, ξ is a root of the poynomia (q r 1 q r+s q r q r+s 1 )X 2 (q r 1 p r+s q r p r+s 1 (6.3) + p r 1 q r+s p r q r+s 1 )X + (p r 1 p r+s p r p r+s 1 ). In particuar, if the continued fraction expansion of ξ is purey periodic, that is, if then ξ is a root of the poynomia ξ = [0; a 1,..., a s ], q s 1 X 2 (p s 1 q s )X p s. (6.4) The poynomias (6.3) and (6.4) may not be primitive, as it is in particuar the case when a r = a r+s. They provide us ony with an upper bound for the height of the rea number ξ defined by (6.2). For the proof of Theorems 4.1 and 4.2 we need a precise estimate of the height of quadratic numbers of a specia form. Lemma 6.3. Let b, c and d be distinct positive integers. Let n 3 be an integer and a 1,..., a n 2 be positive integers. Set Then, the height of ξ satisfies ξ := [0; a 1,..., a n 2, b, c, b]. q 2 n b,c H(ξ) b,c q 2 n, where q n is the denominator of the rationa number p n /q n := [0; a 1,..., a n 2, b, c]. Let m 3 be an integer and set ζ := [0; a 1,..., a n 2, b, c, b, b,..., b, d], where the periodic part b, b,..., b, d has ength m. Then, the height of ζ satisfies q n q n+m b,c,d H(ζ) b,c,d q n q n+m. where q n+m is the denominator of the rationa number [0; a 1,..., a n 2, b, c, b, b,..., b, d]. Proof. We deduce from Lemma 6.2 that H(ξ) b,c q 2 n and H(ζ) b,c,d q n q n+m. 13

14 Let P ξ (X) = AX 2 + BX + C = A(X ξ)(x ξ ) denote the minima defining poynomia of ξ over Z. Since P ξ (p n /q n ) is a non-zero rationa number of denominator dividing qn, 2 we have P ξ (p n /q n ) qn 2. It foows from Lemma 6.1 and the definition of p n /q n that ξ p n /q n < q 2 n, ξ p n /q n b,c q 2 n. Consequenty, we get A b,c q 2 n, thus, H(ξ) b,c q 2 n. This competes the proof of the first assertion of the emma. Since the resutant of the minima defining poynomias of ξ and ζ is a non-zero integer, we get ζ ξ H(ξ) 2 H(ζ) 2 ζ ξ 1 ζ ξ 1 ζ ξ 1 b,c,d H(ξ) 2 H(ζ) 2 q 6 n, by Lemma 6.1. Using ζ ξ b,c,d q 2 n+m and H(ξ) b,c q 2 n, this gives This finishes the proof of the emma. H(ζ) 2 b,c,d q 2 n q 2 n+m. For the proof of Theorem 9.1 we need cassica resuts on continuants which we reca beow (see [33] for a proof). If a 1,..., a m are positive integers, then the continuant K m (a 1,..., a m ) is the denominator of the rationa number [0; a 1,..., a m ]. Lemma 6.4. For any positive integers a 1,..., a m and any integer k with 1 k m 1, we have K m (a 1,..., a m ) = K m (a m,..., a 1 ), and K m (a 1,..., a m ) (1 + max{a 1,..., a m }) m, K m (a 1,..., a m ) max{(min{a 1,..., a m }) m, 2 (m 1)/2 }, K k (a 1,..., a k ) K m k (a k+1,..., a m ) K m (a 1,..., a m ) 2 K k (a 1,..., a k ) K m k (a k+1,..., a m ). 7. Liouvie s inequaity and appications Observe that if α is a rea quadratic number and α denotes its Gaois conjugate, then we have H(α) 1 α α 2H(α). (7.1) 14

15 To see this, it is sufficient to note that, if the minima defining poynomia of α over Z is ax 2 + bx + c, then α α b2 4ac =. a Note that α α can be as sma as 5H(α) 1. discriminant of the poynomia (m 2 + m 1)X 2 (2m + 1)X + 1 Indeed, for any integer m 2, the is equa to 5, thus, the distance between its two roots is equa to 5 divided by its height. Lemma 7.1. Let α and β be rea quadratic numbers. Denote by P α (X) := a(x α)(x α ) and P β (X) := b(x β)(x β ) their minima defining poynomias over Z. Assume that α, α, β, β are distinct. Then we have α β 0.03 max{ α α 1, 1} H(α) 2 H(β) 2. (7.2) Under the assumption of Lemma 7.1, the usua form of Liouvie s inequaity (see e.g. Theorem A.1 of [16]) impies that α β 0.03 H(α) 2 H(β) 2. (7.3) Lemma 7.1 shows that this estimate can be improved if α and its Gaois conjugate are cose to each other. Roughy speaking, in the most favourabe cases, the exponent 2 of H(α) in (7.3) can be repaced by 1. Proof. In view of (7.3), we assume that α α < 1. Since the resutant of P α (X) and P β (X) is a non-zero integer, we get If β 2, then we have ab 2 α β α β P α (β ) 1. (7.4) bβ bββ H(β), if β 1, whie bβ 2 b(β + β ) 2H(β), if β 1. Consequenty, regardess the vaue of β, this gives b 2 P α (β ) 3H(α) b 2 max{1, β } 2 12 H(α) H(β) 2, thus, by (7.4), α β α β 1 H(α) 2 H(β) 2 /12. (7.5) Without any oss of generaity, we may assume that α β α β. 15

16 If α β 2 α β, then (7.5) impies that α β 2 H(α) 2 H(β) 2 /24, which, by (7.1), yieds a much better ower bound than (7.2). If α β 2 α β, then, using the triange inequaity it foows from (7.5) that Combined with (7.3), this gives the emma. α α α β α β α β /2, α β α α 1 H(α) 2 H(β) 2 /24. We point out two important consequences of Lemma 7.1. A first one states that if a rea number ξ is quite cose to a dense (in a suitabe sense) sequence of quadratic numbers having a cose conjugate, then ξ cannot be too we approximated by quadratic numbers. Lemma 7.2. Let ξ be a rea number. Let C > 1 be a rea number and (Q j ) j 1 an increasing sequence of integers such that Q j+1 Q C j for j 1. Let w > 2 and 0 < ε < 1 be rea numbers. Assume that there exists a sequence (α j ) j 1 of quadratic numbers such that, denoting by α j the Gaois conjugate of α j, we have ξ α j < Q 2 ε j max{ α j α j 1, 1}, H(α j ) Q j, and for j 1. Then, w 2(ξ) is finite and ξ α j Q w j, w 2(ξ) w wCε 1. Proof. Set A = 1 + 2Cε 1. Let α be a quadratic rea number and et j be the integer defined by the inequaities Q j H(α) A < Q j+1. By Lemma 7.1, if α α j, then we have α α j 0.03 max{ α j α j 1, 1} Q 2 j H(α) 2. (7.6) For j sufficienty arge, the choice of A impies that H(α) Q 1/A j+1 QC/A j 0.1 Q ε/2 j, 16

17 hence, 0.01 max{ α j α j 1, 1} Q 2 j H(α) 2 max{ α j α j 1, 1} Q 2 ε j > ξ α j. (7.7) Using (7.6) and (7.7), the triange inequaity then gives ξ α α α j ξ α j 0.02 max{ α j α j 1, 1} Q 2 j H(α) Q 2 j H(α) H(α) 4 4C/ε, if j is sufficienty arge. It remains for us to consider the case where α = α j. Then, we have ξ α Q w j H(α) wa H(α) w 2wC/ε. Since w 2, this proves the emma. Let us briefy expain the novety in Lemma 7.2. If nothing is known on the Gaois conjugates of the quadratic approximants α j, then, in order to get an upper bound for w2(ξ), we have to assume that α j is very cose to ξ, namey, roughy speaking, that ξ α j < H(α j ) 2 ε for some positive rea number ε. This is the strong assumption made in [9] (see on page 1370) and in [6] (see on page 896). Fortunatey, we can consideraby weaken it and repace it even by ξ α j < H(α j ) 1 ε (in the most favourabe cases), provided that α j is very cose to its Gaois conjugate α j. This is crucia for the appications we have in mind, especiay for the proof of Theorem 2.2. Lemma 7.3. Let ξ be a rea number. Assume that there exist positive rea numbers c 1, c 2, c 3, δ, ρ, θ and a sequence (α j ) j 1 of quadratic numbers such that and c 1 H(α j ) ρ 1 ξ α j c 2 H(α j ) δ 1. (7.8) H(α j ) H(α j+1 ) c 3 H(α j ) θ, for j 1. Set ε = 0 or assume that there exist c 4 1 and 0 < ε 1 such that α j α j c 4 H(α j ) ε, (7.9) for j 1, where α j denotes the Gaois conjugate of α j. Then we have as soon as Furthermore, if ε > 0, then we have δ w 2(ξ) ρ (ρ 1)(δ 1 + ε) 2θ(2 ε). (7.10) δ w 2(ξ) ρ and w 2 (ξ) = w 2(ξ) + ε, 17

18 as soon as (δ 2 + ε)(δ 1 + ε) 2θ(2 ε) and im j + og α j α j og H(α j ) = ε. (7.11) Proof. Set Q j = H(α j ) for j 1. Let α be either a rationa number, or a quadratic rea number not eement of the sequence (α j ) j 1. Let j be the integer defined by the inequaities Q j 1 < (c 5 H(α)) 2/(ε+δ 1) Q j, where c 5 = (70c 2 c 4 ) 1/2. (7.12) We assume that H(α), hence j, is sufficienty arge. By Lemma 7.2 and (7.12), we have Consequenty, we get α α j 0.03 α j α j 1 Q 2 j 0.03 c 1 4 c2 5 Q 1 δ j 2c 2 Q 1 δ j. H(α) 2 ξ α α α j ξ α j α α j /2. Since Q j c 3 Q θ j 1 c 3 (c 5 H(α)) 2θ/(ε+δ 1), we concude that ξ α 0.01 c 1 4 Q 2+ε j H(α) c 1 4 c 2+ε 3 c 2θ(2 ε)/(ε+δ 1) 5 H(α) 2 2θ(2 ε)/(ε+δ 1), (7.13) thus w 2(ξ) max{ρ, 1 + 2θ(2 ε)/(ε + δ 1)}. Combined with (7.8), this impies that w 2(ξ) ρ when (7.10) is satisfied. Furthermore, the assumption (7.8) ensures that w 2(ξ) δ and w 2 (ξ) w 2(ξ) + ε, by (7.9). This proves the first assertion of the emma. If, moreover, (δ 2 + ε)(δ 1 + ε) 2θ(2 ε), then it foows from (7.13) that ξ α 0.01 c 1 4 c 2+ε 3 c 2θ(2 ε)/(ε+δ 1) 5 H(α) δ ε. (7.14) This means that the best agebraic approximants to ξ of degree at most 2 beong to the sequence (α j ) j 1. Assume that α is irrationa and denote by P α (X) its minima defining poynomia, by α its Gaois conjugate, and by a α 1 the eading coefficient of P α (X). 18

19 Note that a α α α 1. Using (7.14), the fact that δ 2 and the triange inequaity, we then get P α (ξ) a α ξ α ξ α a α ξ α α α / c 1 4 c 2+ε 3 c 2θ(2 ε)/(ε+δ 1) 5 H(α) δ ε. (7.15) For j 1, denoting by P j (X) the minima defining poynomia of α j, we deduce, again from the triange inequaity, that It thus foows from (7.8), (7.11) and (7.16) that P j (ξ) 2 H(α j ) ξ α j α j α j. (7.16) w 2 (ξ) w 2(ξ) + ε δ + ε. (7.17) We concude from (7.15) that the first inequaity in (7.17) is an equaity. This finishes the proof of the emma. 8. Automatic and morphic sequences We reca in this section basic definitions and severa resuts on automatic and morphic sequences. For more information, the reader is advised to consut the monograph [12]. As in [12], but unike in the rest of the present paper, we index the sequences from = 0. Let k 2 be an integer and denote by Σ k the set {0, 1,..., k 1}. A k-automaton is a 6-tupe A = (Q, Σ k, δ, q 0,, τ), where Q is a finite set of states, Σ k is the input aphabet, δ : Q Σ k Q is the transition function, q 0 is the initia state, is the output aphabet and τ : Q is the output function. For a state q in Q and for a finite word W = w 1 w 2... w n on the aphabet Σ k, we define recursivey δ(q, W ) by δ(q, W ) = δ(δ(q, w 1 w 2... w n 1 ), w n ). Let n 0 be an integer and et w r w r 1... w 1 w 0 in (Σ k ) r+1 be the representation of n in base k, meaning that n = w r k r +...+w 0. We denote by W n the word w 0 w 1... w r. Then, a sequence a = (a ) 0 is said to be k-automatic if there exists a k-automaton A such that a = τ(δ(q 0, W )) for a 0. For a finite set A, we denote by A the free monoid generated by A. The empty word ε is the neutra eement of A. Let A and B be two finite sets. An appication from A to B can be uniquey extended to a homomorphism between the free monoids A and B. We ca morphism from A to B such a homomorphism. If there is a positive integer k such that each eement of A is mapped to a word of ength k, then the morphism is caed k-uniform (or, simpy, uniform). Simiary, an appication from A to B can be uniquey extended to a homomorphism between the free monoids A and B. Such an appication is caed a coding (or a etter-to-etter morphism). 19

20 A morphism σ from A into itsef is said to be proongabe if there exists a etter a such that σ(a) = aw, where the word W is such that σ n (W ) is not the empty word for every n 0. In that case, the sequence of finite words (σ n (a)) n 1 converges in A Z 0 (endowed with the product topoogy of the discrete topoogy on each copy of A) to an infinite word a := σ (a). This infinite word is ceary a fixed point for σ and we say that a is generated by the morphism σ. The width of the morphism σ is the maximum of the engths of the words σ(a) for a in A. A morphism σ is primitive if there exists a positive integer n such that a occurs in σ n (a ) for a a, a in A. Cobham [24] estabished that uniform morphisms and automatic sequences are strongy connected. Theorem 8.1. Let k 2 be an integer. A sequence is k-automatic if, and ony if, it is the image under a coding of a fixed point of a k-uniform morphism. Theorem 8.1 means that one can aways associate with a k-automatic sequence a a 5-tupe (φ, σ, i, A, I), where σ is a k-uniform morphism defined over a finite aphabet I, i is a etter of I, φ is a coding from I into A, and such that a = φ(i), with i = σ (i). The set I and the sequence i are respectivey caed the interna aphabet and the interna sequence associated with the 5-tupe (φ, σ, i, A, I). With a sight abuse of anguage, we say that I (respectivey, i) is the interna aphabet (respectivey, interna sequence) associated with a. Indeed, Cobham gives in fact a canonica way to associate with a a 5-tupe (φ, σ, i, A, I). He aso proved ([24], Theorem 2) that p(n, a) k(cardi) 2 n, for n 1. (8.1) The k-kerne of a sequence a = (a ) 0 is the set of a sequences (a ki +j) 0, where i 0 and 0 j < k i. This notion gives rise to another usefu characterization of k- automatic sequences which was first proved by Eienberg [26]. Theorem 8.2. Let k 2 be an integer. A sequence is k-automatic if, and ony if, its k-kerne is finite. We reproduce Lemma 5.1 of [9], which gives an upper bound for the Diophantine exponent of a non-utimatey periodic automatic sequence. Lemma 8.3. Let k 2 be an integer. Let a be a k-automatic sequence which is not utimatey periodic. Let m be the cardinaity of the k-kerne of a. Then the Diophantine exponent of a is ess than k m. We concude this section with three resuts on fixed points of primitive morphisms. Mossé [32] estabished that a fixed point of a primitive morphism either is utimatey periodic, or contains no occurrence of words of the form W x, with W finite and non-empty and x sufficienty arge (independenty of the ength of W ). Her resut immediatey impies the next emma. Lemma 8.4. Let a be a fixed point of a primitive morphism and assume that a is not utimatey periodic. Then the Diophantine exponent of a is finite. A second resut was proved by M. Quefféec [35]. 20

21 Lemma 8.5. Let a = (a ) 1 be a fixed point of a primitive morphism on the aphabet {1, 2,...}. For 1, et q denote the denominator of the -th convergent to [0; a 1, a 2,...]. Then, the sequence (q 1/ ) 1 converges. A third resut, deduced from the proof of Theorem of [12], shows that the compexity of a sequence generated by a primitive morphism is subinear. Lemma 8.6. Let a = (a ) 1 be a fixed point of a primitive morphism σ on an aphabet of cardinaity b 2. Let v denote the width of σ. Then we have p(n, a) 2v 4b 2 b 3 n, for n 1. The reader is referred to [12] for many exampes of sequences generated by an automaton or by a primitive morphism. 9. A combinatoria emma Throughout this section and the next one, we use the foowing notation. If ξ = [0; a 1, a 2,...] is an irrationa rea number whose sequence of convergents is (p /q ) 1, then, for integers r 0 and s 1, we define the integer poynomia P ξ,r,s (X) by P ξ,r,s (X) := (q r 1 q r+s q r q r+s 1 )X 2 (q r 1 p r+s q r p r+s 1 + p r 1 q r+s p r q r+s 1 )X + (p r 1 p r+s p r p r+s 1 ). (9.1) By Lemma 6.2, the quadratic number [0; a 1,..., a r, a r+1,..., a r+s ] is a root of the poynomia P ξ,r,s (X). For the proofs of Theorem 2.2 and 3.2, we need the foowing auxiiary resut. Lemma 9.1. Let A 3 be an integer. Let a = (a ) 1 be an infinite sequence of positive integers at most equa to A and set ξ := [0; a 1, a 2,..., a,...]. Assume that the Diophantine exponent of ξ is finite. Let (p /q ) 1 be the sequence of convergents to ξ. Let M be an upper bound for the sequence (q 1/ ) 1. Assume that there are integers n 0 4 and κ 3 such that, for every integer n n 0, there is a word of ength n having two occurrences in the prefix of ength (κ + 1)n of a. Then, there exist non-negative integers r 1, r 2,... and positive integers s 1, s 2,... such that, for j 1, the word a begins with a word of ength r j foowed by a word of ength s j to the power 1 + 1/κ and (i) r j (2κ + 1)s j ; s j+1 2s j ; 21

22 (ii) a rj a rj +s j ; (iii) (2q rj q rj +s j ) 2 2q rj+1 q rj+1 +s j+1 (2q rj q rj +s j ) 650κ2 (og M) 2 ; (iv) P ξ,rj,s j (ξ) (2q rj q rj +s j ) 1 1/(15κ og M). Furthermore, if α j = [0; a 1,..., a rj, a rj +1,..., a rj +s j ], then α j is the root of P ξ,rj,s j (X) which is the cosest to ξ and, denoting by α j its Gaois conjugate, we have H(α j ) 2q rj q rj +s j, (9.2) ξ α j < (2q rj q rj +s j ) 2 1/(15κ og M) max{ α j α j 1, 1}, (9.3) and ξ α j > (2q rj q rj +s j ) 6(og M)Dio(ξ). (9.4) The estimate (9.2) hods since the quantity 2q r q r+s associated with the poynomia P ξ,r,s (X) defined in (9.1) is an obvious upper bound for its height. Note, however, that the height of P ξ,r,s (X) can be much smaer than 2q r q r+s when q r /q r 1 is cose to q r+s /q r+s 1 ; see Section 12. By Lemma 6.4, we can choose M = A+1 in Lemma 9.1. We have decided to introduce the quantity M since the numerica vaues occurring in Lemma 9.1 heaviy depend on the behaviour of the sequence (q 1/ ) 1, as it is expained after its proof. Moreover, it aows us to adapt more easiy Lemma 9.1 to the case when the sequence (a ) 1 is unbounded; see Section 12. However, in a the appications presented in Sections 2 to 5, the sequence (a ) 1 is bounded. Proof. The first part is purey combinatoria and we argue as in the proof of Lemma 9.1 from [7]. Let 2 be an integer, and denote by A() the prefix of a of ength. By assumption, for n 0, there exists a word W of ength having at east two occurrences in A((κ + 1)). In other words, there exist (possiby empty) words B, D, E and a non-empty word C such that A((κ + 1)) = B W D E = B C W E. We choose these words in such a way that, if B is non-empty, then the ast etter of B differs from the ast etter of C. Assume first that C W. Then, there exists a word F such that A((κ + 1)) = B W F W E. Setting U = B, V = W F, and w = W F W / W F, we observe that U V w prefix of a. Furthermore, we check that w w and is a U V w U V = 1 + W U V κ. 22

23 Assume now that C < W. This means that the two occurrences of W do overap, hence there exists a rationa number d > 1 such that W = C d. Setting U = B and V = C d /2, and noticing that 3 d /2 /2 d + 1, we observe that U V 3/2 is a prefix of a. Furthermore, we check that V ( + 2)/2 and V 1/2 U V V 2( V + κ) κ 1 4κ + 3. To summarize, setting w = 1 + 1/κ, we have shown that, for n 0, there exist two finite words U, V and a rationa number w such that w w 3/2 and (v) U V w is a prefix of a; (vi) U (2κ + 1) V ; (vii) /2 V κ; (viii) If U is not the empty word, then the ast etter of U differs from the ast etter of V ; (ix) U V w U V 4κ + 3. The words U, n 0, constructed above may not be a distinct. For n 0, et ρ and σ denote the engths of U and V, respectivey. The definition of M and Lemma 6.4 impy that 2 (j 1)/2 q j M j, for j 3. (9.5) Set f = 4(og M)/(og 2) + 1. Since we have, for every n 0, 2 2ρ + σ 2(κ + 1), 2f(2ρ + σ ) 4f(κ + 1) 4ρ 4f(κ+1) + 2σ 4f(κ+1) 16f(κ + 1) 2 32f(κ + 1) 2 (2ρ + σ ), by using (vi) and (vii). Setting c = 32f(κ + 1) 2, r j = ρ (4f) j (κ+1) j, and s j = σ (4f) j (κ+1) j, we thus get 2f(2r j + s j ) 2(2r j+1 + s j+1 ) c(2r j + s j ) and s j κ(4f) j (κ + 1) j < (4f)j+1 (κ + 1) j+1 2 It foows from (9.5) that s j+1 2. (2q rj q rj +s j ) 2 2q rj+1 q rj+1 +s j+1 (2q rj q rj +s j ) c(og M)/(og 2). 23

24 Since r j (2κ + 1)s j, Conditions (i) and (iii) of the emma are satisfied. Furthermore, Condition (ii) foows from (viii). Estimate (9.2) directy foows from Lemma 6.1. Set w j = w (4f) j (κ+1) j. Since, by Condition (v), the rea numbers ξ and α j = [0; U (4f) j (κ+1) j, V (4f) j (κ+1) j ] have the same first r j + w j s j partia quotients, we get by Lemma 6.4. The inequaity ξ α j q 2 r j + w j s j q 2 r j +s j 2 s j(w j 1), (9.6) α j α j (q rj q rj +s j ) 1, combined with (9.6) shows that α j is the root of P ξ,rj,s j (X) which is the cosest to ξ, provided that j is arge enough. It foows from (ix) that and, by (9.5) and Lemma 6.4, we get s j (w j 1) (r j + s j )/(4κ + 3), (9.7) 2 r j+s j (q rj q rj +s j ) (r j+s j )(og 2)/((2r j +s j ) og M) (q rj q rj +s j ) (og 2)/(2 og M). (9.8) Putting together (9.6), (9.7) and (9.8), we deduce that ξ α j q 2 r j (2q rj q rj +s j ) 2 (og 2)/(2(4κ+3) og M). (9.9) Furthermore, it foows from Lemma 6.1 that thus α j α j A 3 q 2 r j, (9.10) q 2 r j A 3 max{ α j α j 1, 1}. (9.11) Combining (9.9) and (9.11) with κ 3, we get (9.3) for j arge enough. We aso deduce from (9.9), (9.10) and the triange inequaity that P ξ,rj,s j (ξ) 2 (2q rj q rj +s j ) ξ α j α j α j This shows that (iv) hods for j arge enough. 2A 3 (2q rj q rj +s j ) 1 (og 2)/(2(4κ+3) og M). 24

25 If Dio(ξ) is bounded from above by δ, then the continued fraction expansions of α j and ξ agree at most unti the δ(r j + s j ) partia quotient. Consequenty, using Lemma 5 of [2], we get ξ α j (A + 2) 3 q 2 δ(r j +s j ) (A + 2) 3 M 2 M 2δ(r j+s j ) (A + 2) 3 M 2 (2q rj q rj +s j ) 4δ(r j+s j )(og M)/((2r j +s j )(og 2)) (A + 2) 3 M 2 (2q rj q rj +s j ) 4δ(og M)/(og 2). (9.12) This shows that (9.4) hods for j arge enough. This competes the proof of the emma. It is apparent in the proof of Lemma 9.1 that the estimates can be improved if, instead of (9.5), we use the fact that for some rea numbers m, M with M > m 2 we have m j q j M j, for every sufficienty arge j. The quantity (og M)/(og 2) coming from (9.5) and occurring in the proof of Lemma 9.1 can then be everywhere repaced by (og M)/(og m). In particuar, when the sequence (q 1/ ) 1 converges, then m and M can be taken arbitrariy cose, thus (og M)/(og m) can be taken arbitrariy cose to 1. This shows that, under this assumption, Lemma 9.1 hods with og M repaced by 1 (and, even, by 1/(og 2)) in (iii), (iv), (9.3), and (9.4). 10. Second part of the proof of Theorem 3.2 The proof of Theorem 3.2 party reies on the quantitative version of the Schmidt Subspace Theorem. Theorem E beow statement was proved by Evertse [27]. We refer to his paper for the definition of the height H(L) of the inear form L(x) = α 1 x α m x m, where α 1,..., α m are rea agebraic numbers beonging to a same number fied of degree d. For our purpose, it is sufficient to stress that H(L) can be bounded from above in terms of the heights of the coefficients of L. More precisey, inequaity (6.6) from [5] asserts that H(L) m d/2 (d + 1) m/2 m i=1 H(α i ). (10.1) Theorem E. Let m 2, H and d be positive integers. Let L 1,..., L m be ineary independent (over Q) inear forms in m variabes with rea agebraic coefficients. Assume that H(L i ) H for i = 1,..., m and that the number fied generated by a the coefficients 25

26 of these inear forms has degree at most d. Let ε be a rea number with 0 < ε < 1. Then, the primitive integer vectors x = (x 1,..., x m ) in Z m with H(x) H and such that ie in at most m L i (x) < det(l 1, L 2,..., L m ) (max{ x 1,..., x m }) ε i=1 c m,ε (og 3d) (og og 3d) proper subspaces of Q m, where c m,ε is a constant which ony depends on m and ε. We keep the notation from Section 9. First, we check that the entries of the quadrupe (q r 1 q r+s q r q r+s 1, q r 1 p r+s q r p r+s 1, p r 1 q r+s p r q r+s 1, p r 1 p r+s p r p r+s 1 ) (10.2) are reativey prime. To see this, assume that a positive integer m divides p r 1 q r+s p r q r+s 1 and p r 1 p r+s p r p r+s 1. Then, it aso divides and p r+s 1 (p r 1 q r+s p r q r+s 1 ) q r+s 1 (p r 1 p r+s p r p r+s 1 ) = ±p r 1 p r+s 1 (q r 1 q r+s q r q r+s 1 ) q r+s 1 (q r 1 p r+s q r p r+s 1 ) = ±q r 1. Since p r 1 and q r 1 are reativey prime, the quadrupe (10.2) is primitive. Incidentay, this shows that when q r 1 p r+s q r p r+s 1 is equa to p r 1 q r+s p r q r+s 1, then the tripe (q r 1 q r+s q r q r+s 1, q r 1 p r+s q r p r+s 1, p r 1 p r+s p r p r+s 1 ) is primitive. These resuts wi be used in the seque. Second part of the proof of Theorem 3.2. Let ξ be as in the statement of Theorem 3.2. Assume that the Diophantine exponent of ξ is finite. By Theorem 2.2, this impies that w 2 (ξ) is aso finite. To prove that ξ is either an S- or a T -number, it remains for us to contro the accuracy of the approximation to ξ by agebraic numbers of exact degree d, for every integer d 3. We foow the proof of Theorem 2.1 from [6]. Since there are a few technica difficuties, we give some detais. Let d 3 be an integer. Let α be an agebraic number of degree d. At severa paces in the proof beow, it is convenient to assume that the height of α is sufficienty arge. Let χ be a positive rea number such that Our aim is to prove that ξ α < H(α) χ. χ < exp(c (og 3d) 5 (og og 3d) 4 ) (10.3) 26