Cosine polynomials with few zeros

In a celebrated paper, Borwein, Erd\'elyi, Ferguson and Lockhart constructed cosine polynomials of the form \[ f_A(x) = \sum_{a \in A} \cos(ax), \] with $A\subseteq \mathbb{N}$, $|A|= n$ and as few as $n^{5/6+o(1)}$ zeros in $[0,2\pi]$, thereby disproving an old conjecture of J.E. Littlewood. Here we give a sharp analysis of their constructions and, as a result, prove that there exist examples with as few as $C(n\log n)^{2/3}$ roots.


Introduction
In his 1968 monograph, J.E. Littlewood [18] collected many interesting problems on the behavior of polynomials and trigonometric sums with restricted coefficients, a subject on which he worked extensively throughout his life [13][14][15][16][17][19][20][21][22]. Here we concern ourselves with Problem 22: "If the a j 's are all integral and all different, what is the lower bound on the number of real zeros of n i=1 cos(a j t). Possibly n − 1 or not much less." In 2008, Borwein, Erdélyi, Ferguson and Lockhart [6] showed that there exist examples of cosine polynomials having as few as O(n 5/6 log n) roots, thereby disproving the "n − 1 or not much less" part of the statement and giving a upper bound on the minimum number of zeros of a {0, 1}-cosine polynomial. Despite considerable interest in Littlewood's problem, no progress has been made in the past ten years on the upper bound of [6]. In this paper, we show that there exist {0, 1}-cosine polynomials with as few as n 2/3+o (1) roots. To state this result, define Z(f ) to be the number of zeros of f in the interval [0, 2π]. Obtaining good lower bounds on the minimum number of zeros has also remained a challenging problem. In 2007, Borwein and Erdélyi [5] showed that the number of roots tends to infinity along certain subsequences of functions. In the following year, Borwein, Erdélyi, Ferguson and Lockhart [6] then explicitly conjectured that the number of roots tends to infinity as n → ∞, in general. This conjecture was independently proved by Erdélyi [9] and Sahasrabudhe [24]. In fact, Sahasrabudhe went further and gave the explicit lower bound (log log log n) 1/2+o (1) , for the number of roots. While the exponent of 1/2 + o(1) was later improved to 1 + o(1) by Erdélyi [10], the gap between the upper and lower bounds remains large.
To prove Theorem 1.1, we give a precise analysis of a class of random polynomials introduced by the authors of [6]. For m 0 define f to be the random polynomial where ε 0 , . . . , ε m ∈ {0, 1} are independent Bernoulli random variables such that P(ε i = 0) = P(ε i = 1) = 1/2. This defines a probability measure on the set of {0, 1}-cosine polynomials and we write F m,n for the probability space obtained in this way. Our second main result gives a sharp bound on the expected number of zeros 1 of f ∼ F n,m .
We observe that (n log n) 2/3 is the minimum of this expectation if we optimize over m 0, thus accounting for the bound obtained in Theorem 1.1.
Our main new technical result in this paper (Theorem 1.4) is a tool that gives a sharp, deterministic bound on the number of zeros in short intervals for polynomials of the special form (1). The standard tool for establishing upper bounds on the number of roots in short intervals is the following famous theorem 2 of Erdős and Turán [11]. For this, we let Z I (f ) denote the number of zeros of f in I ⊆ [0, 2π]. While there are various interesting strengthenings of this theorem [1,8,26,27], these results appear to be too weak to improve upon the n 5/6+o(1) bound, obtained by the authors of [6], and it is not even clear that it is possible to overcome this barrier, as Theorem 1.3 is known to be sharp, or close to sharp, in many cases. On the other hand, it is conceivable that the work of Blatt [2] on simple roots could be adapted to this setting to overcome the n 5/6 barrier. However, there are additional complications with this approach and, even if these were to be successfully navigated, it appears that it would still fall short of the optimal bound. Our new technical tool allows us to overcome the n 5/6 barrier and obtain a sharp bound on the number of zeros. Indeed, this result can be viewed a sort of Erdős-Turán result for polynomials of the form (1), which performs much better on short intervals. As an added bonus, our proof of Theorem 1.4 is entirely elementary, whereas the proof of Theorem 1.3 uses methods from Fourier and complex analysis. For the statement of this result, let us write D n (x) = n k=0 cos kx.
where C > 0 is an absolute constant.
Note that Theorem 1.4 does not apply when deg(g) = n 1−o (1) and to deal with this case, we need to develop a slightly more sophisticated version of Theorem 1.4 (See Lemma 5.2 in Section 5).
Taken a little more broadly, Theorem 1.4 can be situated in a group of results and conjectures that say that the roots of polynomials with restricted coefficients can't "clump up" too much [3,8,25,27]. One extreme version of this is the maximum multiplicity of a polynomial at a point; i.e. when the "interval" has length zero. One problem in this domain, perhaps pertinent to our work here, asks for the maximum multiplicity at 1 of a degree n {±1}-polynomial. If we let m(n) be the maximum multiplicity, the best bounds [4,7] to date are log n m(n) c(log n) 2−o(1) .
In a similar vein, the Tarry-Escott Problem, asks for a lower bound on the minimum number of non-zero terms t(n) in a polynomial with coefficients in {0, ±1} and a root at 1 of multiplicity n. Here, the best bounds [4] to date are of the form 2n t(n) n 2+o(1) .

1.1.
Sketch of the proofs of Theorems 1.1 and 1.2. To prove Theorems 1.1 and 1.2, we follow [6] and use the special property of the polynomial D n . This motivates the definition of the following set which we think of as the set of all x where f (x) might have a zero. In the case m < n 1−α , we obtain a close relationship between the number of zeros of f and the measure of the set E(g). To do this, we write E(g) as a union of O(m) intervals and then apply Theorem 1.4 to each of these intervals to obtain an upper bound on the number of zeros in terms of the measure of the set E(g) ⊆ [0, π]. This is the content of Lemma 3.1, where we additionally get a similar lower bound on the number of zeros Up to this point our results have been entirely deterministic and to finish the proof of the upper bound in Theorem 1.1 and Theorem 1.2 (in the case that m < n 1−α ), we simply need to show that E |E(g)| = Θ log m m 1/2 . This we do in Lemma 4.1. To finish the proof of the lower bound on the expectation E Z(f ) in Theorem 1.2, we need to additionally show that f (x) has, on average, at least cm zeros away from x = 0. This is what corrects the "−m" seen in the lower bound in (2).
In the case m > n 1−α , the proof is similar but the analysis becomes more complicated. Here we additionally consider the set E ′ n (g) := {x : |g ′ (x)| 2 7 ns(x)}, and then prove a version of Theorem 1.4 for intervals that are not entirely contained in E ′ n (g) (Lemma 5.2). To deal with roots inside of E ′ n (g), we show that the number of intervals contained in E ′ n (g) (that is, the number of intervals where roots might "clump up") is likely to be small. This is ultimately achieved by appealing to a beautiful anti-concentration result of Halász [12], which we state as Theorem 5.5.

Roots in short intervals
In this section we prove Lemma 2.1, which we will use to show that our polynomial does not have multiple roots in an interval of length ≈ 1/n. Lemma 2.1. For δ = 2 −14 , α ∈ (0, 1), let I ⊆ (2 31 (αn) −1 , π] be an interval with |I| = δ/n, let g be a {0, 1}-cosine polynomial with deg(g) n 1−α and let f = D n − g. Then f has at most 2α −1 roots in I. Theorem 1.4 is immediate from Lemma 2.1: we simply cover the given interval I with intervals of length 2 −14 /n and then apply Lemma 2.1 to deduce that each of these intervals contains at most 2α −1 roots. Now, turning towards the proof of Lemma 2.1, we record the following consequence of the mean value theorem, the proof of which is deferred to the appendix.
We shall also need an elementary lemma concerning "high" derivatives D (r) n of the function D n . To prepare for this, we need the following two basic facts. Again, the proofs of these can be found in the appendix.
. For x ∈ (0, π] and r ∈ N, we have that be an interval of length δ/n. Let C r (x) = (d/dx) r sin(x) and set T = (n + 1/2). Then there exists a point x 0 ∈ I such that We are now able to prove Lemma 2.5. For this, we use the formula for x ∈ (0, 2π), T = n + 1/2.
We are now ready for the key idea behind Lemma 2.1: if f = D n − g contains many roots in an interval of length ≈ n −1 , then some high derivative g (t) of g must be impossibly large.
Lemma 2.6. For t ∈ N and δ = 2 −14 , let I ⊆ [2 30 (t + 2)/n, π] be an interval of length δ/n, let g be a trigonometric polynomial and let f = D n − g. If f has at least t + 2 roots in I then there exists x 0 ∈ I for which Proof. Since I ⊆ [2 30 (t + 2)/n, π] is an interval of length δ/n, Lemma 2.5 guarantees a point for all r ∈ N. Now, for all x ∈ I, and all 1 r t + 2, we may apply Statement 1 in Lemma 2.5 to obtain Since f has t + 2 roots in I, f (t) has at least 2 roots in I. Thus, we may apply Lemma 2.2 to f (t) and (7) to obtain We now specify x = x 0 to learn

By rearranging and using the lower bound |D
Proof of Lemma 2.1. We may assume that Z I (f ) α −1 + 2, otherwise we are done. We write t = α −1 and note that since f has t + 2 roots in I, we may apply Lemma 2.6, to see that On the other hand, for all r ∈ N, we have the crude bound So putting this together with (8), we get δn t /(2x 0 ) 2m t+1 . But this contradicts t = α −1 and m < n 1−α .

A deterministic bound on zeros in terms of |E(g)|
In this section we prove Lemma 3.1, which provides a deterministic bound on the number of zeros of polynomials of the form f = D n −g, where deg(g) n 1−α , in terms of the measure of the set E(g).
where C α > 0 is a constant only depending on α.
We urge the reader to ignore the n 0.6 term in (9), as it is both immaterial in our application and probably not really necessary.
To prove Lemma 3.1, we first show that E(g) can be written as the sum of at most O(m) intervals, a fact that holds simply by the degree of g. Recall that and s(x) = (2 sin(x/2)) −1 .  Proof. If f (x) = 0 then, using the formula (4), we see that and re-arranging yields |g(x) − 1/2| s(x), as desired.
We also have the following companion to this observation; that f is forced to have quite a few roots when |g(x) − 1/2| s(x).
Proof. In any sub-interval J ⊆ I of length 2π/T , there exist x 0 , x 1 ∈ J, such that D n (x 0 ) = 1/2 − s(x 0 ) and D n (x 1 ) = 1/2 + s(x 1 ). Since g is continuous and |g| < s(x) the curves D n and −g cross, resulting in a zero of f . To finish, simply note that there are T 2π |I| pairwise disjoint intervals of length 2π/T in the interval I.
We now turn to finish the proof of Lemma 3.1.
Proof of Lemma 3.1. Since f is a cosine polynomial, it is symmetric about the origin and periodic with period 2π. It is therefore sufficient to count roots in [0, π]. We start by proving the upper bound at (9). First note that by Theorem 1.3 (the Erdős-Turán Theorem) there are at most O(n 0.6 ) roots in the interval [0, n −0.9 ]; so in what follows we may assume that x n −0.9 and, in particular, x ≫ n −1 . Now, by Lemma 3.2, E ′ := [n −0.9 , π] ∩ E(g) can be expressed as the union of t 9m intervals J 1 , . . . , J t . By Observation 3.3, all of the zeros of f lie in E(g). So applying this observation along with Lemma 1.4 to each of these intervals, we have as desired. Putting this together with the bound on Z [0,n −0.9 ] (f ), yields the upper bound in Lemma 3.1.
For the lower bound, we need only to apply Observation 3.4 to each of the intervals J 1 , . . . , J t to obtain as desired.

Two probabilistic calculations
For m ∈ N, we let g = g m be the random polynomial where ε 1 , . . . , ε m ∈ {0, 1} are independent Bernoulli random variables such that P(ε k = 0) = P(ε k = 1) = 1/2 and we let G m denote the corresponding the probability space on cosine polynomials of degree m. We now turn to calculate the expected size of |E(g)|.
Lemma 4.1. Let g ∼ G m . Then To prove Lemma 4.1, we shall use a version of the classical theorem of Berry and Esseen, which roughly says that the random variable (g(a), g(b)) behaves quite a bit like a twodimensional normal distribution with the same mean and covariance matrix. Recall that if (X, Y ) is a random variable taking values in R 2 , then its covariance matrix is defined by The following theorem can be easily derived from the the main theorem of [23]. If Z ∼ N(µ, Σ) then for all measurable convex sets C we have where c > 0 is an absolute constant.
We shall first use a "one dimensional version" of the Theorem 4.2, which says that if the random variables X 1 , . . . , X n take values in R, then where Z ∼ N(µ, σ 2 ), σ 2 = i Var(X i ) and µ = i EX i . To derive this statement from Lemma 4.2, simply apply Lemma 4.2 to the random variablesX i : are independent copies of X i and consider the convex set C t := {x : x t} 2 . Before diving into the proof of Lemma 4.1, let us set out a few of the basic probabilistic quantities in play. As we look to apply Theorem 4.2 to the random sum g(x), we note that and for x such that 3 d(x, πZ) 2π/m, we have that σ 2 (x) := Var(g(x)) is where we have applied the inequality |D m (2x)| m/2 when d(x, πZ) 2π/m. Finally, we see that the sum of the third moments of the summands is Proof of Lemma 4.1. We look to apply the above Berry-Esseen inequality, in one dimension, to each sum g(x) and to the set {y : y t}. Indeed, if we denote by Z(x) the Gaussian random variable with the same first two moments of g(x), we may apply Theorem 4.2 to learn by using (12) and (13). Now, turning to the left-hand-side of (10), we express Since s(x) m/16, when x satisfies d(x, πZ) 2π/m, we have for such x And so, by removing the first removing the intervals J 1 := [0, 2π/m], J 2 := [π − 2π/m, π] from the integral, we have π 0 P (|g(x)| s(x)) dx = as desired.
One can see that Lemma 4.1 along with Lemma 3.1 implies the upper bound on E Z(f ), in Theorem 1.2. For the lower bound we need one further probabilistic calculation. . We note that |D n (x)| < 10, for x > π/4, so the event "g(a) > 10 and g(b) < −10" is enough to guarantee a root of f . For this, we look to apply Theorem 4.2 to the R 2 -valued random variable So we may apply Theorem 4.2 along with our lower bound on the standard deviation at (12) to learn that To prove this Lemma, we introduce a set related to E(g), The following lemma shows the utility of this definition: if a short interval I is not entirely contained in E ′ n (g) then f does not have too many roots in I. We point out that Lemma 5.2 is a sort of companion to Lemma 2.6 and, indeed, a similar idea is applied. Lemma 5.2 will be applied with δ = n −0.1 and so this lemma is telling us that Z I (f ) = O(1).
Rearranging gives, δ −t 3n and so if t = log n this automatically gives a contradiction. Thus t + 1 = Z(f ) and so we arrive at the desired inequality.
We shall also need the following observation.
Observation 5.3. For δ ∈ (0, 1), let I ⊆ [4/n, π] be an interval of length |I| = δ/n and let g be a differentiable function on [0, π]. If I ⊆ E ′ n (g) and I ∩ E(g) = ∅ then I ⊆ E + (g) := {x : |g(x) − 1/2| 4s(x)}. (22) Proof. Let x ∈ I and x 0 ∈ E(g) ∩ I. By the mean value theorem, there exists x 1 ∈ I for which The following Lemma will later be applied to show that |E(g) ∩ E ′ n (g)| is small. Lemma 5.4. For m ∈ N and x with d(x, πZ) To prove Lemma 5.4, we apply the following Theorem of Halász ( [12], Theorem 1), which has been rephrased and simplified slightly for our purposes.
). Now, by Lemma 2.5, we see that for x satisfying d(x, πZ) Since a k , v a k v 1, there must be at least 2 −4 vectors with a k , v 2 −5 . We therefore may apply Theorem 5.5 to finish the proof of the 5.4. Lemma 5.6. For m n, we have Proof. We proceed as we did in Lemma 4.1, except we use Halász's theorem (Theorem 5.5) instead of the Berry-Essen-type theorem, Theorem 4.2. Note that it is enough to only consider x for which d(x, {0, π}) C/m as our final bound n 1.1 /m 2 is larger than 1/m. We want to obtain an upper bound on the quantity To do this, note that we can cover the box {v ∈ R 2 : |v 1 | 1/x, |v 2 | 2 7 n/(mx)} with t = O(n/(mx 2 )) translates B 1 , . . . , B t of {v ∈ R 2 : |v 1 | 2 −3 , |v 2 | 2 −3 }, which has diameter < 2 −5 . Thus we have where we have applied Lemma 5.4 to each summand. To finish, we simply note that as desired.
We now turn to prove Lemma 5.1, the main objective of this section.
Proof of Lemma 5.1. We first notice that Theorem 1.3 tells us that and so we may assume that x n −0.1 in what follows. We set δ = n −0.1 and let I be a partition of [n −0.1 , π] into intervals of length δ/n. From Observation 3.3, we know that if I contains a root then we must have I ∩ E(g) = ∅. We call such an interval dangerous. To count the number of dangerous intervals we notice they come in two types; call a dangerous interval an interior interval if I ⊆ E(g) and call a dangerous interval I a boundary interval if I ⊆ E(g). Lemma 3.2 tells us that there are at most O(m) boundary intervals. On the other hand, if we let a(f ) be the number of interior intervals we see that a(f )(δ/n) |E(g)| and so, applying Lemma 4.1, we have where the last inequality follows from the fact that m > n 0.9 and δ = n −0.1 . Putting these observations together, we see that there are at most m dangerous intervals, in expectation. We now consider two further types of dangerous intervals. We call a dangerous interval, bad if I is contained in E ′ n (g) and call an interval good if I is not contained in E ′ n (g). 13 If I is a good interval, we note that I ⊆ [n −0.1 , π] and so we can apply Lemma 5.2 to see that Z I (f ) = O(1) and therefore the expected number of roots in good intervals is at most O(m).
We now count the number of roots in bad intervals. By Lemma 5.3, we see that if I ⊆ [n −0.1 , π] is bad then 4 I ⊆ E ′ n (g) ∩ E + (g) ∩ [n −0.1 , π]. So, if we let b(f ) be the number of bad intervals in I, we have b(f )(δ/n) |E ′ n (g) ∩ E + (g) ∩ [n −0.1 , π]|. So taking expectations and applying Lemma 5.6 yields where we have applied Lemma 5.6. We can apply Theorem 1.3 to each of these bad intervals to conclude that the number of zeros that f has in bad intervals is at most n|E ′ n (g) ∩ E + (g)| + Cb(f )(n log n) 1/2 , where C > 0 is an absolute constant. And therefore, using Lemma 5.6 and (23), we have that the expected number of roots in bad intervals is at most which is O(m), since n 2.61 δ −1 /m 2 n 2.71 /n 1.8 = n 0.91 m. This finishes the proof.

Proofs of main theorems
All that remains is to put the pieces together and prove Theorem 1.1 and Theorem 1.2.
Proof of Theorem 1.2. We begin with the proof of the following lower bound. If we put Now let us consider the case m < n 0.99 . With this assumption in hand, we may apply Lemma 3.1 and then take expectations to see that E Z(f ) C nE |E(g)| + m + n 0.6 = O n log m m 1/2 + m .
Likewise, for the lower bound, we have E Z(f ) = Ω n log m m 1/2 − m .
In the case that m n 0.99 , the upper bound in Theorem 1.2 follows from Lemma 5.1, while (24) furnishes a matching lower bound.
There is a tiny hiccup in the direct derivation of Theorem 1.1 from Theorem 1.2 as we do not explicitly control the number of terms in the resulting polynomial and therefore we cannot claim the theorem for all sizes of |A|. To get around this, we instead derive Theorem 1.2 from Lemma 3.1.
Proof of Theorem 1.1. Note that it is enough to prove Theorem 1.1 for all sufficiently large N := |A|. Now, given N sufficiently large, put m = (N log N) 2/3 . By Lemma 4.1, we know that there is exists a polynomial with 0 t m terms and |E(g)| = O log m m 1/2 . Choose n so that N = n − t. By Lemma 3.1 we have for an absolute constant C > 0.

acknowledgments
We thank Béla Bollobás and Rob Morris for comments.