Self-inversive polynomials of odd degree - Springer Link

Ramanujan J (2007) 14: 305–320 DOI 10.1007/s11139-007-9029-5

Self-inversive polynomials of odd degree László Losonczi · Andrzej Schinzel

Received: 13 January 2004 / Accepted: 22 February 2006 / Published online: 8 September 2007 © Springer Science+Business Media, LLC 2007

Abstract If the coefficients of a self-inversive polynomial P (z) = C[z] of odd degree m ≥ 3 satisfy the inequality

m

k=0 Ak z

k

∈

π |cAk − d m−k Am |, inf 2(m + 1) c,d∈C m

|Am | ≥ cos

|d|=1 k=0

then all zeros of P are on the unit circle and they are simple. This is an improvement of a recent result of the second author (Ramanujan J. 9, 19–23, 2005) on the zeros of self-inversive polynomials in the case of polynomials of odd degree. A similar improvement in the case of real (reciprocal) polynomials has been given by Lakatos and the first author (J. Inequal. Pure Appl. Math. 4(3), 2003). Keywords Self-inversive polynomials · Zeros Mathematics Subject Classification (2000) Primary 30C15 · Secondary 26C15

Research of László Losonczi supported by Hungarian NFSR grant No. T 047373. L. Losonczi () Faculty of Economics and Business Administration, University of Debrecen, 4028 Debrecen, Kassai u. 26, Hungary e-mail: [email protected] L. Losonczi () e-mail: [email protected] A. Schinzel Institute of Mathematics, Polish Academy of Sciences, Warsaw, Poland e-mail: [email protected]

306

L. Losonczi, A. Schinzel

1 Introduction P. Lakatos [3] proved recently that all zeros of the real reciprocal polynomial P (z) =

m

Ak zk ∈ R[z]

k=0

of degree m ≥ 2 (i.e. Ak ∈ R, Am = 0 and Ak = Am−k for all k = 0, . . . , [ m2 ]) are on the unit circle if |Am | ≥

m−1

|Ak − Am |.

k=1

A. Schinzel[6] generalized this result for self-inversive polynomials. A polynok mial P (z) = m k=0 Ak z ∈ C[z] is called a self-inversive polynomial of degree m if Am = 0,

Am−k = εAk

(k = 0, . . . , m) with a fixed ε ∈ C, |ε| = 1

see e.g. [1, 2]. He proved that all zeros of the self-inversive polynomial P (z) are on the unit circle if |Am | ≥ inf

m

c,d∈C |d|=1 k=0

|cAk − d m−k Am |.

If this inequality is strict, the zeros are simple. P. Lakatos and L. Losonczi [4] noticed that for real reciprocal polynomials of odd degree, Lakatos’ result remains valid even if |Am | ≥ cos2

m−1 π |Ak − Am | 2(m + 1) k=1

and conjectured that for odd degree self-inversive polynomials Schinzel’s result has a similar asymptotic extension. The aim of this paper is to prove this extension.

2 The result and its proof Theorem Let P (z) =

m

Ak zk ∈ C[z]

k=0

be a self-inversive polynomial of odd degree m ≥ 3 i.e. Am = 0,

Ak ∈ C,

and ε A¯ k = Am−k

(k = 0, . . . , m) with ε ∈ C, |ε| = 1.

Self-inversive polynomials of odd degree

307

If the inequality π |cAk − d m−k Am | inf 2(m + 1) c,d∈C m

|Am | ≥ cos

|d|=1 k=0

holds, then all zeros of P are on the unit circle and they are simple. We need two propositions for the proof. Proposition 1 If the coefficients of the self-inversive polynomial P (z) = ∈ C[z] of degree m ≥ 2 satisfy the inequality

m

k=0 Ak z

k

m inf |cAk − d m−k Am |, 2μm c,d∈C m

|Am | ≥

|d|=1 k=0

where

m μm := min kzm−k , |z|=1 k=1

then all zeros of P lie on the unit circle. If the inequality is strict, the zeros are simple. Proof It is enough in the proof given in [6] to replace the right hand side of inequality (5) by 2μm |Am |, the left-hand side of the displayed inequality preceding (6) by 2 μmm |Am |, the left-hand side of the relations 1 = 2|c − 1| and 1 > n−1 n |c − 1| occurring below (6) by 2 μmm . Proposition 2 With μm as in Proposition 1, we have μm =

m 2

if m is even,

m π 1 π m > μm > sec sec +O 3 2 2m + 2 2 2m + 2 m

if m is odd.

Proof By some known identities (see [5, Part 6, Problems 16, 18]) for trigonometric sums we easily get that m Dm (t) := kzm−k k=1

2 m 2 m = k cos(m − k)t + k sin(m − k)t z=eit

k=1

k=1

mt

2 2 m 1 sin m sin t − sin mt 2 2 + = + , t 2 2 sin t 4 sin2 2 2

308


where, in the last expression the fractions at the zeros of sin 2t are defined by their limits. We have μm =

inf Dm (t) ≥

t∈[0,2π]

m 2

and for even m = 2n we have equality here, since Dm (π) = m2 . To prove right hand side of the second inequality of Proposition 2, first we show that for odd m ≥ 9 Dm (t) >

π m sec . 2 2m + 2

(1)

Then we settle the cases m = 1, 3, 5, 7. The proof of (1) requires some work. We get to it through four lemmas. Lemma 1 For all odd m and t ∈ [0, 2π] Dm (t) ≥ νm ,

(2)

where νm is the minimum of Dm on the interval [ m−1 m π, π]. Proof For m = 1 we have Dm (t) = 1 for all t, thus we may assume m ≥ 3. By the identity Dm (π + t) = Dm (π − t)(t ∈ [0, π]) it is enough to prove (2) for t ∈ [0, π]. π If t ∈ [0, m ], then by the inequality sin

t mt t (m − 1)t mt cos − cos sin = sin ≥0 2 2 2 2 2

we have

sin mt 2 sin 2t

2

≥

cos mt 2 cos 2t

2

=

sin m(π−t) 2 sin π−t 2

2 .

Further m sin t − sin mt = m sin(π − t) − sin m(π − t), π −t t . 0 ≤ sin ≤ sin 2 2 Hence Dm (t) ≥ Dm (π − t) ≥ νm . π If t ∈ [ m , π], then t can be written as

t=

k π + εθ, m


309

where 2 ≤ k ≤ m − 1, k is even, ε = ±1, 0 ≤ θ ≤ sin2

mt 2

0 ≤ sin

t 2

m sin t

π m.

Using the relations mεθ mθ m m−1 = sin2 = sin2 = sin2 π +θ , 2 2 2 m 1 m−1 k εθ θ ≤ sin , = sin π+ π+ 2m 2 2 2m 2 k m−1 = m sin π + εθ ≥ m sin π +θ , m m

− sin mt = − sin mεθ = −ε sin mθ m−1 m−1 π + θ ≥ − sin m π +θ , = −ε sin m m m we find that

Dm (t) ≥ Dm

m−1 π + θ ≥ νm . m

In the next lemmas we use several estimates for sin ϕ, cos ϕ, (1 + x)n which follow easily from the Taylor formula by writing the remainder term in Lagrange form. We summarize these here. If ϕ ∈ [0, π2 ], x ∈ [0, 1[, k > 0, then ϕ − S3 = sin ϕ = ϕ −

ϕ3 + S5 6

with 0 ≤ Si ≤

ϕi , i = 3, 5, i!

(3)

ϕ2 ϕ2 ϕ4 ϕi + C4 = cos ϕ = 1 − + − C6 with 0 ≤ Ci ≤ , i = 4, 6, (4) 2 2 24 i! k(k + 1) k(k + 1)(k + 2) 1 − kB1 = (1 + x)−k = 1 − kx + x2 − B3 (5) 2 6

1−

with x i (1 + x)−k−i ≤ Bi ≤ x i , i = 1, 3. We also use the identity 1 = 1 + x + I2 , 1−x

where x = 1, I2 = x 2 (1 − x)−1

(6)

and the inequalities √ x x2 if x ≥ 0, 1+x ≥1+ − 2 8 sin ϕ ≤ ϕ if ϕ ∈ [0, ∞[, π tan ϕ ≥ ϕ if ϕ ∈ 0, . 2 Lemma 2 If m ≥ 9 odd, t =

(m−1)π+ϕ m

with 0 ≤ ϕ ≤ π , then we have

Dm (t)2 ≥ f (ϕ) + A3 ,

(7) (8) (9)

310

where


π − ϕ − sin ϕ 2 f (ϕ) = + 4 (π − ϕ − sin ϕ)(π − ϕ − 3 sin ϕ) π − ϕ 2 + 96 m

m 1 − cos ϕ + 2 4

2

and |A3 |
0, 4 4 4 4 m m 4 4 m m where 0 ≤ S3 ≤

π3 ϕ3 ≤ . 6 6m3

For the second term of f we get

π − ϕ − sin ϕ 4

2

=

π aπ π S3 − + + 4 2m 2m2 4

2 =

π2 aπ 2 π2 π2 − + + + A7 , 2 16 4m 4m 4m2

where π aπ S3 S32 aπ 2 a 2 π 2 π A7 = − 3 + − + + +2 4 2m 2m2 4 16 2m 4m4

2 3 aπ 1 a πm S3 π aπ S3 m3 S32 m3 = 3 − 1− + − − 2 + . 2 2m 8 m m 4 16 m By the inequalities aπ 2 a π2 3 5π 2 3π 2 ≤− 1− ≤− 1− =− , − 2 2 2m 2 18 12 0≤ −

π4 πm3 S3 ≤ , 8 6·8

π4 ϕS3 m3 ≤− ≤ 0, 4·6·9 4 0≤

S32 m3 π6 ≤ 2 16 6 · 16 · 93


315

we conclude that

−15.26 3π 2 π4 1 − ≤ A7 < 3 − 2 4·6·9 m3 m

5π 2 −2.08 1 π4 π6 < ≤ 3 − . + + 2 3 12 6 · 8 6 · 16 · 9 m m3 Next we estimate the third term of f. Write the factors of this term in the form π π − ϕ − sin ϕ = − F1 , 4 4 π π − ϕ − 3 sin ϕ = − F2 , 4 4 π2 (π − ϕ)2 = − F3 . 6m2 6m2 Then, using the estimates for ϕ, S3 π π2 π3 π π ϕ S3 ϕ ≤ 1− − ≤ ≤ , 0< ≤ F1 = − 2 3m 3m 24m3 2 4 2 2m 8·9 2π 3π 2 3π 3 π 2π 3S3 ≤ 1− − ≤ϕ≤ , 0< ≤ F2 = ϕ − 2 3m 3m 4 · 24m3 4 m 64 · 9 1 2π 2 2π 2 π2 2πϕ ϕ2 π2 1 − ≤ 0< − ≤ F3 = − ≤ . 3 3 4 2 2 12 9m 9m 6m 6m 6m 3m3 For the third term we have 2 π π (π − ϕ − sin ϕ)(π − ϕ − 3 sin ϕ) π − ϕ 2 π − F1 − F2 = − F 3 96 m 4 4 6m2 =

π4 + A8 , 96m2

where A8 = −

π 3 (F1 + F2 ) π 2 F1 F2 π 2 F3 π(F1 + F2 )F3 + − F1 F2 F3 . + − 16 4 24m2 6m2

By the estimates for F1 , F2 , F3 we get π 3 (F1 + F2 ) π 2 F3 π4 π4 − F − F F ≥ − − 1 2 3 16 24m2 12m3 6m5 4 2 8.32 π 1+ 2 >− 3 , ≥− 3 12m 9 m

A8 ≥ −

A8 ≤

π4 π 2 F1 F2 π(F1 + F2 )F3 π4 5π 4 2.26 ≤ + + ≤ < 3 . 2 4 4 3 4 6m 12m 8m 24 · 9m m

316


Finally f (ϕ) = Em + A9 , where −

23.37 11.50 < A9 = A6 + A7 + A8 < 3 m m3

proving Lemma 4. Proof of Proposition 2 (continued) Let

2m − 1 (1) π, π , = min Dm (t) : t ∈ νm 2m and (2) νm (1)

m − 1 2m − 1 π, π . = min Dm (t) : t ∈ m 2m (2)

Then νm = min{νm , νm }. If t ∈ [ 2m−1 2m π, π], then Dm (t) ≥

m sin2 mt m 1 − cos mt 2m + 1 2 = + + ≥ . t 2 2 2 2(1 − cos t) 4 2 sin 2

Hence (1) νm ≥

2m + 1 . 4

On the other hand by (4) cos

π π2 4m + 6 2m ≥1− . >1− > 2 2 2m + 2 2m + 1 8(m + 1) 8(m + 1)

Thus (1)

2νm π cos > 1. m 2m + 2 2m−1 If t ∈ [ m−1 m π, 2m π], then by Lemmas 2, 3, 4 we have

Dm (t) ≥

f (ϕ) + A3 ≥

m Em + A3 + A9 = 2

4(Em + A3 + A9 ) m2

m π2 π 4 + 24π 2 4(A3 + A9 ) m √ π2 = − + + = 1 + x, 1+ 2 2 4m2 2m3 24m4 m2 where

π2 2 π 2 + 24 x= + A10 1− + m 4m2 6m2


and

317

14.52 16 16(A3 + A9 ) −40.51 − − 23.37 ≤ A10 = < 3 2 3 9 m π m π2 21.26 14.52 16 + 11.50 < ≤ 2 3 . 9 π m m3

Thus

0.79π 2 2 π2 2 5.64 40.51 5.65 21.26 π2 1 − ≤ x ≤ 1 − + + < − + m m 4m2 4m2 m2 m3 4m2 m2 m3
0.79 (the function value at m = 9) the supremum of the function in the second bracket for m ≥ 9 is 1 (its limit as m → ∞). By the inequality (7) we obtain that m 1+ 2 m 1+ = 2 m 1+ = 2

(2) ≥ Dm (t) ≥ νm

where

π2 π2 π 4 + 24π 2 x 2 − + − 8 8m2 4m3 48m4

π2 π2 π 4 + 24π 2 π4 − + − + A11 8m2 4m3 48m4 128m4 π2 π2 5π 4 + 192π 2 − + + A11 , 8m2 4m3 384m4

2 π4 x2 1 π 2 π = A11 = − −x +x . 8 8 4m2 128m4 4m2

By the estimate for x and by

π 2 + 24 π2 π2 2− − mA10 −x = 6m 4m2 4m3 we have 4.41 1.79π 2 < ≤ m2 4m2

π 2 + 24 21.26 2.73 π2 2 − ≤ − < 6·9 m3 4m3 92

π2 2π 2 4.94 + x ≤ < 2 , 2 2 4m 4m m

2 2 π 40.51 6.17 π 2 + < 3 . − x ≤ 4m2 4m3 92 m

Hence 1.50 4.41 · 2.73 4.94 · 6.17 3.81 < ≤ A11 ≤ < 5 . 5 5 m 8m 8m5 m

318


On the other hand by (4) cos

π π4 π2 + − C6 = 1− 2m + 2 8(m + 1)2 384(m + 1)4

with 0 ≤ C6 ≤

π6 46080(m + 1)6

and by (5) 1+

1 m

−4

= 1 − 4B1 ,

1+

1 m

−2

=1−

2 3 + 2 − 4B3 , m m

where 1 1 −5 1 1 −5 1 0.59 ≤ 1+ 1+ ≤ ≤ B1 ≤ , m m 9 m m m and 0.59 1 1 1 1 −5 1 −5 ≤ ≤ ≤ B3 ≤ 3 . 1 + 1 + 9 m m3 m3 m3 m Thus 1 −2 1 −4 π2 1 + m π π4 1 + m =1− cos + − C6 2m + 2 8 384 m2 m4 2 π4 3 π2 1 − + =1− − 4B (1 − 4B1 ) − C6 3 + 2 2 m m 8m 384m4 =1−

π2 π2 3π 2 π4 + − + + A12 2 3 4 8m 4m 8m 384m4

=1−

π2 π2 144π 2 − π 4 + − + A12 , 8m2 4m3 384m4

where A12 = −C6 +

π 2 B3 2m2

−

π 4 B1 96m4

and

π6 0.59π 2 π 4 4.34 1.89 1 1 π 2 0.59π 4 − + − ≤ A − < 5 . < ≤ 12 46080 · 9 2 96 96 m5 m5 m5 2 m Finally, (2) π π2 2νm π2 5π 4 + 192π 2 cos = 1+ − + + A 11 m 2m + 2 8m2 4m3 384m4 π2 π2 144π 2 − π 4 × 1− + − + A 12 8m2 4m3 384m4 =1−

π4 6π 4 + 48π 2 π2 + + A = 1 + + A13 , 13 64m4 384m4 8m4


319

where 4 π 1 π 4 4π 6 + 336π 4 + m5 (A11 + A12 ) − + m5 A13 = 2 32 m 16 8 · 384 6 1 4π + 336π 4 π 2 5 + m (A12 − A11 ) + 2 4 · 384 8 m 1 (5π 4 + 192π 2 )(144π 2 − π 4 ) π 2 5 m + (A − A ) − 3 12 11 4 m 3842 144π 2 − π 4 5 1 5π 4 + 192π 2 5 m A12 − m A11 + m5 A11 A12 . + 4 384 384 m Omitting the positive terms but the first one, using − m1k ≥ − 91k (k = 1, 2, . . .) and the estimates for A10 , A11 , we get 4 π 1 π 4 4π 6 + 336π 4 5 + m (A11 + A12 ) − + m A13 > 2 32 m 16 8 · 384 2 5 1 (5π 4 + 192π 2 )(144π 2 − π 4 ) π 2 m5 A12 1 π m A11 − 3 − 2 + 8 4 m m 3842 1 144π 2 − π 4 5 m A11 − 4 384 m 4 1 π 4 4π 6 + 336π 4 3.81π 2 π + 1.50 + 1.88 − + − 2 > 16 9 16 8 · 384 9 ·8 3.81(144π 2 − π 4 ) 1 (5π 4 + 192π 2 )(144π 2 − π 4 ) 4.34π 2 − − 3 + 4 9 3842 94 · 384 5

> 7.36. Hence (2) 2νm π π2 7.36 cos >1+ + 5 >1 4 m 2m + 2 8m m

which completes the proof of the inequality μm >

π m sec 2 2m + 2

(11)

if m ≥ 9. For m < 9 we have Dm (t)2 = fm (2 cos t), where f1 (x) = 1, f3 (x) = 3x 2 + 8x + 8, f5 (x) = 5x 4 + 14x 3 + 6x 2 − 2x + 13, and f7 (x) = 7x 6 + 20x 5 − 4x 4 − 40x 3 − 4x 2 + 32x + 32. Put c1 = 1, c3 = 8/3, c5 = 6.70, c7 = 12.74. For m = 1 and 3 we have minx∈R fm (x) = cm , for m = 5 and 7 we have fm (±2) > cm and applying Sturm’s theorem we find that fm (x) − cm has no zeros in the interval ]–2, 2[. Hence Dm (t)2 ≥ √ π , proving (11) for m < 9. cm for all t. However cm > m2 sec 2m+2

320


To prove the inequality π 1 m , +O μm < sec 2 2m + 2 m3 first we note that in the notation of Lemma 2 m sin2 mt m 1 − cos mt m 1 − cos ϕ 2 + = + = + 2 2 2 2 sin2 2t 4 sin2 2t 4 sin2 2t 2 C4 m 1 − cos ϕ ψm − + I2 = + 1+ 2 4 4 2 m 1 − cos ϕ 1 . = + 1+O 2 4 m2 Therefore we have 2 m sin2 mt 1 m 1 − cos ϕ 2 1 − cos ϕ 2 + + O . = + t 2 2 2 4 4 m 2 sin 2 Using this equation instead of (10) in the proof of Lemma 2, we obtain with the notation of Lemma 2, that 1 1 1 − cos ϕ 2 . (12) O +O Dm (t) = f (ϕ) + 4 m m4 π π If ϕ = ϕm = m − 2m 2 , tm = (12) and Lemma 4 that

(m−1)π+ϕm , m

then by

= O( m12 ), we obtain from

1 = Em + O . m3 √ π From the asymptotic expansions given above for Em and for cos 2m+2 it follows that π 1 m . μm ≤ Dm (tm ) < sec +O 2 2m + 2 m3

1 Dm (tm ) = f (ϕm ) + O m3

1−cos ϕm 4

2

This completes the proof of Proposition 2. The theorem follows at once.

References 1. Bonsall, E.M., Marden, M.: Zeros of self-inversive polynomials. Proc. Am. Math. Soc. 3, 471–475 (1952) 2. Cohn, A.: Über die Anzahl der Wurzeln einer algebraischen Gleichung in einem Kreise. Math. Z. 14, 110–148 (1922) 3. Lakatos, P.: On zeros of reciprocal polynomials. Publ. Math. Debrecen 61, 645–661 (2002) 4. Lakatos, P., Losonczi, L.: On zeros of reciprocal polynomials of odd degree, J. Inequal. Pure Appl. Math. 4(3) (2003). Article 60, 8 pp. (electronic, http://jipam.vu.edu.au) 5. Pólya, G., Szeg˝o, G.: Problems and Theorems in Analysis, vol. II. Springer, Berlin (1976) 6. Schinzel, A.: Self-inversive polynomials with all zeros on the unit circle. Ramanujan J. 9, 19–23 (2005)