THE FIRST SIGN CHANGE OF A COSINE POLYNOMIAL

Transcription

1 proceedings of the american mathematical society Volume 111, Number 3, March 1991 THE FIRST SIGN CHANGE OF A COSINE POLYNOMIAL JIANG ZENG (Communicated by Kenneth R. Meyer) Abstract. Nulton and Stolarsky [1] studied the first (i.e., the least positive) sign change of a real cosine polynomial as a function of its smallest frequency. In the present article we will study this problem further, especially to point out that their fundamental proposition is not correct, and that therefore their principal hypothesis is unreasonable. Moreover, various results of Nulton and Stolarsky are improved or corrected and two open questions set in their paper are solved. 1. Introduction Let ax,..., an and 0 < kx < < XN be real numbers, and let N (1.1) f(x) = ^aicosllx, ax 0, ;=1 be the real cosine polynomial. It is well known (see, e.g., [2]) that the number of sign changes of f(x) in the interval (0, t) is kxt/n + 0(\); therefore there exists the first (i.e., the least positive) sign change of f(x). In 1982, Nulton and Stolarsky [1] studied the first sign change of a cosine polynomial as a function of its frequencies A.. In [1], the following proposition was given without proof: Proposition 2.1. If 0 < Xx <l2 (1.2) cosa,x + acosk2x will not increase if Xx is replaced by k', where (1.3) A,<A'<A2. and a is real, the first sign change of From some preliminary work, especially the above proposition, Nulton and Stolarsky proposed the following hypothesis: Hypothesis. Increases in the lowest frequency tend to decrease the position of the first sign change. Although at the end of [1] Nulton and Stolarsky proved that this hypothesis is false in some cases which they thought unusual, they expected that in Received by the editors January 27, 1989 and, in revised form, April 13, Mathematics Subject Classification (1985 Revision). Primary 33A10, 42A05. Key words and phrases. Cosine polynomial, frequencies, first sign changes, zeros American Mathematical Society /91 $1.00+ $.25 per page

2 710 JIANG ZENG (kx,..., kn, ax,..., an) phase space, the set of points where their hypothesis fails is asymptotically very small in relative measure and raised the question of finding the smallest integer N = NQ (difficult to compute by [1]) for which their Hypothesis 2 fails. In fact, there is an explicit counterexample of their Hypothesis 2. We see that the first sign change of cosx -2cos2x is arceos ( 1 + x/33/8) and the first sign change of cos2x - 2 cos 2.x is n/4 i.e., arceos (\/32/8). Therefore their Proposition 2.1 is false. Their Hypothesis 2 is consequently not valid even for N = N0 = 2. We will give a corrected version of their Proposition 2.1 in our Remark 2 and show that the sign of f(0) plays a crucial role in this problem. The object of the present paper is to give a further study of the relation between the first sign change and the smallest frequency of a real cosine polynomial. Some necessary lemmas will be given in 2 and the main results in 3. Finally, in 4, a number of hypotheses are proposed and some special cases are discussed. 2. Some lemmas Lemma 1. Let m>3n>0 be real. Then there exists a point x e (n/(2n), n/n] such that cos mx = -1. Lemma 2. Let m > 3«> 0 be real. Then there exists a point x e [n/(2n), n/n) such that cosmx = 0. These facts are simple, so we omit their proofs here. Lemma 3. Let a>0, change of b > 0, and 0 < kx < k2 < A3 be real. Then the first sign (2.1) f(x) = coskxx + acosk2x + bcosk3x belongs to (0, n/kx). Proof. At first, we have (2.2) /(0) = H-a + 6>0. Therefore, to prove the lemma, it is sufficient to prove that there exists a point x0 in (0, n/kx] such that f(x0) < 0. We will prove this fact by dividing the set {(a, b)\a >0,b>0} into five cases. Case 1. 0 < a < 1/2 and 0 < b < 1/2. Let x0 = n/kx. Then k k 11 f(x0) = -1 + acosy-n + bcosy-n <-l = 0. A. A, L L Case 2. a + 1/2 < b. If k3 < 3kx, take x0 = n/k3. Then k k 1 f(x0) = cos y-n + acos-yn -b < - + a-b <0. If A3 > 3A,, by Lemma 1 there exists x0e (n/(2kx), n/kx] such that cosk3x0 = -1. Hence /(x0) = coskxx0 + acosk2x0 - b < a- b <0.

3 THE FIRST SIGN CHANGE OF A COSINE POLYNOMIAL 711 Case 3. b > 1/2 and 0 < b - a < 1/2. If k3 < 2kx, take x0 = n/k3. Then k k f(x0) = cos y-n + a cos yn - b < a- b <0. A-* A-i If k3 > 3kx, by Lemma 1 there exists x0 e (n/(2kx), n/kx] such that cosk3x0 = -1. Therefore, f(x0) = cos-i.,x0 + acosa2x0 - b < a- b <0. If 2kx < k3 < 3kx and k3 < 2k2, take x0 = n/k3. Then k k 1 f(x0) = cos-pjr + acosyn - b < - -b <0. A-i A-) Z. If 2kx < k3< 3kx and k3 > 2k2, then 9 3kx <k2 + k3< -kx. It follows that Let x0 = 3n/(k2 + k3). Then we have 2n 3n n JA. A-, + A-, A, f(xa) 3k,n... 3k,n 1 vo^ - = ""a cos 3!i ' +(b-a)cos + ( > - cos -f- ^ <-- <-^ + e-a<0. A-) ^2 T" ~r ^-i A~\ A-) ^2 ~r i A-, A-* Z. Case 4. a > 1/2 and 0 < a - b < 1/2. If k2 < 2/1,, let x0 = n/k2. Then k k f(xq) = cos y1 n -a + b cos y*-n < -a + b < 0. A') Ay If k2 > 3/1,, by Lemma 1 there exists x0 e (n/(2kx), n/kx] such that cosk2xq -1. Therefore, f(x0) = cosa,x0 - a + bcosk3x0 < -a + b < 0. If 2A, < k2< 3kx and k3 > Akx, then 2n Aj T" A? So there exists x0 = (2n + l)n/(k2 + k}) e[2n/(3kx), n/kx] suchthat cosk2x0 + cosk3x0 = 0, from which it follows that f(x0) = cosa,x0 + (a - b)cosk2x0 < -- + a - b <0. If 2kx < k2< 3kx and k3 < 3kx, then k3 < 3k2/2. Taking xq = n/k2, we have < it jai f(xq) = cos yn - a + b cos y-n < - a < 0. A-y A-> 2. If 2kx < k2 < 3kx and 3kx < k} < 4kx, then 2n 5n n ja. A2 + A, A.

4 712 JIANG ZENG Taking x0 = 5n/(k2 + k3), we have 5k 5k 1 f(x0) = cos-:-y-n + (a- b)cos--~n < -- + a- b <0. Ay ~r A-* Ay ~r A-î. Case 5. a > b + 1/2. If k2 < 3kx, let x0 = n/k2. Then k k 1 f(x0) = cos yn - a + b cos y-n <? - a + b < 0. k2 k2 I If k2 > 3kx, by Lemma 1 there exists x0 e (n/(2kx), n/kx] such that cosa2x0 = -1. Hence f(x0) = coskxxq - a + bcosk3x0 < -a + b < 0. Combining these five cases and (2.2), the proof is completed. Lemma 4. Let a < 0, b < 0, and 0 < kx < k2< k3 be real. Then the first sign change of (2.3) f(x) = coskxx + acosk2x + bcosk3x belongs to (0, n/kx). Since the proof is similar to that of Lemma 3, we omit it here. 3. Main results Theorem 1. Let 0 < kx < k2< k3. If a and b have the same sign, the first sign change of (3.1) f(x) = coskxx + acosk2x + bcosk3x D belongs to (0, n/kx). Proof. This proof follows from Lemmas 3 and 4. D Theorem 2. Under the same conditions as in Theorem 1, if a + b > -1, the first sign change of f(x) is a decreasing function of kx; if a + b < -1, the first sign change of f(x) is an increasing function of kx. Proof. Let f(x ; k) = cos Ax + a cos A2x + b cos k3x. For kx < k' < k2, note x0 (resp. x ) the first sign change of f(x; kx) (resp. f(x ; k')). It follows from Theorem 1 that / n 0 < x0 < and 0 < -<o<jj x0 <. Noting that x'0 < n/k' < 2n/(kx + k') and coskxx > cos/x for x e (0, 2n/ (kx + k')), we have (3.2) f(x;kx)>f(x;k') fotxe(0,x'].

5 THE FIRST SIGN CHANGE OF A COSINE POLYNOMIAL 713 We will distinguish three cases according to the value of a + b : Case 1. a + b > -1. Then f(0; kx) = /(0; k') > 0. By (3.2), we have x0 < x0. Case 2. a + b = -1. Then /(0;A,) = f(0;k') = 0 and f(0;kx) = f'(0;k') = 0. But /'(0 ; k') = -k'2 - ak\ - bk] > k\ - k'2 > 0. Hence f(x ; k') > 0, for x e (0, x ). By (3.2), we have x'0 < x0. Case 3. a + b < -1. Then f(0;kx) = f(0;k') < 0. By (3.2), we have Remark 1. Let a = b = 1. Theorem 2 shows that Hypothesis 2 of [ 1 ] is true for all trinomials coskxx + cosk2x + cosa3x, which is an open question in [1]. Remark 2. Let b = 0. Theorem 2 gives a corrected version of Proposition 2.1 of[l](cf. 1). Remark 3. If a and b do not have the same sign, the above theorems may not be valid. This can be seen by the fact that the first sign change of g(x, 1) is not in (0, n) and the first sign change increases with k e (1,1.005), where g(x; k) = cos Ax - - cosxx + -cos2x. Finally, as a consequence of Theorem 2, we have the following result which improves the main theorem of [1]. Theorem 3. Let m > 0 and 1 < kx < k2 < k3 be real. Let T(x ; kx, k2, k3) = coskxx + a cosk2x + b cosa3x, with a and b nonpositive and a + b < -1. Then, whenever 0 < e < k2 - kx, there exists a point k0 e (0,1] such that the first sign change of T(x ; k0 + e, kqk2, k0k3) exceeds the first sign change of T(x ; kq, k0k2, k0k3) by more than m. Proof. Let x, and x2 be the first sign change of T((m + l)x; 1, k2, k3) and r((m+l)x; 1 +, k2, k3), respectively. By Theorem 2, we have f5 = x2-x1 > 0. Let (1 ifs>l, "U if «f < i. Clearly, (m + l)xj/a0 and(m + l)x2/a0 are, respectively, the first sign change of T(x ; k0, k0k2, k0k3) and T(x ; k0 + ek0, k0k2, kqk3). By Theorem 2, the first sign change of T(x ; k0 + e, kqk2, kqk3), say x3, is not less than that of T(x ; k0 + skq, kqk2, k0k3). Therefore (m-f-l)x, (m+l)x7 (w-l-l)x, (m + 1)0 x _.1_Í_L > j_í_ _ 2_Í_L ^_L- > m n a0 a0 a0 a0 4. Some conjectures It is plausible that the previous results are also valid in more general cases. From Remark 1, we obtain naturally the following two conjectures.

6 714 JIANG ZENG Hypothesis 1. Let N > 1, be an integer and 0 < kx < < kn. Then the first sign change of coskxx + cosa2x + + cosa^x belongs to (0, n/kx Hypothesis 2. Let N > 1 be an integer and 0 < kx < < kn. Then the first sign change of cosa,x + cosa2x -l-h cosa^x is a decreasing function of A,. It is easy to see that if Hypothesis 1 turns out to be true, Hypotheses 2 is true also. So far, the above two hypotheses are verified only for N < 4. Indeed, for N = 4 we have the following propositions: Proposition 1. Let 0 < kx < k2 < k3 < A4 and Then the first sign change of f(x) f(x) = cosa.x + cosa2x + cosa3x + cosa4x. is in (0, n/kx). Proposition 2. Hypothesis 2 is true for N = 4. 1 and leave the details to in- Here we only sketch the proof of Proposition terested readers. Proof. Since (4.1) /(0) = 4>0, we need to show only that there exists a point xqe (0, n/kx] such that f(x0) < 0. Note that if A3 < A, + A2, then if k4 < k2 + k3, then r(lt\ -, ^1+^-7 ^?~^1 ^d / I y I = 2cos-Lry -ncos-^ l-n + cos y-n - 1 < 0; r ( n \ ^1 -. k? + k, k3 k7 f = cosy^n + 2cos^r1 ^cos^ j -n - 1 < 0. Therefore we can assume in the sequel that A3>A,- -A2 and k4>k2 + k3. We will distinguish among six cases. In every case, we only indicate the point x0 we wish but omit the details. Case 1. A2 > 8A,. It is not difficult to see that there exist two integers n > 1 and k > 1 such that 2n + 1 (4k + l 4k+ 3 \ ( n n -n e ^ n, Z-. n c /-, + A. \ 2.A-. ZÁy J \ LA. A, Take x0 = (2n + l)n/(k3 + A4).

7 THE FIRST SIGN CHANGE OF A COSINE POLYNOMIAL 715 Case 2. 10A,/3 < A2 < 8A,. We show that there exists an integer m > 1 such that 2m + 1 /8?r 107r\ / n n_\ 3 + 4ne V3A2' 3A2J C V3A, ' kx) ' Take x0 = (2m + l)n/(k3 + A4). Case 3. 3A, < A2 < 10A,/3. We show that there exists an integer 5 > 1 such that 2s + 1 ( In 3n\ ( 2n_ n\ k^+t4n 6 \3T2 ' k2 ) C \3kx ' A, ) ' Take x0 = (25 + l)n/(k3 + A4). Case 4. 3A,/2 < A2 < 3kx and k3 + A4 > 4A2. We show that there exists an integer / > 1 such that 2t+ 1 in 3n_\ in n\ i~nr4n \t2 ' 2k2)c [wx ' tx) Take x0 = (2t+ l)n/(k3 + A4). Case 5. 3A,/2 < A2 < 3A, and k3+k4< 4k2. Take x0 = 3t:/(A3 + A4). Case 6. 0 < A2 < 3A,/2. We show that there exists an integer i > 1 such that 2/ + 1 in n\ An 3n \ k^+t4n \2TX' Tx) c \2V 2~k2) Take x0 = (2i + l)n/(k3 + A4). The conclusion follows by combining the above six cases and (4.1). Some other special cases of the above hypotheses are also verified. For example, it is not difficult to establish the following more general result, which both improves and corrects Proposition 3.1 of [1]. Proposition 3. Let g(x) be an even, real-valued function of period 2n that is strictly decreasing on [0, n]. Assume that (4.2) g(0) = l, *(f)=0. Let 0 < Aj < A2 < < kn and (4.3) G(x) = g(kxx) + a2g(k2x + c2) + + ang(knx + cn), where a, c are real numbers. Then whenever 0<4A, j <k,.lt j+i Kj<N-l, j if G(0) > 0, the first sign change of G(x) is a decreasing function of A, ; // G(0) < 0, the first sign change of G(x) is an increasing function of kx ; if G(0) = G'(0) = = Gin~X)(0) = 0 but G{n)(0) 0 for some integer n > 1 (only here we assume that g(x) is differentiable up to the nth derivative), then if C7("'(0) > 0, that sign change is a decreasing function of kx ; if t7(,!)(0) < 0, that sign change is an increasing function of A,. D

8 716 JIANG ZENG More ambitiously, the problems are the following: Hypothesis 3. Let ax,..., an and 0 < A, < < kn be real numbers. If all a have the same sign, then the first sign change of belongs to (0, n/kx). f(x) = cosajx + a2cosa2x + + an cos knx Hypothesis 4. Under the same condition as Hypothesis 3, if f(0) > 0 (resp. f(0) < 0), the first sign change of f(x) is a decreasing (resp. increasing) function ofkx. Acknowledgment The author is indebted to Professors D. Foata and S. P. Zhou for their helpful suggestions on the original version of this paper. References 1. J. D. Nulton and K. B. Stolarsky, The first sign change of a cosine polynomial, Proc. Amer. Math. Soc. 84(1982), G. Polya, On polar singularities of power series, and Dirichlet series, Proc. London Math. Soc. 33(1932), département de mathématique, université louis-pasteur, 7 rue rene-descartes, f Strasbourg, France