On Zeros of Sums of Cosines
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eros of Sums of Cosines S. V. Konyagin1* 1
Steklov Mathematical Institute of Russian Academy of Sciences, Moscow, 119991 Russia
Received April 29, 2020; in final form, April 29, 2020; accepted May 14, 2020
Abstract—It is shown that there exist arbitrarily large natural numbers N and distinct nonnegative integers n1 , . . . , nN for which the number of zeros on [−π, π) of the trigonometric polynomial N 2/3 log2/3 N ). j=1 cos(nj t) is O(N DOI: 10.1134/S0001434620090254 Keywords: trigonometric polynomials, Dirichlet kernel.
1. INTRODUCTION Littlewood [1, Problem 22] posed the following question. Is it true that, for any different nonnegative integers n1 , . . . , nN , the number of real zeros on [−π, π) of the trigonometric polynomial N
cos(nj t)
(1)
j=1
is at least N − 1 (or slightly less)? For a long time, there were no results concerning this problem. In [2], by using a computer, examples of polynomials with a considerably smaller number of zeros were found and the following theorem was proved. Theorem A. There exists a sequence of numbers Nm ≥ 2, m ≥ 2, such that Nm /m → 1 as m → ∞ m and trigonometric polynomials N j=1 cos(nj t) with different nonnegative integers n1 , . . . , nN for which the number of real zeros on [−π, π) is 5/6 log Nm ) = O(m5/6 log m). O(Nm
Using the same construction as in [2], we improve the upper bound for the number of zeros. Theorem 1. There exists a sequence of numbers Nm ≥ 2, m ≥ 2, such that Nm /m → 1 as m → ∞ m and trigonometric polynomials N j=1 cos(nj t) with different nonnegative integers n1 , . . . , nN for which the number of real zeros on [−π, π) is 2/3 log2/3 Nm ) = O(m2/3 log2/3 m). O(Nm
Note that it has been unknown for a long time that the number of zeros of trigonometric polynomials of the form (1) tends to infinity as N → ∞. This fact was discovered recently in [3]–[5]. *
E-mail: [email protected]
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ON ZEROS OF SUMS OF COSINES
539
2. LOCAL BEHAVIOR OF THE ZEROS OF SPECIAL TRIGONOMETRIC POLYNOMIALS Just as in [2], for a sufficiently large m, we shall seek the required trigonometric polynomial among polynomials of the form Sm (t) = Dm (t) − Pn (t),
(2)
where Dm (t) =
m
cos(jt) =
j=0
1 sin((m + 1/2)t) + , 2 2 sin(t/2)
Pn (t) =
n
aj cos(jt),
aj ∈ {0, 1}.
j=0
In contrast to [2], we take n = n(m) = [m2/3 log2/3 m]. For a sufficiently large m, the inequality n ≤ m holds. Therefore, the polynomial Sm has the required form and, at the same time, the number Nm of its nonzero terms satisfies the inequalities m − n ≤ Nm ≤ m + 1, whence it follows that Nm /m → 1 as m → ∞. Lemma 1. For a sufficiently large m and any integer l, the polynomial Sm cannot have more than four zeros on the closed interval π πl 3π πl + , + . [u, v] = m + 1/2 4m + 2 m + 1/2 4m + 2 Proof of the lemma. Multiplying (2) by 2 sin(t/2), we obtain n+1 1 1 t + sin(t/2) − t , bj sin j + 2Sm (t) sin(t/2) = sin m + 2 2
(3)
j=−1
where |bj | ≤ 2 for j = −1, . . . , n + 1. After differentiating equality (3) four times, we obtain (4) t = R1 (t) − R2 (t), 2Sm (t) sin 2
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