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)
and a is real, the first sign change of
cosA,x + acosk2x
will not increase if Xx is replaced by k', where
(1.3)
A,<A'<A2.
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, 1989.
1980 Mathematics Subject Classification (1985 Revision). Primary 33A10, 42A05.
Key words and phrases. Cosine polynomial, frequencies, first sign changes, zeros.
©1991 American Mathematical Society
0002-9939/91 $1.00+ $.25 per page
709
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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
such that cos mx = -1.
be real. Then there exists a point x e (n/(2n),
n/n]
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
(2.1)
b > 0, and 0 < kx < k2 < A3 be real. Then the first sign
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
A.
A,
f(x0) = -1 + acosy-n + bcosy-n <-l
11
+ - + - = 0.
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.
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711
THE FIRST SIGN CHANGE OF A COSINE POLYNOMIAL
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
f(x0) = cos-pjr + acosyn
A-i
1
- b < - -b <0.
A-)
Z.
If 2kx < k3< 3kx and k3 > 2k2, then
9
3kx <k2 + k3< -kx.
It follows that
2n
3n
n
JA.
A-, + A-,
A,
Let x0 = 3n/(k2 + k3). Then we have
3k,n
..
.
3k,n
f(xa)
+
(¿>- a) cos ^-fvo^-= cos
""a 3 !i'
+(b-a)cos
^2 T"
A-)
~r ^-i
A~\
^2
i A-*
A-) ~r
A-,
<-^
<--
1
Z.
+ e-a<0.
Case 4. a > 1/2 and 0 < a - b < 1/2 . If k2 < 2/1,, let x0 = n/k2. Then
k
k
A')
Ay
f(xQ) = cos y1 n -a + b cos y*-n < -a + b < 0.
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?
<
it
jAi
So there exists x0 = (2n + l)n/(k2 + k}) e[2n/(3kx),
cosk3x0 = 0, from which it follows that
n/kx] suchthat
f(x0) = cosA,x0 + (a - b)cosk2x0 < --
cosk2x0 +
+ a - b <0.
If 2kx < k2< 3kx and k3 < 3kx , then k3 < 3k2/2 . Taking xQ = n/k2, we have
f(xQ) = cos yn - a + b cos y-n < - —a < 0.
A-y
A->
If 2kx < k2 < 3kx and 3kx < k} < 4kx, then
2n
5n
ja.
A2 + A,
2.
n
A.
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712
JIANG ZENG
Taking x0 = 5n/(k2 + k3), we have
5k
5k
f(x0) = cos-:-y-n
Ay ~r A-*
+ (a- b)cos--~n
< --
Ay ~r A-î
1
¿.
+ a- b <0.
Case 5. a > b + 1/2. If k2 < 3kx , let x0 = n/k2. Then
k
k
1
k2
k2
I
f(x0) = cos yn - a + b cos y-n < ■?- a + b < 0.
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.
D
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
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
0 < x0 < —
and
/
n
0 <-<o<JJ
x0 < — .
Noting that x'0 < n/k' < 2n/(kx + k') and coskxx > cos/x
(kx + k')), we have
(3.2)
f(x;kx)>f(x;k')
foTxe(0,x'].
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for x e (0, 2n/
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.
f'(0;k') = 0. But
Then /(0;A,) = f(0;k') = 0 and f(0;kx) =
/'(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
4
4
1
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
°"U
ifS>l,
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
a0
> j_Í_£
a0
_ 2_Í_L
a0
— ^_La0
•> m
n
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.
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JIANG ZENG
714
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
f(x) = cosA.x + cosA2x + cosA3x + cosA4x.
Then the first sign change of f(x)
is in (0, n/kx).
Proposition 2. Hypothesis 2 is true for N = 4.
Here we only sketch the proof of Proposition
terested readers.
1 and leave the details to in-
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
r(lt\
-,
^1+^-7
^?~^1
^d
/ I y I = 2cos-Lry—-ncos-^—l-n
+ cos y-n - 1 < 0;
if k4 < k2 + k3, then
r ( n \
f
—
^1
-.
k? + k,
k3—k7
= 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
/-, + A.
\
-n e
—^—n,
2.A-.
4k+ 3 \
—Z-.—n
ZÁy
J
c
( n
n
\ LA.
A,
Take x0 = (2n + l)n/(k3 + A4).
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THE FIRST SIGN CHANGE OF A COSINE POLYNOMIAL
Case 2.
such that
715
10A,/3 < A2 < 8A, . We show that there exists an integer m > 1
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).
D
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,
Let 0 < Aj < A2<
(4.3)
*(f)=0.
< kN and
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,.
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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
f(x) = cosAjX + a2cosA2x + • • ■+ aN cos kNx
belongs to (0, n/kx).
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), 55-59.
2. G. Polya, On polar singularities of power series, and Dirichlet series, Proc. London Math.
Soc. 33(1932), 85-101.
département
de mathématique,
67084 Strasbourg, France
université
louis-pasteur,
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7 rue rene-descartes,
f-