polynomials with all zeros on the unit circle - Springer Link

Acta Math. Hungar., 125 (4) (2009), 341356.

DOI: 10.1007/s10474-009-9028-7 First published online June 10, 2009

POLYNOMIALS WITH ALL ZEROS ON THE UNIT CIRCLE P. LAKATOS 1

1 ∗

and L. LOSONCZI

2 †

Institute of Mathematics, Debrecen University, 4010 Debrecen, pf. 12, Hungary e-mail: [email protected] 2 Faculty of Economics, Debrecen University, 4028 Debrecen, pf. 82, Hungary e-mail: [email protected] (Received January 29, 2009; accepted February 24, 2009)

Abstract. We give a new sucient condition for all zeros of self-inversive polynomials to be on the unit circle, and nd the location of zeros. This generalizes some recent results of Lakatos [7], Schinzel [17], Lakatos and Losonczi [9], [10]. By this sucient condition the mentioned results can be treated in a unied way.

1. Introduction

By a classical theorem of A. Cohn [4] (see also [13], p. 18, Theorem 2.1.6)

all zeros of a polynomial

(1)

P (z) =

m

Ak z k ∈ C[z]

k=0

of degree m lie on the unit circle if and only if (i) P is self-inversive, i.e.

(2)

∗ †

Am−k = εAk

(k = 0, . . . , m)

where ε ∈ C,

|ε| = 1,

Research supported by Hungarian NFSR (OTKA) grants No. NK-68040, NK-72523. Research supported by Hungarian NFSR (OTKA) grant No. NK-68040. Key words and phrases: self-inversive polynomial, zeros on the unit circle. 2000 Mathematics Subject Classication: primary 26C10, secondary 30C15.

02365294/$ 20.00 c 2009 Akadémiai Kiadó, Budapest

P. LAKATOS and L. LOSONCZI

!"

all zeros of P lie in or on this circle. the coecients Ak (k = 0, . . . , m) are real

(ii)

If then in (2) real, thus such polynomials are self-inversive if and only if (j) either ε = 1 and Am−k = Ak (k = 0, . . . , m) i.e. P is

ε

should also be

reciprocal,

(jj) or ε = −1 and Am−k = −Ak (k = 0, . . . , m) i.e. P is anti-reciprocal. Recently several authors found sucient conditions for self-inversive or reciprocal polynomials to have all their zeros on the unit circle. Lakatos [7] proved that all zeros of the reciprocal polynomial

P (z) =

m

Ak z k

(z ∈ C)

k=0

m 2 with real coecients Ak ∈ R (i.e. Am = 0 k = 0, . . . , [ m 2 ]) are on the unit circle, if

of degree for all

(3)

|Am |

m−1

and

Ak = Am−k

|Ak − Am |

k=1

holds. A. Schinzel [17] generalized this result for self-inversive polynomials proving that all zeros of the self-inversive polynomial (1) are on the unit circle if m cAk − dm−k Am . |Am | inf c,d∈C |d|=1 k=0

If the inequality is strict the zeros are simple. The authors proved [9] that if the coecients of a self-inversive polynomial P satisfy the inequality

(4)

m−1 1 |Am | |Ak | 2 k=1

then all zeros of P are on the unit circle. Moreover, they found the approximate location of the zeros. For reciprocal polynomials we recently proved [10] that if there is a B ∈ R such that

(5)

Am B 0, |Am | |B|

and

|Am + B|

m−1 k=1

Acta Mathematica Hungarica 125, 2009

|B + Ak − Am |

POLYNOMIALS WITH ALL ZEROS ON THE UNIT CIRCLE then all zeros of the zeros of

P

P

!"!

are on the unit circle. If the inequality (5) is strict then

have the form

e±iuj (j = 1, . . . , [ m 2 ])

2(j − 1)π 2jπ < uj < m m

where

j = 1, . . . ,

m 2

and they are simple. In case of odd m in addition to these, −1 = e−iπ is a zero too. In [7], [8], [10] the method of proof was Chebyshev transformation. It was shown that the Chebyshev transform of the polynomial has one more change of sign than the degree of the polynomial. In [9] from the self-inversive poly-

¯0 nomial (1) the real polynomial ε−1/2 e−imϕ/2 P (eiϕ ) (ϕ ∈ [0, 2π[, ε = Am /A was constructed and again the number of sign changes of this real polynomial was estimated. Schinzel [17] used an indirect proof based on a result of Cohn. There are however other methods to give sucient conditions for polynomials to have all, or some of their zeros on the unit circle. One possibility is reducing the problem (by Chebyshev transformation in case of reciprocal polynomials, or by the result mentioned above in case of self-inversive polynomials) to determining the number of real zeros of suitable real polynomials in an interval. Concerning this there are several results, the reader may consult the book [15], whose full third chapter is devoted to the latter problem. Another method is using known bounds (see e.g. [12], Chs. VII, VIII) for the zeros of the derivative of a polynomial and Cohn's theorem. Applying e.g. the well-known Cauchy estimate (see [12], Theorem (27,1)) for the derivative P (z) = mAm z m−1 + (m − 1)Am−1 z m−2 + · · · + 2A2 z + A1 of the polynomial (1), we conclude that all zeros of

|z| r

where

r

P

lie in the closed disk

is the positive root of the equation

−m|Am |z m−1 + (m − 1)|Am−1 |z m−2 + · · · + 2|A2 |z + |A1 | = 0. Hence, if

r1

then all zeros of (the self-inversive)

A third possibility is using

matrix methods.

P

Let

lie on the unit circle.

P (z) = z m +

m−1 k=0

Ak z k

be an mth degree monic polynomial. Any m × m matrix whose characterism tic polynomial is (−1) P is called a companion matrix of P . The simplest companion matrix is the Frobenius companion matrix:

⎛

0

1

...

0

⎞

.. .. .. ⎟ ⎜ .. . . . ⎟. ⎜ . ⎝ 0 0 ... 1 ⎠ −A0 −A1 . . . −Am−1



!""

Companion matrices give a link between matrix analysis and polynomials. This connection has been used to locate zeros of polynomials using matrix methods, see [3], [12], [16]. If P has only unimodular zeros then unitary companion matrix of P can be constructed, see [5], where two algorithms are given producing this matrix. Unimodular zeros of some special polynomials were studied in [6], [14]. It is also worth to mention that polynomials with unimodular zeros are used in signal processing, see [1] and the references therein. The aim of this paper is to generalize the result [10] for self-inversive polynomials. In this way the mentioned results of [7], [8], [9], [10], [17] can be treated in a unied way. The main advantage of our sucient condition is its simplicity and generality: an inequality, nearly linear in the coecients, containing three parameters. With suitable choice of the parameters previous results can be obtained from it. 2. The main result

Theorem 1. Let P (z) = Aj zj ∈ C[z] be a self-inversive polynomial of j=0 degree m 2. If there are B, c, d ∈ C such that m

B Am

(6)

and (7)

is real,

0

B 1, Am

c = 0,

|d| = 1

m−1 cAk + (B − Am )dm−k + |cAm − Am |, |Am + B| cA0 − dm Am + k=1

then all zeros of P are on the unit circle. If the inequality (7) is strict then the zeros are simple. Remark 1. If m = 1 then the only zero of the self-inversive polynomial

P1 (z) = A1 z + A0 = A1 z + εA¯1

natural.

is on the unit circle, thus assuming m 2 is

Remark 2. For real coecients Ak and B = 0,

Theorem 1 gives the result of Lakatos [7], for B = 0 it goes over into the theorem of Schinzel [17]. Namely, in this case (7) clearly implies (8)

|Am |

inf

c,d∈C, |d|=1

c=d=1

m−1 cA0 − dm Am + cAk − Am dm−k + |cAm − Am |


k=1

POLYNOMIALS WITH ALL ZEROS ON THE UNIT CIRCLE

!"#

Schinzel's condition. On the other hand from (8) it follows the existence of c, d ∈ C, |d| = 1 for which m−1 cAk − Am dm−k + |cAm − Am | |Am | cA0 − dm Am + k=1

holds. Here however c = 0, otherwise |Am | (m + 1)|Am | would follow which is impossible. For B = Am condition (7) goes over into m−1 |cAk | + |cAm − Am |. 2|Am | cA0 − dm Am + k=1 1/m

With c = 1, d = ( AAm0 ) this inequality is equivalent to (4) thus from Theorem 1 the result [9] follows. Finally, for real reciprocal polynomials the condition (7) with c = d = 1 gives (5). Remark 3. From Cohn's theorem it follows easily that both zeros of the ¯ self-inversive polynomial A2 z 2 + A1 z + AA1¯A1 2 (A2 = 0) are unimodular if and only if |A2 | 12 |A1 |. For this polynomial of second degree condition (7) (with the choice B = A2 , c = 1, d = (A1 A¯2 /A¯1 A2 )1/2 ) is necessary and sucient. For m 3 degrees however, we believe that (7) is not necessary, no matter how we choose the parameters B, c, d. Namely, assume that our self-inversive polynomial P satises (10), (11) with strict inequality in (11). Then by our Theorem 2 there is only one zero eiu2 of P satisfying 2π/m < u2 < 4π/m. It is clear that there are polynomials with all zeros unimodular (by Cohn's theorem necessarily self-inversive) which do not satisfy this condition. Remark 4. How should we choose the parameters B , c, d ? Writing B = βAm the inequality (7) can be replaced by

(9) |Am |

m−1 cAk + (β − 1)Am dm−k + |cAm − Am | cA0 − dm Am + k=1

1+β

where β ∈ [0, 1], c = 0, |d| = 1. For a given polynomial (i.e. for a given xed set of coecients) clearly the best choice of the parameters is the one which minimizes R(β, c, d), the right hand side of (9). Since R is non-negative, its inmum taken over the parameters does exist, however, as the set of parameters β ∈ [0, 1], c = 0, |d| = 1 is not compact, the inmum may not be



!"$

assumed, the function R may not have a global minimum. Even the calculation of local minimum is not easy at all, due to the complicated form of the function R. Without local or global minimizing points we can just experiment with some values e.g. taking c = d = 1, β = 0, 1/2, 1, etc. For real reciprocal polynomials however the situation is simpler, see the rst part of the last section. To prove Theorem 1 we need two lemmas. Lemma 1. If P is self-inversive and η = 0 then ηP is also self-inversive (of the same degree).

Proof. The coecients of Bk of ηP

satisfy

η Bm−k = ηAm−k = ηεAk = ε B k = ε1 B k η where |ε1 | = 1, which shows the statement. Lemma 2. If Am , B ∈ R; Am B 0, Am > 0

then

m m m−k inf (Am − B) kz + Bm (Am + B). 2 |z|1 k=1

Proof. For B = 0 our statement follows from the inequality m m m−k inf kz 2 |z|1 k=1

see [17] or [8]. If B > 0 then let X(z) :=

m

kz m−k . We have

k=1

(Am − B)X(z) + Bm = Bm Am − B X(z) + 1 . Bm With the notation a :=

Am −B Bm ,

Y (z) := aX(z) + 1 we have

Y (z) 2 = aX(z) + 1 2 = a2 X(z) 2 + 2aX(z) + 1. As X is analytic, its real part X is harmonic, thus by the min-max theorem for harmonic functions it follows that 2 m 1 sin mt m 2 , + inf X(z) = inf X(z) = inf t 2 2 sin 2 2 |z|=1 t∈[0,2π] |z|1



!"%

where z = r(cos t + i sin t) and in the last equality we used the representation of [8] for the real part of X . Therefore, for |z| 1 2 Y (z) 2 a m + 1 , 2

and

Y (z) a m + 1 2

(Am − B)X(z) + Bm = Bm Y (z) Bm a m + 1 2 m m = (Am − B) + Bm = (Am + B). 2 2

Proof of Theorem 1. Suppose that (7) holds with m 2. The polynomial P (dz) is also self-inversive, its coecients Bk = dk Ak satisfy the inequality |Bm + β| |cB0 − Bm | +

m−1

cBk + (β − Bm ) + |cBm − Bm |,

k=1

where β = dm B . If the zeros of P (dz) are on the unit circle, then so are zeros of P (z) and by the denition of β the condition (6) is also satised if Am , B are replaced by Bm , β respectively. By Lemma 1 and the homogeneity of condition (7) it suces to prove the assertion concerning the zeros of P under the assumption that (10)

Am , B ∈ R; Am > 0, Am B 0,

c = 0

and (11)

Am + B |cA0 − Am | +

m−1

|cAk + B − Am | + |cAm − Am |.

k=1

We adopt Schinzel's indirect proof [17] for our case. Suppose that P has a zero inside the unit circle. By another theorem of Cohn ([4], p. 113, Theom rem IV, see also [2], Theorem 1) the polynomial z m−1 P (z −1 ) = kAk z m−k k=1 also has a zero |z0 | < 1, thus m

kAk z0m−k = 0.

k=1



!"&

Multiplying this by c + c¯AA¯ = c + c¯ε¯ (ε = AA¯ , c = 0) and subtracting from m both sides the expression 2(Am − B) kz0m−k + 2Bm we obtain m 0

m 0

k=1

m k (c + c¯ε¯)Ak − 2(Am − B) z0m−k − 2Bm

(12)

k=1

= −2 (Am − B)

m

kz0m−k

+ Bm .

k=1

By Lemma 2 for the right hand side R of (12) we have (13) |R| m(Am + B). Let L denote the left hand side of (12) then L=

m−1

k(cAk − Am + B)z0m−k + m(cAm − Am )

k=1

+

m−1

k(¯ cε¯Ak − Am + B)z0m−k + m(¯ cε¯Am − Am ).

k=1

Using the self-inversiveness condition Ak = εA¯m−k the second line can be written as m−1

k(¯ cε¯εA¯m−k − Am + B)z0m−k + m(¯ cε¯εA¯0 − Am )

k=1

=

m−1

(m − k)(¯ cA¯k − Am + B)z0k + m(¯ cA¯0 − Am )

k=1

=

m−1

(m − k)(cAk − Am + B)z0k + m(cA0 − Am ).

k=1

Therefore we have m−1 |L| m |cAk − Am + B| + |cAm − Am | + |cA0 − Am | , k=1



!"'

where strict inequality holds if (14)

there exists k,

1k m−1

for which

cAk − Am + B = 0.

Thus, if (14) is satised then from (12) we conclude that m−1

m

|cAk − Am + B| + |cAm − Am | + |cA0 − Am |

k=1

> |L| = |R| m(Am + B),

which contradicts (11) proving that P cannot have a zero inside the unit circle. As P is self-inversive it cannot have a zero outside the unit circle either, thus all zeros are on the unit circle. Let us now consider the case when (14) is not satised. Then c = 0,

cAk − Am + B = 0

for

k = 1, . . . , m − 1,

therefore (11) takes the form (15)

Am + B |cA0 − Am | + |cAm − Am |.

If strict inequality holds here then we can get a contradiction and complete the proof, as above. Thus the only case left to discuss is when (14) is not satised and (15) holds with equality, i.e. (16) Am + B = |cA0 − Am | + |cAm − Am | and cAk − Am + B = 0 for k = 1, . . . , m − 1. Hence Ak = A c−B (k = 1, . . . , m − 1) and from A1 = εA¯m−1 we get that ε = cc¯ . Therefore the rst equation of (16) can be written as m

A0 B 1+ = c − 1 + |c − 1| = 2|c − 1| Am Am

since c AA

0 m

m − 1 = cε A Am − 1 = c − 1.

¯

P (z) = Am

Thus, in the case under discussion

m−1 1 − b k c¯ z + z + , c c m

k=1



!#

where b = AB , a quantity between 0 and 1. Dene the sequence of polynomials Pn by m

m−1 1−b k u ¯n z + Pn (z) := Am z m + (n ∈ N), un un k=1

where un = c − c−1 n and c is a complex number satisfying 1 + b = 2|c − 1|. We claim that Pn (n ∈ N) satises (11) with strict inequality (where now in (11) the role of c is played by un , and B is the same as earlier). Indeed, substituting the coecients of Pn into (11), after dividing it by Am we get or 2|c − 1| 2|un − 1|. 1 + b |¯ un − 1| + |un − 1| As |un − 1| = |c − 1|

n−1 < |c − 1|, n

we conclude that the previous inequality is strict, justifying our claim. Therefore all zeros of Pn (n ∈ N) are on the unit circle. The zeros of Pn tend to the zeros of P as n → ∞ thus all zeros of P lie on the unit circle as well. If the inequality (7) is strict then so is (11). The above argument shows that P (z −1 ) has no zeros satisfying |z| 1 hence all zeros of P (z) are inside the unit circle. By Cohn's theorem all zeros of P (z) are on or outside the unit circle. Thus P and P have no common zeros, proving the second part of our theorem. 3. Location of the zeros

We study the location of the zeros of the self-inversive polynomial P (z) = Ak z k by assuming that conditions (10), (11) hold. As we have seen this k=0 does not restrict the generality. First we prove (cf. [18], p. 229, (7.4)) m

Lemma 3. If c = 0 ε =

m

P (z) =

Ak z k is a self-inversive polynomial of degree m,

k=0 cAm , then the function cA0 1

m

1

R(ϕ) := ε− 2 z − 2 cP (z)|z=eiϕ = ε− 2 e− (where we x one of the two values of for real ϕ.


1

imϕ 2

cP (eiϕ )

ε− 2 and use it in the sequel) is real


Proof. With Bk = ε−

2n + 1

1 2

is odd, z = then eiϕ

R(ϕ) =

n

cAk

we have B

k

!#

. If m =

= B m−k (k = 0, . . . , m)

m

m

m

m

(Bm−k z 2 −k + Bm−k z 2 −k ),

k=0

if m = 2n is even, we have R(ϕ) =

n−1

(Bm−k z 2 −k + Bm−k z 2 −k ) + Bn

k=0

where B is real. In both cases R(ϕ) is real. It is easy to check that cP (z) = cP (z) − (A + B)S(z) + (A (17) n

m

= (cAm − Am )z + m

+ B)S(z)

(cAk + B − Am )z k + (cA0 − Am ) + (Am + B)S(z)

k=1

where Am S(z) =

m−1

m

m

zk − B

k=0

Am + B

m−1

Lemma 4. Let P (z) =

(10) (11)

zk

k=1

=

m k=0

⎧ Am (z m+1 − 1) − Bz(z m−1 − 1) ⎪ ⎪ ⎪ ⎪ ⎪ (Am + B)(z − 1) ⎨ ⎪ ⎪ ⎪ A (m + 1) − B(m − 1) ⎪ ⎪ ⎩ m Am + B

Ak z k

if if

z = 1, z = 1,

be a self-inversive polynomial of degree m

(11)

1

satisfying , such that in strict inequality holds. Then ε− 2 = 0 m . where ε = cA cA0 If ϕ± are solutions of the equation e−

imϕ± 2

S eiϕ± = ±1

then

(18)

1

sgn R(ϕ± ) = ± sgn ε− 2 ,

where R(ϕ) is the function dened in Lemma 3. Thus P has a zero eiϕ with ϕ between ϕ+ and ϕ− . Acta Mathematica Hungarica 125, 2009


!#

Proof. By Lemma 3 and (17) we have (19)

1 1 R(ϕ± ) = Q(ϕ± ) ± (Am + B)ε− 2 = Q(ϕ± ) ± (Am + B)ε− 2 , 1

0 = Q(ϕ± ) ± (Am + B)ε− 2 ,

where denotes the imaginary part and − 12 − m 2

Q(ϕ) := ε +

m−1 k=1

(cAm − Am )z m

z

(cAk + B − Am )z + (cA0 − Am ) k

. z=eiϕ

By (11) we have m−1 Q(ϕ) |cAm − Am | + |cAk + B − Am | + |cA0 − Am | < Am + B, k=1

therefore by (19) (Am + B)2

1

(ε− 2 )

2

1

+ (ε− 2 )

2

= (Am + B)2

2 2 2 > Q(ϕ± ) = Q(ϕ± ) + Q(ϕ± ) 2 1 2 = Q(ϕ± ) + (Am + B)2 (ε− 2 ) .

From this we get that (Am + B)|ε− | > Q(ϕ± ) showing that ε− and by (19) 1 2

1 2

= 0,

1

sgn R(ϕ± ) = sgn (Q(ϕ± ) ± (Am + B)ε− 2 ) 1

1

= ±(Am + B) sgn ε− 2 = ± sgn ε− 2

proving (18). The last statement of Lemma 4 follows from the intermediate value theorem applied for R(ϕ). m Lemma 5. Let P (z) = Ak zk be a self-inversive polynomial of degree k=0 m 2 satisfying (10) and (11) with strict inequality. There exists a ϕ0 ∈ B [0, 2π m [, which depends only on m and b = Am , such that

e−

imϕ0 2


S(eiϕ0 ) = 1.


Let ϕj =

then

2jπ m

e−

(j = 1, . . . , m − 1),

imϕj 2

S(eiϕj ) = (−1)j

Proof. The equation e form

− imϕ 2

!#!

ϕm = 2π − ϕ0 ,

(j = 0, . . . , m).

S(eiϕ ) = (−1)j

can easily be rewritten in the

⎧ m−1 1 m+1 j ⎪ ⎪ ⎨sin 2 ϕ − b sin 2 ϕ − (−1) (1 + b) sin 2 ϕ = 0 Tj (ϕ) := if 0 < ϕ < 2π, ⎪ ⎪ ⎩ m + 1 − b(m − 1) − (−1)j (1 + b) = 0 if ϕ = 0.

0 b 1, therefore T0 (0) = (1 − b)m 0 and T0 ( 2π m) = −2(1 + b) sin < 0. The intermediate value theorem implies the existence m of a ϕ0 with 0 ϕ0 < 2π m , T0 (ϕ0 ) = 0, further Tm (ϕm ) = (−1) T0 (ϕ0 ) = 0. j jπ A simple calculation shows that sin m±1 2 ϕj = ±(−1) sin m (j = 1, . . . , m − 1),

By (10) we have π m

hence

Tj (ϕj ) = (−1)j sin

jπ (1 + b) − (1 + b) = 0 (j = 1, . . . , m − 1), m

completing the proof. A consequence of Lemmas 35 is m Theorem 2. Let P (z) = Ak z k be a self-inversive polynomial of dek=0 gree m 2 satisfying (10), (11) such that in (11) strict inequality holds. Let 0 ϕ0