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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

Annulus containing all the zeros of a polynomial Aseem Dalal a, N.K. Govil b,⇑ a b

Department of Mathematics, Indian Institute of Technology Delhi, New Delhi 110016, India Department of Mathematics & Statistics, Auburn University, Auburn, AL 36849-5108, USA

a r t i c l e

i n f o

Keywords: Polynomials Location of zeros of polynomials Eigenvalues MATLAB

a b s t r a c t Recently Dalal and Govil (2013) proved that for any sequence of positive numbers fAk gnk¼1 P P such that nk¼1 Ak ¼ 1, a complex polynomial PðzÞ ¼ nk¼0 ak zk with ak – 0; 1 6 k 6 n has all its zeros in the annulus C ¼ fz : r1 6 jzj 6 r 2 g, where

1=k 1=k a0 1 ank r 1 ¼ min16k6n Ak and r 2 ¼ max16k6n : ak Ak an They also showed that their result includes as special cases, many known results in this direction. In this paper we prove that the bounds obtained by making choice of different fAk gnk¼1 ’s cannot be in general compared, that is one can always construct examples in which one result gives better bound than the other and vice versa. Also, we provide a result which gives better bounds than the existing results in all cases. Finally, using MATLAB, we compare the result obtained by our theorem with the existing ones to show that our theorem gives sharper bounds than many of the results known in this direction. Ó 2014 Elsevier Inc. All rights reserved.

1. Introduction The properties of polynomials have been studied since the time of Gauss and Cauchy, and have played an important role in many scientiﬁc disciplines (see [11–13]). Problems involving location of their zeros ﬁnd important applications in many areas of applied mathematics such as control theory, signal processing, communication theory, coding theory, cryptography, combinatorics, and mathematical biology. Since Abel and Rufﬁni proved that there is no general algebraic solution to polynomial equations of degree ﬁve or higher, the problem of ﬁnding an annulus containing all the zeros of a polynomial became much more interesting and over a period a large number of results have been provided in this direction. An accurate estimate of the annulus containing all the zeros of a polynomial can considerably reduce the amount of work needed to ﬁnd exact zeros, and so there is always a need for better and better estimates for the annulus containing all the zeros of a polynomial. The earliest result concerning the location of the zeros of a polynomial is probably due to Gauss, who showed that a polynomial

pðzÞ ¼ an zn þ an1 zn1 þ . . . þ a1 z þ a0 : with all ak real, has no zeros outside certain circle jzj ¼ R , where

⇑ Corresponding author. E-mail addresses: [email protected] (A. Dalal), [email protected] (N.K. Govil). http://dx.doi.org/10.1016/j.amc.2014.10.038 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.

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A. Dalal, N.K. Govil / Applied Mathematics and Computation 249 (2014) 429–435 1=k

R ¼ max16k6n ðn21=2 jak jÞ

:

Cauchy [2] improved the above result of Gauss by proving the following Theorem 1.1. Let pðzÞ ¼ zn þ

Pn1

k¼0 ak z

k

, be a complex polynomial. Then all the zeros of pðzÞ lie in the disc

fz : jzj < gg fz : jzj < 1 þ Ag; where

A ¼ max06k6n1 jak j; and g is the unique positive root of the real-coefﬁcient equation

zn jan1 jzn1 jan2 jzn2 . . . ja1 jz ja0 j ¼ 0: The above result of Cauchy has been sharpened, among others, by Joyal, Labelle and Rahman [9], Datt and Govil [4], Sun and Hsieh [14], Jain [8], and Affane-Aji, Biaz and Govil [1]. The following result, which provides an annulus containing all the zeros of a polynomial is due to Diaz-Barrero [5]. P Theorem 1.2. Let pðzÞ ¼ nk¼0 ak zk ðak – 0; 1 6 k 6 nÞ be a non-constant complex polynomial. Then all its zeros lie in the annulus C ¼ fz : r1 6 jzj 6 r 2 g , where

r1 ¼

1=k n 3 2 F k Cðn; kÞ a0 min16k6n a 2 F 4n k

ð1:1Þ

r2 ¼

ank 1=k 2 F 4n max16k6n n : 3 2 F k Cðn; kÞ an

ð1:2Þ

and

th

n! Here F k is the k Fibonacci number, namely, F 0 ¼ 0; F 1 ¼ 1 and for k P 2; F k ¼ F k1 þ F k2 . Furthermore, Cðn; kÞ ¼ k!ðnkÞ! are the binomial coefﬁcients. Another result in this direction providing annulus containing all the zeros of a polynomial is the following, and is due to Kim [10].

P Theorem 1.3. Let pðzÞ ¼ nk¼0 ak zk ðak – 0; 1 6 k 6 nÞ be a non constant polynomial with complex coefﬁcients. Then all the zeros of p(z) lie in the annulus A ¼ fz : r 1 6 jzj 6 r 2 g where,

r1 ¼ min16k6n

1=k Cðn; kÞ a0 n 2 1 ak

ð1:3Þ

and

1=k 2n 1 ank : Cðn; kÞ an

r2 ¼ max16k6n

ð1:4Þ

Here Cðn; kÞ are the binomial coefﬁcients. The following two results by Diaz-Barrero and Egozcue [7], also provide annuli containing all the zeros of a polynomial. P Theorem 1.4. Let pðzÞ ¼ nk¼0 ak zk ðak – 0; 1 6 k 6 nÞ be a non-constant complex polynomial. Then for j P 2, all its zeros lie in the annulus C ¼ fz : r1 6 jzj 6 r 2 g where,

( ) nk 1=k Cðn; kÞAk Bkj ðbBj1 Þ a0 r1 ¼ min16k6n a Ajn k

ð1:5Þ

and

( r2 ¼ max16k6n

)1=k ank A : nk jn a n Cðn; kÞAk Bkj ðbBj1 Þ Ajn

ð1:6Þ

P n k n1k Here, Bn ¼ n1 and An ¼ cr n þ ds , where c, d are real constants and r,s are the roots of the equation x2 ax b ¼ 0 in k¼0 r s P nk which a; b are strictly positive real numbers. For j P 2; nk¼0 Cðn; kÞðbBj1 Þ Bkj Ak ¼ Ajn . Furthermore, Cðn; kÞ are the binomial coefﬁcients.

A. Dalal, N.K. Govil / Applied Mathematics and Computation 249 (2014) 429–435

431

P Theorem 1.5. Let pðzÞ ¼ nk¼0 ak zk ðak – 0; 1 6 k 6 nÞ be a non-constant polynomial with complex coefﬁcients. Then, all its zeros lie in the ring shaped region C ¼ fz : r1 6 jzj 6 r2 g where,

r1 ¼ min16k6n

1=k n 2 P k Cðn; kÞ a0 a P 3n k

ð1:7Þ

and

r2 ¼ max16k6n

ank 1=k P3n : 2n Pk Cðn; kÞ an

ð1:8Þ

th

n! Here Pk is the k Pell number, namely, P0 ¼ 0; P1 ¼ 1 and for k P 2; Pk ¼ 2Pk1 þ Pk2 . Furthermore, Cðn; kÞ ¼ k!ðnkÞ! are the binomial coefﬁcients. Finally, we present the following result in this direction, which is due to Diaz-Barrero [6], also providing an annulus containing all the zeros of a polynomial.

P Theorem 1.6. Let pðzÞ ¼ nk¼0 ak zk ðak – 0; 1 6 k 6 nÞ be a complex monic polynomial. Then, all its zeros lie in the disks C 1 ¼ fz : jzj 6 r1 g or C 2 ¼ fz : jzj 6 r 2 g where

r1 ¼ max16k6n

( 2n1 Cðn þ 1; 2Þ 2

k Cðn; kÞ

)1=k jank j

ð1:9Þ

and

( r2 ¼ max16k6n

F 3n

Cðn; kÞ2k F k

)1=k jank j

ð1:10Þ

:

Here Cðn; kÞ is the binomial coefﬁcient. Recently, Dalal and Govil [3] presented a uniﬁed result which includes all the above-mentioned results as special cases. More precisely Dalal and Govil [3] proved the following. P P Theorem 1.7. Let nk¼1 Ak ¼ 1 be an identity with Ak > 0 for 1 6 k 6 n, and let PðzÞ ¼ nk¼0 ak zk ðak – 0; 1 6 k 6 nÞ be a non constant polynomial with complex coefﬁcients. Then all the zeros of P(z) lie in the annulus C ¼ fz : r 1 6 jzj 6 r2 g where,

1=k a0 r1 ¼ min16k6n Ak ak

ð1:11Þ

and

r2 ¼ max16k6n

1=k 1 ank : Ak an

ð1:12Þ

Theorem 1.7 implies that inﬁnitely many annuli containing all the zeros of a complex polynomial PðzÞ ¼ Pn a zk ðak – 0 for 1 6 k 6 nÞ can be obtained from inﬁnitely many sequences of positive numbers fAk gnk¼1 such that k¼0 Pn k Pn n k¼1 Ak ¼ 1. Then, it is natural to ask which sequence of positive numbers fAk gk¼1 with k¼1 Ak ¼ 1 gives the best result. Pn ak k Pn ank 1 It is easy to see that the best possible result in fact comes when k¼1 a0 r 1 ¼ 1 and k¼1 an rk ¼ 1, which is proved in the 2 following result of Cauchy. Theorem 1.8. All the zeros of a polynomial f ðzÞ ¼ unique positive root of the equation

Pn

k¼0 ak z

k

; an – 0 lie in the annulus C ¼ fz : r1 6 jzj 6 r2 g, where r1 is the

jan jjzjn þ jan1 jjzjn1 þ . . . þ ja1 jjzj ja0 j ¼ 0 and r2 is the unique positive root of the equation

ja0 j þ ja1 jjzj þ . . . þ jan1 jjzjn1 jan jjzjn ¼ 0: Although, the result of Cauchy gives the best bound but it is implicit in the sense that in order to ﬁnd the annulus containing all the zeros of a polynomial, we need to compute a zero of another polynomial while Theorems 1.2–1.7 mentioned above are explicit, that is, the annulus can be obtained by only using the polynomial coefﬁcients. P Now coming back to the question that for a given polynomial PðzÞ ¼ nk¼0 ak zk , what set of positive numbers Pn n fAk gk¼1 ; k¼1 Ak ¼ 1 should be chosen out of inﬁnitely many possible values that would give a sharp bound, we consider for example the polynomial PðzÞ ¼ z3 þ 0:1z2 þ 0:1z þ 0:7. For this polynomial, Theorems 1.2 and 1.6 gives the annulus with the area 3:4730 and 8:3820 respectively, but for randomly generated sequences f0:0184; 0:1576; 0:8240g and f0:1586; 0:3659; 0:4755g of fAk g3k¼1 , the area comes out to be 93:9685 and 2:5568 respectively. Clearly, the bound with area

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A. Dalal, N.K. Govil / Applied Mathematics and Computation 249 (2014) 429–435

of 93:9685 is poor while the bound with the area of 2:5568 is desirable. The question that now arises is how to get a sharp bound, because putting fAk gnk¼1 randomly does not ensure the improvement in the bound. Thus, there is a need for a result that can provide bounds with monotonically decreasing area, and in this paper we study this problem and obtain a result in this direction. 2. Main results For any r 1 > 0 and r 2 > 0, we deﬁne Dr1 ¼ f ðr1 ; a0 ; . . . ; an Þ ¼ 1 P 1 Dr2 ¼ f ðr 2 ; a0 ; . . . ; an Þ ¼ 1 nk¼1 aank rk , well-deﬁned when an – 0. n

Pn ak k k¼1 a0 r 1 , which is well-deﬁned when a0 – 0; and

2

It is easily seen from Theorem 1.8 that, if for any positive numbers r 1 and r 2 , we have Dr1 P 0 and Dr2 P 0, then all the P zeros of the polynomial PðzÞ ¼ nk¼1 ak zk ; ðak – 0 for 1 6 k 6 nÞ lie in the region ðr1 6 jzj 6 r2 Þ. Also, it is clear that for r 1 and r2 in Theorem 1.7, we have Dr1 P 0 and Dr2 P 0. The following lemma states the necessary and the sufﬁcient condition for Dr1 to be equal to zero. P Lemma 2.1. Let nk¼1 Ak ¼ 1; with Ak > 0 for 1 6 k 6 n, be an identity used to ﬁnd the inner radius r 1 in Theorem 1.7. Then, Dr1 ¼ 0 if and only if ja10 j ; jaA11 j ; . . . jaAnn j are in geometric progression with common ratio r1 ¼ aa01 A1 . Proof. Suppose,

1 ; A1 ; . . . jaAnn j ja0 j ja1 j

are in geometric progression with common ratio r1 ¼ aa01 A1 . Then,

Ak rk ¼ 1 for 1 6 k 6 n; jak j ja0 j which is equivalent to

ak k r ¼ Ak for 1 6 k 6 n: a 1 0 Therefore,

Dr1 ¼ 1

n n X X ak rk ¼ 1 Ak ¼ 1 1 ¼ 0: a0 1 k¼1

k¼1

Now suppose, Dr1 ¼ 0. Then from the deﬁnition of Dr1 , we have

1

n X ak rk ¼ 0; a 1 k¼1

0

which implies n n X X ak r k ¼ 0; Ak a 1 k¼1 k¼1 0

that is,

n X ak Ak r k1 ¼ 0: a0 k¼1

ð2:1Þ

Since by (1.11), we have Ak aa0k rk1 P 0, therefore this when combined with (2.1) gives

ak Ak rk1 ¼ 0 for 1 6 k 6 n; a0

which clearly implies ratio r1 ¼ aa10 A1 . h

Ak jak j

rk

¼ ja10 j for 1 6 k 6 n. Thus,

1 ; A1 ; . . . jaAnn j ja0 j ja1 j

are in geometric progression with common

Similarly, the necessary and the sufﬁcient condition for Dr2 to be equal to zero can be found, and this is given in the following lemma. P Lemma 2.2. Let nk¼1 Ak ¼ 1; with Ak > 0 for 1 6 k 6 n, be an identity used to ﬁnd the outer radius r 2 in Theorem 1.7. Then, an An 1 1 Dr2 ¼ 0 if and only if ja1n j ; jaAn1 j ; . . . ja0 j are in geometric progression with common ratio r 2 ¼ an1 A1 . We omit the proof, since the proof is on the same lines as of Lemma 2.1. As mentioned earlier, Theorem 1.7 can generate inﬁnitely many results giving annulus containing all the zeros of a polynomial, and over the years, mathematicians have compared their bounds with the existing bounds in the literature by generating some examples, and hence showing their bound is better in special cases.

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A. Dalal, N.K. Govil / Applied Mathematics and Computation 249 (2014) 429–435

In the following two theorems, we show that no-matter what result you obtain as a corollary of Theorem 1.7, you can always generate examples in which your bound is better than the existing ones. P P Theorem 2.3. Let fAk gnk¼1 and fBk gnk¼1 be sequences of positive numbers such that nk¼1 Ak ¼ 1 and nk¼1 Bk ¼ 1. Then, there A B A B always exists a polynomial for which r1 > r 1 and vice versa, where r1 and r 1 are the inner radii of the annulus obtained from the Theorem 1.7 by using the sequences fAk gnk¼1 and fBk gnk¼1 respectively. Proof. Let us consider the ¼ 1 for 1 6 k 6 n, therefore

Bk ja0 j jak j

polynomial

PðzÞ ¼

Pn

k k¼0 ak z ;

ðak – 0 for 0 6 k 6 nÞ

such

that

ak ¼ Bk a0 .

1=k a0 rB1 ¼ min16k6n Bk ¼ 1: ak

Because

ð2:2Þ

0j Also, ja10 j ; jaB11 j ; . . . jaBnn j are in geometric progression with common ratio jaBk j jaBk1 j ¼ Bk ja ¼ 1, and since by (2.2) we have rB1 ¼ 1 jak j k k1 B1 Bn 1 hence we get that ja0 j ; ja1 j ; . . . jan j are in geometric progression with common ratio r B1 . Therefore, by Lemma 2.1 we have P DrB ¼ 0. Now, note that Ak < Bk for at least one k; 1 6 k 6 n, because otherwise nk¼1 Ak would be greater than 1, which is 1

a contradiction. For this k, we get by Theorem 1.7, that

1k 1k a0 a0 rA1 6 Ak < Bk ¼ rB1 ; ak ak

because Bk aa0 ¼ 1 for each k; 1 6 k 6 n. Hence r A1 < r B1 . k P Similarly, for the polynomial PðzÞ ¼ nk¼0 ak zk , with ak – 0 for 0 6 k 6 n and ak ¼ Ak a0 , we will have r B1 < r A1 , and the proof of this theorem is thus complete. h Pn Pn Theorem 2.4. Let fAk gnk¼1 and fBk gnk¼1 be sequences of positive numbers such that k¼1 Ak ¼ 1 and k¼1 Bk ¼ 1. Then, there always exists a polynomial for which r A2 > rB2 and vice versa, where rA2 and rB2 are the outer radii of the annulus obtained from the Theorem 1.7 by using sequences fAk gnk¼1 and fBk gnk¼1 respectively. We omit the proof, as it follows on the same lines as that of Theorem 2.3. Thus the above discussion implies that the bounds computed by substituting different fAk gnk¼1 ’s in Theorem 1.7 cannot be in general compared. However, the following theorem improves any bound r 1 ; r 2 obtainable from any theorem if Dr1 > 0 and Dr2 > 0. Pn Pn k Theorem 2.5. Suppose k¼1 Bk ¼ 1; with Bk > 0 for 1 6 k 6 n and PðzÞ ¼ k¼0 ak z ðak – 0; 1 6 k 6 nÞ be a non-constant polynomial with complex coefﬁcients. Let r ; r be any positive numbers such that D P 0 and Dr2 P 0. Then, all the zeros of PðzÞ r1

1 2 lie in the annulus C ¼ z : r01 6 jzj 6 r 02 where

1=k a0 r01 ¼ min16k6n r k1 þ Dr1 Bk ak

ð2:3Þ

and

r02 ¼ max16k6n

an 1=k 1 þ D B : r k 2 a rk2 nk

ð2:4Þ

Remarks. Before we proceed to the proof of the above theorem we would like to make the following remarks. 1. Since, it is clear that r01 P r 1 and r02 6 r 2 therefore the above Theorem 2.5 always gives a bound sharper than any bound that has been obtained in terms of r1 and r 2 . 2. Since we know that Dr1 P 0 and Dr2 P 0 for r 1 and r2 obtained from any theorem that is obtainable from Theorem 1.7 therefore the Theorem 2.5 will always give a bound that is sharper than the bound obtainable from any result that is obtained from Theorem 1.7. In particular Theorem 2.5 will always give a bound sharper than obtainable from any of the Theorems 1.2–1.6. 3. In order to obtain better and better bound, Theorem 2.5 can be applied multiple times by substituting newly computed r 01 and r 02 as r 1 and r2 in the formula, and making corresponding substitutions in Dr1 and Dr2 respectively.

Proof of Theorem 2.5. If a0 ¼ 0, then r1 ¼ 0 which implies r01 ¼ r1 ¼ 0, and therefore we can suppose without loss of generality that a0 – 0. Let z be such that jzj < r 01 . Then,

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A. Dalal, N.K. Govil / Applied Mathematics and Computation 249 (2014) 429–435

! X n n n n X X X k ak ðr 0 Þk : jPðzÞj ¼ ak zk P ja0 j jak jjzjk > ja0 j jak jðr01 Þ ¼ ja0 j 1 a 1 k¼0 k¼1 k¼1 k¼1 0

ð2:5Þ

Since by (2.3), we have

ak 0 k ak k ðr Þ 6 r þ Dr Bk ; 1 a0 1 a0 1

for 1 6 k 6 n;

ð2:6Þ

hence on substituting (2.6) in (2.5) we get

jPðzÞj > ja0 j 1

! ! ! n n n n X X X X ak rk ¼ a ¼ a ¼ 0; D B D D B 1 B j j j jD r1 k 0 r1 r1 0 r1 k k a 1 k¼1

0

k¼1

k¼1

k¼1

because

n X Bk ¼ 1; k¼1

and consequently PðzÞ does not have any zeros in the region jzj < r 01 . It is known that all the zeros of PðzÞ have modulus less than or equal to the unique positive root of the equation

GðzÞ ¼ jan jzn jan1 jzn1 þ . . . þ ja1 jz þ ja0 j ¼ 0:

In view of this, the second part of the theorem will follow if we show that Gðr02 Þ P 0. Since by (2.4) we have for 1 6 k 6 n,

ank 1 ank 1 6 a 0 k a r k þ Dr2 Bk n ðr Þ n 2 2

hence

"

Gðr02 Þ

# " # n n X X X ank 1 ank 1 0 n ¼ 1 nBk a 0 k P jan jðr 2 Þ 1 a r k Dr 2 n ðr Þ n 2 k¼1 k¼1 k¼1 2 n X n P jan jðr 02 Þ Dr2 Dr2 ; because Bk ¼ 1 ¼ 0; n jan jðr 02 Þ

k¼1

and the proof of the theorem is thus complete. h It may be remarked that r02 in (2.4) can also be obtained by applying the formula of r01 in (2.3) to the polynomial zn Pð1z Þ. 3. Comparison of bounds In this section we will compare using MATLAB, the results obtained by Theorem 2.5 with the existing results in the literature and subsequently show the importance of such a theorem. In fact as proved above, our Theorem 2.5 will always give a bound sharper than any bound obtainable from a theorem that has been obtained from Theorem 1.7, therefore for this reason, we compare the bound obtained by our theorem with the bounds obtained by only Theorems 1.2 and 1.6. Example. Let pðzÞ ¼ z3 þ 0:1z2 þ 0:1z þ 0:7. k Fk In the following table, we use Theorem 1.2 to get inner radii r 1 and outer radii r 2 (with Ak ¼ Cðn;kÞ2 ), and use Theorem 2.5 F 3n 0 0 with Bk ¼ 1=3; 1 6 k 6 3 to compute the corresponding r1 and r2 . Result

Inner Radii (r1 or r 01 )

Outer Radii (r2 or r 02 )

Area of Annulus

Theorem 1.2 Theorem 2.5 Actual Bound

0.6402 0.6576 0.8840

1.2312 1.094 0.8899

3.4730 2.4027 0.0328

2

k Cðn;kÞ In the table below, we use Theorem 1.6 to get the inner radii r 1 and r2 . (with Ak ¼ 2n1 ), and use Theorem 2.5 with Cðnþ1;2Þ Bk ¼ 1=3; 1 6 k 6 3 to compute the corresponding r 01 and r 02 .

Result

Inner Radii (r1 or r 01 )

Outer Radii (r2 or r 02 )

Area of Annulus

Theorem 1.6 Theorem 2.5 Actual Bound

0.4641 0.5002 0.8840

1.6984 1.2076 0.8899

8.382 3.7957 0.0328

It is clear from the above tables that Theorem 2.5 can give signiﬁcant improvements over the bounds obtainable from Theorem 1.2 and Theorem 1.6.

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