ON MAHLER MEASURES OF A SELF-INVERSIVE POLYNOMIAL AND ITS DERIVATIVE

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1 ON MAHLER MEASURES OF A SELF-INVERSIVE POLYNOMIAL AND ITS DERIVATIVE Abstract. Let M(f) be the Mahler measure of a polynomial f C[z]. An ol inequality (which is ue to Mahler himself) asserts that M(f ) M(f) for each f C[z] of egree. In contrast, we prove that if f C[z] is a self-inversive polynomial of egree then M(f ) > M(f). We also show that this inequality is best possible for even, namely, that the quotient M(f )/M(f) takes every value in the interval (/, ] as f runs through reciprocal polynomials f R[z] of egree. It seems likely that for o the constant / is not optimal. For instance, for = 3, the optimal value of the constant is instea of 3/. For each o 5, we prove that there exists a monic reciprocal polynomial f Z[z] of egree such that M(f ) < +1 M(f). A corresponing problem for L p norms of a self-inversive polynomial an its erivative is also consiere. Recall that, for any polynomial 1. Introuction f(z) = a z + a 1 z a = a (z α 1 )... (z α ) C[z] of egree, its Mahler s measure is efine as M(f) = a max{1, α j }. j=1 By Jensen s formula, the logarithm of the Mahler measure of f C[z] can be also expresse by the formulae log M(f) = 1 log f(e πit ) t (1) (see [14]). There is a classical inequality between the Mahler measure of a polynomial f C[z] of egree an its erivative M(f ) M(f). () It was prove by Mahler himself [15]. See also Section D in [1] for another proof of () an two recent papers of Pereira [17] an Pritsker [18], where they showe that () can be erive from an earlier result of e Bruijn an Springer [9]. The example f(z) = z + 1 shows that the constant in () is best possible. Mathematics Subject Classification. 1C8. Key wors an phrases. Mahler s measure, self-inversive polynomial, reciprocal polynomial. 1

2 By the efinition of Mahler s measure, we have M(f ) a (3) for every polynomial f(z) = a z + +a C[z] of egree. The example f(z) = a z +a with a an a C shows that we can have M(f ) = a, so inequality (3) is sharp. Furthermore, by aing a large positive integer n to any polynomial f we see that lim n M(f +n) =, whereas the erivative of f +n is equal to f. So one cannot replace the right-han sie of (3), a, by, say, εm(f) with some small positive ε, unless there is some restriction on the constant term of a polynomial. The most natural restriction is f() =. Then a result of Storozhenko [7] asserts that M(f ) s()m(f) for each f C[z] of egree satisfying f() =, where s() = /6<k<5/6 = ( sin(πk/)) M((1 + z) 1). The example f(z) = (1 + z) 1 shows that the boun of Storozhenko is best possible. However, s() is approximately 1.4 for large, so the constant s() is very small. For which f C[z] there is a lower boun for M(f ) given in terms of M(f) an such that the epenence on is similar to ()? An important class of such polynomials is given in Theorem 1. Let f enote the reciprocal polynomial of f C[z], efine by f (z) = z eg f f(1/z), where the bar enotes complex conjugation. More precisely, for f(z) = a z + + a, where a, we have f (z) = a z + +a. Generally speaking, a polynomial f C[z] is calle self-inversive if f (z) = θf(z) for some θ C. Eviently, θ = a a /(a a ), so such a number θ must be of moulus 1. In particular, a polynomial f R[z] is calle reciprocal if its reciprocal polynomial f is ±f, i.e., f (z) = ±f(z). In other wors, f is reciprocal if it is a real self-inversive polynomial. The set of roots of a self-inversive polynomial is invariant uner the map z 1/z, i.e., the multiset {α 1,..., α } coincies with the multiset {1/α 1,..., 1/α }. Reciprocal polynomials play a very important role in the theory of Mahler measure of algebraic numbers. Recall that the Mahler measure of an algebraic number is the Mahler measure of its minimal polynomial in Z[z]. In 1933, Lehmer [13] aske whether for every ε > there is a polynomial f Z[z] satisfying 1 < M(f) < 1 + ε? In 1971, Smyth [5] showe a result which implies that such a polynomial f, if exists, must be reciprocal. It supports the opinion that the expecte answer to Lehmer s question is negative. One shoul also mention the papers of Schinzel [1], Amoroso an Dvornicich [1], Borwein, Dobrowolski an Mossinghoff [6], where Lehmer s problem was resolve for some other classes of polynomials with integer coefficients. However, Smyth s result remains one of the most important results in this area, because it reuces Lehmer s problem to a small class of reciprocal polynomials. Clearly, our next theorem, which gives a kin of reverse inequality to (), is applicable to reciprocal polynomials in Z[z] an more generally to reciprocal polynomials in R[z]:

3 ON MAHLER MEASURES OF A SELF-INVERSIVE POLYNOMIAL AND ITS DERIVATIVE 3 Theorem 1. Suppose that f C[z] is a self-inversive polynomial of egree. Then M(f ) > M(f). We remark that two trivial cases = an = 1 are exclue from Theorem 1. If = 1 then f(z) = a 1 z +a C[z] is self-inversive if an only if a = a 1. Then M(f) = a 1 an M(f ) = a 1, giving M(f ) = M(f) for each linear self-inversive polynomial f. A key tool in the proof of Theorem 1 is Corollary 6. Although several versions of Corollary 6 (for example, Corollary 8) have alreay been use in several earlier papers (e. g., [3], [4]), it seems that neither Lemma 5 nor Corollary 6 were state in this form an, moreover, were never applie in the context of Mahler measures of polynomials. (One can fin many references on Mahler s measure in a recent survey of Smyth [6].) Note that if M(f) a then the trivial inequality (3), which hols for every f C[z] of egree, is stronger than the inequality of Theorem 1. In particular, Theorem 1 is of no use if f Z[z] is a monic reciprocal polynomial whose Mahler measure M(f) satisfies 1 M(f). In this case, the trivial inequality M(f ) is stronger. Moreover, the inequality is strict M(f ) > in case a self-inversive polynomial f of egree has at least one root of moulus 1. Inee, by an ol result of Maren (see Theorem (45,) in [16]), a self-inversive polynomial f an its erivative f have the same number of zeros in z > 1, so f has at least one root in z > 1 provie that f has a root of moulus 1. Combining Maren s inequality with Theorem 1 we thus obtain M(f ) > max{m(f)/, 1} for each self-inversive polynomial f of egree having at least one root of moulus 1. Self-inversive polynomials with all zeros on the unit circle have been stuie by Bonsall an Maren [5], Lakatos an Losonczy [1], Schinzel [3], Sinclair an Vaaler [4]. (For those polynomials of egree we obviously have M(f ) = M(f).) Combining Theorem 1 with Mahler s inequality (), we have < M(f ) M(f) for each self-inversive polynomial f C[z] of egree. How close is the number inf M(f ) M(f), where the infimum is taken over every self-inversive polynomial f C[z] of egree, to the value /? Our next theorem shows that for even the lower boun M(f )/M(f) > / is best possible even we are restricte to the class of monic integer reciprocal polynomials. Theorem. Let be an even integer. Then there is a sequence of monic integer reciprocal polynomials f n, n = 1,, 3,..., of egree such that M(f n)/m(f n ) / as n. Furthermore, for every w (/, ] there is a monic reciprocal polynomial f R[z] of egree such that M(f )/M(f) = w. For 3 o we prove the following: Theorem 3. Let 3 be an o integer. Then there is a monic integer reciprocal polynomial f of egree such that M(f ) < +1M(f).

4 4 For = 3 we shall fin the exact value of inf M(f ), where the infimum is taken over M(f) every cubic self-inversive polynomial f C[z]: Theorem 4. Let f C[z] be a cubic self-inversive polynomial. Then M(f )/M(f) 3( ) 1/ ( ) + 9 1/3 3 = (4) The equality in (4) is attaine for the polynomial f(z) = z 3 B z B z + 1 R[z], where B = ( ) 1/3 8 One can also express B as B = t + 1/t 1 with Then where t satisfies The minimal polynomial of B is ( = (5) ) 1/3 8 t = (3 + ) 1/3 + (3 ) 1/3 + = f(z) = z 3 B z B z + 1 = (z + 1)(z t)(z 1/t), t 3 6t + 9t 8 =. 8z 3 33z + 18z 9. (6) By (5), the right han sie of (4) is equal to 3(B 1)/4. The minimal polynomial of the cubic number 3(B 1)/4 is 3z 3 7z 54z 7. (7) Let S be the set of complex self-inversive polynomials of egree, an let R S be M(f the subset of (real) reciprocal polynomials. Theorems 1, an 3 imply that inf ) = f S M(f) ( + τ )/, where τ = for even an τ < 1 for 3 o. Theorem 4 asserts M(f that min ) f S3 = Their proofs will be given in Sections 3 an 4. In M(f) Section 5 we shall stuy the quotients of L p norms f p / f p for f S an f R. We also give some numerical examples of polynomials of o egree an low value of the quotient M(f )/M(f) in Section 6.. Some lemmas Lemma 5. Let f C[z] be a self-inversive polynomial of egree 1, an let z be a complex number of moulus 1. Then ( ) f(z f (z ) = ) + z f (z ) f(z ). (8) If, in aition, z is not a root of f then R(z f (z )/f(z )) = /.

5 ON MAHLER MEASURES OF A SELF-INVERSIVE POLYNOMIAL AND ITS DERIVATIVE 5 Proof: Since f C[z] is self-inversive, f (z) = z f(1/z) = θf(z) for some θ C of moulus 1. Calculating the erivative of θf, we obtain θf (z) = z 1 f(1/z) z f (1/z) = θz 1 f(z) z f (1/z) for every non-zero complex number z. Diviing both sies of the above equality by θz 1 f(z), we euce that = zf (z) f(z) + z 1 f (1/z) θf(z) = zf (z) f(z) + f (1/z) zf(1/z) if z is non-zero an not a root of f. Fix any z of moulus 1 which is not a root of f. Then 1/z = z, because z = 1. Setting z = z into the right han sie of (9), we see that the secon term is equal to the complex conjugate z f (z )/f(z ) of the first term z f (z )/f(z ). It follows that = R(z f (z )/f(z )), giving R(z f (z )/f(z )) = /. This proves the secon part of the lemma. It is clear that equality (8) hols for each z of moulus 1 which is a root of f. Suppose that z = 1, where z is not a root of f. Then R(z f (z )/f(z )) = / implies that z f (z )/f(z ) = / + ii(z f (z )/f(z )). Hence z f (z )/f(z ) / = I(z f (z )/f(z )). Since z = 1, we have f (z )/f(z ) = z f (z )/f(z ) = R(z f (z )/f(z )) + I(z f (z )/f(z )) Multiplying by f(z ) yiels (8). = (/) + z f (z )/f(z ) /. We remark that ientity (8) of Lemma 5 implies that if f C[z] is a self-inversive polynomial then its erivative f has no zeros on the unit circle z = 1 except for the multiple zeros of f (if any). This is an ol result of Bonsall an Maren [5] (see Lemma (45,) in [16]) which was prove by an entirely ifferent argument. Below, we shall use the following corollary: Corollary 6. Let f C[z] be a self-inversive polynomial of egree 1, an let z be a complex number of moulus 1. Then f (z ) f(z ), where equality hols if an only if z is one of at most points of the circle z = 1 satisfying z f (z ) = f(z ). Proof: From (8) we see that f (z ) f(z ), where equality hols if an only if z f (z ) = f(z ). Note that zf (z) f(z) is a polynomial of egree, so it has at most complex roots. In particular, the equality z f (z ) = f(z ) hols for at most complex numbers z lying on the unit circle z = 1. Lemma 7. Let D be a fixe positive integer, an let g 1, g, g 3, C[z] be a sequence of polynomials satisfying eg g n D an lim n L(g n ) =. Then lim n M(g +g n ) = M(g) for each g C[z]. (9)

6 6 Here an below, for any polynomial f(z) = j= a jz j C[z], its length L(f) is efine by the formulae L(f) = j= a j. The inequality for the ifference of two Mahler measures M(f) 1/ M(g) 1/ L(f g) 1/, (1) where f, g C[z] are any polynomials of egree, was obtaine by Chern an Vaaler [8]. Setting = max(d, eg g) an f = g + g n into (1) yiels Lemma 7. The fact that lim n M(f n ) = M(f) if lim n L(f f n ) = an the polynomials f, f 1, f,... have the same egree was earlier prove by Boy [7]. See also Corollary 14 on p. 51 in []. 3. Proofs of Theorems 1, an 3 Proof of Theorem 1: Suppose that f C[z] is a self-inversive polynomial of egree 1. By Corollary 6, there exists a finite set Λ of at most + 1 points of the interval I = [, 1) which inclues an all the points t I for which either f (e πit ) = f(eπit ) or f(e πit ) =. Note that, by Lemma 5, every t I satisfying f (e πit ) = also belongs to Λ. Fix a small positive number ε. Let I ε = λ Λ I λ,ε be a finite union of intervals I λ,ε = (λ ε/, λ + ε/), where I,ε is efine as I,ε = [, ε) (1 ε, 1). All those intervals I λ,ε are isjoint for sufficiently small ε. So I ε is the union of at most + intervals of length ε, an I \ I ε is the union of at most + 1 close intervals. By (1), we have 1 log M(f) = log f(e πit ) t = log f(e πit ) t + log f(e πit ) t. I\I ε I ε Similarly, log M(f ) = 1 log f (e πit ) t = log f (e πit ) t + I\I ε log f (e πit ) t. I ε Note that, by Corollary 6, for every t I \ I ε, we have f (e πit ) > f(eπit ) >. Thus log f (e πit ) t log f(e πit ) t > (1 I ε ) log(/). (11) I\I ε I\I ε Also, f (e πit ) f(eπit ) for t I ε. Hence log f (e πit ) t log f(e πit ) t I ε log(/). (1) I ε I ε (In fact, the singularities of the integrans in (1) are logarithmic, because log e πit e πiλ = log sin(π(t λ)) log(π t λ ) as t λ. So, using the fact that I ε is a finite union of intervals of length ε, as on p. 8 of [1], one can show that there is a constant C(f) inepenent of ε such that I ε log f (e πit ) t < C(f)ε log(1/ε) an I ε log f(e πit ) t < C(f)ε log(1/ε).)

7 ON MAHLER MEASURES OF A SELF-INVERSIVE POLYNOMIAL AND ITS DERIVATIVE 7 Aing (11) with (1) an using the formulas for log M(f ) an log M(f), we euce that log M(f ) log M(f) > (1 I ε ) log(/) + I ε log(/) > log(/). This proves the inequality M(f )/M(f) > /. The referee observe that by using Lemma 5 one can prove a slightly stronger inequality, namely, ( ) ( M(f ) M(f) + M zf (z) ) f(z) (13) for every self-inversive polynomial f C[z]. We give a sketch of the proof. Suppose that the polynomials g, h C[z] have no zeros on the unit circle. The Mahler s inequality n n n (x k + y k ) 1/N x 1/N k + y 1/N k, k=1 where x k, y k >, k = 1,..., N, for the corresponing Riemann integral sums implies that ( 1 ) exp log ( g(e πiθ ) + h(e πiθ ) )θ k=1 ( 1 ) ( 1 ) exp log g(e πiθ ) θ + exp log h(e πiθ ) θ = M(g) + M(h). Selecting here g(z) = (f(z)/) an h(z) = (zf (z) f(z)/) for z = e πiθ an using (8) we obtain ( 1 M(f ) = exp ) ( 1 log f (e πiθ ) θ = exp k=1 ) log ( g(e πiθ ) + h(e πiθ ) )θ ( ) ( M(g) + M(h) = M(f) + M zf (z) ) f(z), which proves (13). The case when the polynomial gh has some zeros on the unit circle can be treate by introucing I ε as in the proof of Theorem 1. Proof of Theorem : Put f n (z) = z nz / + 1. The Mahler measure of f n is equal to the Mahler measure of the polynomial z nz + 1, so M(f n ) = (n + n 4)/ if n. From f n(z) = z / 1 (z / n/) it follows that M(f n) = n/ for n. Thus M(f n) M(f n ) = /n as n, which proves the first part of the theorem. For the secon part, fix w (/, ] an select a real number u such that /w 1 = 1 4/u. By the above, for the polynomial f(z) = z uz / + 1 R[z], we have M(f )/M(f) = u/(u + u 4) = w.

8 8 Proof of Theorem 3: For = 3 we can take f(z) = z 3 4(z + z) + 1 an fin that M(f )/M(f) = is smaller than. Suppose that 5. Set f n (z) = z ng(z) + 1 with g(z) = (z + 1)(z 4z + 1)z ( 3)/. Using L(f n /n + g) = /n as n, by Lemma 7, we obtain Note that M(f n ) lim n n = M(g) = M(z 4z + 1) = + 3. g (z) = (( + 3)z 3 3( + 1)z 3( 1)z + 3)z ( 5)/ /. Hence, using L(f n/n + g ) = /n as n, we fin that M(f lim n) = M(g ) = 1 n n M(( + 3)z3 3( + 1)z 3( 1)z + 3) = 1 M(h), where h(z) = ( + 3)z 3 3( + 1)z 3( 1)z + 3. In case the limit lim n M(f n)/m(f n ) = ( 3)M(h)/ is smaller than ( + 1)/, we can take f = f n with sufficiently large n. Inee, if ( 3)M(h) = + 1 κ() with some positive number κ() then M(f n)/m(f n ) < ( + 1 κ()/)/ for each sufficiently large n, say, n n. Taking f = f n, where n n, we woul obtain M(f )/M(f) = M(f n)/m(f n ) < ( + 1 κ()/)/ < ( + 1)/, as claime. So it remains to prove the inequality M(h) < ( + 3)( + 1). Note that h( 1) = 1, h() = 3 > an h(1) = 4 <. So h has a root in ( 1, ), a root in (, 1), an a root β > 1. Hence M(h) = ( + 3)β. The require inequality is equivalent to the inequality β < (+ 3)(+1)/(+3). For this, it suffices to show that h((+ 3)(+1)/(+3)) >. Inee, by a simple computation, we have h(( + 3)( + 1)/( + 3)) = 4(3( 3 + 1) )/( + 3) >, which completes the proof. 4. The cubic case: Proof of Theorem 4 If all three roots of f lie on the unit circle then, by the Gauss-Lucas theorem, all roots of f lie in the unit circle z 1, hence M(f )/M(f) = 3. This is more than we claim in (4). Next, since M(uf) = u M(f) an M(uf ) = u M(f ), u C, we can assume that f is monic. The Mahler measure of f(z) an that of f(ze iφ ), φ R, are equal. The same hols for their erivatives. So we may assume that f has a real root t greater than 1. Since the polynomial f is self-inversive, its two other roots are 1/t an e iφ, where φ [, π). However, as M(f) = M(f) an M(f ) = M(f ), without loss of generality, we can assume that φ [, π].

9 ON MAHLER MEASURES OF A SELF-INVERSIVE POLYNOMIAL AND ITS DERIVATIVE 9 It remains to minimize the quotient M(f )/M(f), where f runs through the polynomials f(z) = (z e iφ )(z t)(z 1/t) with t > 1 an φ [, π]. Eviently, M(f) = t. Set w = e iφ an λ = t + 1/t. Since f(z) = (z w)(z λz + 1) = z 3 (w + λ)z + (wλ + 1)z w, we fin that f (z) = 3z (w + λ)z + wλ + 1. By the above mentione theorem of Maren, f has a unique root outsie the unit circle, say, β = β(φ, t). Hence M(f )/M(f) = 3 β /t. (14) Clearly, β = β(φ, t) = λ + eiφ + λ λe iφ + e iφ 3, (15) 3 where 3β(φ, t) (t + 1/t + e iφ )β(φ, t) + (t + 1/t)e iφ + 1 =. (16) The sign of the square root in (15) is chosen so that its real part is positive. We claim that then β(φ, t) > 1. Inee, this is the case if λ = t + 1/t is large enough. Suppose for some t > 1 an φ [, π] we have β(φ, t) < 1. Then, by continuity, there exist t > 1 an φ [, π] such that β(φ, t ) = 1. However, this is impossible, because f has no multiple roots on z = 1 (see a remark after Lemma 5). Set γ = γ(φ, t) = β(φ, t) = λ + e iφ + λ λe iφ + e iφ 3. 3 Then 3γ(φ, t) (t + 1/t + e iφ )γ(φ, t) + (t + 1/t)e iφ + 1 =. (17) Note that if t 3/ then M(f )/M(f) 3/t. Also, β(φ, t) /t /3 as t, hence M(f )/M(f) tens to as t The extremal polynomial f(z) = z 3 B z B z+1 given in the statement of the theorem shows that M(f )/M(f) attains smaller value. We can thus restrict t to the interval 3/ t t with some absolute constant t. By (14), we have M(f ) /9M(f) = β /t = βγ/t, so we nee to fin the minimum of the function h(φ, t) = β(φ, t)γ(φ, t)/t in the rectangle φ [, π], t [3/, t ]. Suppose that the minimum is attaine at the point (φ, t), where < φ < π an 3/ t t. (It will be clear from the context when the same letters φ, t are use to enote the variables of the functions β(φ, t), γ(φ, t) an h(φ, t) an when the pair (φ, t) is use to enote the point, where the minimum of h is attaine.) Then this point is a critical point of the function h(φ, t), so h φ = β φ γ + γ φ β = an h t = 1 t ( β t γ + γ t β ) βγ t 3 =. (18) Our goal is to prove that this is not the case. Suppose for a moment that this is alreay establishe. Then, as we know, the minimum is not attaine at t = 3/ an at t = t, so

10 1 it must be attaine at some point (φ, t), where φ {, π} an 3/ < t < t. Moreover, it is easy to check that h(, t) > h(π, t). Inee, t h(, t) = β(, t) = λ λ λ 3 > λ 1 + λ + λ 3 = β(π, t) = t h(π, t), because by squaring the inequality + λ λ > λ + λ we have λ λ > λ. By squaring again, we see that this inequality hols, because λ >. Thus the minimum of h must be attaine at some point (π, t) with 3/ < t < t. Then M(f ) M(f) = 3 3β(π, t) h(π, t) = = (λ 1 + λ + λ ) t λ +. λ 4 Setting x = λ 1, we thus nee to fin the minimum of the function Ψ(x) = (x + x + 3x) x (x + 1) 4 for x > λ 1 = t + 1/t 1 > 7/6. A simple computation with Maple shows that the function Ψ(x) has only one critical point in [1, ). From Ψ (x) = + (x + 3)/ x + 3x x x + x 3 (x + x + 3x)(1 + (x + 1)/ x + x 3) (x = x + x 3) we fin that the minimum of Ψ(x) in [1, ) is attaine at the point x = B = ( ) 1/ ( ) 1/ = , where Ψ (B ) =. It is equal to Ψ(B ) = One of the extremal polynomials at which the equality M(f )/M(f) = Ψ(B ) is attaine is the polynomial f(z) = z 3 B z B z +1. The minimal polynomials for B an Ψ(B ), namely, (6) an (7) have been foun with Maple. This proves the theorem. In the remainer of this section we will show that for no point (φ, t), where < φ < π an 3/ < t < t all four equalities given in (16), (17), (18) can hol. Differentiating (16) with respect to the variable φ, we obtain Hence Analogously, (17) yiels thus 6β β φ (λ + eiφ ) β φ iβeiφ + ie iφ λ =. (6β λ e iφ ) β φ = ieiφ (β λ). (19) 6γ γ φ (λ + e iφ ) γ φ + iγe iφ ie iφ λ =, (6γ λ e iφ ) γ φ = ie iφ ( γ + λ). ()

11 ON MAHLER MEASURES OF A SELF-INVERSIVE POLYNOMIAL AND ITS DERIVATIVE 11 Since, by (18), β γ γ+ β =, multiplying (19) by φ φ (3γ λ e iφ )γ an () by (3β λ e iφ )β an then aing, we obtain = (3γ λ e iφ )γie iφ (β λ) + (3β λ e iφ )βie iφ ( γ + λ). Hence (β λ)(3γ (λ + e iφ )γ)e iφ = (γ λ)(3β (λ + e iφ )β)e iφ. By (16) an (17), we know that 3β (λ+e iφ )β = (λ+e iφ )β λe iφ 1 an 3γ (λ+e iφ )γ = (λ + e iφ )γ λe iφ 1. So (β λ)((λ + e iφ )γ λe iφ 1)e iφ = (γ λ)((λ + e iφ )β λe iφ 1)e iφ. Here, the left han sie is equal to βγ(1 + λe iφ ) λ(1 + λe iφ )γ β(λ + e iφ ) + λ(λ + e iφ ), an the right han sie is equal to βγ(1 + λe iφ ) λ(1 + λe iφ )β γ(λ + e iφ ) + λ(λ + e iφ ). Subtracting one sie from another an multiplying by w = e iφ yiels λβγ(w 1) + β( λw w + λ ) + γ(λw + λ w ) + λ(w 1) =. (1) Differentiating (16) with respect to the variable t an using λ = 1 t 1/t, we obtain 6β β (λ + w) β t t (1 1/t )β + w(1 1/t ) =. Since, by (16), β(3β λ w) = (λ + w)β (λw + 1), this yiels 1 β β t = (β w)(1 1/t ) ((λ + w)β λw 1). Taking the complex conjugates of both sies, we obtain By (18), their sum 1 β β + 1 γ t γ t 1 γ γ t = (γ w)(1 1/t ) ((λ + w)γ λw 1). must be equal to /t, hence (β w)(1 1/t ) ((λ + w)β λw 1) + (γ w)(1 1/t ) ((λ + w)γ λw 1) = t Expressing w = 1/w, λ = t+1/t an multiplying by common enominators, we can rewrite () in the form p 3 (w, t)βγ + p (w, t)β + p 1 (w, t)γ + p (w, t) =, (3) where p j (w, t) Z[w, t] for j =, 1,, 3. Clearly, (16) can be written as an (17) can be written as () 3β (λ + w)β + λw + 1 = (4) 3wγ (λw + 1)γ + λ + w =. (5)

12 1 Observe that (1), (4), (5) are three equations in four variables β, γ, λ, w. Here, β an γ are the values of the corresponing functions at the critical point (φ, t) of h. By computing the resultant of (1) an (4) with respect to β we shall obtain a polynomial F in γ, λ, w. Then, by computing the resultant of this polynomial F an the polynomial on the right han sie of (5) with respect to γ, we shall obtain the polynomial F 1 (w, λ) = 3w (w 1) (w + 1) (λ ) (λ + ) (w λw + 1) ((λ 4 λ 3)w (λ 5 6λ λ)w + λ 4 λ 3). Similarly, we can substitute λ = t + 1/t into (4) an (5), multiply them by t an then obtain a polynomial G 1 (w, t) as a result of first computing the resultant G of (3) (obtaine from ()) an (4) with respect to β an then the resultant of this polynomial G in γ, w, t an the polynomial (5) with respect to γ. The polynomial G 1 (w, t) = 8(w 7 + w 5 )t 3 + (1w w 6 + 1w 4 )t + 888(w 7 + w 5 )t + 115w 6 is to large to be reprouce here. It has 1 nonzero terms, its total egree is 3, its egree in w is 1, an its egree in t is 3. (Of course, the polynomial G 1 (w, t) can be quickly compute with Maple exactly as it is escribe above.) Since w ±1 an λ >, only factor of F 1 which may vanish at points w with moulus 1 is F (w, λ) = (λ 4 λ 3)w (λ 5 6λ λ)w + λ 4 λ 3. We substitute λ = t+1/t into F an eliminate t by computing the resultant of t 5 F (w, t+ 1/t) an G 1 (w, t) with respect to t. This resultant F 3 (w) factors in Z[w] into irreucible factors as follows: where F 3 (w) = (w 1) 3 (w + 1) 3 F 4 (w)f 5 (w)f 6 (w)f 6 (w), F 4 (w) =11w w w w w w w w + 11, F 5 (w) =763w w w 9444w w w w w w w w w + 763, F 6 (w) =5887w w w w w w w w w w The polynomials F 6, F6 are irreucible in Z[w] an non-reciprocal, so they have no zeros of moulus 1. A simple numerical computation with Maple with extene precision shows that neither F 4 nor F 5 have such zeros. It follows that the function h(φ, t) has no critical points if < φ < π. This proves our claim an completes the proof of Theorem 4.

13 ON MAHLER MEASURES OF A SELF-INVERSIVE POLYNOMIAL AND ITS DERIVATIVE 13 Another way to compute the extremal value t is to use (). Since it is alreay establishe that the minimum is attaine at w = 1 (an so β = γ), by (), we have β + 1 (λ 1)(β + 1) = t t 1. Thus β = (t t+3)/(t ). Substituting this expression into 3β (t+1/t 1)(β+1) = (which is obtaine from (4) with w = 1), we fin that 3t(t t+3) = 4(t t+1) (t ). Hence (t 3 6t + 9t 8)(t + 1) =, giving t 3 6t + 9t 8 =. Thus t = (3 + ) 1/3 + (3 ) 1/3 + = is the critical point of h(π, t) an f(z) = (z + 1)(z t)(z 1/t) is a corresponing extremal polynomial. 5. L p norms of a polynomial an its erivative Suppose that < p. It is well-known that the Mahler measure of a polynomial f C[z] is the limit of its L p norm efine by namely, f p = ( 1 f(e πit ) p t ) 1/p, lim f p = M(f) p + (see, e.g., [11]). Now, a irect analogue of () is the upper boun f p f p, (6) which hols for every p > an every f C[z] of egree. For p = inequality (6) is a classical inequality of Bernstein. Later, Zygmun [8] prove (6) for 1 p < an Arestov [] for < p < 1. Fix any p >. Which values the quotient f p / f p takes as f runs through self-inversive polynomials of egree? For p =, we have f = max z =1 f(z). Theorem of [] asserts that if f C[z] is a self-inversive polynomial of egree 1 then f = f. So the quotient f / f, where f S, takes only one value /. The upper boun for f p / f p, where f C[z] is a self-inversive polynomial of egree, was establishe by Rahman an Schmeisser [19]. Combining Theorem an Remark of [], we have f p f p z + 1 p = z + 1 p = ( πγ(p/ + 1) ) 1/p (7) Γ(p/ + 1/) for every f S. Equality in (7) is attaine for the polynomial f(z) = z + 1 R.

14 14 Since the integrans f(e πit ) p an f (e πit ) p, where f is a polynomial, are efine for every point t [, 1], Corollary 6 immeiately implies the following lower boun for L p -norms of a self-inversive polynomial an its erivative: Corollary 8. Suppose that < p <. If f C[z] is a self-inversive polynomial of egree then f p > f p. The boun of Corollary 8 is best possible for even. Inee, one can take the same example f n (z) = z nz / +1 with n large as in the proof of Theorem. Then, for every fixe p >, we have f n p n an f n p n/ as n. Hence f n p / f n p / as n. Combining with (7), by continuity, we euce that for every < p < the quotient f p / f p takes every value in the interval (/, ( πγ(p/ + 1)/Γ(p/ + 1/) ) 1/p/] as f runs through S (or R ), where is even. For example, one can take f(z) = z uz / + 1 R with u [, + ). Corollary 8 for p 1 was alreay establishe in [4]: see Theorem an Remark in [4], an also an earlier paper [3]. The question of fining inf f S f p / f p an inf f R f p / f p remains open for 3 o. One case when both infimums can be establishe is p =. For f(z) = a z + a 1 z a we have f = a + a a. Thus if f C[z] is a self-inversive polynomial of egree = 1 then f(z) = a(z + θ) with a an θ = 1. Hence f = 1 f. For 3 we prove the following: Theorem 9. If f C[z] is a self-inversive polynomial of o egree 3 then f > +1 f. This inequality is best possible. Proof: Without loss of generality we may assume that f is monic. Then f(z) = z + a 1 z a ( 1)/ z (+1)/ + θa ( 1)/ z ( 1)/ + + θa 1 z + θ with some θ C of moulus 1. Hence an f = (1 + a 1 + a + + a ( 1)/ ) f = + (( 1) + 1 ) a 1 + (( ) + ) a a ( 1)/. The coefficients ( j) + j are greater than ( + 1)/ for each j in the range j < ( 1)/. Therefore, f > + 1 (1 + a 1 + a + + a ( 1)/ ) = + 1 f 4. This yiels the require inequality f > +1 f for 3 o. In orer to show that this inequality is tight let us take the reciprocal polynomial f n (z) = z +n(z (+1)/ +z ( 1)/ )+1 Z[z]. We have f n = (1+n ) an f n = +( +1)n /. So lim n f n / f n = + 1/.

15 ON MAHLER MEASURES OF A SELF-INVERSIVE POLYNOMIAL AND ITS DERIVATIVE 15 Theorem 9 implies that inf f / f = inf f / f = + 1/ f S f R for 3 o. Note that the right han sie of (7) for p = is equal to /. By continuity, i.e., taking f(z) = z + u(z (+1)/ + z ( 1)/ ) + 1 R, where u [, + ), we euce that if f runs through every reciprocal polynomial f of o egree 3 then the quotient f / f takes every value in the interval ( + 1/, / ]. 6. Numerical examples By the above results, it seems that the constant / in the L p norm inequality f p > f p, where p <, is not optimal for self-inversive polynomials f(z) of o egree. In general, the computation of the infimum f p / f p over self-inversive polynomials f of o egree, is non-trivial except in two cases p = (Theorem 9) an p =. We give several examples of integer monic self-reciprocal polynomials with low value of M(f )/M(f) (corresponing to the case p = ), compute with Maple. Table 1. Monic reciprocal polynomials f Z[z] of small o egree an low value M(f )/M(f). f(z) M(f) M(f ) M(f )/M(f) 3 z 3 4z 4z z 5 z 4 + 7z 3 + 7z z z 7 z 6 + z 5 8z 4 8z 3 + z z z 9 z 8 + z 7 3z z z 4 3z 3 + z z Theorem 4 gives a precise answer for cubic self-reciprocal polynomials f(z) C[z]. The cubic polynomial in Table 1 is the monic polynomial with integer coefficients closest to the optimal cubic polynomial in Theorem 4. The value is the minimal value of M(f )/M(f) among all monic cubic reciprocal polynomials with integer coefficients. For = 5, we searche for optimal monic reciprocal polynomial with integer coefficients in the interval [ 1, 1]. For = 7 an = 9, the search interval was reuce to [, ] an to [ 15, 15], respectively. We must note that the values of M(f )/M(f) in Table 1 are not far from (+1)/ (see also the Theorem 3), but the istance from those values to (+1)/ is increasing with. In the case = 5, we have also compute the polynomial g(z) = z z z z 1.73z + 1 with M(g )/M(g) = It seems that this value is close to the infimum of M(f )/M(f), where f R 5.

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17 ON MAHLER MEASURES OF A SELF-INVERSIVE POLYNOMIAL AND ITS DERIVATIVE April, 6, (es. James McKee an Chris Smyth), LMS Lecture Note Series 35 (8), pp [5] C.J. Smyth, On the prouct of the conjugates outsie the unit circle of an algebraic integer, Bull. Lonon Math. Soc., 3 (1971), [6] C.J. Smyth, The Mahler measure of algebraic numbers: a survey, in Number theory an polynomials, Proceeings of the conference hel at the University of Bristol, 3-7 April, 6, (es. James McKee an Chris Smyth), LMS Lecture Note Series 35 (8), pp [7] E.A. Storozhenko, A problem of Mahler on the zeros of a polynomial an its erivative, Sb. Math., 187 (1996), [8] A. Zygmun, A remark on conjugate series, Proc. Lonon Math. Soc., 34 (193), Department of Mathematics an Informatics, Vilnius University, Naugaruko 4, Vilnius LT-35, Lithuania aress: arturas.ubickas@mif.vu.lt jonas.jankauskas@gmail.com