One-sided power sum and cosine inequalities

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1 Available online at Indagationes Mathematicae 24 (2013) One-sided power sum and cosine inequalities Frits Beukers a, Rob Tideman b, a Mathematical Institute, University Utrecht, P.O.Box 80010, 3508 TA Utrecht, The Netherlands b Mathematical Institute, Leiden University, P.O.Box 9512, 2300 RA Leiden, The Netherlands Received 13 January 2012; received in revised form 6 November 2012; accepted 29 November 2012 Communicated by J. Top Abstract In this note we prove results of the following types. Let be given distinct complex numbers z satisfying the conditions z = 1, z 1 for = 1,..., n and for every z there exists an i such that z i = z. Then k z k 1. If, moreover, none of the ratios z i /z with i is a root of unity, then k z k 1 log n. π4 The constant 1 in the former result is the best possible. The above results are special cases of upper bounds for n k b z k obtained in this paper. c 2012 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved. Keywords: Power sums; One-sided inequality; Sums of cosines; Littlewood conecture; Multistep methods 1. Introduction Our colleague Marc N. Spiker asked the following question in view of an application in numerical analysis [6]: Corresponding author. Tel.: ; fax: addresses: f.beukers@uu.nl (F. Beukers), tideman@math.leidenuniv.nl (R. Tideman) /$ - see front matter c 2012 Royal Dutch Mathematical Society (KWG). Published by Elsevier B.V. All rights reserved. doi: /.indag

2 374 F. Beukers, R. Tideman / Indagationes Mathematicae 24 (2013) Problem 1. Is it true that for given real numbers b 1 and distinct complex numbers z satisfying the conditions z = 1, z 1 for = 1,..., n and for every z there exists an i such that z i = z, b i = b we have lim k n b z k 1? Note that by the conugacy conditions on b i, z i the sum n b z k is real for all k. In Section 2 we answer Spiker s question in a slightly generalized and sharpened form (see Theorem 1 and Corollary 1). The solution of Problem 1 has an application to numerical analysis, more particularly Linear multistep methods (LMMs). They form a well-known class of numerical step-by-step methods for solving initial-value problems for certain systems of ordinary differential equations. In many applications of such methods it is essential that the LMM has specific stability properties. An important property of this kind is named boundedness and has recently been studied by Hundsdorfer, Mozartova and Spiker [3]. In that paper the stepsizecoefficient γ is a crucial parameter in the study of boundedness. In [6] Spiker attempts to single out all LMMs with a positive stepsize-coefficient γ for boundedness. By using Corollary 1 below he is able to nicely narrow the class of such LMMs. As a fine point we can remark that the bound 1 in Spiker s problem is the optimal one. Namely, take z = ζ where ζ = e 2πi/(n+1) and b = 1 for all. Then the exponential sum equals n if k is divisible by n + 1 and 1 if not. If, moreover, none of z /z i with i is a root of unity, then the upper bound in Problem 1 can be improved to log n/π 4. We deal with this question in Theorem 3 and more particularly Corollary 2. The obtained results can easily be transformed into estimates for m b cos(2πα k) where b, α are real numbers and α 1,..., α n are strictly between 0 and 1/2. Theorem 4 states that this imum is equal to m t R b cos(2πα t), provided that the Q-span of α 1,..., α n does not contain The general case We provide an answer to Problem 1. Theorem 1. Let n be a positive integer. Let b 1,..., b n be nonzero complex numbers such that b n+1 i = b i for all i = 1, 2,..., n. Let z 1,..., z n be distinct complex numbers with absolute value 1, not equal to 1, such that z n+1 i = z i for all i = 1, 2,..., n. Then b 2 lim k b z k. b Note that n b z k is real because of the conugacy conditions. By applying the Cauchy Schwarz inequality we immediately obtain the following consequence. Corollary 1. Let n, b, z be as in Theorem 1. Then lim b z k 1 b. k n

3 F. Beukers, R. Tideman / Indagationes Mathematicae 24 (2013) Obviously this answers Problem 1, since in that case b R 1 for all. In the special case when the b are positive real numbers we can even drop the distinctness condition on the z. Theorem 2. Let n, b, z be as in Theorem 1 with the additional condition b R >0 for all and let the distinctness condition on the z be dropped. Then the conclusions of Theorem 1 and Corollary 1 still hold. Furthermore, since n b z k is almost periodic we can apply Dirichlet s Theorem on simultaneous Diophantine approximation, and find that the lim coincides with the imum in the above theorems and corollary. Proof of Theorem 1. Put s k = b 1 z k b nz k n. Put c = lim k s k. Note that the z k equal 0 on average for each. Hence the same holds for the values s k and thus we see that c 0. Let ε > 0. Choose N so large that s k > (c + ε) for all k N. For any positive integer K consider the sum Σ 1 := N+K 1 k=n s k. Since none of the z i is 1, we have Σ 1 = N+K 1 k=n b z k = z N z N+K b. 1 z Thus there exists C 1, independent of N and K, such that 2 b Σ 1 1 z = C 1. Define Σ + 1 to be the subsum of Σ 1 of all nonnegative s k and Σ 1 to be minus the subsum of Σ 1 of all negative s k. Let P be the number of nonnegative s k for k = N,..., N + K 1. Then and Σ + 1 Σ 1 + C 1 (K P)(c + ε) + C 1 Σ 1 Σ C 1 P b i + C 1. Consider the sum Σ 2 := N+K 1 k=n s 2 k. Then, using s k = s k, Σ 2 = = K N+K 1 k=n s k 2 = i, b i 2 + i N+K 1 k=n b i b z k i z k b i b (z i /z ) N (z i /z ) N+K 1 z i /z. Hence there exists C 2, independent of N and K, such that Σ 2 K b i 2 2 b i b 1 z i i /z = C 2.

4 376 F. Beukers, R. Tideman / Indagationes Mathematicae 24 (2013) The terms sk 2 in Σ 2 can be estimated above by ( n b i )Σ + 1 when s k 0 and by (c + ε)σ 1 when s k < 0. So we get the upper bound Σ 2 b i Σ (c + ε)σ 1. Now use the upper bounds for Σ ± 1 we found above to get Σ 2 (c + ε)(k P) b i + (c + ε)p b i + C 3, where C 3 = C 1 (c+ε+ n b i ). Combine this with the lower bound Σ 2 C 2 +K n b i 2 to get C 2 + K b i 2 (c + ε)k b i + C 3. Dividing on both sides by K and letting K yields b i 2 (c + ε) b i. Since ε can be chosen arbitrarily small, the assertion follows. Proof of Theorem 2. Take the distinct elements from {z 1,..., z n } and write them as w 1,..., w m. Denote for any r the sum of the b over all such that z = w r by B r. Then s k = m r=1 B r wr k for every k. We can apply Theorem 1 to this sum to obtain Br 2 r=1 lim s k. k B r r=1 Since the b are positive reals, we get that m r=1 B r = n b and m r=1 Br 2 n b 2. Our theorem now follows. 3. The non-degenerate case In the next theorem we make an additional assumption on the z i, which allows us to improve on the upper bound in Theorem 1 considerably. Theorem 3. Let b C and let z C be as in Theorem 1. Assume in addition that z i 1 for all i and that none of the ratios z /z i with i is a root of unity. Then k b z k < 1 π 4 (min b ) log n. If the z i satisfy the conditions of Theorem 3 we say that we are in the non-degenerate case. Notice in particular that z i /z n+1 i = z 2 i and hence none of the z i are roots of unity in the nondegenerate case. For the proof of Theorem 3 we use the following result.

5 F. Beukers, R. Tideman / Indagationes Mathematicae 24 (2013) Lemma 1. Let b 1,..., b n C. Let q 1,..., q n be distinct integers. Set f (t) = n b e iq t. Then π f (t) dt 4 π 3 (min b ) log n. Proof. See Stegeman, [7]. Lemma 1 is a refinement of a result independently obtained by McGehee, Pigno, Smith [5] and Konyagin [4] who thereby established a conecture of Littlewood [1]. Already Littlewood noticed that the constant in Lemma 1 cannot be better than 4/π 2, cf. [7] p. 51. Stegeman expects that the optimal constant in Lemma 1 is 4/π 2 indeed. As a fine point we mention that the choice of 4/π 3 by Stegeman is for esthetical reasons only, the best possible value with his method happens to lie close to this value. See also [8]. The following lemma connects Littlewood s conecture with minima of sums of exponentials. Lemma 2. Let b 1,..., b n C. Let q 1,..., q n be distinct nonzero integers. Suppose that f (t) = n b e iq t is real-valued for all real t. Then min f (t) < 1 t R π 4 (min b ) log n. Proof. Denote the minimum of f (t) by c. Define f + (t) = max( f (t), 0) and f (t) = min( f (t), 0). Then f = f + f. Since the exponents q are nonzero, we have that π f (t)dt = 0, hence that π f + (t)dt = π f (t)dt and π f (t) dt = π f + (t)dt + π f (t)dt = 2 π f (t)dt < 4πc. Now combine this upper bound with the lower bound from Lemma 1 to find the assertion of our lemma. Proof of Theorem 3. Consider the subgroup G of C \ {0} generated by z 1,..., z n. By the fundamental theorem of finitely generated abelian groups, G is isomorphic to T Z d for some d and some finite group T consisting of roots of unity. More concretely this means that there exist w 1,..., w d G and µ T such that w 1,..., w d are multiplicatively independent and every z can be written in the form z = µ r w a 1 1 w a d d, r, a i Z, 0 r < T. It follows from the condition in Theorem 1 that a n+ 1,h = a,h for all = 1,..., n and h = 1,..., d. Our exponential sum can be rewritten as s k := b µ kr w ka 1 1 w ka d d. By Kronecker s approximation theorem, the closure of the set of points (w1 k,..., wk d ) for k Z 0 equals the set (S 1 ) d consisting of points (ω 1,..., ω d ) with ω = 1 for = 1,..., d. The same holds true if we restrict ourselves to values of k that are divisible by T. Hence s k min ω 1 = = ω d =1 b ω a 1 1 ω a d d.

6 378 F. Beukers, R. Tideman / Indagationes Mathematicae 24 (2013) Because there are no roots of unity among the z, for every at least one coefficient a i is nonzero. Since the ratios z i /z are not a root of unity for every i, the vectors (a 1,..., a d ) are pairwise distinct. Hence we can choose p 1,..., p d Z such that the numbers q = a 1 p a d p d, = 1,..., n are distinct and nonzero. Let us now restrict to the points with ω l = e iplt, t R, l = 1,..., d. Then we get s k min t R b e iq t, where the sum on the right-hand side is real for all t in view of b n+1 = b, q n+1 = q for = 1,..., n. By Lemma 2 the right-hand side is bounded above by 1 (min π 4 b ) log n. In the special case b = 1 for all we have the following corollary. Corollary 2. Let z 1,..., z n be as in Theorem 1. Suppose in addition that none of the ratios z i /z with i is a root of unity. Then z k < 1 log n. π 4 Proof. Note that we have not excluded the possibility that z i = 1 for some i. When z i 1 for all i we are in the non-degenerate case and can apply Theorem 3 immediately. When z i = 1 for some i we can take z n = 1. We now consider the subsequence of sums for odd k. Put k = 2κ + 1. Note that z k n κ z 2κ+1. Apply Theorem 3 to the numbers b = z for = 1,..., n 1 and z 2 instead of z for = 1,..., n 1. Then we find z k log(n 1) < log n. π 4 π 4 4. The continuous case The conditions on b, z in Theorem 1 can be seen as an invitation to write the power sum as a cosine sum. We consider the easier case when b R for all. Then we have b z k = Re b z k = b cos(2πα k), where we have written z = exp(2πiα ) for all. To make things simpler assume that z 1 for all. Then n is even and the arguments α come in pairs which are opposite modulo Z. Letting m = n/2 we rewrite our sum as 2 b cos(2πα k), where we can assume that 0 < α < 1/2 for all.

7 F. Beukers, R. Tideman / Indagationes Mathematicae 24 (2013) Corollary 1 immediately implies the following. Corollary 3. Let b 1,..., b m, α 1,..., α m be real numbers such that the α are distinct and strictly between 0 and 1/2 for all. Then b cos(2πα k) < 1 2m b. Proof. Apply Corollary 1 with n = 2m and b = b m when > m and z = exp(2πiα ) when m and z = exp( 2πiα m ) when > m. Similarly Theorem 3 implies the following. Corollary 4. Let b 1,..., b m, α 1,..., α m be real numbers such that the α are distinct and strictly between 0 and 1/2 for all. Suppose in addition that none of the differences α i α with i and none of the sums α i + α is rational. Then b cos(2πα k) < log(2m) 2π 4 We introduce the notation c S = b cos(2πα k), c T = t R b cos(2πα t). min b. In the notation c S, c T we suppress the dependence on the α s and b s. Of course, c S c T for all numbers α and b. Problem 2. What are the corresponding upper bounds for c T? The following result shows that c T = c S under a general condition. Theorem 4. Let b 1,..., b m be real numbers and let α 1,..., α m be real numbers such that their Q-span does not contain 1. Then c S = c T. In the proof we use the following consequence of Kronecker s theorem on simultaneous Diophantine approximation. Lemma 3. Let α 1,..., α n be numbers such that their Q-span does not contain 1. Let t 0 R. Given δ > 0 there exist integers k, k 1,..., k n such that α t 0 α k k < δ for = 1,..., n. Proof. Let β 1,..., β d be a basis of the Q-vector space spanned by the α. Choose λ i Q such that d α = λ i β i.

8 380 F. Beukers, R. Tideman / Indagationes Mathematicae 24 (2013) By a convenient choice of the β i we can see to it that λ i Z for all i,. Put Λ = max ( i λ i ). By Kronecker s theorem ([2], Theorem 442) there exist integers k, m 1,..., m d such that β i t 0 β i k m i < δ/λ for i = 1,..., d. Here we use the ormation that the Q-span of the β i s does not contain 1. Put k = d λ i m i. Then we get, for = 1,..., n, α t 0 α k k d d λ i β i t 0 β i k m i < λ i δ Λ δ. Proof of Theorem 4. It remains to prove that c T c S. Let ε > 0. Choose t 0 such that 0 c T + b cos(2πα t 0 ) < ε. We apply Lemma 3 with a δ which is so small that there exists an integer k 0 with b cos(2πα t 0 ) b cos(2πα k 0 ) < ε. Hence 0 c T + We deduce that b cos(2πα k 0 ) < 2ε. c T c S = c T + c T + b cos(2πα k) b cos(2πα k 0 ) < 2ε. Since ε can be chosen arbitrarily close to zero, we conclude c S = c T. Acknowledgments We thank Marc Spiker for making valuable comments on drafts of the paper and providing the ormation about the application in numerical analysis. We are much indebted to the referee for his careful reading of the manuscript and for saving us from two embarrassing mistakes. References [1] G.H. Hardy, J.E. Littlewood, A new proof of a theorem on rearrangements, J. Lond. Math. Soc. 23 (1948) [2] G.H. Hardy, E.M. Wright, An Introduction to the Theory of Numbers, fourth ed., Oxford at the Clarendon Press, [3] W. Hundsdorfer, A. Mozartova, M.N. Spiker, Stepsize restrictions for boundedness and monotonicity of multistep methods, J. Sci. Comput. 50 (2012) [4] S.V. Konyagin, On a problem of Littlewood, Izv. Akad. Nauk SSSR Ser. Mat. 45 (1981) (in Russian); transl.: Math. USSR-Izv. 18 (1981)

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