# Annular Bounds for the Zeros of a Polynomial

Authors: Le Gao; N. K. Govil
Journal: International Journal of Mathematics and Mathematical Sciences
 (2018)

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## Abstract

The problem of obtaining the smallest possible region containing all the zeros of a polynomial has been attracting more and more attention recently, and in this paper, we obtain several results providing the annular regions that contain all the zeros of a complex polynomial. Using MATLAB, we construct specific examples of polynomials and show that for these polynomials our results give sharper regions than those obtainable from some of the known results.


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## Body

## 1. Introduction

The Fundamental Theorem of Algebra states that every nonzero, single-variable, polynomial(1)pz=a0+a1z+a2z2+a3z3+⋯+anzn,of degree n with complex coefficients, has exactly n complex zeros. The zeros may however be coincident. Although the above theorem tells about the number of zeros, it does not mention anything about the location of these zeros. The problem of obtaining the smallest possible region containing all the zeros of a polynomial has been attracting more and more attention recently, since the results are very useful in engineering applications as well as in many areas of applied mathematics such as control theory, cryptography, mathematical biology, and combinatorics [1–4].For any second-degree polynomial equation,ax2+bx+c=0,  a≠0, the roots can be found by using the familiar quadratic formula, x=(-b±b2-4ac)/2a, and for polynomials of third and fourth degree, there are analogous formulas to find the zeros. However, for polynomials of degree five or higher with arbitrary coefficients, the Abel-Ruffini theorem states that there is no algebraic solution. Several methods, for example, Aberth-Ehrlich method [5, 6], have been proposed for the simultaneous determination of zeros of algebraic polynomials and there are studies [7, 8] to accelerate convergence and increase computational efficiency of these methods. Approximations to the zeros of a polynomial can be drawn by these methods, and these methods can become more efficient when an annulus containing all the zeros of the polynomial is provided. So this paper is focused on finding new theorems that can provide smaller annuli containing all the zeros of a polynomial.


## 2. Preliminaries

The first contributor to this subject was probably Gauss, who proved the following.Theorem 1 (Gauss).
A polynomial(2)pz=a0+a1z+a2z2+a3z3+⋯+anzn,with all ak real, has no zeros outside the circle z=R, where (3)R=max1≤k≤n⁡n2ak1/k.Later in 1850, Gauss [9] proved that, for any real or complex ak, the radius R may be taken as the positive root of the equation (4)zn-21/2a1zn-1+⋯+an=0.Theorem1 of Gauss was improved in 1829 by Cauchy [10] who derived more exact bounds for the moduli of the zeros of a polynomial than those given by Gauss, by proving the following.Theorem 2 (Cauchy).
If(5)pz=a0+a1z+a2z2+a3z3+⋯+anznis a complex polynomial of degree n, then all the zeros lie in the disc(6)z:z<η⊂z:z<1+A,where A=max1≤k≤n-1⁡ak and η is the unique positive root of the real-coefficient equation (7)zn-an-1zn-1-an-2zn-2-⋯-a1z-a0=0.By applying Theorem 2 to the polynomial pz=znp(1/z), we easily get Theorem 3.Theorem 3.
All the zeros of the polynomial(8)pz=a0+a1z+a2z2+a3z3+⋯+anzn,an≠0,lie in the annulus r1≤z≤r2, where r1 is the unique positive root of the equation(9)anzn+an-1zn-1+an-2zn-2+⋯+a1z-a0=0,and r2 is the unique positive root of the equation(10)a0+a1z+⋯+an-1zn-1-anzn=0.Diaz-Barrero [11] gave the following results, providing circular regions containing all the zeros of a polynomial in terms of the binomial coefficients and Fibonacci’s numbers. Note that the binomial coefficients are defined by (11)Cn,k=n!k!n-k!,0!=1,and Fibonacci’s numbers are defined by (12)F0=0,F1=1,Fj=Fj-1+Fj-2,forj≥2.Theorem 4.
Letpz=∑j=0najzj(aj≠0,0≤j≤n) be a nonconstant complex polynomial. Then all its zeros lie in the disc C=z:r1≤z≤r2, where (13)r1=32min1≤j≤n⁡2nFjCjnF4na0aj1/k,r2=23max1≤j≤n⁡F4n2nFjCjnan-jan1/k.Kim [12] provides another annulus containing all the zeros of a polynomial with the binomial coefficients.Theorem 5.
Letpz=∑k=0nakzk(ak≠0,0≤k≤n) be a nonconstant polynomial with complex coefficients. Then all the zeros of p(z) lie in the annulus A=z:r1≤z≤r2, where(14)r1=min1≤k≤n⁡Cn,k2n-1a0ak1/k,r2=max1≤k≤n⁡2n-1Cn,kan-kan1/k.There are in fact many results in this direction, and for some more results, see, for example, [13, 14].Recently, Dalal and Govil [15] unified the above theorems and proved the following theorem.Theorem 6.
LetAk>0 for 1≤k≤n such that ∑k=1nAk=1. If pz=a0+a1z+a2z2+a3z3+⋯+anzn is a nonconstant complex polynomial of degree n, with ak≠0 for 1≤k≤n, then all the zeros of p(z) lie in the annulus C=z:r1≤z≤r2, where (15)r1=min1≤k≤n⁡Aka0ak1/k,r2=max1≤k≤n⁡1Akan-kan1/k.Using the above theorems, Dalal and Govil [15] also gave the following.Theorem 7.
Letpz=a0+a1z+a2z2+a3z3+⋯+anzn be a nonconstant complex polynomial of degree n, with ak≠0 for 1≤k≤n. Then all the zeros of p(z) lie in the annulus C=z:r1≤z≤r2, where(16)r1=min1≤k≤n⁡Ck-1Cn-kCna0ak1/k,r2=max1≤k≤n⁡CnCk-1Cn-kan-kan1/k,where (17)Ck=C2k,kk+1is the kth Catalan number in which C(2k,k) are the binomial coefficients.Theorem6 of Dalal and Govil [15] can generate infinitely many results, including Theorems 4 and 5, giving annulus containing all the zeros of a polynomial, and over the years, mathematicians have shown the usefulness of their results by comparing their bounds with the existing bounds in the literature by giving some examples and thus showing that their bounds are better in some special cases. In this connection, Dalal and Govil in [16] have shown that no matter what result you obtain as a corollary to Theorem 6, one can always generate polynomials for which the corollary so obtained gives better bound than the existing ones, implying that every result obtained by Theorem 6 can be useful. Since the results obtained as corollaries of Theorem 6 cannot in general be compared, more recently Dalal and Govil [17] have given results that help to compare the bounds for a subclass of polynomials. For this, they provide a class of polynomials with some conditions on degree or absolute range of coefficients of the polynomial, and for this class of polynomials, the bound obtained by one corollary is always better than the bound obtained from the other.


## 3. Main Results and Their Proofs

In this section, we obtain some new results which provide annuli containing all the zeros of a polynomial. These results have been obtained by making use of some identities and Theorem6. Then, in Section 4, we use MATLAB to obtain examples of polynomials for which our results give bounds that are sharper than those obtainable from Theorems 5 and 7. In fact, it is not difficult to construct polynomials for which our theorems also give better bounds than those obtainable from Theorem 4.Theorem 8.
Letpz=a0+a1z+a2z2+a3z3+⋯+anzn be a nonconstant complex polynomial of degree n, with ak≠0 for 1≤k≤n. Then all the zeros of pz lie in the annulus C=z:r1≤z≤r2, where(18)r1=min1≤k≤n⁡kCn,k2nn-k+1Cna0ak1/k,r2=max1≤k≤n⁡nn-k+1CnkCn,k2an-kan1/k.Theorem 9.
Letpz=a0+a1z+a2z2+a3z3+⋯+anzn be a nonconstant complex polynomial of degree n, with ak≠0 for 1≤k≤n. Then all the zeros of p(z) lie in the annulus C=z:r1≤z≤r2, where(19)r1=min1≤k≤n⁡Cn,2kCk2n-2kCn+1-2na0ak1/k,r2=max1≤k≤n⁡Cn+1-2nCn,2kCk2n-2kan-kan1/k.Theorem 10.
Letpz=a0+a1z+a2z2+a3z3+⋯+anzn be a nonconstant complex polynomial of degree n, with ak≠0 for 1≤k≤n. Then all the zeros of p(z) lie in the annulus C=z:r1≤z≤r2, where(20)r1=min1≤k≤n⁡F2kF2n+1-1a0ak1/k,r2=max1≤k≤n⁡F2n+1-1F2kan-kan1/k.Theorem 11.
Letpz=a0+a1z+a2z2+a3z3+⋯+anzn be a nonconstant complex polynomial of degree n, with ak≠0 for 1≤k≤n. Then all the zeros of p(z) lie in the annulus C=z:r1≤z≤r2, where(21)r1=min1≤k≤n⁡Fk2FnFn+1a0ak1/k,r2=max1≤k≤n⁡FnFn+1Fk2an-kan1/k.Theorem 12.
Letpz=a0+a1z+a2z2+a3z3+⋯+anzn be a nonconstant complex polynomial of degree n, with ak≠0 for 1≤k≤n. Then all the zeros of p(z) lie in the annulus C=z:r1≤z≤r2, where(22)r1=min1≤k≤n⁡k2Cn,knn+12n-2a0ak1/k,r2=max1≤k≤n⁡nn+12n-2k2Cn,kan-kan1/k.Theorem 13.
Letpz=a0+a1z+a2z2+a3z3+⋯+anzn be a nonconstant complex polynomial of degree n, with ak≠0 for 1≤k≤n. Then all the zeros of p(z) lie in the annulus C=z:r1≤z≤r2, where(23)r1=min1≤k≤n⁡kCn,k2n/2C2n,na0ak1/k,r2=max1≤k≤n⁡n/2C2n,nkCn,k2an-kan1/k.Theorem 14.
Letpz=a0+a1z+a2z2+a3z3+⋯+anzn be a nonconstant complex polynomial of degree n, with ak≠0 for 1≤k≤n. Then all the zeros of p(z) lie in the annulus C=z:r1≤z≤r2, where(24)r1=min1≤k≤n⁡k2Cn,k2n2C2n-2,n-1a0ak1/k,r2=max1≤k≤n⁡n2C2n-2,n-1k2Cn,k2an-kan1/k.


### 3.1. Lemmas

For the proofs of the above theorems, we will need the following lemmas.Lemma 15.
If(25)Cn=C2n,nn+1is the nth Catalan number in which C(2k,k) and C(n,k) are binomial coefficients, then for n≥1,(26)1n∑k=1nkCn,k2n-k+1=Cn.Proof of Lemma15.
Note that(27)1n∑k=1nkCn,k2n-k+1=1n∑k=1nCn,kkn-k+1Cn,k=1n∑k=1nCn,kkn-k+1n!k!n-k!=1n∑k=1nCn,kn!k-1!n-k+1!=1n∑k=1nCn,kCn,k-1=Cn.Lemma 16.
If(28)Cn=C2n,nn+1is the nth Catalan number in which C(2k,k) and C(n,k) are binomial coefficients, then for n≥1,(29)Cn+1=∑k=0nCn,2kCk2n-2k.Proof of Lemma16.
Note that, by [18, p. 292], we have(30)Cn+1=∑j=0nCn,jMj,where Mj is the jth Motzkin number defined by(31)M0=M1=M-1=1;Mj+1=2j+3j+3Mj+3jj+3Mj-1,j≥1.Also, by [18, p. 292], we have(32)Mj=∑k=0jCj,2kCk.Therefore, (33)Cn+1=∑j=0nCn,j∑k=0jCj,2kCk=∑k=0nCn,2kCk2n-2k.Lemma 17.
IfFn is the nth Fibonacci number, then for n≥1,(34)∑k=1nF2k=F2n+1-1.Proof of Lemma17.
SinceF0=0,F1=1, and Fj=Fj-1+Fj-2 for j≥2, hence(35)∑k=1nF2k=F2+F4+F6+⋯+F2n-2+F2n=F1+F2+F4+F6+⋯+F2n-2+F2n-F1=F3+F4+F6+⋯+F2n-2+F2n-F1⋮=F2n-1+F2n-F1=F2n+1-F1.Lemma 18.
IfFn is the nth Fibonacci number, then for n≥1,(36)∑k=1nFk2=FnFn+1.Proof of Lemma18.
One has(37)FnFn+1=FnFn+Fn-1=Fn2+FnFn-1=Fn2+Fn-1Fn-1+Fn-2=Fn2+Fn-12+Fn-1Fn-2⋮=Fn2+Fn-12+⋯+F12=∑k=1nFk2.Lemma 19.
If(38)Cn=C2n,nn+1is the nth Catalan number in which C(2k,k) and C(n,k) are binomial coefficients, then for n≥1,(39)∑k=0nk2Cn,k=n1+n2n-2.Proof of Lemma19.
Note that(40)∑k=0nk2Cn,k=∑k=1nk2Cn,k=∑k=1nknCn-1,k-1sincekCn,k=nCn-1,k-1=n∑k=0n-1k+1Cn-1,k=n∑k=0n-1kCn-1,k+n∑k=0n-1Cn-1,k=n∑k=1n-1kCn-1,k+n∑k=0n-1Cn-1,k=nn-12n-2+n2n-1,since∑k=1n-1kCn-1,k=n-12n-2=n1+n2n-2.Lemma 20.
If(41)Cn=C2n,nn+1is the nth Catalan number in which C(2k,k) and C(n,k) are binomial coefficients, then for n≥1,(42)∑k=0nkCn,k2=n2C2n,n.Proof of Lemma20.
One has(43)∑k=0nkCn,k2=∑k=0nnCn-1,k-1Cn,k=∑k=0nnCn-1,k-1Cn,n-k=nC2n-1,n-1=n2n-1!n-1!n!=n2C2n,n.Lemma 21.
If(44)Cn=C2n,nn+1is the nth Catalan number in which C(2k,k) and C(n,k) are binomial coefficients, then for n≥1,(45)∑k=0nk2Cn,k2=n2C2n-2,n-1.Proof of Lemma21.
One has(46)∑k=0nk2Cn,k2=∑k=0nn2Cn-1,k-12=n2∑k=0nCn-1,k-1Cn-1,n-k=n2C2n-2,n-1.


### 3.2. Proofs of the Theorems

Proof of Theorem8.
By Lemma15, we have that(47)1n∑k=1nkCn,k2n-k+1=Cn.If we take(48)Ak=1nkCn,k2n-k+1Cn,then Ak>0 and ∑k=1nAk=1. Hence applying Theorem 6 for this set of values of Ak, we get our desired result.Proof of Theorem9.
From Lemma16, we have that(49)Cn+1=∑k=0nCn,2kCk2n-2k.So Cn+1-2n=∑k=1nC(n,2k)Ck2n-2k. If we take(50)Ak=Cn,2kCk2n-2kCn+1-2n,then Ak>0 and ∑k=1nAk=1. Hence by applying Theorem 6 for this set of values of Ak, Theorem 9 can be proved.Proof of Theorem10.
From Lemma17, we have that(51)∑k=1nF2k=F2n+1-1.If we take(52)Ak=F2kF2n+1-1,then Ak>0 and ∑k=1nAk=1. Hence by applying Theorem 6 for this set of values of Ak, Theorem 10 can be proved.Proof of Theorem11.
From Lemma18, we have that(53)∑k=1nFk2=FnFn+1.If we take(54)Ak=Fk2FnFn+1,then Ak>0 and ∑k=1nAk=1. Hence by applying Theorem 6 for this set of values of Ak, Theorem 11 can be proved.Proof of Theorem12.
From Lemma19, we have that(55)∑k=0nk2Cn,k=n1+n2n-2.If we take(56)Ak=k2Cn,kn1+n2n-2,then Ak>0 and ∑k=1nAk=1. Hence by applying Theorem 6 for this set of values of Ak, Theorem 12 can be proved.Proof of Theorem13.
From Lemma20, we have that(57)∑k=0nkCn,k2=n2C2n,n.If we take(58)Ak=kCn,k2n/2C2n,n,then Ak>0 and ∑k=1nAk=1. Hence by applying Theorem 6 for this set of values of Ak, Theorem 13 can be proved.Proof of Theorem14.
From Lemma21, we have that(59)∑k=0nk2Cn,k2=n2C2n-2,n-1.If we take(60)Ak=k2Cn,k2n2C2n-2,n-1,then Ak>0 and ∑k=1nAk=1. Hence by applying Theorem 6 for this set of values of Ak, Theorem 14 can be proved.


## 3.1. Lemmas

For the proofs of the above theorems, we will need the following lemmas.Lemma 15.
If(25)Cn=C2n,nn+1is the nth Catalan number in which C(2k,k) and C(n,k) are binomial coefficients, then for n≥1,(26)1n∑k=1nkCn,k2n-k+1=Cn.Proof of Lemma15.
Note that(27)1n∑k=1nkCn,k2n-k+1=1n∑k=1nCn,kkn-k+1Cn,k=1n∑k=1nCn,kkn-k+1n!k!n-k!=1n∑k=1nCn,kn!k-1!n-k+1!=1n∑k=1nCn,kCn,k-1=Cn.Lemma 16.
If(28)Cn=C2n,nn+1is the nth Catalan number in which C(2k,k) and C(n,k) are binomial coefficients, then for n≥1,(29)Cn+1=∑k=0nCn,2kCk2n-2k.Proof of Lemma16.
Note that, by [18, p. 292], we have(30)Cn+1=∑j=0nCn,jMj,where Mj is the jth Motzkin number defined by(31)M0=M1=M-1=1;Mj+1=2j+3j+3Mj+3jj+3Mj-1,j≥1.Also, by [18, p. 292], we have(32)Mj=∑k=0jCj,2kCk.Therefore, (33)Cn+1=∑j=0nCn,j∑k=0jCj,2kCk=∑k=0nCn,2kCk2n-2k.Lemma 17.
IfFn is the nth Fibonacci number, then for n≥1,(34)∑k=1nF2k=F2n+1-1.Proof of Lemma17.
SinceF0=0,F1=1, and Fj=Fj-1+Fj-2 for j≥2, hence(35)∑k=1nF2k=F2+F4+F6+⋯+F2n-2+F2n=F1+F2+F4+F6+⋯+F2n-2+F2n-F1=F3+F4+F6+⋯+F2n-2+F2n-F1⋮=F2n-1+F2n-F1=F2n+1-F1.Lemma 18.
IfFn is the nth Fibonacci number, then for n≥1,(36)∑k=1nFk2=FnFn+1.Proof of Lemma18.
One has(37)FnFn+1=FnFn+Fn-1=Fn2+FnFn-1=Fn2+Fn-1Fn-1+Fn-2=Fn2+Fn-12+Fn-1Fn-2⋮=Fn2+Fn-12+⋯+F12=∑k=1nFk2.Lemma 19.
If(38)Cn=C2n,nn+1is the nth Catalan number in which C(2k,k) and C(n,k) are binomial coefficients, then for n≥1,(39)∑k=0nk2Cn,k=n1+n2n-2.Proof of Lemma19.
Note that(40)∑k=0nk2Cn,k=∑k=1nk2Cn,k=∑k=1nknCn-1,k-1sincekCn,k=nCn-1,k-1=n∑k=0n-1k+1Cn-1,k=n∑k=0n-1kCn-1,k+n∑k=0n-1Cn-1,k=n∑k=1n-1kCn-1,k+n∑k=0n-1Cn-1,k=nn-12n-2+n2n-1,since∑k=1n-1kCn-1,k=n-12n-2=n1+n2n-2.Lemma 20.
If(41)Cn=C2n,nn+1is the nth Catalan number in which C(2k,k) and C(n,k) are binomial coefficients, then for n≥1,(42)∑k=0nkCn,k2=n2C2n,n.Proof of Lemma20.
One has(43)∑k=0nkCn,k2=∑k=0nnCn-1,k-1Cn,k=∑k=0nnCn-1,k-1Cn,n-k=nC2n-1,n-1=n2n-1!n-1!n!=n2C2n,n.Lemma 21.
If(44)Cn=C2n,nn+1is the nth Catalan number in which C(2k,k) and C(n,k) are binomial coefficients, then for n≥1,(45)∑k=0nk2Cn,k2=n2C2n-2,n-1.Proof of Lemma21.
One has(46)∑k=0nk2Cn,k2=∑k=0nn2Cn-1,k-12=n2∑k=0nCn-1,k-1Cn-1,n-k=n2C2n-2,n-1.


## 3.2. Proofs of the Theorems

Proof of Theorem8.
By Lemma15, we have that(47)1n∑k=1nkCn,k2n-k+1=Cn.If we take(48)Ak=1nkCn,k2n-k+1Cn,then Ak>0 and ∑k=1nAk=1. Hence applying Theorem 6 for this set of values of Ak, we get our desired result.Proof of Theorem9.
From Lemma16, we have that(49)Cn+1=∑k=0nCn,2kCk2n-2k.So Cn+1-2n=∑k=1nC(n,2k)Ck2n-2k. If we take(50)Ak=Cn,2kCk2n-2kCn+1-2n,then Ak>0 and ∑k=1nAk=1. Hence by applying Theorem 6 for this set of values of Ak, Theorem 9 can be proved.Proof of Theorem10.
From Lemma17, we have that(51)∑k=1nF2k=F2n+1-1.If we take(52)Ak=F2kF2n+1-1,then Ak>0 and ∑k=1nAk=1. Hence by applying Theorem 6 for this set of values of Ak, Theorem 10 can be proved.Proof of Theorem11.
From Lemma18, we have that(53)∑k=1nFk2=FnFn+1.If we take(54)Ak=Fk2FnFn+1,then Ak>0 and ∑k=1nAk=1. Hence by applying Theorem 6 for this set of values of Ak, Theorem 11 can be proved.Proof of Theorem12.
From Lemma19, we have that(55)∑k=0nk2Cn,k=n1+n2n-2.If we take(56)Ak=k2Cn,kn1+n2n-2,then Ak>0 and ∑k=1nAk=1. Hence by applying Theorem 6 for this set of values of Ak, Theorem 12 can be proved.Proof of Theorem13.
From Lemma20, we have that(57)∑k=0nkCn,k2=n2C2n,n.If we take(58)Ak=kCn,k2n/2C2n,n,then Ak>0 and ∑k=1nAk=1. Hence by applying Theorem 6 for this set of values of Ak, Theorem 13 can be proved.Proof of Theorem14.
From Lemma21, we have that(59)∑k=0nk2Cn,k2=n2C2n-2,n-1.If we take(60)Ak=k2Cn,k2n2C2n-2,n-1,then Ak>0 and ∑k=1nAk=1. Hence by applying Theorem 6 for this set of values of Ak, Theorem 14 can be proved.


## 4. Computational Results and Analysis

In this section, we present two examples of polynomials in order to compare our theorems with some of the above stated known theorems and show that for these polynomials our theorems give better bounds than those obtainable by these known theorems.Example 1.
Consider the polynomialpz=z3+0.1z2+0.1z+0.7.Table1 suggests that our Theorem 11 gives significantly better bounds than those obtained from any result, including Theorems 5 and 7. As can be seen, the area of the annulus obtained by Theorem 11 is about 16.17% of the area of the annulus obtained by Theorem 5 and about 42.14% of the area obtained by Theorem 7. The inner radius obtained from Theorem 11 is almost 167% of the inner radius obtained from Theorem 5, and similarly the outer radius obtained from Theorem 11 is almost 59.84% of the outer radius obtained from Theorem 5.Table 1 Theorems r 1 r 2 Area of the annulus 5 0.4641 1.6984 8.382 7 0.6542 1.205 3.2159 8 0.5192 1.5183 6.395 10 0.5833 1.20 3.4549 11 0.7757 1.0164 1.3553 12 0.6403 1.2313 3.4748 14 0.4886 1.6134 7.428Example 2.
Consider the polynomial(61)pz=z5+0.006z4+0.29z3+0.29z2+0.29z+0.001.As shown in Table2, the best upper bound of the annular region containing all the zeros of the polynomial p(z) comes from Theorem 8, which also gives the smallest area of the annulus containing all the zeros. The area of the annulus obtained by Theorem 8 is about 90.84% of the area of the annulus obtained by Theorem 5 and about 50% of the area of the annulus obtained by Theorem 7.Table 2 Theorems r 1 r 2 Area of the annulus 5 5.56 × 10−4 1.158 4.2125 7 0.0011 1.5608 7.6529 8 8.21 × 10−5 1.1036 3.8265 12 7.18 × 10−5 1.3191 5.4664 13 1.37 × 10−4 1.1626 4.2464 14 4.93 × 10−5 1.1264 3.9859



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