Journal of Inequalities in Pure and
Applied Mathematics
ON ZEROS OF RECIPROCAL POLYNOMIALS OF ODD DEGREE
PIROSKA LAKATOS AND LÁSZLÓ LOSONCZI
Institute of Mathematics,
Debrecen University,
4010 Debrecen, pf.12, Hungary.
EMail: lapi@math.klte.hu
volume 4, issue 3, article 60,
2003.
Received 12 December, 2002;
accepted 30 July, 2003.
Communicated by: A. Sofo
EMail: losi@math.klte.hu
Abstract
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2000
Victoria University
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Abstract
The first author [1] proved that all zeros of the reciprocal polynomial
m
X
Pm (z) =
Ak z k (z ∈ C),
k=0
of degree m ≥ 2with real coefficients Ak ∈ R (i.e. Am 6= 0 and Ak = Am−k for
all k = 0, . . . , m2 ) are on the unit circle, provided that
m
m−1
X
X
|Am | ≥
|Ak − Am | =
|Ak − Am |.
k=0
k=1
Moreover, the zeros of Pm are near to the m + 1st roots of unity (except the root
1). A. Schinzel [3] generalized the first part of Lakatos’ result for self–inversive
polynomials i.e. polynomials
m
X
Pm (z) =
Ak z k
Piroska Lakatos and
László Losonczi
Title Page
k=0
for which Ak ∈ C, Am 6= 0 and Āk = Am−k for all k = 0, . . . , m with a fixed
∈ C, || = 1. He proved that all zeros of Pm are on the unit circle, provided that
m
X
|Am | ≥ inf
|cAk − dm−k Am |.
c,d∈C, |d|=1
On Zeros of Reciprocal
Polynomials of Odd Degree
k=0
If the inequality is strict the zeros are single. The aim of this paper is to show
that for real reciprocal polynomials of odd degree Lakatos’ result remains valid
m−1
X
even if
π
2
|Am | ≥ cos
|Ak − Am |.
2(m + 1)
k=1
We conjecture that Schinzel’s result can also be extended similarly: all zeros of
Pm are on the unit circle if Pm is self-inversive and
m
X
π
|Am | ≥ cos
inf
|cAk − dm−k Am |.
2(m + 1) c,d∈C, |d|=1
k=0
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2000 Mathematics Subject Classification: Primary 30C15, Secondary 12D10, 42C05
Key words: Reciprocal, Semi-reciprocal polynomials, Chebyshev transform, Zeros
on the unit circle.
The authors had their research supported by Hungarian NFSR grant No.s T 043034
and T 043080 respectively.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2
The Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3
Remarks on Schinzel’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 13
References
On Zeros of Reciprocal
Polynomials of Odd Degree
Piroska Lakatos and
László Losonczi
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1.
Introduction
Studying the spectral properties of the Coxeter transformation Lakatos [1]
found that all zeros of the reciprocal polynomial
Pm (z) =
m
X
Ak z k
(z ∈ C)
k=0
of degree m ≥ 2 with real coefficients, i.e.
(1.1)
Am 6= 0,
Ak ∈ R, and Ak = Am−k
k = 0, . . . ,
m 2
Piroska Lakatos and
László Losonczi
are on the unit circle, provided that
|Am | ≥
(1.2)
m−1
X
|Ak − Am |.
k=1
inf
c,d∈C, |d|=1
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are on the unit circle, provided that
|Am | ≥
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Contents
She used Chebyshev transformation to prove this result.
The manuscript on this was sent to A. Schinzel for his comments. He generalized the above theorem [3] for self-inversive
polynomials by proving that all
Pm
k
zeros of the polynomial Pm (z) = k=0 Ak z where
(1.3)
Am 6= 0, Ak ∈ C, and and Āk = Am−k (k = 0, . . . , m) with ∈ C, || = 1
(1.4)
On Zeros of Reciprocal
Polynomials of Odd Degree
m
X
k=0
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|cAk − dm−k Am |
J. Ineq. Pure and Appl. Math. 4(3) Art. 60, 2003
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holds.
Schinzel’s proof was based on a theorem of Cohn [4] and on the estimate
m
X
m
m−k (1.5)
min kz
≥ .
z∈C,|z|=1 2
k=1
Learning of Schinzel’s generalization, the first author made an attempt to
improve her result. Although the method of Chebyshev transformation does not
seem to work for self-inversive polynomials, it can be used to obtain information
about the location of the zeros on the unit circle. She could prove [1] that
condition (1.2)
that the distribution of the zeros eiuj (j = 1, . . . , m)
Pensures
m
k
of Pm (z) =
k=0 Ak z satisfying (1.1), (1.2) is quite regular. They can be
arranged such that
j − eiuj < π
(1.6)
(j = 1, . . . , m)
m+1
holds, where
j
j = ei m+1 2π (j = 1, 2, . . . , m)
are the m + 1st roots of unity except 1.
If m = 2n + 1 is odd then −1 = eiun+1 is always a zero and all zeros of P2n+1
are single.
If m = 2n is even, (1.2) holds with equality and
On Zeros of Reciprocal
Polynomials of Odd Degree
Piroska Lakatos and
László Losonczi
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sgn A2n = sgn(−1)k+1 (Ak − A2n )
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for all k = 1, 2, . . . , n with Ak − A2n 6= 0, then un = un+1 = π, the number
−1 = eiun = eiun+1 is a double zero of P2n . Otherwise all zeros of P2n are
single.
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The aim of this paper is to show that for polynomials of odd degree both results can be improved. For even degree polynomials this is not possible. Since,
for first degree reciprocal or self-inversive polynomials the only zero of the
polynomial has modulus one we may assume that the degree m ≥ 2.
On Zeros of Reciprocal
Polynomials of Odd Degree
Piroska Lakatos and
László Losonczi
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2.
The Main Result
The theorem of Lakatos can be improved as follows.
Theorem 2.1. All zeros of the reciprocal polynomial
P2n+1 (z) =
(2.1)
2n+1
X
Ak z k
(z ∈ C)
k=0
of odd degree 2n + 1 ≥ 3 with real coefficients i.e.
(2.2)
A2n+1 6= 0,
Ak ∈ R, and Ak = A2n+1−k (k = 0, . . . , n)
On Zeros of Reciprocal
Polynomials of Odd Degree
Piroska Lakatos and
László Losonczi
are on the unit circle, provided that
2n
(2.3)
X
π
|Ak − A2n+1 |.
|A2n+1 | ≥ cos
2(2n + 2) k=1
2
Moreover, if (2.2), (2.3) hold then all zeros eiuj (j = 1, 2, . . . , 2n + 1) of P2n+1
are single, −1 = eiun+1 is always a zero and the zeros can be arranged such
that
j − eiuj < π
(2.4)
(j = 1, . . . , 2n + 1),
2n + 2
where
j
j = ei 2n+2 2π
(j = 1, 2, . . . , 2n + 1)
are the 2n + 2 nd roots of unity except 1.
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Proof. With the notation
l := A2n+1 = A0 , a1 := A2n − l = A1 − l, . . . , an := An+1 − l = An − l
we have
h2n+1 (z) : = l(z 2n+1 + z 2n + · · · + z + 1) +
n
X
ak z 2n+1−k + z k
k=1
= P2n+1 (z) =
2n+1
X
Ak z k
(z ∈ C).
On Zeros of Reciprocal
Polynomials of Odd Degree
k=0
(2.2) goes over into
l 6= 0, l, ak ∈ R (k = 1, . . . , n)
while (2.3) goes over into
(2.5)
Piroska Lakatos and
László Losonczi
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|l| ≥ 2 cos2
π
2(2n + 2)
n
X
|ak |.
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k=1
We have h2n+1 (z) = (z + 1)h̄2n (z) with
h̄2n (z) = lv̄2n (z) +
n
X
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k=1
where
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v̄2n (z) = z
w̄2n−2k (z) =
2n
z
+z
2n−2
2n+1−2k
z+1
2
+ ··· + z + 1
+1
,
ek (z) = z k .
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The Chebyshev transform T h̄2n of h̄2n was calculated in [1]:
T h̄2n (x) = lT v̄2n (x) +
= l Un
x
2
+
n
X
k=1
n
X
ak T (ek · w̄2n−2k )(x)
h
ak Un−k
x
k=1
2
− Un−k−1
x i
2
,
where Un is the Chebyshev polynomial of degree n of the second kind defined
by Un (cos x) := sin(n+1)x
(n = 1, 2, . . . ) and U−1 (x) := 0. Evaluating of T h̄2n
sin x
at the points
xj = 2 cos yj
with
yj =
j + 12
2π
2n + 2
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n
yj
l(−1)j X cos 2n−2k+1
2
T h̄2n (xj ) =
+
ak
yj
sin yj
cos 2
k=1
P
y
l(−1)j + 2 nk=1 ak sin 2j cos 2n−2k+1
yj
2
.
=
sin yj
We have for j = 0, . . . , n − 1
(2.6)
Piroska Lakatos and
László Losonczi
(j = 0, . . . , n)
of the open interval ] − 2, 2[ gives that (see [1])
yj
yn
(0 <) sin < sin
= sin
2
2
On Zeros of Reciprocal
Polynomials of Odd Degree
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π
π
−
2 2(2n + 2)
= cos
π
,
2(2n + 2)
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while for j = n there is equality here. The absolute value of the factor
cos
2n − 2k + 1
yj
2
2(n − k)j + j + (n − k) + 12
= cos
π
2n + 2
(k = 1, . . . , n; j = 0, . . . , n)
takes its maximum if the fraction
2(n − k)j + j + (n − k) +
2n + 2
1
2
is nearest to an integer. Clearly the nearest possible value of this fraction to an
1
integer is 2(2n+2)
(this value is attained at j = 0, k = n). Thus we have shown
that
π
2n − 2k + 1 (2.7) cos
yj ≤ cos
(k = 1, . . . , n; j = 0, . . . , n).
2
2(2n + 2)
Let for j = 0, . . . , n
n
X
yj
2n − 2k + 1 Sj := 2 ak sin cos
yj .
2
2
k=1
Then, for j = 0, . . . , n − 1 by (2.6), (2.7) we have
n
n
X
π
π
yj X
2
Sj ≤ 2 sin
|ak | cos
< 2 cos
|ak |
2 k=1
2(2n + 2)
2(2n + 2) k=1
On Zeros of Reciprocal
Polynomials of Odd Degree
Piroska Lakatos and
László Losonczi
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P
unless nk=1 |ak | = 0.
Thus, by (2.3) or (2.5) we have
n
Sj < 2 cos2
X
π
|ak | ≤ |l|
2(2n + 2) k=1
and the resulting inequality
Sj < |l|
Pn
remains valid even if
For j = n we have
k=1
|ak | = 0.
On Zeros of Reciprocal
Polynomials of Odd Degree
n
1 2(n
−
k)n
+
n
+
(n
−
k)
+
yn X
2 Sn ≤ 2 sin
|ak | cos
π
2 k=1
2n + 2
n
1 X
k
+
π
2
|ak | cos n − k +
= 2 cos
π 2(2n + 2) k=1
2n + 2
n
= 2 cos
X
k + 12
π
|ak | cos
π
2(2n + 2) k=1
2n + 2
n
Pn
k=1
π
2(2n + 2)
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|ak | = 0. Thus, by (2.3) or (2.5) we have
Sn < 2 cos2
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X
π
< 2 cos
|ak |
2(2n + 2) k=1
2
unless
Piroska Lakatos and
László Losonczi
n
X
k=1
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|ak | ≤ |l|.
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and the inequality
Sn < |l|
Pn
remains valid even if k=1 |ak | = 0.
Looking again at the Chebyshev transform we can see that by the inequalities
Sj < |l| (j = 0, . . . , n) we have
sgn T h̄2n (xj ) = sgn l sgn(−1)j (j = 0, 1, . . . , n)
therefore T h̄2n has n different zeros in ] − 2, 2[. Writing these zeros in the form
2 cos vj with 0 ≤ v1 ≤ v2 ≤ · · · ≤ vn ≤ π and applying Lemma 1 of [1] we
conclude that all zeros of h̄2n are single, and of the form e±ivj , where
(2.8)
On Zeros of Reciprocal
Polynomials of Odd Degree
Piroska Lakatos and
László Losonczi
yj−1 < vj < yj (j = 1, . . . , n).
Title Page
Let uj := vj for j = 1, . . . , n, un+1 := π and un+1+j := 2π − un+1−j (j =
1, . . . , n), then we obtain that all zeros of P2n+1 = h2n+1 are eiuj (j = 1, . . . , 2n+
1) and by (2.8) the condition (2.4) holds.
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3.
Remarks on Schinzel’s Theorem
Schinzel’s result can be generalized as follows.
P
k
Theorem 3.1. Let Pm (z) = m
k=0 Ak z be a self–inversive polynomial of degree m, i.e. let Ak ∈ C, Am 6= 0 and Āk = Am−k for all k = 0, . . . , m with a
fixed ∈ C, || = 1. If
(3.1)
m
|Am | ≥
2µm
inf
c,d∈C, |d|=1
m
X
|cAk − dm−k Am |,
k=0
where
(3.2)
On Zeros of Reciprocal
Polynomials of Odd Degree
Piroska Lakatos and
László Losonczi
m
X
µm := min kz m−k |z|≤1 k=1
then all zeros of Pm are on the unit circle. If the inequality is strict the zeros are
single.
Apart from minor changes Schinzel’s proof [3] is valid for this more general
result hence we omit the proof.
By Cohn’s theorem [4] any self–inversive polynomial Pm and the polynomial
m
X
m−1 0
−1
z
Pm (z ) =
kAk z m−k (z ∈ C)
k=1
have the same
of zeros inside
circle. Applying this for the
P number
Pmthe unit
k
m−k
polynomial m
z
we
obtain
that
kz
has no zeros inside the unit
k=0
k=1
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circle, thus the modulus of the latter takes its positive minimum in the unit disk
on the unit circle:
m
X
µm = min kz m−k > 0.
|z|=1 k=1
By some known identities (see [2, Part 6, Problems 16, 18]) for trigonometric
sums we easily get that
m
X
m−k Dm (t) : = kz
it
k=1
z=e
v"
#2 " m
#2
u m
u X
X
k cos(m − k)t +
k sin(m − k)t
=t
k=1
k=1
v"
#
u
u m 1 sin mt 2 2 m sin t − sin mt 2
2
+
+
.
=t
2
2 sin 2t
4 sin2 2t
From this it follows that µm = min Dm (t) ≥
t∈[0,2π]
m
2
and for even m = 2n we
have equality here, since for t = π
Dm (π) =
On Zeros of Reciprocal
Polynomials of Odd Degree
Piroska Lakatos and
László Losonczi
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m
.
2
This means that for even m Theorem 3.1 coincides with Schinzel’s result.
For odd m however
m
µm > .
2
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Let for t ∈ [0, 2π]
2
m 1 sin mt
2
xm (t) :=
+
,
2
2 sin 2t
m sin t − sin mt
ym (t) :=
,
4 sin2 2t
zm (t) := xm (t) + iym (t),
2
then Dm
(t) = |zm (t)|2 = xm (t)2 + ym (t)2 . As Dm (π + t) = Dm (π − t) it is
enough to consider Dm on the interval [0, π].
The next figure shows the graph of z9 in the complex plane and an enlargement of the part which is nearest to the origin. In the latter the distance of
the origin from the graph of z9 is also shown. The point of the graph with
t = tm = m−1
π is distinguished by a small circle.
m
On Zeros of Reciprocal
Polynomials of Odd Degree
Piroska Lakatos and
László Losonczi
Title Page
Contents
Our numerical experiments give base to the following conjecture.
Conjecture 1. For odd m we have
m
π
µm = min Dm (t) ≥
sec
.
t∈[0,2π]
2
2m + 2
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A simple calculation shows that for odd m = 2n + 1
m
π
sec
.
Dm (tm ) =
2
2m
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40
4
30
3
20
2
10
1
0
10
20
30
0
40
1
2
3
4
5
Figure 1: Graph of z9 .
It is clear that µm is the distance of the graph of zm from the origin. The minimum of Dm is attained near to tm . It is relatively easy to show that the minimum
is attained in the interval [tm , π]. Numerical calculations seem to justify that the
minimum point is in the smaller interval
2π
tm , tm + 2 .
m
Piroska Lakatos and
László Losonczi
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Theorem 3.1 and Conjecture 1 give
Conjecture 2. All zeros of the self–inversive polynomial
P2n+1 (z) =
On Zeros of Reciprocal
Polynomials of Odd Degree
2n+1
X
k=0
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Ak z k
(z ∈ C)
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of odd degree 2n + 1, i.e.
A2n+1 6= 0, Ak ∈ C, and Āk = A2n+1−k (k = 0, . . . , 2n+1) with ∈ C, || = 1
are on the unit circle, provided that
2n+1
|A2n+1 | ≥ cos
X
π
|cAk − d2n+1−k A2n+1 |.
inf
2(2n + 2) c,d∈C, |d|=1 k=0
holds. If the inequality is strict here then the zeros are single.
On Zeros of Reciprocal
Polynomials of Odd Degree
Piroska Lakatos and
László Losonczi
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References
[1] P. LAKATOS, On zeros of reciprocal polynomials, Publ. Math. (Debrecen)
61 (2002), 645–661.
[2] G. PÓLYA AND G. SZEGŐ, Problems and Theorems in Analysis, Vol. II,
Springer 1976.
[3] A. SCHINZEL, Self-inversive polynomials with all zeros on the unit circle,
Ramanujan Journal, to appear.
[4] A. COHN, Über die Anzahl der Wurzeln einer algebraischen Gleichung in
einem Kreise, Math. Zeit., 14 (1922), 110–148.
On Zeros of Reciprocal
Polynomials of Odd Degree
Piroska Lakatos and
László Losonczi
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