aa r X i v : . [ m a t h . N T ] J u l On some power sum problems of Montgomeryand Tur´an
Johan Andersson ∗ October 24, 2018
Abstract
We use an estimate for character sums over finite fields of Katz to solveopen problems of Montgomery and Tur´an. Let h ≥ | z k | =1 max ν =1 ,...,n h | P nk =1 z νk | ≤ ( h − o (1)) √ n. This givesthe right order of magnitude for the quantity and improves on a bound ofErd˝os-Renyi by a factor of the order √ log n . One of the simplest and most striking problems in the Tur´an power sum theoryis to determine the quantity( ⋆ ) = inf | z k | =1 max ν =1 ,...,m (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) for various choices of integers n, m . The purpose of this paper is to find thecorrect order of magnitude for this quantity when m ∼ n B . For any fixed B > m = ⌊ n B ⌋ we prove that ( ⋆ ) ≍ √ n . This solves open problems of HughMontgomery and Paul Tur´an.We first present some well known results about ( ⋆ ). By letting z k = e (cid:18) kn (cid:19) , ( e ( x ) = e πix )we see that ( ⋆ ) = 0 if 1 ≤ m ≤ n −
1. Tur´an [13] proved that ( ⋆ ) = 1 if m = n .In general it is difficult to determine the quantity ( ⋆ ) exactly but in [2] we provedthat ( ⋆ ) = √ n − m = n − n and n − m = n − j for 2 ≤ j ≤ n − n is a prime power then ( ⋆ ) = √ n . ∗ Email:[email protected] ⋆ ) when n, m → ∞ . In the special case n δ < m ≤ n we have proved in [3] that ( ⋆ ) ∼ √ n .Finding asymptotic estimates can be quite difficult as well and the next ques-tion is to ask for choices of m, n where we can find the order of magnitude for( ⋆ ). Montgomery proved ([12], page 100, Theorem 10) that √ nB ≪ max ν =1 ,...,n B (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (1)uniformly for 1 + δ ≤ B ≤ n when | z k | = 1. Erd˝os-Renyi [9] have used proba-bilistic reasoning to show that there exists an n -tuple ( z , . . . , z n ) of unimodularcomplex numbers such that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ p n log( m + 1) ( ν = 1 , . . . , m ) (2)thus giving a general upper bound for ( ⋆ ). Let us now consider the case m = ⌊ n B ⌋ for a fixed constant B >
1. Leenman-Tijdeman [11] have given an explicitconstruction which yields the same order of magnitude as Eq. (2). Unfortunatelythese upper estimates differ by a essentially a √ log n factor from the lower boundEq. (1). This inspired Hugh Montgomery to state the following problem Problem 1. (Montgomery, [12], page 197, Problem 13) Show that for any posi-tive B there exist complex numbers z , . . . , z n such that | z k | = 1 for all k and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≪ B √ n. ( ν = 1 , . . . , ⌊ n B ⌋ )In contrast Tur´an had previously stated the following problem. Problem 2. (Tur´an, [14] page 197, Problem 54) Does there exist an w ( x ) ր ∞ such that b j > , | z j | = 1 , j = 1 , . . . , n implies for g ( ν ) = P nj =1 b j z νj the inequalitymax ≤ ν ≤ n | g ( ν ) | > w ( n ) √ n | g (0) | ?If we can solve Montgomery’s problem for B = 100 it is clear that there doesnot exist such a function w ( x ) in Tur´an’s problem. These conflicting guesses anda general lack of understanding of the situation has meant that until very recentlyit has not been clear to us what the correct order of magnitude in the problemshould be. See the discussion in arXiv version 2 of [3].2 Main results
The purpose of this paper is to solve Montgomery’s problem and give the rightorder of magnitude for ( ⋆ ) if m = ⌊ n B ⌋ . We will use an estimate of character sumsover finite fields of Katz. Before we state our theorems we state the followingLemma which we will prove in Section 3. Lemma 1.
Let h ≥ be an integer and let q be a prime power. Then there existsunimodular complex numbers z , . . . z q such that max ν =1 ,...,q h − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( h − √ q. We first state a non explicit version of our result which has a somewhat easierproof than our sharpest version.
Theorem 1.
Let δ > . Then we have uniformly for δ ≤ B ≤ n the followingestimate √ Bn ≪ inf | z k | =1 max ν =1 ,..., ⌊ n B ⌋ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≪ B √ n. If | z k | = 1 is replaced by | z k | ≥ then the result still holds for ≤ B ≤ n . This Theorem gives a direct solution to Problem 1 of Montgomery and im-proves on Erd˝os-Renyi’s result Eq. (2) for 2 < B ≪ log n by removing a factor p (log n ) /B . Proof.
The lower bound is Eq. (1). For the upper bound it is sufficient to consider B = h for integers h ≥
2. Since 2 m is a prime power we can choose { z k,m } m k =1 inLemma 1 such that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) m X k =1 z νk,m (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) ⌈ Nm ⌉ − (cid:19) m/ . ( ν = 1 , . . . , N − a m = 0 , n , i.e. n = P Mm =0 a m m , thenthe sum P Mm =0 a m P m k =1 z νk,m is a power sum of n elements and can be written as P nk =1 z νk . By the triangle inequality it is clear that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ M X m =0 a m (cid:18) ⌈ Nm ⌉ − (cid:19) m/ ≪ NM M/ for ν = 1 , . . . , N −
2. By the binary expansion of n we have that n ≤ M +1 − n h ≤ N − n, h ≥ N = h ( M + 1). We obtain thefollowing estimate. (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≪ h √ n. ( ν = 1 , . . . , n h )3heorem 1 implies the following Corollary. Corollary 1.
It is not possible to find a function w ( x ) in Tur´an’s problem, Prob-lem 2.Proof. Choose B = 100 in Theorem 1 and b = · · · = b n = 1 in Tur´an’s problem.We will now state a somewhat sharper version of Theorem 1, which we willchoose to state for | z k | ≥ | z k | = 1. Theorem 2.
Let h ≥ be an integer and ǫ > . One then has that C h √ n − O ( n − / ) ≤ inf | z k |≥ max ν =1 ,...,n h (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( h − √ n + O (cid:0) n . ǫ (cid:1) . where C m = ( m !) / m and C m +1 = C m . We will prove this Theorem in Section 4.
Remark 1.
Theorem 2 still holds if | z k | ≥ | z k | = 1 since theconstruction that yields the upper bound uses unimodular complex numbers. Inthis case the lower bound in Theorem 2 can be improved slightly by using themethod from [4], e.g. for h = 3 we get the lower bound √ n instead of √ n . Atpresent there is however no h > h = 3 in Theorem 2 (with | z k | = 1) we get the following Corollary. Corollary 2.
Let α > be a real number. One then has that inf | z k | =1 max ν =1 ,..., ⌊ αn ⌋ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ √ n + O (cid:0) n . ǫ (cid:1) . ( ǫ > α > Remark 2.
In the proof of Theorem 2 we use the methods of [3], Section 6. Thesimpler methods of [3], Subsection 4.1 can be used to prove a weaker form ofTheorem 2 that still implies the Corollary.We will here suggest the following open problems (Compare with [3], Problem3). 4 roblem 3.
Find an increasing function Λ( x ) such that for each B > | z k | =1 max ν =1 ,..., ⌊ n B ⌋ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∼ Λ( B ) √ n. Question 1.
Is the upper bound in Theorem 2 sharp? Can we choose Λ( h ) = h − h ≥ Problem 4.
Let h ≥ h ( α ) suchthat for each α > | z k | =1 max ν =1 ,..., ⌊ αn h ⌋ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∼ Λ h ( α ) √ n. For h = 2 this is given as Conjecture 4 in [3]. Question 2.
Assume that we can solve Problem 4. Is it true that lim α →∞ Λ h ( α ) =lim α → + Λ h +1 ( α )?If Question 2 is true and Problem 3 solvable then we have that the functionΛ( x ) in Problem 3 is piece wise constant for n < x < n + 1. Proof.
Let F be a finite field of order q and let E be an extension field of F oforder q h . Let x , . . . , x q denote the elements of F . Let ω be an element thatgenerates the multiplicative group E ∗ , and let χ be a multiplicative character on E of order q h −
1. Choose z k = χ ( ω + x k ) . ( k = 1 , . . . , q )Then q X k =1 z νk = X x ∈ F χ ν ( ω + x ) , (3)where χ ν = χ ν is a non trivial character on E unless ( q h − | ν . By Theorem 1of Katz [10] its absolute value can be estimated from above by ( h − √ n for ν = 1 , . . . , q h −
2. 5 emark 3.
The character sum Eq. (3) first occurred as eigenvalues of adjacencymatrices of graphs in Chung [8]. Katz proof uses Weil’s estimates [15], althoughhe prefers to use the language of Deligne.
Remark 4.
For h = 2 the construction in Lemma 1 coincides with our con-struction in Eq. (7) of [2], where we used a result of Bose [6] on B sequences(mod q − B h Sidon sequences (mod q h −
1) in Bose-Chowla [7], and this was how we originally arrived at the constructionin Lemma 1. We will therefore describe the construction of Bose-Chowla. Let F and E be a finite fields and ω an element in E which is a multiplicative generatorfor E ∗ as as in the proof of Lemma 1. Choose the discrete logarithm so thatlog ω = 1. Let a k = log( ω + x k ) . ( k = 1 , . . . , q ) (4)Then { a k } is a Sidon set of order h , or B h set (mod q h − a j + · · · + a j h ≡ a i + · · · + a i h (mod q h −
1) has no non-trivialsolution. Since the case h = 2 had been successfully used in the problem it wasnatural to expect that this construction could give good estimates for a general h ≥ ν = 1 , . . . , p h −
2. Indeed, numerical investigation (with h = 3 , z k = e (cid:18) a k q h − (cid:19) , ( k = 1 , . . . , q )then we should have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( h − √ n. ( ν = 1 , . . . , q h − χ by χ ( ω n ) = e (cid:18) nq h − (cid:19) , χ (0) = 0 . Then z k = χ ( ω + x k ) , where χ is a character of order q h − E ∗ as in the proof of Lemma 1. The lower estimate follows from Theorem 2 in our paper [1] (cid:18) n ! m !( n − m )! (cid:19) / (2 m ) ≤ max ν =1 ,...,n m (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . ( | z k | ≥ Lemma 2.
Let ǫ > , < θ < , C ≥ and let h ≥ be an integer. Supposethat ( z , . . . , z n ) is an n − tuple of unimodular complex numbers, m ∼ n θ , (5) and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C √ n. ( ν = 1 , . . . , n h ) Let N = { , . . . , n } . Then there exist a subset M ⊂ N , with M = m suchthat (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X k ∈M z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≪ ǫ m / ǫ . ( ν = 1 , . . . , n h ) Proof.
The Lemma is the same as Lemma 9 from [3] except for the fact that ⌊ αn ⌋ is replaced by n h . The proof is the same, but we need the choice N > h θǫ instead of N > θǫ .By the Baker-Harman-Pintz theorem [5] we can choose a prime n < p suchthat p − n ≍ p . . By the construction given in Lemma 1, we can choose a p − tuple ( z , . . . , z p ) of unimodular complex numbers such that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( h − √ p. ( ν = 1 , . . . , p h − m = p − n . By Lemma 2 with θ = 0 .
525 and C = h − M ⊂ { , . . . , p } with M = m such that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X k ∈M z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n . ǫ . ( ν = 1 , . . . , p h −
2) ( ǫ > N = { , . . . , p } \ M . It is clear that N = n and by the triangle inequalityit follows for 1 ≤ ν ≤ n h ≤ p h − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X k ∈N z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p X k =1 z νk − X k ∈M z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p X k =1 z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + O (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X k ∈M z νk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)! , ≤ ( h − p n + O ( n . ) + O (cid:0) n . ǫ (cid:1) , ≤ ( h − √ n + O (cid:0) n . ǫ (cid:1) . which finishes the proof of our Theorem.7 eferences [1] J. Andersson. On some power sum problems of Tur´an and Erd˝os. Acta Math. Hungar. , 70(4):305–316, 1996.[2] J. Andersson. Explicit solutions to certain inf max problems fromTur´an power sum theory, 2006 arXiv:math/0607238 . To Appearin Indagationes Mathematicae.[3] J. Andersson. Tur´an’s problem 10 revisited, 2007. arXiv:math/0609271 .[4] J. Andersson. Lower bounds in some power sum problems, 2007. arXiv:0704.1879 .[5] R. C. Baker, G. Harman, and J. Pintz. The difference betweenconsecutive primes. II.
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