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ЧЕБЫШЕВСКИЙ СБОРНИК
Том 20. Выпуск 1.
УДК 512.62 DOI 10.22405/2226-8383-2019-20-1-197-203
О многочленах Нюмена без корней на единичном круге
А. Дубицкас
А рту рас Дубицкас — доктор математических наук, ведущий научный сотрудник, Институт математики Вильнюсского университета, г. Вильнюс (Литва). e-mail: [email protected]
Аннотация
В настоящей заметке мы получим необходимое и достаточное условие на тройку неотрицательных целых чисел а < b < с при выполнении которого многочлен Нюмена YTj=о х3 + Cj=b х3 имеет корень на единичном круге. Изпользуя это условие мы докажем, что для каждого d > 3 существует такое целое положительное число п > d, что многочлен Нюмена 1 + х + • • • + xd-2 + хп длины d не имеет корней на единичном круге.
Ключевые слова: многочлен Нюмена, корень из единицы.
Библиография: 11 названий.
Для цитирования:
А. Дубицкас. О многочленах Нюмена без корней на единичном круге // Чебышевский сборник, 2019, т. 20, вып. 1, с. 197-203.
CHEBYSHEVSKII SBORNIK Vol. 20. No. 1.
UDC 512.62 DOI 10.22405/2226-8383-2019-20-1-197-203
On Newman polynomials without roots on the unit circle
A. Dubickas
Arturas Dubickas — habilitated doctor of mathematics, research professor, Institute of Mathematics, Vilnius University, Vilnius (Lithuania). e-mail: [email protected]
Abstract
In this note we give a necessary and sufficient condition on the triplet of nonnegative integers a < b < c for which the Newman polynomial Ylj=o x3 + ^ j=b x3 a root 011 the unit circle. From this condition we derive that for each d > 3 there is a positive integer n > d such that the Newman polynomial 1 + x + ■ ■ ■ + xd-2 + xn of length d has no roots on the unit circle.
Keywords: Newman polynomial, root of unity.
Bibliography: 11 titles.
For citation:
A. Dubickas, 2019, "On Newman polynomials without roots on the unit circle" , Chebyshevskii sbornik, vol. 20, no. 1, pp. 197-203.
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In honor of Professor Antanas Laurincikas on the occasion of his 70th birthday
1. Introduction
For a polynomial f £ let throughout
m(f) := min |/^
be the minimal value of f on the unit circle. Motivated bv some questions raised by Campbell, Ferguson, Forcade [2], and Smyth [11] Bovd in [1] studied the behavior of m(f) for Newman polynomials f. (Recall that f (x) = <j=o aj'IS a Newman polynomial if aj £ {0,1} for each j = 0,1,..., d.) In particular, in [1] it was shown that for each d > 12 there is a Newman polynomial f of degree d satisfying m(f) > 1 and that for each sufficiently large d £ N there is a Newman polynomial f of degree d satisfying m(f) > d0A37. Here, 12 is the smallest possible such degree. For example,
m(1 + x + + + + x7 + + x10 + x12) = 1.36237....
See also [9] for some results on the irreducibilitv of Newman polynomials.
One can raise similar questions with degree of f replaced by its length (see [1], [4] and [8]). Such questions turn out to be more difficult. More precisely, let Nd be the set of Newman polynomials of length d, namely,
Md := {xkl + ■ ■ ■ + xkd, where k1 < ■ ■ ■ < kd are nonnegative integers}.
As in [8], we put
a(d) := max m(f).
Evidently, ^(1) = 1 and ^(2) = 0. The values of ^(4) have been calculated in [11] and [4],
respectively. (They come from polynomials 1 + x2 + x3 and 1 + x2 + x3 + x4.) It is conjectured that y(d) ^ ^ as d ^ ^ (see but even much weaker inequality y(d) > 1 is still not established
for each d > 5 (see [8]).
In [8] Mercer proved that y(d) > 0 for each d > 3. This is equivalent to the fact that for each d > 3 there is a Newman polynomial of length d which has no roots on the unit circle. In this note we describe which Newman polynomials of the form
c
f (x) = ^ Xj + xj = 1+-----h xa + xb +-----h
j=0 j=b
where 0 < a < b < c, have roots on the unit circle and which do not have. These polynomials are naturally obtained by taking a degree c polynomial with all coefficients 1 and then replacing in it a string of consecutive coefficients by zeros. Note that Theorem 3 immediately implies the above mentioned result y(d) > 0 for everv d > 3.
For a positive integer m and a prime number p, let up(m) be the power of p in the prime factorization of m, md vp(0) = 0.
Theorem 1. Let a,b,c be integers satisfying 0 < a < b < c. Then the Newman polynomial 1 + ■ ■ ■ + xa + xb + ■ ■ ■ + xc has a root on the unit circle if a nd only if c = a + b or at least one of the inequalities
gcd(a + 1,c - b +1) > 1, (1)
gcd(c + 1,c - b + a + 2) > 1, (2)
(|c - a - bl) >p2(b) (3)
holds.
In particular, selecting a = d — 2 > 1 and b = c = n > a, we see that c = a + b and that gcd(a + 1,c — b + 1) = gcd(d — 1,1) = 1, so (1) does not hold. Inserting c + 1 = n + 1, c — b + a + 2 = d, |c — a — 6| = d — 2 and b = n into (2) and (3), we obtain the following special case of Theorem 1:
Theorem 2. Let, d and n be integers satisfying n > d — 2 > 1. Then the Newman polynomial 1 + x + ■ ■ ■ + xd~2 + xn £ Md has a root on the unit circle if and only if at least one of the inequalities
gcd(n + 1,d) > 1, (4)
U2(d — 2) >V2(n) (5)
holds.
Let f(m) be the Euler totient function. From Theorem 2 we shall derive the following:
Theorem 3. Let d > 3 be an integer. Write d in the form d = 2m(2l +1) with integers m,l > 0. If k £ N satisfies
k> lQg d (Q)
k > <p(2l + 1) log 2 (6}
then the Newman polynomial 1 + x + ■ ■ ■ + xd-2 + xn £ Nd, with degree n = 2^(2i+1)fc; has no roots on the unit circle.
The proof of Theorem 1 is based on the following very simple lemma which was proved in [3]. (Subsequently, it was used in a different context in [6], [7], [10].)
Lemma 1. Suppose z1, z2, z3, z4 are complex numbers of modulus 1 satisfying
Zi + Z2 + z3 + Z4 = 0.
Then z1 + Zj = 0 for some j £ {2, 3, 4}.
In the next section we give the proof of Theorem 1. In Section 3 we prove Theorem 3. Finally, in Section 4 we present one more construction of Newman polynomials without roots on the unit circle.
2. Proof of Theorem 1
For c = a + b, the polynomial
1 + ■ ■ ■ + xa + xb + ■■■ + xc = (1 + ■ ■ ■ + xa)(1+ xb)
is a product of cvclotomic polynomials, so all of its roots are roots of unity. In all what follows we will assume that c = a + b.
If a complex number ( of modulus 1 is a root of
f (x) := 1 +-----+ xa + xb +-----+ xc,
then ( = 1 and
o = (1 — C )f «) = 1 — (a+1 + (b — (c+1.
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Since the four numbers on the right hand side of this equality, namely, 1, -(a+1, (b, -(c+1, are 1
1 - (a+1 = (b - Cc+1 = 0, 1 - Cc+1 = -<a+1 + (b = 0, 1 + Cb = Ca+1 + Cc+1 = 0.
In particular, these equalities imply that if ( of modulus 1 is a root of /, then ( must be a root of unity.
Now, will show that in the first (resp. second and third) case the inequality (1) (resp. (2) and (3)) holds and, conversely, if (1), (2) or (3) holds then f ({) = 0 to some root of unitv (Evidently, l£l = 1)
In the first case, we have (a+1 = (c-b+1 = 1. Set g := gcd(a + 1,c — b + 1). Then, there are some u,v £ Z such th at g = u(a + 1) + v(c - b + 1) and therefore (9 = 1. This is impossible if g = 1, because ( = 1. Hence g > 1 which proves (1). On the other hand, we will show that £ := e2m/a = 1 is a root of /.Indeed, then, as g divides a + 1 and c - b + 1, we have £a+1 = {c-6+1 = 1, which yields £c+1 = Consequently,
(1 - of (o = 1 - r+1 - ec+1+e6 = 1 -1 + 0 = 0,
giving f ) = 0-
In the second case, we have (c+1 = 1 and ^b-a-1 = 1. Therefore, (c+1 = (c-b+a+2 = 1. As above, putting g1 := gcd(c + 1, c - b + a + 2) we derive that (9l = 1. Hence g1 > 1 which proves (2). On the other hand, we will show that £ := e?m/gi = 1 is a root of /.Indeed, then £c+1 = ^c-b+a+2 = 1 which yields (b = £a+1. Consequently,
(1 - of (o = 1 - ec+1 - r+1+e6 = 1 -1 + 0 = 0,
giving f ) =
In the third case, we obtain (b = -1 and (c-a = -1. Hence (c-a-b = 1, and so (\c-a-b\ = 1, From (b = -1 we tod that ( = ¿Kl(2k+1)/b for some k £ Z. So, using (\c-a-b\ = 1, we obtain
eKi(2k+1)\c-a-b\/b = 1
It follows that (2k + 1)|c - a - bl/b must be an even integer, which is only possible when u2(|c - a - b\) > v2(b). This implies (3). To prove that the condition (3) is sufficient, we assume that V2(b) = t > 0 and ^(|c - a - b\) = s > t +1. Then b = 2\2q + 1) and |c - a - b\ = 2s(2l + 1), where q,£ > 0 are integers. (Here, we use the fact that c = a + b.) Putting £ := em/2 , we deduce that
£b = e2t(2q+1)ni/2t = &(2q+1)^i = -
and
^\c-a-6\ = p2s(2e+1)ni/2t = e2s-t(2£+1)ni = 1.
Thus 1 + = 0 and £c-a-b = 1, which yields £c+1 = -£°+1. It follows that £ is a root of (1 - x)f (x) = 1 - xa+1 + xb - xc+1. Since ( = 1, it is a root of f. This completes the proof of Theorem 1.
3. Proof of Theorem 3
To derive Theorem 3 from Theorem 2 we first observe that (6) implies
n = 2<p(2l+1)k = e<p(2l+1)k log 2 > elog d = d,
so 1 + x +-----+ xd-2 + xn is indeed a Newman polynomial of length d. Next, from (6),n = 2v(2l+1">k,
d > 3 and the trivial inequality v2(c) < log2 c for c £ N, it follows that
V2(n) = <p(2l + 1)k > log2 d > log2(d — 2) > V2(d — 2),
so (5) does not hold. To show that (4) does not hold as well, we need to prove that the numbers n + 1 = 2^(2l+1)k + 1 and d = 2m(2l + 1) are coprime. By Euler's theorem, 2^(2l+1)k = 1 (mod 21 + 1). Consequently, n + 1 = 2v(2l+1">k + 1 modulo 21 + 1 is 2. Combining this with the fact that n + 1 is odd, we derive that
gcd(n + 1,d)= gcd(n + 1, 21 + 1) = gcd(2,21 + 1) = 1. This completes the proof of Theorem 3.
4. Another SGFIGS of Newman polynomials without unimodular roots
Finally, in order to give one more alternative (and very short) proof of the fact that ^(d) > 0 for each d > 3 we recall the following result of Filaseta, Finch and Nicol (see the proof of Theorem 4.1 in [5]):
Lemma 2. There is an infinite sequence of nonnegative integers S := ^ < s2 < s3 < ...} such that for every finite set T C S the polynomial 1 + teT x4 irreducible over the rationals.
Fix d > 3 and take any T C S with d — 1 elements, for instance, T = {i1 < ■ ■ ■ < td-1} C S. Then
f (x) := 1 + £ x4t £Nd. teT
We claim that f has no roots on the unit circle. Indeed, if ( is a root of f ^^fefving | = 1 then so is ( = 1/^, and hence the minimal polynomial g of ( over Q is reciprocal. Since and, by Lemma 2, / is irreducible, we must have g = f. ft follows that f is reciprocal, namely, f (x) = xD f (1/x), with D = 4td-1. However, the identity
xD + ^ xD-4t = xD f (1/x) = f (x) = 1 + ^ x4t teT teT
does not hold in view of
D — (d — 4ti) = 4ti < 3 ■ 4td-2 < 4td-i — 4td-2 = p — 4td-2.
Hence f has no roots on the unit circle, as claimed.
Acknowledgement. This research was funded by a grant (No. S-MIP-17-66/LSS-110000-1274)
from the Research Council of Lithuania.
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СПИСОК ЦИТИРОВАННОЙ ЛИТЕРАТУРЫ
1. Boyd D. W. Large Newman polynomials // in: Diophantine analysis (Kensigton, 1985), London Math. Soc. Lecture Note Ser. Vol. 109, Cambridge Univ. Press, Cambridge, 1986. P. 159-170.
2. Campbell D.M., Ferguson H.R. P., Forcade R. W. Newman polynomials on |z| = 1 // Indiana Univ. Math. J. 1983. Vol. 32. P. 517-525.
3. Dubickas A. Nonreciprocal algebraic numbers of small measure // Comment. Math. Univ. Carolin. 2004. Vol. 45. P. 693-697.
4. Goddard B. Finite exponential series and Newman polynomials // Proc. Amer. Math. Soc. 1992. Vol. 116. P. 313-320.
5. Filaseta M., Finch C., Nicol C. On three questions concerning 0,1-polvnomials //J. Théorie des Nombres Bordx. 2006. Vol. 118. P. 357-370.
6. Finch C., Jones L. On the irreducibilitv of {—1, 0,1}-quadrinomials // Integers. 2006. Vol. 6, A16. 4 P.
7. Mercer I. Newman polynomials, reducibilitv and roots on the unit circle // Integers. 2012. Vol. 12, A6. 16 P.
8. Mercer I. Newman polynomials not vanishing on the unit circle // Integers. 2012. Vol. 12, A67. 7 P.
0, 1
1999. Vol. 88. P. 333-350.
10. Luo J. J., Ruan H.-J., Wang, Y.-L. Lipschitz equivalence of Cantor sets and irreducibilitv of polynomials // Mathematika. 2018. Vol. 64. P. 730-741.
11. Smyth C.J. Some results on Newman polynomials // Indiana Univ. Math. J. 1985. Vol. 34. P. 195-200.
REFERENCES
1. Bovd, D.W. 1986, "Large Newman polynomials", in: Diophantine analysis (Kensigton, 1985), London Math. Soc. Lecture Note Ser., vol. 109, Cambridge Univ. Press, Cambridge, pp. 159170.
2. Campbell, D.M., Ferguson, H.R. P. k, Forcade, R. W. 1983, "Newman polynomials on |z| = 1", Indiana Univ. Math. J., vol. 32, pp. 517-525.
3. Dubickas, A. 2004, "Nonreciprocal algebraic numbers of small measure", Comment. Math. Univ. Carolin., vol. 45, pp. 693-697.
4. Goddard, B. 1992, "Finite exponential series and Newman polynomials", Proc. Amer. Math. Soc., vol. 116, pp. 313-320.
0, 1
Théorie des Nombres Bordx., vol. 118, pp. 357-370.
{-1, 0, 1}
A16, 4 pp.
7. Mercer, I. 2012, "Newman polynomials, reducibilitv and roots on the unit circle", Integers, vol. 12, A6, 16 pp.
8. Mercer, I. 2012, "Newman polynomials not vanishing on the unit circle", Integers vol. 12, A67, 7 pp.
0, 1
Arith., vol. 88, pp. 333-350.
10. Luo, J. J., Ruan, H.-J. k, Wang, Y.-L., 2018, "Lipschitz equivalence of Cantor sets and irreducibilitv of polynomials", Mathematika, vol. 64, pp. 730-741.
11. Smyth, C.J. 1985, "Some results on Newman polynomials", Indiana Univ. Math. ,J., vol. 34, pp. 195-200.
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