On Two Exponents of Approximation Related to a Real Number and Its Square

Transcription

1 Canad. J. Math. Vol. 59 (1), 2007 pp On Two Exponents of Approxmaton Related to a Real Number and Its Square Damen Roy Abstract. For each real number ξ, let λ 2 (ξ) denote the supremum of all real numbers λ such that, for each suffcently large X, the nequaltes x 0 X, x 0 ξ x 1 X λ and x 0 ξ 2 x 2 X λ admt a soluton n ntegers x 0, x 1 and x 2 not all zero, and let ω 2 (ξ) denote the supremum of all real numbers ω such that, for each suffcently large X, the dual nequaltes x 0 + x 1 ξ + x 2 ξ 2 X ω, x 1 X and x 2 X admt a soluton n ntegers x 0, x 1 and x 2 not all zero. Answerng a queston of Y. Bugeaud and M. Laurent, we show that the exponents λ 2 (ξ) where ξ ranges through all real numbers wth [Q(ξ) : Q] > 2 form a dense subset of the nterval [1/2, ( 5 1)/2] whle, for the same values of ξ, the dual exponents ω 2 (ξ) form a dense subset of [2, ( 5 + 3)/2]. Part of the proof rests on a result of V. Jarník showng that λ 2 (ξ) = 1 ω 2 (ξ) 1 for any real number ξ wth [Q(ξ) : Q] > 2. 1 Introducton Let ξ and η be real numbers. Followng the notaton of Y. Bugeaud and M. Laurent [3], we defne λ(ξ, η) to be the supremum of all real numbers λ such that the nequaltes x 0 X, x 0 ξ x 1 X λ and x 0 η x 2 X λ admt a non-zero nteger soluton (x 0, x 1, x 2 ) Z 3 for each suffcently large value of X. Smlarly, we defne ω(ξ, η) to be the supremum of all real numbers ω such that the nequaltes x 0 + x 1 ξ + x 2 η X ω, x 1 X and x 2 X admt a non-zero soluton (x 0, x 1, x 2 ) Z 3 for each suffcently large value of X. An applcaton of Drchlet box prncple shows that we have 1/2 λ(ξ, η) and 2 ω(ξ, η). Moreover, n the (non-degenerate) case where 1, ξ and η are lnearly ndependent over Q, a result of V. Jarník, kndly ponted out to the author by Yann Bugeaud, shows that these exponents are related by the formula (1) λ(ξ, η) = 1 1 ω(ξ, η), wth the conventon that the rght-hand sde of ths equalty s 1 f ω(ξ, η) = (see [7, Theorem 1]). Receved by the edtors July 23, 2004; revsed October 12, Work partally supported by NSERC and CICMA AMS subject classfcaton: Prmary: 11J13; secondary: 11J82. c Canadan Mathematcal Socety

2 212 D. Roy In the case where η = ξ 2, we use the shorter notaton λ 2 (ξ) := λ(ξ, ξ 2 ) and ω 2 (ξ) := ω(ξ, ξ 2 ) of [3]. The condton that 1, ξ and ξ 2 are lnearly ndependent over Q smply means that ξ s not an algebrac number of degree at most 2 over Q, a condton whch we also wrte as [Q(ξ):Q] > 2. Under ths condton, t s known that these exponents satsfy (2) 1 2 λ 2 (ξ) 1 γ = and 2 ω 2(ξ) γ 2 = , where γ = (1 + 5)/2 denotes the golden rato. By vrtue of W. M. Schmdt s subspace theorem, the lower bounds n (2) are acheved by any algebrac number ξ of degree at least 3 (see [12, Ch. VI, Corollares 1C, 1E]). They are also acheved by almost all real numbers ξ, wth respect to Lebesgue s measure (see [3, Theorem 2.3]). On the other hand, the upper bounds follow respectvely from [5, Theorem 1a] and from [2]. They are acheved n partcular by the so-called Fbonacc contnued fractons (see [8, 2] or [9, 6]), a specal case of the Sturman contnued fractons of [1]. Now, thanks to Jarník s formula (1), we recognze that each set of nequaltes n (2) can be deduced from the other one. Generalzng the approach of [8], Bugeaud and Laurent have computed the exponents λ 2 (ξ) and ω 2 (ξ) for a general (characterstc) Sturman contnued fracton ξ. They found that, after 1/γ and γ 2, the next largest values of λ 2 (ξ) and ω 2 (ξ) for such numbers ξ are, respectvely, and , and they asked f there exsts any transcendental real number ξ whch satsfes ether 2 2 < λ 2 (ξ) < 1/γ or < ω 2 (ξ) < γ 2 (see [3, 8]). Our man result below shows that such numbers exst. Theorem The ponts ( λ 2 (ξ), ω 2 (ξ)) where ξ runs through all real numbers wth [Q(ξ):Q] > 2 form a dense subset of the curve C = {(1 ω 1, ω) ; 2 ω γ 2 }. Snce ( λ 2 (ξ), ω 2 (ξ)) = (1/2, 2) for any algebrac number ξ of degree at least 3, t follows n partcular that (1/γ, γ 2 ) s an accumulaton pont for the set of ponts ( λ 2 (ξ), ω 2 (ξ)) wth ξ a transcendental real number. Because of Jarník s formula (1), ths theorem s equvalent to ether one of the followng two assertons. Corollary The exponents λ 2 (ξ) attached to transcendental real numbers ξ form a dense subset of the nterval [1/2, 1/γ]. The correspondng dual exponents ω 2 (ξ) form a dense subset of [2, γ 2 ]. The proof s nspred by the constructons of [9, 6] and [11, 5]. We produce countably many real numbers ξ of Fbonacc type (see 7 for a precse defnton) for whch we show that the exponents ω 2 (ξ) are dense n [2, γ 2 ]. By (1), ths mples the theorem. One may then reformulate the queston of Bugeaud and Laurent by askng f there exst transcendental real numbers ξ not of that type whch satsfy ω 2 (ξ) > The work of S. Fschler announced n [6] should shed some lght on ths queston.

3 On Two Exponents of Approxmaton Notaton and Equvalent Defntons of the Exponents We defne the norm of a pont x = (x 0, x 1, x 2 ) R 3 as ts maxmum norm x = max 0 2 x. Gven a second pont y R 3, we denote by x y the standard vector product of x and y, and by x, y ther standard scalar product. Gven a thrd pont z R 3, we also denote by det(x, y, z) the determnant of the 3 3 matrx whose rows are x, y and z. Then we have the well-known relaton det(x, y, z) = x, y z and we get the followng alternatve defnton of the exponents λ(ξ, η) and ω(ξ, η). Lemma 2.1 Let ξ, η R, and let y = (1, ξ, η). Then λ(ξ, η) s the supremum of all real numbers λ such that, for each suffcently large real number X 1, there exsts a pont x Z 3 wth 0 < x X and x y X λ. Smlarly, ω(ξ, η) s the supremum of all real numbers ω such that, for each suffcently large real number X 1, there exsts a pont x Z 3 wth 0 < x X and x, y X ω. In the sequel, we wll need the followng nequaltes. Lemma 2.2 (3) (4) For any x, y, z R 3, we have x, z y x, y z 2 x y z, y x z z x y + 2 x y z. Proof Wrtng y = (y 0, y 1, y 2 ) and z = (z 0, z 1, z 2 ), we fnd x, z y x, y z = max =0,1,2 x, y z z y 2 x y z, whch proves (3). Smlarly, one fnds y x z z x y 2 x y z for = 0, 1, 2, and ths mples (4). For any non-zero pont x of R 3, let [x] denote the pont of P 2 (R) havng x as a set of homogeneous coordnates. Then (4) has a useful nterpretaton n terms of the projectve dstance defned for non-zero ponts x and y of R 3 by dst([x], [y]) = dst(x, y) = x y x y. Indeed, for any trple of non-zero ponts x, y, z R 3, t gves (5) dst([x], [z]) dst([x], [y]) + 2 dst([y], [z]).

4 214 D. Roy 3 Fbonacc Sequences n GL 2 (C) A Fbonacc sequence n a monod s a sequence (w ) 0 of elements of that monod such that w +2 = w +1 w for each ndex 0. Clearly, such a sequence s entrely determned by ts frst two elements w 0 and w 1. We start wth the followng observaton. Proposton 3.1 There exsts a non-empty Zarsk open subset U of GL 2 (C) 2 wth the followng property. For each Fbonacc sequence (w ) 0 wth (w 0, w 1 ) U, there exsts N GL 2 (C) such that the matrx (6) y = { w N w t N f s even, f s odd, s symmetrc for each 0. Any matrx N GL 2 (C) such that w 0 N, w 1 t N and w 1 w 0 N are symmetrc satsfes ths property. When w 0 and w 1 have nteger coeffcents, we may take N wth nteger coeffcents. Proof Let (w ) 0 be a Fbonacc sequence n GL 2 (C) and let N GL 2 (C). Defnng y by (6) for each 0, we fnd y +3 = y t +1 Sy Sy +1 wth S = N 1 f s even and S = t N 1 f s odd. Thus, y s symmetrc for each 0 f and only f t s so for = 0, 1, 2. Now, for any gven pont (w 0, w 1 ) GL 2 (C) 2, the condtons that w 0 N, w t 1 N and w 1 w 0 N are symmetrc represent a system of three lnear equatons n the four unknown coeffcents of N. Let V be the Zarsk open subset of GL 2 (C) 2 consstng of all ponts (w 0, w 1 ) for whch ths lnear system has rank 3. Then, for each (w 0, w 1 ) V, the 3 3 mnors of ths lnear system convenently arranged nto a 2 2 matrx provde a non-zero soluton N of the system, whose coeffcents are polynomals n those of w 0 and w 1 wth nteger coeffcents. Then the condton det(n) 0 n turn determnes a Zarsk open subset U of V. To conclude, we note that U s not empty as a short computaton shows that t contans the pont formed by w 0 = ( ) and w 1 = ( ) Defnton 3.2 Let M = Mat 2 2 (Z) GL 2 (C) denote the monod of 2 2 nteger matrces wth non-zero determnant. We say that a Fbonacc sequence (w ) 0 n M s admssble f there exsts a matrx N M such that the sequence (y ) 0 gven by (6) conssts of symmetrc matrces. Snce M s Zarsk dense n GL 2 (C), Proposton 3.1 shows that almost all Fbonacc sequences n M are admssble. The followng example s an llustraton of ths. Example 3.3 Fx ntegers a, b, c wth a 2 and c b 1, and defne w 0 = ( ) ( ) 1 b 1 c, w a a(b + 1) 1 = a a(c + 1)

5 On Two Exponents of Approxmaton 215 and ( ) 1 + a(b + 1)(c + 1) a(b + 1) N =. a(c + 1) a These matrces belong to M snce det(w 0 ) = det(w 1 ) = a and det(n) = a. Moreover, one fnds that and w 0 N = ( ) 1 + a(c + 1) a, w a 0 t 1 N = w 1 w 0 N = ( ) 1 + a a a a 2 ( 1 + a(b + 1) ) a a 0 are symmetrc matrces. Therefore, the Fbonacc sequence (w ) 0 constructed on w 0 and w 1 s admssble wth an assocated sequence of symmetrc matrces (y ) 0 gven by (6), the frst three matrces of ths sequence beng the above products y 0 = w 0 N, y 1 = w 1 t N and y 2 = w 1 w 0 N. 4 Fbonacc Sequences of 2 2 Integer Matrces In the sequel, we dentfy R 3 (resp., Z 3 ) wth the space of 2 2 symmetrc matrces wth real (resp., nteger) coeffcents under the map x = (x 0, x 1, x 2 ) ( ) x0 x 1. x 1 x 2 Accordngly, t makes sense to defne the determnant of a pont x = (x 0, x 1, x 2 ) of R 3 by det(x) = x 0 x 2 x 2 1. Smlarly, gven symmetrc matrces x, y and z, we wrte x y, x, y and det(x, y, z) to denote respectvely the vector product, scalar product and determnant of the correspondng ponts. In ths secton we look at arthmetc propertes of admssble Fbonacc sequences n the monod M of Defnton 3.2. For ths purpose, we defne the content of an nteger matrx w Mat 2 2 (Z) or of a pont y Z 3 as the greatest common dvsor of ther coeffcents. We say that such a matrx or pont s prmtve f ts content s 1. Proposton 4.1 Let (w ) 0 be an admssble Fbonacc sequence of matrces n M and let (y ) 0 be a correspondng sequence of symmetrc matrces n M. For each 0, defne z = det(w ) 1 y y +1. Then, for each 0, we have (a) tr(w +3 ) = tr(w +1 ) tr(w +2 ) det(w +1 ) tr(w ), (b) y +3 = tr(w +1 )y +2 det(w +1 )y, (c) z +3 = tr(w +1 )z +1 + det(w )z, (d) det(y, y +1, y +2 ) = ( 1) det(y 0, y 1, y 2 ) det(w 2 ) 1 det(w +2 ), (e) z z +1 = ( 1) det(y 0, y 1, y 2 ) det(w 2 ) 1 y +1. Proof For each ndex 0, let N denote the element of M for whch y = w N. Accordng to (6), we have N = N f s even and N = t N f s odd. We frst prove

6 216 D. Roy (b) followng the argument of the proof of [10, Lemma 2.5()]. Multplyng both sdes of the equalty w +2 = w +1 w on the rght by N +2 = N, we fnd (7) y +2 = w +1 y, whch can be rewrtten as y +2 = y +1 N 1 +1 y. Takng the transpose of both sdes, ths gves y +2 = y N 1 y +1 = w y +1. Replacng by + 1 n the latter dentty and combnng t wth (7), we get (8) y +3 = w +1 y +2 = w 2 +1y. Then (b) follows from (7) and (8), usng the fact that, by the Cayley Hamlton theorem, we have w 2 +1 = tr(w +1)w +1 det(w +1 )I. Multplyng both sdes of (b) on the rght by N 1 and takng the trace, we deduce that tr(y +3 N 1 ) = tr(w +1 ) tr(w +2 ) det(w +1 ) tr(w ). Ths gves (a) because tr(y +3 N 1 ) = tr( t y t +3 N 1 ) = tr(w +3 ). Takng the exteror product of both sdes of (b) wth y +1, we also fnd y +1 y +3 = tr(w +1 ) det(w +1 )z +1 + det(w +1 ) det(w )z. Smlarly, replacng by + 1 n (b) and takng the exteror product wth y +3 gves det(w +3 )z +3 = det(w +2 )y +1 y +3. Then (c) follows upon notng that det(w +3 ) = det(w +2 ) det(w +1 ). The formula (d) s clearly true for = 0. If we assume that t holds for some nteger 0, then usng the formula for y +3 gven by (b) and takng nto account the multlnearty of the determnant we fnd det(y +1, y +2, y +3 ) = det(w +1 ) det(y, y +1, y +2 ) = ( 1) +1 det(y 0, y 1, y 2 ) det(w +3) det(w 2 ). Ths proves (d) by nducton on. Then (e) follows snce, for any x, y, z Z 3, we have (x y) (y z) = det(x, y, z) y whch, n the present case, gves z z +1 = det(w +2 ) 1 det(y, y +1, y +2 ) y +1. Corollary 4.2 The notaton beng as n the proposton, assume that tr(w ) and det(w ) are relatvely prme for = 0, 1, 2, 3 and that det(y 0, y 1, y 2 ) 0. Then for each 0, (a) the ponts y, y +1, y +2 are lnearly ndependent, (b) tr(w ) and det(w ) are relatvely prme, (c) the matrx w s prmtve,

7 On Two Exponents of Approxmaton 217 (d) the content of y dvdes det(y 2 )/ det(w 2 ), (e) the pont det(w 2 ) z belongs to Z 3 and ts content dvdes det(y 2 ) det(y 0, y 1, y 2 ). Proof The asserton (a) follows from Proposton 4.1(d). Snce (b) holds by hypothess for = 0, 1, 2, 3, and snce det(w 2 ) and det(w ) have the same prme factors for each 2, the asserton (b) follows, by nducton on, from the fact that Proposton 4.1(a) gves tr(w +1 ) tr(w ) tr(w 1 ) modulo det(w 2 ) for each 3. Then (c) follows snce the content of w dvdes both tr(w ) and det(w ). Let N M such that y 2 = w 2 N. For each, we have y = w N where N = N f s even and N = t N f s odd. Ths gves y Adj(N ) = det(n)w where Adj(N ) M denotes the adjont of N. Thus, by (c), the content of y dvdes det(n) = det(y 2 )/ det(w 2 ), as clamed n (d). The fact that det(w 2 ) z belongs to Z 3 s clear for = 0, 1, 2 because det(w 0 ) and det(w 1 ) dvde det(w 2 ). Then Proposton 4.1(c) shows, by nducton on, that det(w 2 ) z Z 3 for each 0. Moreover, the content of that pont dvdes that of det(w 2 ) 2 z z +1 whch, by (d) and Proposton 4.1(e), dvdes det(y 0, y 1, y 2 ) det(y 2 ). Ths proves (e). Example 4.3 Let (w ) 0, N and (y ) 0 be as n Example 3.3. Snce w 0, w 1 and N are congruent to matrces of the form ( ) ±1 0 0 modulo a and have determnant ±a, all matrces w and y are congruent to matrces of the same form modulo a and ther determnant s, up to sgn, a power of a. Thus these matrces have relatvely prme trace and determnant, and so are prmtve for each 0. Snce det(y 0, y 1, y 2 ) = a 4 (c b), Proposton 4.1(e) shows that the ponts z = det(w ) 1 y y +1 satsfy z z +1 = ( 1) a 2 (c b)y +1 for each 0. Moreover, we fnd that a 1 z 0 = (0, 0, b c), a 1 z 1 = (a, 1 + a(b + 1), b) and a 1 z 2 = (a, 1 + a(c + 1), c) are nteger ponts. Then Proposton 4.1(c) shows, by nducton on, that a 1 z Z 3 for each 0. In partcular, f c = b + 1, we deduce from the relaton a 1 z a 1 z +1 = ±y +1 that a 1 z s a prmtve nteger pont for each 0. 5 Growth Estmates Defne the norm of a 2 2 matrx w = (w k,l ) Mat 2 2 (R) as the largest absolute value of ts coeffcents w = max 1 k,l 2 w k,l, and defne γ = (1 + 5)/2 as n the ntroducton. In ths secton, we provde growth estmates for the norm and determnant of elements of certan Fbonacc sequences n GL 2 (R). We frst establsh two basc lemmas. Lemma 5.1 Let w 0, w 1 GL 2 (R). Suppose that, for = 0, 1, the matrx w s of the form ( ) a b c d wth 1 a mn{b, c} and max{b, c} d. Then all matrces of the Fbonacc sequence (w ) 0 constructed on w 0 and w 1 have ths form and for each 0, they satsfy (9) w w +1 < w +2 2 w w +1.

8 218 D. Roy Proof The frst asserton follows by recurrence on and s left to the reader. It mples that w s equal to the element of ndex (2, 2) of w for each 0. Then (9) follows by observng that, for any 2 2 matrces w = (w k,l ) and w = (w k,l ) wth postve real coeffcents, the product w w = (w k,l ) satsfes w 2,2w 2,2 < w 2,2 2 w w. Lemma 5.2 Let (r ) 0 be a sequence of postve real numbers. Assume that there exst constants c 1, c 2 > 0 such that c 1 r r +1 r +2 c 2 r r +1 for each 0. Then there also exst constants c 3, c 4 > 0 such that c 3 r γ r +1 c 4 r γ for each 0. Proof Defne c 3 = c γ 1 /(cc 2) and c 4 = cc γ 2 /c 1, where c 1 s chosen so that the condton c 3 r +1 /r γ c 4 holds for = 0. Assumng that the same condton holds for some ndex 0, we fnd r +2 r γ +1 c 1 r r 1/γ +1 c 1 c 1/γ 4 = c 1/γ2 c 3 c 3, and smlarly r +2 /r γ +1 c 4. Ths proves the lemma by recurrence on. Proposton 5.3 Let (w ) 0 be a Fbonacc sequence n GL 2 (R). Suppose that there exst real numbers c 1, c 2 > 0 such that (10) c 1 w w +1 w +2 c 2 w w +1 for each 0. Then there exst constants c 3, c 4 > 0 for whch the nequaltes (11) c 3 w γ w +1 c 4 w γ, c 3 det(w ) γ det(w +1 ) c 4 det(w ) γ hold for each 0. Moreover, f there exst α, β 0 such that (12) (c 2 w ) α det(w ) (c 1 w ) β holds for = 0, 1, then ths relaton extends to each 0. Proof The frst asserton of the proposton follows from Lemma 5.2 appled once wth r = w and once wth r = det(w ). To prove the second asserton, assume that for some ndex j 0 the condton (12) holds both wth = j and = j + 1. We fnd det(w j+2 ) = det(w j+1 ) det(w j ) (c 2 w j+1 ) α (c 2 w j ) α (c 2 w j+2 ) α and smlarly det(w j+2 ) (c 1 w j+2 ) β. Therefore, (12) holds wth = j + 2. By recurrence on, ths shows that (12) holds for each 0 f t holds for = 0, 1.

9 On Two Exponents of Approxmaton 219 Example 5.4 Let the notaton be as n Example 3.3. Snce w 0 and w 1 satsfy the hypotheses of Lemma 5.1, the Fbonacc sequence (w ) 0 that they generate fulflls for each 0 the condton (10) of Proposton 5.3 wth c 1 = 1 and c 2 = 2. As det(w 0 ) = det(w 1 ) = a, we also note that for ths choce of c 1 and c 2 the condton (12) holds for = 0, 1 wth α = log a log(2a(c + 1)) and β = log a log(a(b + 1)). Then, for an approprate choce of c 3, c 4 > 0, both (11) and (12) hold for each 0. Moreover, the estmates (9) of Lemma 5.1 mply that the sequence (w ) 0 s unbounded. 6 Constructon of a Real Number Gven sequences of non-negatve real numbers wth general terms a and b, we wrte a b or b a f there exsts a real number c > 0 such that a cb for all suffcently large values of. We wrte a b when a b and b a. Wth ths notaton, we now prove the followng result (cf. [11, 5]). Proposton 6.1 Let (w ) 0 be an admssble Fbonacc sequence n M and let (y ) 0 be a correspondng sequence of symmetrc matrces n M. Assume that (w ) 0 s unbounded and satsfes the condtons (13) w +1 w γ, det(w +1 ) det(w ) γ and det(w ) w β for a real number β wth 0 < β < 2. Vewng each y as a pont n Z 3, assume that det(y 0, y 1, y 2 ) 0 and defne z = (det(w )) 1 y y +1 for each 0. Then we have (14) y w, det(y ) det(w ), z w 1, and there exsts a non-zero pont y of R 3 wth det(y) = 0 such that (15) y y det(w ) w and z, y det(w +1) w +2. If β < 1, the coordnates of such a pont y are lnearly ndependent over Q and we may assume that y = (1, ξ, ξ 2 ) for some real number ξ wth [Q(ξ):Q] > 2. Proof For each 0, let N denote the element of M for whch y = w N. Puttng N = N 0, we have by hypothess N = N when s even and N = t N otherwse. Ths mples that y w and det(y ) det(w ). In the sequel, we wll repeatedly use these relatons as well as the hypothess (13). We clam that we have (16) y y +1 det(w ) w 1.

10 220 D. Roy To prove ths, we defne J = ( ) and note that for each 0 the coeffcents of the dagonal of y Jy +1 concde wth the frst and thrd coeffcents of y y +1 whle the sum of the coeffcents of y Jy +1 outsde of the dagonal s the mddle coeffcent of y y +1 multpled by 1. Ths gves (17) y y +1 2 y Jy +1. Snce y +1 = y N 1 y 1 and snce x Jx = det(x) J for any symmetrc matrx x, we also fnd that y Jy +1 = det(y ) JN 1 y 1 and therefore y Jy +1 det(w ) w 1. Combnng ths wth (17) proves our clam (16), whch can also be wrtten n the form (18) z w 1. As y w and y +1 w γ, the estmate (16) shows, n the notaton of 2, that (19) dst([y ], [y +1 ]) cδ, where δ = det(w ) w 2 and where c s some postve constant whch does not depend on. Snce by hypothess we have det(w ) w β wth β < 2, we fnd that lm δ = 0. Snce moreover, we have δ +1 δ γ, we deduce that there exsts an ndex 0 1 such that δ +1 δ /4 for each 0. Then, usng (5), we deduce that j 1 (20) dst([y ], [y j ]) 2 k dst([y k ], [y k+1 ]) c 2 k δ k 2cδ k= for each choce of and j wth 0 < j. Thus the sequence ([y ]) 0 converges n P 2 (R) to a pont [y] for some non-zero y R 3. Snce the rato det(y ) / y 2 depends only on the class [y ] of y n P 2 (R) and tends to 0 lke δ as, we deduce by contnuty that det(y) / y 2 = 0 and thus that det(y) = 0. By contnuty, (20) also leads to dst([y ], [y]) 2cδ for each 0, and so (21) y y det(w ) w. Applyng (3) together wth the above estmates (18) and (21), we fnd j 1 k= z, y y +2 z, y +2 y 2 z y +2 y w 1 det(w +2) w +2 det(w +1 ) δ. Usng Proposton 4.1(d), we also get (22) z, y +2 y = det(y, y +1, y +2 ) y det(w +1 ). det(w )

11 On Two Exponents of Approxmaton 221 Combnng the above two estmates, we deduce that z, y y +2 det(w +1 ) and therefore that z, y det(w +1 ) / w +2. The latter estmate s the second half of (15). It mples z +1, y y det(w +2) w +3 w det(w ) δ +1. Snce z +1, y = det(w 1 ) 1 z, y +2, the estmate (22) can also be wrtten n the form z +1, y y det(w ). Then, applyng (3) once agan, we fnd 2 z +1 y y z +1, y y z +1, y y det(w ). Snce, by (18) and (21), we have z +1 w and y y det(w ) / w, we conclude from ths that z +1 w and y y det(w ) / w, whch completes the proof of (14) and (15). Now, assume that β < 1, and let u Z 3 such that u, y = 0. By (3), we have (23) 2 u y y u, y y u, y y = u, y y for each 0. Snce y y det(w ) / w w β 1 tends to 0 as, we deduce from (23) that the nteger u, y must vansh for all suffcently large values of. Ths mples that u = 0 because t follows from the hypothess det(y 0, y 1, y 2 ) 0 and the formula n Proposton 4.1(d) that any three consecutve ponts of the sequence (y ) 0 are lnearly ndependent. Thus the coordnates of y must be lnearly ndependent over Q. In partcular, the frst coordnate of y s non-zero and, dvdng y by ths coordnate, we may assume that t s equal to 1. Then, upon denotng by ξ the second coordnate of y, the condton det(y) = 0 mples that y = (1, ξ, ξ 2 ) and thus [Q(ξ):Q] > 2. 7 Estmates for the Exponent ω 2 We frst prove the followng result and then deduce from t our man theorem n 1. Proposton 7.1 Let (w ) 0 be an admssble Fbonacc sequence n M, and let (y ) 0 be a correspondng sequence of symmetrc matrces n M. Assume that (w ) 0 s unbounded and satsfes (24) w +1 w γ, det(w +1 ) det(w ) γ, w α det(w ) w β for real numbers α and β wth 0 α β < γ 2. Assume moreover that tr(w ) and det(w ) are relatvely prme for = 0, 1, 2, 3 and that det(y 0, y 1, y 2 ) 0. Then the real number ξ whch comes out from the last asserton of Proposton 6.1 satsfes γ 2 βγ ω 2 (ξ) γ 2 αγ.

12 222 D. Roy Proof Put y = (1, ξ, ξ 2 ) and defne the sequence (z ) 0 as n Proposton 4.1. Snce y 1, the nequalty (3) combned wth the estmates of Proposton 6.1 shows that, for any pont z Z 3 and any ndex 1, we have { (25) z, y y z, y +2 z y y < c 5 max w z, y, z det(w } ), w wth a constant c 5 > 0 whch s ndependent of z and. Suppose that a pont z Z 3 satsfes (26) 0 < z Z := c 6 w and z, y det(w +1) w +2, where c 6 = c 1 5 det(y 2 ) 1. Usng (25) wth replaced by + 1, we fnd z, y +1 det(w ) γ w 1/γ. Snce det(w ) w β wth β < γ 2, ths gves z, y +1 < 1 provded that s suffcently large. Then the nteger z, y +1 must be zero and, by Proposton 4.1(e), we deduce that z = az + bz +1 for some a, b Q where b s gven by z z = bz z +1 = ( 1) b det(y 0, y 1, y 2 ) det(w 2 ) 1 y +1. Snce det(w 2 )z z Z 3 and snce, by Corollary 4.2(d), the content of y +1 dvdes det(y 2 )/ det(w 2 ), ths mples that b det(y 0, y 1, y 2 ) det(y 2 )/ det(w 2 ) s an nteger. So, f b s non-zero, t satsfes the lower bound b det(w 2 )/(det(y 0, y 1, y 2 ) det(y 2 )). We note that z, y = 0 and by Proposton 4.1(d) that z +1, y = det(y, y +1, y +2 ) det(w +1 ) = ( 1) det(y 0, y 1, y 2 ) det(w 2 ) Therefore, f b 0, the pont z = az + bz +1 satsfes det(w ). z, y = b z +1, y det(y 2 ) 1 det(w ) = c 5 c 6 det(w ). However, (25) and (26) gve { det(w+1 ) w } z, y < c 5 max, c 6 det(w ) = c 5 c 6 det(w ) w +2 f s suffcently large, because the rato det(w +1 ) w / w +2 w βγ γ tends to 0 as. Comparson wth the prevous nequalty then forces b = 0, and so we get z = az wth a 0. Snce det(w 2 )z s, by Corollary 4.2(e), an nteger pont whose content dvdes det(y 2 ) det(y 0, y 1, y 2 ), we deduce that a det(y 2 ) det(y 0, y 1, y 2 )/ det(w 2 )

13 On Two Exponents of Approxmaton 223 s a non-zero nteger and therefore, usng the second part of (15) n Proposton 6.1, we fnd that z, y = a z, y det(w 2 ) det(y 2 ) det(y 0, y 1, y 2 ) z, y det(w +1) w +2. Snce ths holds for any pont z satsfyng (26) wth suffcently large, we deduce that for any ndex 0 and any pont z Z 3 wth 0 < z Z we have z, y det(w +1) w +2 w γα γ2 Z γα γ2. Ths shows that ω 2 (ξ) γ 2 γα. Fnally, for any real number Z z 0, there exsts an ndex 0 such that z Z < z +1 and, for such choce of, we fnd by Proposton 6.1 that z, y det(w +1) w +2 w βγ γ2 z +1 βγ γ2 Z βγ γ2, showng that ω 2 (ξ) γ 2 γβ. Let us say that a real number ξ s of Fbonacc type f there exst an unbounded Fbonacc sequence (w ) 0 n M and a real number θ wth θ > 1/γ such that (ξ, 1)w w θ for each suffcently large ndex. There are countably many such numbers, and any real number ξ obtaned from Proposton 6.1 wth β < γ 2 s of ths type. The followng corollary shows that the exponents ω 2 (ξ) attached to transcendental numbers of Fbonacc type are dense n the nterval [2, γ 2 ]. By Jarník s formula (1), ths mples our man theorem n 1. Corollary 7.2 Let t and ǫ be real numbers wth 0 < t < γ 2 and ǫ > 0. Then there exst a transcendental real number ξ and an unbounded Fbonacc sequence (w ) 0 n M whch satsfy (a) (ξ, 1)w w 1+t for each suffcently large, (b) γ 2 tγ ω 2 (ξ) γ 2 (t ǫ)γ. Proof Snce t < 1, there exst ntegers k and l wth 0 < l < k and t ǫ l/(k+2) l/k < t. For such a choce of k and l, consder the Fbonacc sequence (w ) 0 of Example 3.3 wth parameters a = 2 l, b = 2 k l 1 and c = 2 k l. Accordng to Example 4.3, w has relatvely prme trace and determnant for each 0 and the correspondng sequence of symmetrc matrces (y ) 0 satsfes det(y 0, y 1, y 2 ) = 2 4l 0. Moreover, Example 5.4 shows that (w ) 0 s unbounded and satsfes the estmates (24) of Proposton 7.1 wth α = l/(k + 2) and β = l/k (note that the example provdes a slghtly larger value for α). So, Proposton 7.1 apples and shows that the correspondng real number ξ constructed by Proposton 6.1 satsfes the above condton (b). In partcular, ξ s transcendental snce ω 2 (ξ) > 2. Moreover, snce (ξ, 1)w (ξ, 1)y y y, the frst estmate n (15) leads to (a).

14 224 D. Roy Acknowlegments The author warmly thanks Yann Bugeaud for pontng out the results of Jarník n [7] whch brought a notable smplfcaton to the present paper. References [1] J.-P. Allouche, J. L. Davson, M. Queffélec, L. Q. Zambon, Transcendence of Sturman or morphc contnued fractons. J. Number Theory 91(2001), no. 1, [2] B. Arbour and D. Roy, A Gel fond type crteron n degree two. Acta Arth. 11(2004), no. 1, [3] Y. Bugeaud and M. Laurent, Exponents of Dophantne approxmaton and sturman contnued fractons. Ann Inst. Fourer (Grenoble) 55(2005), no. 3, [4] H. Davenport and W. M. Schmdt, Approxmaton to real numbers by quadratc rratonals. Acta Arth. 13(1967), [5], Approxmaton to real numbers by algebrac ntegers. Acta Arth. 15(1969), [6] S. Fschler, Spectres pour l approxmaton d un nombre réel et de son carré. C. R. Acad. Sc. Pars 339(2004), no. 10, [7] V. Jarník, Zum Khntchneschen Übertragungssatz. Trudy Tblsskogo mathematcheskogo nsttuta m. A. M. Razmadze = Travaux de l Insttut mathématque de Tblss 3(1938), [8] D. Roy, Approxmaton smultanée d un nombre et de son carré. C. R. Acad. Sc., Pars 336(2003), no. 1, 1 6. [9], Approxmaton to real numbers by cubc algebrac ntegers. I. Proc. London Math. Soc. 88(2004), no. 1, [10], Approxmaton to real numbers by cubc algebrac ntegers. II. Ann. of Math. 158(2003), no. 3, [11], Dophantne approxmaton n small degree. In: Number Theory, CRM Proceedngs and Lecture Notes 36, Amercan Mathematcal Socety, Provdence, RI, 2004, pp [12] W. M. Schmdt, Dophantne Approxmaton, Lecture Notes n Mathematcs 785, Sprnger-Verlag, Berln, Département de Mathématques Unversté d Ottawa 585 Kng Edward Ottawa, ON K1N 6N5 e-mal: droy@uottawa.ca