Minimal polynomials of some beta-numbers and Chebyshev polynomials

Minimal polynomials of some beta-numbers and Chebyshev polynomials DoYong Kwon Abstract We consider the β-expansion of 1 which encodes a rational rotation on R/Z under a certain partition Via transforming its β- polynomial into a linear combination of Chebyshev polynomials, we specify the minimal polynomial of β over the field of rationals At the end, some number theoretical conjectures are posed with computational evidence to support them 1 Introduction For β > 1, the β-transformation T β : x βx (mod 1) is multiplication by β modulo 1 Then the β-expansion d β (x) = (x i ) i 1 of x [0, 1] is defined by x i = βt i 1 β (x) While Rényi [14] introduced the β-transformation and proved that it is ergodic, Parry [1] found its invariant measure and characterized possible sequences that can be a β-expansion It was shown that the β-expansion of 1 specifies any other β-expansion of x [0, 1) in terms of lexicographic order To be more precise, let us fix β > 1 and assume that s = x 1 x is an infinite word over an alphabet {0, 1,, β } In addition, putting d β(1) := lim d β (1 ɛ), ɛ 0 we have d β (1) = (e 1 e m 1 (e m 1)) ω if d β (1) = e 1 e m 0 ω and d β (1) = d β (1) otherwise Here a ω := aa Then Parry showed that s = d β (x) for some x [0, 1) if and only if σ n (s) < d β (1) for all n 0, where σ is the shift If d β (1) is eventually periodic, say d β (1) = e 1 e n (e n+1 e n+p ) ω, then we call β a beta-number In this case, β is the dominant root of ( ) ( ) n+p x n+p e i x n+p i x n e i x n i = 0 000 Mathematics Subject Classification: Primary 11R06, 11R09; Secondary 37B10, 68R15 Key words: beta-numbers, Chebyshev polynomials, Pisot numbers, Mahler measure 1

In particular if d β (1) is finite, ie, d β (1) = e 1 e n 0 ω, then we call β a simple beta-number and β is the dominant root of x n e i x n i = 0 These two polynomials above are called the beta-polynomial of each β, or simply the β-polynomial if β is clear from the context The term characteristic polynomial is also used in the literature The special class of β-expansions will be focused on We study a real β > 1 for which d β (1) encodes a rational rotation on R/Z In this case, β is an algebraic integer For the case of irrational rotation, see [4, 6] Arithmetic study was also pursued there and the current work is in fact motivated by these previous papers Let α > 0 and ρ [0, 1] We consider the following two infinite sequences called lower and upper mechanical words with slope α and intercept ρ: for n 0, s α,ρ (n) = α(n + 1) + ρ αn + ρ, s α,ρ(n) = α(n + 1) + ρ αn + ρ Note that these are infinite words over the alphabet A = { α 1, α } For an irrational α, the words are aperiodic and called Sturmian words On the other hand a rational α produces purely periodic words, whose shortest period words are called Christoffel words Our main concern is the latter We need to describe this in more detail Let the slope be α = p/q > 0 with gcd(p, q) = 1 and b = α Then the working alphabet is A = {b 1, b} and the lengths of the Christoffel words, say, t p,q, t p,q are q So one has s α,0 = t ω p,q and s α,0 = t p,qω They are also represented by t p,q = (b 1)z p,q b, t p,q = bz p,q (b 1), for some word z p,q, called a central word We see that z p,q is a palindrome, ie, z p,q is equal to its reversal See [11] The motivation of this paper originated from the next proposition For any word w = a 0 a 1 a n 1 with a i Z, we denote by w, a vector (a 0,, a n 1 ) Z n Proposition 11 ([4, 6]) For α > 0, there exists a unique β > 1 for which d β (1) = s α,0 So define : (0, ) (1, ) by (α) = β Then (a) At an irrational α > 0, then is continuous and (α) is a transcendental number

(b) is left-continuous but not right-continuous at every rationals (c) If α = p/q with gcd(p, q) = 1 and b = α, then the (α)-polynomial is x q bz p,q b (x q 1, x q,, 1), (d) With the same rational α as above, let (α+) := lim x α+ (x) Then the (α+)-polynomial is x q+1 bz p,q b (x q, x q 1,, x) x + 1 For a rational α > 0, (α) is called a lower self-christoffel number, and the right limit (α+) an upper self-christoffel number We are now in a position to state our main theorem THEOREM Let p 0 /q 0 be a fixed rational with 0 < p 0 q 0 and gcd(p 0, q 0 ) = 1 For a positive integer b, suppose p/q = b 1 + p 0 /q 0 Then there exists an effectively computable B such that for all b B, we have: (a) x q bz p,q b (x q 1, x q,, 1) is irreducible over Q, (b) x q+1 bz p,q b (x q, x q 1,, x) x + 1 is irreducible over Q Thus the polynomials mentioned are eventually the minimal polynomials of (p/q) and (p/q+) The above theorem was in fact proved in [7] but we did not know from which value b on, the polynomials are irreducible We give here another proof involving Chebyshev polynomials The new proof enables us to find B effectively for which b B implies the corresponding polynomials are irreducible over Q The effective procedure for finding B is demonstrated at page 10 This quantitative results accordingly lead us to some connection with other fields of mathematics, which will be discussed in Section 4 Preliminaries In this section we briefly review some concepts and known results to be used in the main proof First we recall some definitions in number theory Among algebraic integers, a Pisot number is α > 1 all of whose conjugates lie inside the unit circle Suppose g(x) = a n x n + + a 1 x + a 0 = a n n (x α i) Z[x] with a n 0 Then the Mahler measure of g is a positive number defined by n M(g) = a n max{1, α i } 3

In particular, cyclotomic polynomials have 1 as their Mahler measures The most famous problem on Mahler measures, posed by Lehmer [9], is whether or not 1 is an accumulation point of the set {M(g) g Z[x]} Though it is still open, a certain answer can be given for a special subclass of polynomials Let f R[x] and deg(f) = n If f satisfies f(x) = x n f(x 1 ) then we say that f is reciprocal Now the Mahler measures of nonreciprocal polynomials cannot be less than the smallest Pisot number as stated below Theorem 1 ([16]) Let p(x) Z[x] and let θ 0 = 1347 be the smallest Pisot number, ie, the real zero of x 3 x 1 If M(p) < θ 0, then p(x) is a reciprocal polynomial Let us denote by T n and U n the n th Chebyshev polynomials of the first and the second kinds respectively Among many equivalent definitions we adopt the following simple equations, which are suitable for our proof T n (cos θ) := cos nθ, U n (cos θ) := sin(n + 1)θ sin θ For instance, T (x) = x 1 and U 3 (x) = 8x 3 4x In analysis of our polynomials, we encounter some reciprocal polynomials, and they will be converted into more convenient ones to handle via some transformation introduced in [3] Suppose that p(z) = n i=0 a iz i R[z] is a nonzero reciprocal polynomial with a i = a n i, i = 0,, n If a n = a n 1 = = a n+k+1 = 0 but a n+k 0 hold, then we observe that [ ( p(z) = a i z i = z n a n+k z k + 1 ) ( + + a z k n+1 z + 1 ) ] + a n z i=0 = a n+k z n k ( z + 1 ) z α i k = a n+k z n k (z α i z + 1), for some α i C, i = 1,, k Given a reciprocal polynomial p as above, the Chebyshev transform T of p is defined by T p(x) := a n+k k (x α i ) Theorem ([3]) The Chebyshev transform T is a linear isomorphism of the space of real reciprocal polynomials, {p(z) = n i=0 a iz i R[z] : a i = a n i, i = 0,, n} into the space of real polynomials of degree at most n One notes that p(e iθ ) = T p( cos θ) Chebyshev polynomials now naturally appear via Chebyshev transforms 4

Lemma 3 ([8]) If p(z) = z n + 1 and q(z) = z n + z n + + z + 1, then T p(x) = T n (x/) and T q(x) = U n (x/) The following geometric study on beta-polynomials will be useful below The reader should note that all the bounds given in the proposition tend to 1 as β increases Proposition 4 ([7]) (a) If β is an upper self-christoffel number and γ β is a zero of the β-polynomial, then γ β + β + 4β β (b) If β is a lower self-christoffel number and γ β is a zero of the β-polynomial, then β + 1 8β + 1 β γ β + 1 + 8β + 1 β 3 Proof Throughout this section, notations appearing in THEOREM will be used without explicit mention First we consider lower self-christoffel numbers Put f(x) = x q bz p,q b (x q 1, x q,, 1) and suppose that f(x) = g(x)h(x) over Q and g(β) 0 = h(β) Then Theorem 1 together with Proposition 4 implies that g is eventually reciprocal as b increases If γ is a zero of g, then γ 1 is also a zero of g Thus one has γ q = bz p,q b (γ q 1, γ q,, 1) = γ q 1 bz pb,qb (γ 1 q, γ q,, 1) = γ q 1 γ q = γ 1, where we use the fact that bz p,q b is a palindrome So we conclude γ q+1 = 1 Let us assume that q = n + 1 and γ j = e iθ j = e jπi/(q+1) is a zero of g for some j = 1,, q We then consider a polynomial r(z) := z q f(z) = bz p,q b (z n, z n 1,, 1) = b(z n + z n 1 + + 1) a k z k (z n k + 1), a k {0, 1/, 1} Together with an equation z n +z n 1 + +1 = (z n +z n + +z +1)+z(z n +z n 4 + +z +1), 5

Lemma 3 guarantees that the Chebyshev transform of r(z) is ( ( x ) ( x )) T r(x) = b U n + U n 1 ( x ) a k T n k To compute T r( cos θ j ), we use the fact that T n k (cos θ j ) = cos(n k)θ j and that U n (cos θ j ) + U n 1 (cos θ j ) = sin(n + 1)θ j + sin nθ j sin θ j Thus one finds that 1 = T r( cos θ j ) b = sin (n+1)θ j cos θ j sin θ j sin (n+1)θ j sin θ j = a k sin (n+1)θj sin θ j (1) Since 0 < θ j / < π and (n + 1)θ j / = (n + 1)jπ/(n + ) can not be an integer multiple of π for any j = 1,, n+1, this lead us to a contradiction as b increases If q = n, then the polynomial r(z) above is r(z) := z q f(z) = bz p,q b (z n 1, z n,, 1) n 1 = b(z n 1 + z n + + 1) a k z k (z n k 1 + 1), a k {0, 1} If n = 1 here, then we read the summation from 1 to 0 as zero Define a polynomial r 1 (z) by r 1 (z) := r(z) z + 1 = b(zn + z n 4 + + z + 1) n 1 a k z k (z n k z n k 3 + z + 1) Now we find its Chebyshev transform as ( x ) n 1 ( ( x ) ( x )) T r 1 (x) = bu n 1 a k U n k 1 U n k If γ j = e iθ j see that = e jπi/(q+1) is a zero of g for some j = 1,, q, then one can U n k 1 (cos θ j ) U n k (cos θ j ) = sin(n k)θ j sin(n k 1)θ j sin θ j = cos (n k 1)θ j sin θ j sin θ j = cos (n k 1)θj cos θ j () 6

Since γ j 1, we can fix some constant M so that M T r 1 ( cos θ j ) b sin nθ j sin θ j a k cos θ j For any j = 1,, n, the number nθ j = njπ/(n + 1) is never an integer multiple of π and the number θ j / = jπ/(n + 1) can not be π/ Now we get a contradiction for every sufficiently large b This completes Part (a) of THEOREM The crucial ingredient in the previous proof is from splitting given a betapolynomial into a reciprocal polynomial and a monomial This technique also works for upper self-christoffel numbers Put f(x) = x q+1 bz p,q b (x q, x q 1,, x) x + 1 and suppose that f(x) = g(x)h(x) over Q and g(β) 0 = h(β) Then the same reasoning as before shows that g is eventually reciprocal as b increases If γ is a zero of g, then γ 1 is also a zero of g Hence one has γ = γ q+1 bz p,q b (γ q, γ q 1,, γ) + 1 = γ q+1 [γ q 1 bz p,q b (γ q, γ q+1,, γ 1 ) + 1] = γ q+1 γ 1 = γ q, which gives γ q 1 = 1 First, we suppose that q = n 1 and γ j = e iθ j g for some j = 1,, q We then put = e jπi/(q 1) is a zero of r(z) := f(z) + z = z n bz p,q b (z n 1, z n,, z) + 1 = z n + z n 1 + + 1 (b + 1)z(z n + + 1) + a k z k (z n k + 1), a k {0, 1/, 1} The similar arguments to the above derive the Chebyshev transform of r(z) by T r(x) = U n ( x ) ( x ) ( ( x ) ( x )) +U n 1 (b+1) U n 1 + U n + ( x a k T n k Using T n k (cos θ j ) = cos(n k)θ j and Equation (1) one thus finds that sin (n 1)θ j sin (n+1)θ j 1 = T r( cos θ j ) (b + 1) sin θ j sin θ j a k (3) Since 0 < θ j / < π and (n 1)θ j / = (n 1)jπ/(n ) can not be an integer multiple of π for any j = 1,, n 3, we eventually get a 7 )

contradiction as b increases If q = n, then we have r(z) := f(z) + z = z n+1 bz p,q b (z n, z n 1,, z) + 1 = z n+1 + z n + + 1 (b + 1)z(z n 1 + + 1) + a k z k (z n+1 k + 1), a k {0, 1} This can be rewritten in the form, r 1 (z) := r(z) z + 1 = z n + z n + + z + 1 (b + 1)z(z n + z n 4 + + z + 1) + a k z k (z n k z n k 1 + z + 1) Hence we find ( x ) ( x ) T r 1 (x) = U n (b + 1)U n 1 + ( ( x ) ( x )) a k U n k U n k 1 If γ j = e iθ j = e jπi/(q 1) is a zero of g for some j = 1,, q, then one can verify as in Equation () that U n k (cos θ j ) U n k 1 (cos θ j ) = cos (n k+1)θ j cos θ j Since γ j 1, we can fix some constant M so that M T r 1 ( cos θ j ) (b + 1) sin nθ j sin θ j sin(n + 1)θ j sin θ j a k cos θ j For any j = 1,, n, the number nθ j = njπ/(n 1) is never an integer multiple of π and the number θ j / = jπ/(n 1) can not be π/ Now we get a contradiction for every sufficiently large b 4 Discussion and further studies Suppose that an integer b 1 is divisible by a prime p but not by p Then the Eisenstein Criterion shows that the beta-polynomial of ( bq 1 ) is q irreducible for any q 1 Now we introduce another interesting connection between prime numbers and irreducibility for self-christoffel numbers This is related to Mersenne primes, ie, primes of the form n 1 One of famous open questions in number theory is: 8

Are there infinitely many Mersenne primes? We need a classical irreducibility criterion for polynomials in Z[x], which appears for example in [13] Theorem 41 Let β 1,, β n be the zeros of some f(x) Z[x] of degree n If there exists an integer b such that f(b) is prime, f(b 1) 0 and the real parts of zeros satisfy R(β i ) < b 1 for 1 i n, then f(x) is irreducible over Q We consider the beta-polynomials of ( 1) and ( 1 +) Let us denote q q them by fq(x) l and fq u (x) respectively, and take b = Then fq(1) l = fq u (1) = 1 If q 3 then ( 1) < 3, and if q 4 then ( 1+) < 3 Since q q beta-numbers are Perron numbers [10], Condition (4) holds for lower (resp upper) self-christoffel numbers whenever q 3 (resp q 4) We also note that fq() l = q 1 1, fq u () = q 3 Now one can say the following: 1 If there exist infinitely many Mersenne primes, then f l q(x) is irreducible for infinitely many q 3 (4) If there exist infinitely many primes of the form n 3, then f u q (x) is irreducible for infinitely many q 4 Using a computer the author checked the irreducibility of beta-polynomials of self-christoffel numbers To be more precise, suppose α = p/q + b 1, where 1 p < q 00, gcd(p, q) = 1 and 1 b 150 For given 1 p 199, the irreducibility of beta-polynomials of (α) and (α+) q 00 was checked as b runs over 1 b 150 When some beta-polynomial is reducible, the case was recorded as (p, q, b) in a row As a result, a table was obtained, which comprises 564 (resp 34) reducible cases for lower (resp upper) self-christoffel numbers This is too huge to be included here We refer to [17] instead For example, if α = 6 + b 1 then the 11 beta-polynomial of (α) is irreducible for all b = 1,, 4 b 150, and the beta-polynomial of (α+) is irreducible for all b 150 On the other hand, if α = 4 + b 1 then both beta-polynomials of (α) and (α+) are 9 irreducible for all 1 b 150 We will use below this table implicitly 9

Suppose α = p/q +b 1 If we follow the proof of THEOREM described in Section 3, then we can find effectively the constant B smallest possible for which b B implies that the beta-polynomial of (α) or (α+) is irreducible This procedure is demonstrated in the next example Example 4 Let us consider the case α = 4 + b 1 and find B for upper 7 self-christoffel number (α+) First we must determine B 1 so that the Mahler measure of g is less than θ 0 given in Theorem 1 One can see that such B 1 is enough for our purpose if it satisfies ( B 1 + ) 5 B1 + 4B 1 < θ 0 B 1 So B 1 17 Second we also find B which contradicts Inequality (3) Computation shows that B 11 Let B 3 = max{b 1, B } = 17 Then b B 3 implies that the beta-polynomial of (( 4 + b 1)+) is irreducible 7 One readily notes that the constant B 3 works not only for 4 but for all 7 1,,, 6 To find the smallest B instead of B 7 7 7 3, we consult the table [17] It suffices to check whether or not (1, 7, b), (, 7, b),, (6, 7, b) for 1 b 16 are in the list But there are none even for 1 b 150 Now we can state the following (ie, in this case we have the beta-polynomial of (α+) is irreducible for every b B = 1): For any integer p 1 which is not a multiple of 7, the betapolynomial of ( p +) is irreducible 7 Some calculations tempt us to conjecture that all lower self-christoffel numbers are Pisot numbers But this is false as explained below Flatto et al [5] considered the dynamical zeta-function ζ β (z) = exp ( P n n=1 n z n) for the β-transformation T β, where P n is the number of fixed points of Tβ n They formulated it in terms of the β-expansions of 1: 1 If β is not a simple beta-number with d β (1) = e 1 e, then ζ β (z) = 1 1 e iz i If β is a simple beta-number with d β (1) = e 1 e n, then ζ β (z) = 1 z n 1 n e iz i In both cases, ζ β (z) is meromorphic in the open unit disk It has a simple pole at z = 1/β and no other pole in {z : z 1/β} Then M(β) was defined by the second smallest modulus of the poles of ζ β (z) If no pole 10

other than 1/β exists in the open unit disk, then M(β) = 1 The main study was the behavior of the function M(β) The next theorem alludes to an abundance of non-pisot lower self-christoffel numbers Theorem 43 ([5]) There exists ɛ > 0 such that M(β) < 1 for every β (1, 1 + ɛ) In particular, if the Mersenne Prime Conjecture is true then there are infinitely many non-pisot lower self-christoffel numbers There still remains a question in this direction Question 1 Let p/q be a fixed rational with 0 < p q and gcd(p, q) = 1 If α = b 1 + p/q for b N, then (α) is eventually a Pisot number as b increases Together with THEOREM, the affirmative answer to Question 1 would produce many concrete examples of irreducible Pisot type beta-substitutions See for details [1] While our main interest in this paper is the irreducible beta-polynomial, Boyd [] focused on the reducible beta-polynomial Specifically, he dealt with the situation where β is a Pisot number It is well known that every Pisot number is a beta-number [15] Let β be a Pisot number and f(x) be the β-polynomial Now we suppose that f(x) = g(x)h(x) is a nontrivial factorization over Q, where h(x) is the minimal polynomial of β Boyd called g(x) the complementary factor Although considerable calculations made some mathematicians suspect that g(x) should be a product of cyclotomic polynomials, Boyd showed by systematic computation that there exist many Pisot numbers β for which their complementary factors are noncyclotomic or even nonreciprocal Motivated by Boyd s work, the author checked the complementary factors of all reducible cases listed in [17] for both lower (564) and upper (34) self-christoffel numbers Surprisingly enough, all the complementary factors are (either cyclotomic polynomials or) products of cyclotomic polynomials For lower self-christoffel numbers in [17], irreducibles of the complementary factors consist of, 3, 4, 5, 6, 7, 8, 9, 10, 1, 14, 15, 16, 18, 0, 1,, 4, 6, 8, and 30 th cyclotomic polynomials only For upper self-christoffel numbers in [17], irreducibles of the complementary factors consist of 1,, 3, 4, 5, 6, 8, 9, 10, 1, and 14 th cyclotomic polynomials only A natural question arises Question For β a lower or an upper self-christoffel number, if the β- polynomial is reducible, then its complementary factor is a product of cyclotomic polynomials 11

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[15] K Schmidt On periodic expansions of Pisot numbers and Salem numbers Bull London Math Soc 1 (1980) 69 78 [16] CJ Smyth On the product of the conjugates outside the unit circle of an algebraic integer Bull London Math Soc 3 (1971) 169 175 [17] DY Kwon Table of reducible beta-polynomials of lower and upper self-christoffel numbers available at http://mathyonseiackr/doyong/paper/reduciblepdf Department of Mathematics, Yonsei University, 134 Shinchon-dong, Seodaemun-gu, Seoul 10-749, Republic of Korea E-mail: doyong@yonseiackr 13