Approximation by polynomials with bounded coefficients

Journal of Number Theory 27 (2007) 03 7 www.elsevier.com/locate/jnt Approximation by polynomials with bounded coefficients Toufik Zaimi Département de mathématiques, Centre universitaire Larbi Ben M hidi, Oum El-Bouaghi 04000, Algérie Received 24 June 2006; revised 2 December 2006 Available online 30 March 2007 Communicated by David Goss Abstract Let θ be a real number satisfying <θ<2, and let A(θ) be the set of polynomials with coefficients in {0, }, evaluated at θ. Using a result of Bugeaud, we prove by elementary methods that θ is a Pisot number when the set (A(θ) A(θ) A(θ)) is discrete; the problem whether Pisot numbers are the only numbers θ such that 0 is not a limit point of (A(θ) A(θ)) is still unsolved. We also determine the three greatest limit points of the quantities inf{c, c > 0, c C(θ)}, wherec(θ) is the set of polynomials with coefficients in {, }, evaluated at θ, and we find in particular infinitely many Perron numbers θ such that the sets C(θ) are discrete. 2007 Elsevier Inc. All rights reserved. MSC: R06; Y60; C08 Keywords: Polynomial approximation; Beta-expansion; Pisot numbers. Introduction For a real number θ> and a rational integer m, let A m = A m (θ) = { ε 0 + ε θ + +ε n θ n,n N, ε k {0,,...,m} }, where N is the set of positive rational integers, and let E-mail address: toufikzaimi@yahoo.com. 0022-34X/$ see front matter 2007 Elsevier Inc. All rights reserved. doi:0.06/j.jnt.2007.0.04

04 T. Zaimi / Journal of Number Theory 27 (2007) 03 7 B m = B m (θ) = A m A m = { ε 0 + ε θ + +ε n θ n,n N, ε k { m, m +,...,m} }. Several authors have studied the distribution of the elements of the sets above in the real line R (see for instance [2 7,5]), and considered in this context the quantities β m = β m (θ) = inf{b, b B m,b>0}. It is clear that A m is uniformly discrete if and only if β m > 0, or equivalently if and only if 0 is not a limit point of B m. Recall that a subset X of R is uniformly discrete if the usual distance between two distinct elements of X is greater than a positive constant depending only on X; a uniformly discrete set is a discrete set, that is a set with no finite limit point. Notice also by the relation B 2m = B m B m, that the two propositions: the sets B m are uniformly discrete for all m, and the sets B m are discrete for all m, are equivalent, since we have β 2m > 0 when B 2m is discrete, and B m is uniformly discrete if and only if β 2m > 0. As usual, Pisot numbers were the first algebraic integers θ which had been considered in such a problem. A Pisot number is a real algebraic integer greater than whose other conjugates over the field of the rationals Q are of modulus less than. By the Pigeon-hole principle, it is easy to check (see also [3,8] and [6]) that the sets B m are discrete for all m when θ is a Pisot number. Bugeaud [3] was the first to show that the converse of the last sentence is also true: Theorem A. If B m is discrete for each m, then θ is a Pisot number. In fact Bugeaud proved that θ is a Pisot number, when the equivalent condition: β m > 0 for all m, holds. The proof of Theorem A uses a result of Frougny [9] from automata theory, and does not provide any estimation for m. Generalizing some former results of Frougny [9] and Erdős, Joó and Schnitzer [7], Erdős and Komornik [8] obtained an improvement of Theorem A: Theorem B. Let m be the smallest positive rational integer satisfying m θ θ.ifb m is discrete, then θ is a Pisot number. In these pages we denote by x and {x} the integer and the fractional parts of a real number x, respectively ( x is the greatest rational integer less than or equal to x and {x}=x x ). The first aim of this paper is to show: Theorem. If (B θ A θ ) is discrete, then B m is discrete for each m. Clearly, we can infer by Theorem and Theorem A that θ is a Pisot number when the set (B θ A θ ) is discrete; in particular, if θ<2and (A(θ) A(θ) A(θ)) is discrete, where A(θ) is the set of polynomials with coefficients in {0, }, evaluated at θ, then θ is a Pisot number. It is worth noting that Theorem (together with Theorem A) improves Theorem B for θ ] + 5 2, 2[, as the inclusion B A B 2 is strict for a.e. θ ] + 5 2, 2[ (for instance we have that 2 B 2, and if 2 B A then θ is a root of a non-zero polynomial with rational integers coefficients of modulus at most 4). We prefer to state Theorem in the general case for the simplicity of its proof. In Section 2, we shall also show the following weaker form of Theorem A without using automata theory: If B m is discrete for each m, then θ is a Pisot or a Salem number. Recall that a Salem number is an algebraic integer greater than, whose other conjugates over Q are of modulus at most and with a conjugate of modulus.

T. Zaimi / Journal of Number Theory 27 (2007) 03 7 05 Now, assume θ<2 and consider the set C = C(θ) = { ε 0 + ε θ + +ε n θ n,n N, ε k {, } }. Then, we have C = C B and the set C is uniformly discrete when θ is a Pisot number. It has been proved in [2] that C is dense in R for a.e. θ ] 2, 2[. Further, if θ ], 2] and is not a root of a polynomial with coefficients in {, 0, }, then the set C is also dense in R. In somewhat the opposite direction, Borwein and Hare [2] found a family of Salem numbers θ such that the corresponding sets C are discrete. They also exhibit a finite set of Perron numbers that are not Pisot nor Salem numbers, with the same property. A Perron number is a real algebraic integer θ>whose other conjugates over Q are of modulus less than θ. In their proof Borwein and Hare used a simple algorithm to determine the elements of the set C ]0, θ [ (the same algorithm was used in [2] and in [6] to determine the elements of B m ]0, θ m [ for some Pisot numbers θ), and the following discreteness test: Theorem C. Let θ be a real number satisfying <θ<2. Then, the set C (respectively, B m ) is discrete if and only if C [0, θ ] (respectively, B m [0, θ m ]) is finite. We shall use the same arguments in the proof of the next result to determine whether the sets C(θ) are discrete for some classes of Perron numbers θ, and also to compute the quantities when the corresponding sets C(θ) are discrete. γ = γ(θ)= inf { c, c C(θ), c > 0 }, Theorem 2. Let θ be a real number satisfying <θ<2. Then, the possible values of γ greater than are θ ( 0 i i ), θ, θ +, n+ 0 i n ( θ 2 i ) and n+ + ( θ 2 i ), 0 i n where n is a non-negative rational integer. Moreover, each of the equalities γ(θ)= θ, γ(θ)= θ θ+, γ(θ)= and γ(θ)= θ hold for infinitely many Perron numbers θ for which the sets + C(θ) are discrete. From the proof of Theorem 2 we easily deduce: Corollary. The three greatest limit points of the set {γ(θ), θ ], 2[} are 2, 3 and 4 ( 2 and 3 are both right hand limit points and 4 is a left hand limit point).

06 T. Zaimi / Journal of Number Theory 27 (2007) 03 7 The solutions θ ], 2[ of the inequality β (θ) > 2 have been determined in Theorem 3 of [5]. The same result asserts also that the implication β (θ) 2 β (θ) < 2 5, is true. By Theorem 2 we have: Corollary 2. If β (θ) 2, then β (θ) θ. The proofs of Theorems and 2 appear in Sections 2 and 3, respectively. The proof of Theorem uses elementary properties of the beta-expansion of a real number, and the proof of Theorem 2 is inductive. We also show in Section 4: Theorem 3. The set B m is discrete if and only if B m [0, θ+ ] is finite. It is clear when θ<2 that Theorem 3 improves Theorem C for the sets B m. The proof of Theorem 3 follows essentially from the ones of Theorems and 2. All the computations in the paper were performed using the computer system Pari []. 2. Proof of Theorem It is clear that B m is discrete when θ N, since a subset of a discrete set is discrete and B m Z, the ring of the rational integers. So, assume that B θ A θ is discrete and θ/ N. Notice also by the relations B θ = B θ {0} B θ A θ that B m is discrete when m θ, as B m B θ. To prove the result for m> θ, we shall only use the relations B k θ B θ + F k, () where k is rational integer greater than, and F k is a finite subset of R depending on k and θ.to show that the inclusion () is true for k = 2, we first recall the definition of the beta-expansion of a real number. Following [0], let x be a positive real number, and let p = p(x) Z be such that θ p x< θ p+. Then, the beta-expansion of x in base θ, or simply the beta-expansion of x, is the sequence (ε k ) k p = (ε k (x)) k p defined by the relations ε p = θ x p, r p = r p (x) ={ θ x p }, and ε k = θr k+ and r k = r k (x) ={θr k+ } for k running through the set of the rational integers less than p.inthis casewehave x = ε p θ p + ε p θ p + +ε 0 + ε θ + ε 2 θ 2 +, ε k {0,,..., θ } and r k [0, [. Now, let b B θ [, [ and let p = p(b). If the beta-expansion of b is the sequence (ε k ) k p, then b = ε p θ p + ε p θ p + +ε 0 + ε θ + ε 2 θ 2 +, ε p θ p + ε p θ p + +ε 0 A θ and so the number b (ε p θ p + ε p θ p + +ε 0 ) belongs to the finite set E := (B θ A θ ) [0, [,

T. Zaimi / Journal of Number Theory 27 (2007) 03 7 07 since (B θ A θ ) is discrete and b (ε p θ p + ε p θ p + +ε 0 ) = r 0 (b). Consider an element d B 2 θ. It is clear that d can be written as d = b b, for some elements b and b of the set B θ.letn be a sufficiently large rational integer so that θ N + b and θ N + b belong to B θ [, [. By the above, there are a A θ, r E, a A θ and r E such that b + θ N = a + r and b + θ N = a + r. It follows that d = b + θ N ( b + θ N ) = ( a a ) + ( r r ) (A θ A θ ) + (E E) = B θ + (E E) and so () is satisfied with F 2 := E E. Assume that () is true for some k 2. Then, by the relations B (k+) θ = B k θ + B θ B θ + F k + B θ = B 2 θ + F k B θ + (F k + F 2 ), we see that the inclusion () is true for k + with F k+ := F k + F 2. Now, by () we have immediately that B k θ is discrete for each k 2. Indeed, otherwise there is a convergent sequence of distinct elements of B k θ,say(d n ) n N. Since F k is finite and each term of (d n ) n N can be written as d n = b n + f n, where b n B θ and f n F k, we can extract from (d n ) n N a subsequence of the form (b n + f) n I, where f {f n, n N} and I is an infinite subset of N, and so we obtain a convergent sequence of distinct elements of B θ, namely the sequence (b n ) n I ; this is absurd because B θ is discrete. Hence, B k θ is discrete and so are all the sets B m, since for each m there is k N such that m k θ. Remark. Next we give two simple proofs of the following weaker form of Theorem A: If B m is discrete for each m, then θ is a Pisot or a Salem number. The first proof uses a result due to Schmidt [4] and the second one a theorem of Parry s []. Recall that if the beta-expansion (to base θ) of a positive real number x is the sequence (ε k ) k p, then ( ε k ) k p is the betaexpansion of x, and we say that the real x has a periodical expansion if its beta-expansion is eventually periodic. Let Per(θ) be the set of numbers having periodical expansions. Then, Per(θ) Q(θ), and by Theorem 2.4 of [4] we have that if Q(θ) = Per(θ) then θ is a Pisot or a Salem number. Recall also that the number θ is said to be a beta-number when {θ} Per(θ). In [] it has been shown that a beta-number is an algebraic integer and the other conjugates of a beta-number over Q are of modulus less than 2. To make the notation easier (as in [] and [3]), we let the beta-expansion of a real number α satisfying θ α< to be the sequence (ε n) n N, where ε n =[θr n ] and r n ={θr n } when n N, and r 0 = α. Then, r n = ε n+ θ +ε n+2 θ 2 + ε n+3 θ 3 +, α = ε θ + ε 2 θ 2 + +ε n θ n + r n θ n (2) and the sequence (ε n ) n N is eventually periodic if and only if the set {r n, n 0} is finite. Finally, suppose that θ/ N, since θ is a Pisot number when it is a rational integer.

08 T. Zaimi / Journal of Number Theory 27 (2007) 03 7 The first proof. Let k N, k 2, and let l be the greatest rational integer such that θ l <k. Then, θ l k [ θ l θ, [, and by (2) we have for α := k, kr n = θ l+n k ( ε θ n + ε 2 θ n 2 + +ε n ). Hence, there is an m N independent of n (we may choose m = k θ ) such that kr n B m (θ) for all n, and so {kr n, n 0} is a subset of the finite set B m (θ) [0,k[. It follows that {r n, n 0} is finite and α Per(θ). Moreover, there are u and v such that u>vand kr u = kr v, and so θ is a root of the polynomial x l+u k ( ε x u + +ε u ) x l+v + k ( ε x v + +ε v ). Thus, θ is an algebraic integer of degree, say d, over Q and Q(θ) ={P(θ), P Q[X],deg(P ) d }. To prove the inclusion Q(θ) Per(θ), it suffices to show that Q(θ) [ θ, [ Per(θ), since by definition 0 Per(θ), x Per(θ) when x Per(θ), and if θ p x<θ p+ for some p Z then θ θ p x<, θ p x Q(θ) when x Q(θ) and the beta-expansion of x and θ p x are identical. Let α Q(θ) [ θ, [. Then, α can be written as α = n 0 + n θ + +n d θ d k for some n 0,n,...,n d Z and k N, and so by (2) we have kr n = ( n 0 + n θ + +n d θ d ) θ n k ( ε θ n + ε 2 θ n 2 + +ε n ) ; thus kr n B m (θ), where m = max{ n 0, n,..., n d, k θ }, and similarly as for the case where α = θ a k we easily obtain that α Per(θ). After this we use Schmidt s result to infer that θ is a Pisot or a Salem number. Finally, recall that the question whether Pisot numbers are the only numbers θ satisfying the relation Per(θ) = Q(θ), remains open. The second proof. Let k N. Then, the set B θ k (θ k ) is discrete, since B θ k (θ k ) B θ k (θ) and B θ k (θ) is discrete. Considering the beta-expansion of {θ k } to base θ k, we obtain identically as in the first proof that θ k is a beta number. Hence, θ is an algebraic integer with no other conjugate over Q of modulus greater than, since otherwise we obtain a contradiction with Parry s result when k is large. 3. Proof of Theorem 2 and its corollaries Proof of Theorem 2. To make the proof clear we consider the cases corresponding to the greatest limit points of the set {γ(θ), θ ], 2[} separately. Step. We show that C ]0, θ ], and solve the equation γ(θ)= θ.

T. Zaimi / Journal of Number Theory 27 (2007) 03 7 09 Consider the function T 0 (x) = θx in the real variable x. It is clear that T 0 (C) C and T 0 (] n k= n+ θ k, k= ]) ] n θ k k= n θ k, k= ] θ k, where n N. Since ] [ C θ, θ and θ = k θ k, by iterating the map T 0 we obtain that there exists n N such that T (n) 0 () C ]0, θ ], and so C ]0, θ ]. To find the numbers θ satisfying the equation γ(θ)= θ, we shall use the polynomials f n (x) = x n x n x, where n 2. It is known (see [5] and [5]) that f n is the minimal polynomial of a Pisot number, say q n, the sequence (q n ) n 2 is increasing towards 2 with q 2 =.68..., and the real function f n (x) is increasing on the interval [ q n, [ ; by convention q := and f (x) := x. Let θ be such that γ(θ)= θ. Then, C ]0, θ [= and so θ q 2, since θ C (and θ θ ). Further, if θ [q n,q n+ [ for some n 2, then by the relations f n (x) = xf n (x), f n (θ) C and 0 f n (θ) < θ, we have f n (θ) = 0. Hence, θ = q n and so C is uniformly discrete. Now, we use the algorithm of [2] to determine the elements of C ]0, θ [, where θ = q n. Clearly, by the inequalities q k <θ, where k {, 2,...,n }, wehavef k (θ) > 0 and f k (θ) + 2 > 2 > θ ; thus the only element of C ]0, θ [ with degree k (as a polynomial in θ) is f k(θ) and so the set C ] θ, θ [ contains exactly one element with degree n, namely f n (θ) = 0. Hence, ] [ C 0, = {,f (θ), f 2 (θ),..., f n (θ) }, θ and γ(θ)= θ, since >f (θ) > f 2 (θ) > >f n (θ) = θ. Step 2. We prove that C ]0, γ(θ)= θ+. θ+ ] when θ/ F 0 := {q n, n 2}, and solve the equation

0 T. Zaimi / Journal of Number Theory 27 (2007) 03 7 Assume θ/ F 0. Then, C ]0, θ [. By the relations T 0 (C) C and (] T 0 θ +, [) ] [ 0,, θ θ + we find that C ]0, θ+ ], and γ(θ)= θ+ if and only if C ]0, θ [={ θ+ }.Now,letθ be such that γ(θ)= θ+. Then, θ>q 2, because the inequality θ < θ yields θ = θ+, and the set C( 2) is dense in R. Further, if θ ]q n,q n+ [ for some n 2, then we have 0 <f n (θ) < θ and so f n (θ) = θ+ ; thus θ is a root of the polynomial g n+ (x) = (x + ) ( x n x n x ) = x n+ 2 ( x n + x n 2 + +x + ). It is clear that the polynomial g n+ is irreducible over Q, as it is a 2-Einstein polynomial, and can also be written, for x, as since g n+ (x) = f n (x) + f n+ (x) = xn (x 2)(x + ) + 2, x f n (x) = xn (x 2) +. x It follows by the relations g n+ (q n+ ) = f n (q n+ ) = q n+ > 0 and g n+ (q n ) = f n+ (q n ) = < 0 that g n+ has a real root, say r n+, such that q n <r n+ <q n+ ; so the sequence (r n ) n 3 is increasing towards 2 with r 3 =.769... Moreover, if we fix δ ], 2[ and choose N N so that δ N (2 δ)(δ )>2, then we have on the circle z =δ (in the complex plane) that z n (z 2)(z + ) > 2 when n N. It follows by Rouché s theorem that the polynomial x n (x 2)(x + ) + 2 = (x )g n+ (x) has n + roots in the disc z <δ, and so the polynomial g n+ (x) has n roots with modulus less than δ. A short computation shows that we can choose N = 5forδ = q 2 ; thus the polynomial g n+ has exactly n roots with modulus less than q 2 when n 5, and the remaining root, which is r n+, satisfies r n+ >q n q 2. Hence, the conjugates of the algebraic integer r n are of modulus less than r n for each n 6. Directly we verify that r 3, r 4 and r 5 are also Perron numbers (it is easy to see that for any <δ<2, there is N N such that the conjugates of the Perron number r n are in the annulus δ < z <δfor all n N). Note also that if u and v are two positive roots of the polynomial g n+, where u v, then ( u = 2 u + u 2 + + ) ( u n 2 v + v 2 + + ) v n = v

T. Zaimi / Journal of Number Theory 27 (2007) 03 7 and so u = v; thus r n+ is the only root of g n+ in ], 2[. Similarly as in the case above, by the relations g n (x) = f n (x) + f n (x), where n 3, and r n >q n q 2, we easily obtain for θ = r n that ] C 0, [ = {,f (θ), f 2 (θ),..., f n (θ) } ; θ thus γ(θ)= f n (θ) = θ+, and by Theorem C we have that the set C is discrete (we will see in the proof of the corollaries that β (θ) = 0; so by the relation 2B C C, thesetc is not uniformly discrete). Finally, notice that the number r n has at least a conjugate of modulus > (respectively, has no conjugate of modulus ), because r n < 2 and r n has norm 2 (respectively, because r n is not a unit); thus r n is not a Pisot nor a Salem number. Step 3. Let F 0 ={r n, n 3}. We show that C ]0, θ ] when θ/ F 0 F 0, and prove that the equation γ(θ)= θ holds at least for two families of Perron numbers θ. Let θ/ F 0 F 0. Then, C ]0, θ+ [. Similarly as in Step, by iterating the real function when it is necessary and when C ] θ, T (] n k= n+ θ k, k= T (x) = x θ + θ+ θ [, we obtain that C ]0, ], since T (C) C, ]) θ k ] n k= θ n k, k= ] θ k, where n N, and k θ k = θ +. Moreover, we have as γ(θ)= θ ] if and only if C 0, [ { n θ + k= } θ k,n N, T (x) n k= θ k n+ when x k= θ k. Now, let θ be such that γ(θ)= θ. It is clear when θ< 2 that θ < θ +, θ = n k= θ k

2 T. Zaimi / Journal of Number Theory 27 (2007) 03 7 for some n 2, and θ = q n.ifθ> 2, then there is n 2 such that θ ]r n,q n [, where r 2 := 2, or θ ]q n,r n+ [, and by the same arguments as in the above cases we obtain that there is m N such that θ is respectively a root of one of the polynomials or h + m,n (x) = x2m f n (x) + x2m x + h m,n (x) = x2m f n (x) x2m x +. By Theorems 5.3 and 5.4 of [2], we see that the polynomial h,n has only one root of modulus greater than, say s,n, and s,n is a Salem number such that γ(s,n ) = s,n (s,n and A(s )2,n ) is discrete. Notice also that Now, consider the polynomial q n <s,n <r n+. h +,n (x) = x2 f n (x) + x. It is clear that h +,n (q n) = q n > 0. Further, by the identities xg n (x) = xf n (x) + xf n (x) = (x + )f n (x) + wehaveh +,n (r n) = r 2 n f n(r n ) + r n = r n + < 0; thus the polynomial h+,n has a real root, say s +,n, satisfying r n <s +,n <q n, and the sequence (s +,n ) n 2 is increasing towards 2. Writing h +,n (x) = x2 f n (x) + x = xn+2 (x 2) + (2x 2 2x + ), x we obtain identically as for the polynomials g n that the roots of h +,n other than s+,n are of modulus less than r 2 ; so the other conjugates of s +,n over Q are of modulus less than s+,n, and s +,n is a Perron number. Similarly as for the case where θ = q n, a short computation shows when θ = s +,n and n 3 that ] C 0, [ { = θ thus C is discrete and γ(θ)= θ, since θ>r 3 and,f (θ),...,f n (θ), f n (θ) = θ, θ θ >f n (θ) = f n(θ) + = θ + θ θ 3 > θ. } ;

T. Zaimi / Journal of Number Theory 27 (2007) 03 7 3 Finally, notice when θ = s +,2 =.52... that 0 <θ6 θ 5 θ 4 θ 3 + + θ + = 0....< θ = 0.2... Step 4. Let F ={θ ], 2[, γ(θ)= θ }. We prove that C ]0, θ + ] when θ/ F 0 F 0 F, and show that there are infinitely many Perron numbers θ satisfying γ(θ)= θ +. Let θ/ F 0 F 0 F. Then, C ]0, θ [. By the relations T (C) C and (] θ T +, θ [) ] 0, θ [, + we obtain C ]0, θ θ ], and γ(θ)= + + such that γ(θ)= θ + and θ ]q n,s,n θ if and only if C ]0, [={ θ }.Letθ be + [, where n 3 (we will see that such a θ exists). Then, h,n (θ) < 0, as h,n () = f n()<0 and h,n has no root in ],s,n [, 0 <f n(θ) < θ f n (θ) = θ ; thus θ is a root of the polynomial + From the identities l n (x) = x n+ x n 2 ( x n 2 + x n 3 + +x + ). xl n (x) = ( x 2 + ) f n (x) (x ) = f n (x) + h,n (x), we have l n (q n )<0, l n (s,n ) = f n(s,n )>0 and so l n has a real root, say t n, satisfying q n <t n <s,n ; and so thus the sequence (t n ) n 3 is increasing towards 2 with t 3 =.873...By the same arguments as in the above cases, we obtain that t n is a Perron number and ] [ { C(t n ) 0, =,f (t n ),...,f n (t n ), f n+ (t n ) = t } n + t n tn 2 +. Hence, C(t n ) is discrete and γ(t n ) = f n (t n ) = t n t 2 n +. Step 5. We use induction to complete the proof. Let F n and F n be the sets of the numbers θ satisfying γ(θ)= 0 i n (θ 2i ) 0 i n and γ(θ)= (i ), n n +

4 T. Zaimi / Journal of Number Theory 27 (2007) 03 7 where n N, respectively. It suffices to show that the following two propositions ] n ( P n : ifθ/ Fi F i ) 0 i n Fn, then C(θ) 0, (θ ] 2i ) n + i=0 and P n : n ] if θ/ ( Fi F ) 0 i n, then C(θ) 0, (θ ] 2i ) i=0 i n+ are true for all n.fromstep4,p is true. Further, if θ/ F 0 F 0 F F By iterating the map T 2 (x) = θ 4 x (θ ) ( ) = θ 4 x θ 3 + + θ when C ] (θ )( ), θ θ 4 + [, we obtain that P is true, since T 2(C) C, T 2 (] n k= (θ )( ) n+ θ 4k, k= ]) (θ )( ) θ 4k then C ]0, θ + [. is contained in ] n k= (θ )( ) θ 4k, n k= ] (θ )( ) θ 4k, where n N, and Identically, by considering the map k T n (x) = n+ x + ( ) θ 4k = +. 0 i n ( θ 2 i ), we easily show that the propositions P n+ and P n+ are true, when P n and P n are so. The relation T n (C) C follows from the fact that the polynomial 0 i n (x2i ) has its coefficients in {, } and is of degree 2 n+. Proof of the corollaries. From the proof of Theorem 2, we have when θ/ F 0 F 0 F that C ]0, θ ] and so + β (θ) γ(θ) θ { + < min 5, θ }.

T. Zaimi / Journal of Number Theory 27 (2007) 03 7 5 Moreover, if θ F 0 (respectively, θ F 0 ) then θ = q n for some n 2 and γ(q n ) = q n (respectively, θ = r n for some n 3 and γ(r n ) = r n ) tends to 2 (respectively, to 3 ) when n tends to infinity; for θ F,wehaveγ(θ)= θ < 4, and in particular when θ = s+,n (or θ = s,n ), γ(θ) tends to 4 when n tends to infinity (I am not able to determine whether 4 belongs to the second derived set of {γ(θ), θ ], 2[}). Recall also that the equality β (θ) = θ when θ F 0, has been proved in many places and firstly in [5]. Finally, if θ F 0 then by Remark 2 of [3] we have β (θ) = 0, since θ is not a root of a polynomial with coefficients in {, 0, }. Remark 2. Let θ =.7548... be the Pisot number root of x 3 2x 2 + x (θ is the square of the smallest Pisot number). Then, β (θ) = θ and so Corollary 2 is optimal. From the proof of Theorem 2 we have that the solutions of the equality β (θ) = θ are among the numbers q n, where n 2, and the roots, say s m,n ±,ofh+ m,n and h m,n which belong to ], 2[. To determine whether 4 is a limit point of {β (θ),θ ], 2[} it suffices to consider the numbers s m,n ±, since we have β ( q n ) = 0 by the following proposition: If p N and θ p / Q(θ), then β (θ p ) = 0. Indeed, with the notation of Remark, let (ε n ) n N be the beta-expansion to base θ of α, where α = θ p, and let r n = ε n+ θ + ε n+2 θ 2 + ε n+3 θ 3 +, where n 0. Then, α = + ε θ + ε 2 θ 2 + + ε n θ n + r n θ n, r n [0, [, and the set {r n, n 0} is not finite because α/ Q(θ) and Per(θ) Q(θ). It follows from the last equality that r n = α np+ α np ε α np p ε 2 α np 2p ε n α p ε n B (α) and so B (α) has a limit point, say l. Let(r nk ) k N be a subsequence of (r n ) such that n <n 2 < n 3 <,r ni r nj when i j, and lim r nk = l. Then, lim(r nk+ r nk ) = 0, and r nk+ r nk B (α), asp 2 and p does not divide in Z the numbers n k p +. 4. Proof of Theorem 3 Note first by Lemma 2.(b) of [8] (or by Proposition (i) of [6]) that Theorem 3 is true when m θ, since in this case the set B m is discrete. So, assume m θ. By definition we have that B m [0, θ+ ] is finite when B m is discrete. To prove the converse, we shall first show that B m is discrete when the set B m [0, ] is finite. Since B m = B m, it suffices to prove that each finite subinterval, say [0,ε], of[0, [ contains at most a finite number of elements of B m.let B m ]0, ]={b,b 2,...,b k }. We will show that each b B m [0, [ can be written as b = i k n i b i, (3) for some non-negative rational integers n,n 2,...,n k. Indeed, if (3) is true, then for each i {, 2,...,k} we have n i n i b i β m b β m,

6 T. Zaimi / Journal of Number Theory 27 (2007) 03 7 as β m = min{b,b 2,...,b k }, and so n i {0,,..., β ε m } when b [0,ε] B m ; thus there are at most ( + β ε m ) k elements of B m in [0,ε] (this is a quantitative version of what we want to prove). Now, let b B m ], [. Since B m = B m θ + B θ = B m θ + A θ A θ, there exist z B m θ + A θ and a A θ such that b = z a; by convention B 0 + A θ = A θ. Considering the beta-expansion of the number a +, which satisfies a + <z, we deduce that there is a A θ such that a + a [0, [; thus a A θ ]a,a + ] and so a A θ ]a,z[. Letb = a a and b = z a. Then, and b = b + b, b B θ ]0, ] B m ]0, ] b B m ] 0,b b [ B m ]0,b β m [. (4) It follows when b that b is a sum of two elements of B m ]0, ]; otherwise we repeat the same process for b instead of b, and by induction we obtain (3). By the relation (4), the process must terminate. To complete the proof of Theorem 3 it is enough to verify that the following two assertions are true: B m has a limit point in [ θ, ] B m has a limit point in [0, θ ], and B m [ θ+, θ ] is not finite B m [0, θ+ ] is not finite. The first proposition follows easily by iterating the continuous real function T 0 defined in the proof of Theorem 2 (Step ), since T 0 (B m ) B m and T 0 (l) is a limit point of B m when l is so, and the second implication is immediate by considering the injective map T 0. Remark 3. The constant θ+ in Theorem 3 is not certainly the best one (in fact the initial aim was to prove that B m is discrete when 0 is not a limit point of B m ). For example by considering the functions T 2 (x) (defined in the proof of Theorem 2) and f(x)= x in the real variable x, we obtain when θ<2that B m is discrete if and only if B m [0, θ ] is finite. Finally, notice + that Theorem 3 is also true for sets of the form (A ± A ± ±A ) A θ (like the one in Theorem ). Acknowledgment The author would like to thank the referee for careful reading of the paper. References [] C. Batut, D. Bernardi, H. Cohen, M. Olivier, Pari Calculator, copyright 989, 992. [2] P. Borwein, K.G. Hare, Some computations on the spectra of Pisot and Salem numbers, Math. Comp. 7 (238) (2002) 767 780. [3] Y. Bugeaud, On a property of Pisot numbers and related questions, Acta Math. Hungar 73 (996) 33 39. [4] P. Erdős, I. Joó, M. Joó, On a problem of Tamás Varga, Bull. Soc. Math. France 20 (992) 507 52. [5] P. Erdős, I. Joó, V. Komornik, Characterization of the unique expansion = i qn i and related problems, Bull. Soc. Math. France 8 (990) 377 390.

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