For a polynomial of degree n, we have obtained some results, which generalize and improve upon the earlier well known results (under certain conditions). A similar result is also obtained for analytic function

Hans, Sunil

Kumar, Susheel

Lal, Roshan

English

University of Maria Curie-Skłodowska (UMCS): Scientific e-Journals / Uniwersytet Marii Curie-Skłodowskiej: e-czasopisma naukowe

doi: 10.2478/v10062-011-0008-3ANNALESU N I V E R S I T A T I S M A R I A E C U R I E - S K Ł O D O W S K AL U B L I N – P O L O N I AVOL. LXV, NO. 1, 2011 SECTIO A 97–108ROSHAN LAL, SUSHEEL KUMAR1 and SUNIL HANS2 3On the zeros of polynomialsand analytic functionsAbstract. For a polynomial of degree n, we have obtained some results,which generalize and improve upon the earlier well known results (under cer-tain conditions). A similar result is also obtained for analytic function.1. Introduction and statement of results. The following theorem isdue to Pellet ([6], [5, p. 128]).Theorem A. Let q(z) = a0 + a1z + . . . + apzp + . . . + anzn, ap 6= 0, bea polynomial of degree n. If the polynomialQp(z) = |a0|+ |a1|z + . . .+ |ap−1|zp−1 − |ap|zp + |ap+1|zp+1 + . . .+ |an|zn,has two positive zeros r and R, r < R, then q(z) has exactly p zeros in thedisc|z| ≤ rand no zero in the annular ringr < |z| < R.The next result is due to Jayal, Labelle and Rahman [4].1The research of the author is supported by UGC, New Delhi; F. No. 17-52/98(SA-I).2The research of the author is supported by CSIR, New Delhi; F. No. -9/466(95)/2007-EMR-I.3Corresponding author.2000 Mathematics Subject Classification. 30C15, 30A10.Key words and phrases. Polynomial, analytic function, zeros.98 R. Lal, S. Kumar and S. HansTheorem B. Let p(z) =∑nv=0 avzv be a polynomial of degree n such thatan ≥ an−1 ≥ . . . ≥ a1 ≥ a0.Then p(z) has all its zeros in(1) |z| ≤ an − a0 + |a0||an| .Gardner and Govil [1] improved Theorem B as follows.Theorem C. Let p(z) =∑nv=0 avzv be a polynomial of degree n such thatan ≥ an−1 ≥ . . . ≥ a1 ≥ a0.Then p(z) has all its zeros in the annular ring(2)|a0|an − a0 + |an| ≤ |z| ≤an − a0 + |a0||an| .Recently Jain [3] proved the following result for the upper bound involvingcoefficients of the polynomial.Theorem D. Let q(z) = a0 + a1z + . . .+ ap−1zp−1 + apzp + . . .+ anzn, bea polynomial of degree n such that ap 6= ap−1 for some p ∈ {1, 2, . . . , n}. SetM =Mp :=n∑j=p+1|aj − aj−1|+ |an| (1 ≤ p ≤ n− 1), Mn := |an|andm = mp :=p−1∑j=1|aj − aj−1| (2 ≤ p ≤ n), m1 := 0.Suppose thatpM|ap − ap−1|p+ 1< 1and that|a0|+m pM|ap − ap−1|p+ 1<( pM)p( |ap − ap−1|p+ 1)p+1.Then q(z) has at least p zeros in|z| < pM|ap − ap−1|p+ 1.Jain [3] again, in the same paper proved the following.Theorem E. Let q(z) = a0 + a1z + . . .+ ap−1zp−1 + apzp + . . .+ anzn, bea polynomial of degree n such that ap 6= ap−1 for some p ∈ {1, 2, . . . , n− 1},|arg ak − β| ≤ α ≤ pi2,On the zeros of polynomials and analytic functions 99k = 0, 1, 2 . . . , n, for some real β and α and|an| ≥ |an−1| ≥ . . . ≥ |a1| ≥ |a0|.LetL = Lp := |an|+ (|an| − |ap|) cosα+n∑j=p+1(|aj |+ |aj−1|) sinαandl = lp := (|αp−1|−|a0|) cosα+p−1∑j=1(|aj |+|aj−1|) sinα (2 ≤ p ≤ n−1), l1 := 0.Suppose that|a0|+ l pL|ap − ap−1|p+ 1<(pL)p( |ap − ap−1|p+ 1)p+1.Then q(z) has at least p zeros in|z| < pL|ap − ap−1|p+ 1.In this paper, firstly we prove the following.Theorem 1. Let q(z) = a0+a1z+a2z2+ . . .+ap−1zp−1+apzp+ . . .+anznbe a polynomial of degree n such that ap 6= ap−1 for some p ∈ {1, 2, . . . , n},with coefficients aj, j = 0, 1, 2, . . . , n, satisfying(3) an ≥ an−1 ≥ . . . ≥ ap > ap−1 ≥ . . . ≥ a1 ≥ a0and(4)(pM1)p(ap − ap−1p+ 1)p+1> |a0|+ pM1(ap − ap−1p+ 1)(ap−1 − a0)where M1 = an + |an| − ap.Then q(z) has at least p zeros in(5)|a0|an − a0 + ρn1 |an|≤ |z| < ρ1 = p(|an| − an + ap)(ap − ap−1)(p+ 1).Remark 1. In Theorem 1, we haveM1 = |an|+ an − apfor 1 ≤ p ≤ (n− 1) and M1 = |an| for p = n. The value M1 = |an|+an−apserves the purpose for 1 ≤ p ≤ n (see also equality (1.6) of Jain [3]).For the case p = n, in Theorem 1, we have the following.100 R. Lal, S. Kumar and S. HansCorollary 1. Let q(z) = a0 + a1z + . . .+ anzn be a polynomial of degree nsuch that(6) an > an−1 ≥ an−2 ≥ . . . ≥ a1 ≥ a0,and(7)(n|an|)n(an − an−1n+ 1)n+1> |a0|+ nn+ 1(an − an−1|an|)(an−1 − a0)then q(z) has all its zeros in(8)|a0|an − a0 + ρn2 |an|≤ |z| < ρ2 =nn+ 1(an − an−1)|an| .Remark 2. Corollary 1 is a refinement of Theorem B due to Joyal, Labelleand Rahman [4] as well as Theorem C due to Gardner and Govil [1] underthe conditions (6) and (7).As it can be shown easily from (8) and (2) thatnn+ 1(an − an−1)|an| <an − a0 + |a0||an|is always true.And also|a0|an − a0 + ρn2 |an|>|a0|an + |an| − a0forρ2 =nn+ 1an − an−1|an| .Remark 3. If we take a0 > 0, then Corollary 1 gives a refinement of a resultdue to Jain [3, Corollary 1].Instead of proving Theorem 1, we prove the following result. Theorem 1can be proved in a similar way as the next result (Theorem 2) except theonly change that is in Theorem 1 p ∈ {1, 2, . . . , n}.Theorem 2. Let q(z) = a0 + a1z + . . . + ap−1zp−1 + apzp + . . . anzn bea polynomial of degree n such that ap 6= ap−1 for some p ∈ {1, . . . , n − 1},with the coefficients aj, j = 0, 1, . . . , n, for some K ≥ 1, satisfying(9) Kan ≥ an−1 ≥ . . . ≥ ap > ap−1 ≥ . . . ≥ a1 ≥ a0and(10)(pM2)p(ap − ap−1p+ 1)p+1> |a0|+ pM2(ap − ap−1p+ 1)(ap−1 − a0),where(11) M2 = K(an + |an|)− ap.On the zeros of polynomials and analytic functions 101Then q(z) has at least p zeros in(12)|a0|Kan + (K − 1)|an| − a0 + ρn3 |an|≤ |z| ≤ ρ3 = pM2(ap − ap−1)p+ 1where we assume that ρ3 < 1.For the case a0 > 0, we have the following.Corollary 2. Let q(z) = a0+a1z+a2z2+ . . .+ap−1zp−1+apzp+ . . .+anznbe a polynomial of degree n, with the condition ap 6= ap−1 and for somep ∈ {1, 2, . . . , n− 1}, K ≥ 1 satisfying(13) Kan ≥ an−1 ≥ . . . ≥ ap > ap−1 ≥ . . . ≥ a0 > 0and(14)(pM3)p(ap − ap−1p+ 1)p+1> a0 +pM3(ap − ap−1p+ 1)(ap−1 − a0),where(15) M3 = 2Kan − ap.Then q(z) has at least p zeros in(16)a0(2K − 1 + ρn4 )an − a0≤ |z| < ρ4 = pp+ 1(ap − ap−1)(2Kan − ap) .Remark 4. As ap−ap−12Kan−ap < 1 (by (13)), we havepp+1(ap−ap−12Kan−ap)= ρ4 < 1.For the polynomials with complex coefficients, we have been able to provethe following.Theorem 3. Let q(z) = a0 + a1z + . . . + ap−1zp−1 + apzp + . . . + anzn bea polynomial of degree n such that ap 6= ap−1 for some p ∈ {1, 2, . . . , n− 1},for some real β and α|arg aj − β| ≤ α ≤ pi2, j = 0, 1, . . . , n,and for some K ≥ 1,(17) K|an| ≥ |an−1| ≥ . . . ≥ |a1| ≥ |a0|and(18)(pM4)p( |ap − ap−1|p+ 1)p+1> |a0|+ pM4|ap − ap−1|(p+ 1)m′where(19)M4 = K|an|+ (K|an| − |ap|) cosα+ (K|an|+ |an−1|) sinα+n−1∑j=p+1(|aj |+ |aj−1|) sinα,102 R. Lal, S. Kumar and S. Hans(20) m′ = (|ap−1| − |a0|) cosα+p−1∑j=1(|aj |+ |aj−1|) sinα.Then q(z) has at least p zeros in(21) |z| < ρ5 = pp+ 1(|ap − ap−1|)M4.Remark 5. In the case K = 1, the above theorem reduces to Theorem Edue to Jain [3].Remark 6. ρ5 < 1, as can be verified by using (19), (17) and Lemma 1.Now we turn to the study of zeros of an analytic function. In this direc-tion, we have been able to prove the following.Theorem 4. Let the function f(z) =∑∞j=0 ajzj ( 6≡ 0) be analytic in |z| ≤ρ6, for some p ∈ N such that ap−1ap< 2 +1p. Assume that(22) a0 ≥ a1 ≥ . . . ≥ ap−1 > ap ≥ ap+1 ≥ . . .and(23)(pap)p(ap−1 − app+ 1)p+1> |a0|+ pap(ap−1 − app+ 1)(a0 − ap−1).Then the function f(z) has at least p zeros in(24) |z| < ρ6 = pp+ 1(ap−1 − apap).2. Lemma. For the proof of the theorems, we need the following lemma.Lemma 1. If aj is any complex number with|arg aj − β| ≤ α ≤ pi2,for certain real β and α, then|aj − aj−1| ≤ ||aj | − |aj−1|| cosα+ (|aj |+ |aj−1|) sinα.This lemma is due to Govil and Rahman (proof of Theorem 2 of [2]).3. Proofs of Theorems.Proof of Theorem 2. ConsiderOn the zeros of polynomials and analytic functions 103(25)g(z) = (1− z)q(z)= (1− z)(a0 + a1z + . . .+ ap−1zp−1 + apzp + . . .+ anzn)= a0 +p−1∑j=1(aj − aj−1)zj + (ap − ap−1)zp+n∑j=p+1(aj − aj−1)zj − anzn+1.= φ(z) + ψ(z),whereφ(z) = a0 +p−1∑j=1(aj − aj−1)zj ,andψ(z) = (ap − ap−1)zp +n∑j=p+1(aj − aj−1)zj − anzn+1.Now for |z| = ρ3 (ρ3 < 1 (as assumed)) and p ≤ n− 1,|ψ(z)| ≥ |ap − ap−1|ρp3 −{n∑j=p+1|aj − aj−1|ρj3 + |an|ρn+13}≥ (ap − ap−1)ρp3 − ρp+13{|an|ρn−p3 + |an − an−1|ρn−p3+n−1∑j=p+1|aj − aj−1|ρj−(p+1)3}≥ (ap − ap−1)ρp3 − ρp+13{|an|+ |Kan − an−1|+ (K − 1)|an|+n−1∑j=p+1|aj − aj−1|}= (ap − ap−1)ρp3 − ρp+13{|an|+Kan − an−1+ (K − 1)|an|+ an−1 − ap}= (ap − ap−1)ρp3 − ρp+13 {K(an + |an|)− ap}=(pM2)p(ap − ap−1p+ 1)p+1104 R. Lal, S. Kumar and S. Hans> |a0|+ pM2(ap − ap−1p+ 1)(ap−1 − a0) by (10).= |a0|+ ρ3(ap−1 − a0).Thus for |z| = ρ3,(26) |ψ(z)| > |a0|+ ρ3(ap−1 − a0).On the other hand, for |z| = ρ3,(27)|φ(z)| ≤ |a0|+p−1∑j=1|aj − aj−1|ρj3≤ |a0|+ ρ3(ap−1 − a0).From (26) and (27) it follows that |ψ(z)| > |φ(z)| for |z| = ρ3. By Rouche’stheorem, g(z) = φ(z)+ψ(z) and ψ(z) has same number of zeros in |z| < ρ3.But ψ(z) has at least p zeros in |z| < ρ3. Therefore g(z) and hence q(z) hasat least p zeros in|z| < ρ3.This proves one part of Theorem 2.Now it remains to prove that there are no zeros of q(z) in|z| < |a0|Kan + (K − 1)|an| − a0 + ρn3 |an|.Let(28)g(z) = (1− z)q(z) = a0 +n∑j=1(aj − aj−1)zj − anzn+1= a0 + h(z).Now for |z| = ρ3, (ρ3 < 1) we havemax|z|=ρ3|h(z)| ≤n∑j=1|aj − aj−1|ρj3 + |an|ρn+13≤ ρ3{|an − an−1|+n−1∑j=1|aj − aj−1|+ |an|ρn3}≤ ρ3{|Kan − an−1 + an −Kan|+ an−1 − a0 + |an|ρn3}≤ ρ3{Kan + (K − 1)|an| − a0 + ρn3 |an|}.Since h(0) = 0, h(z) is analytic in |z| ≤ ρ3, by Schwarz lemma we have|h(z)| ≤ {Kan + (K − 1)|an| − a0 + ρn3 |an|}|z|On the zeros of polynomials and analytic functions 105if |z| ≤ ρ3. Now from (28) we see that for |z| ≤ ρ3,|g(z)| ≥ |a0| − |h(z)|≥ |a0| − {Kan + (K − 1)|an| − a0 + ρn3 |an|}|z|> 0,if|z| < |a0|Kan + (K − 1)|an| − a0 + ρn3 |an|.This proves the theorem completely. Proof of Theorem 3. Consider the polynomial(29)g(z) = (1− z)q(z)= a0 +p−1∑j=1(aj − aj−1)zj + (ap − ap−1)zp+n∑j=p+1(aj − aj−1)zj − anzn+1= φ(z) + ψ(z),whereφ(z) = a0 +p−1∑j=1(aj − aj−1)zjandψ(z) = (ap − ap−1)zp +n∑j=p+1(aj − aj−1)zj − anzn+1.Now for |z| = ρ5 (ρ5 < 1),|ψ(z)| ≥ |ap − ap−1|ρp5 − ρp+15{|an|ρn−p5 +n∑j=p+1|aj − aj−1|ρj−(p+1)5}≥ |ap − ap−1|ρp5 − ρp+15{|an|+n∑j=p+1|aj − aj−1|}= |ap − ap−1|ρp5 − ρp+15{|an|+ |an − an−1|+n−1∑j=p+1|aj − aj−1|}= |ap − ap−1|ρp5 − ρp+15{|an|+ |Kan − an−1 + an −Kan|+n−1∑j=p+1|aj − aj−1|}106 R. Lal, S. Kumar and S. Hans≥ |ap − ap−1|ρp5 − ρp+15{|an|+ |Kan − an−1|+ (K − 1)|an|+n−1∑j=p+1|aj − aj−1|}≥ |ap − ap−1|ρp5 − ρp+15{K|an|+ (K|an| − |an−1|) cosα+ (K|an|+ |an−1|) sinα+n−1∑j=p+1[(|aj | − |aj−1|) cosα+ (|aj |+ |aj−1|) sinα]}(by Lemma 1)= |ap − ap−1|ρp5 − ρp+15{K|an|+ (K|an| − |ap|) cosα+ (K|an|+ |an−1|) sinα+n−1∑j=p+1(|aj |+ |aj−1|) sinα}= |ap − ap−1|ρp5 − ρp+15 M4=(pM4)p( |ap − ap−1|p+ 1)p+1, (by definition of M4).Thus on |z| = ρ5,(30)|ψ(z)| ≥(pM4)p( |ap − ap−1|p+ 1)p+1> |a0|+ pM4( |ap − ap−1|p+ 1)m′= |a0|+ ρ5m′ (by (18)).Now for |z| = ρ5,|φ(z)| ≤ |a0|+p−1∑j=1|aj − aj−1|ρj5 ≤ |a0|+ ρ5p−1∑j=1|aj − aj−1|≤ |a0|+ ρ5{ p−1∑j=1(|aj | − |aj−1|) cosα+p−1∑j=1(|aj |+ |aj−1|) sinα}= |a0|+ ρ5{(|ap−1| − |a0|) cosα+p−1∑j=1(|aj |+ |aj−1|) sinα}= |a0|+ ρ5m′ (by (20)).On the zeros of polynomials and analytic functions 107Thus for |z| = ρ5,(31) |φ(z)| ≤ |a0|+ ρ5m′.From (30) and (31), we see that on |z| = ρ5, |φ(z)| < |ψ(z)|, therebyimplying by Rouche’s theorem that g(z) = φ(z) + ψ(z) and ψ(z) have thesame number of zeros in |z| < ρ5. Since ψ(z) has at least p zeros in |z| < ρ5,this implies that g(z) and hence q(z) has at least p zeros in |z| < ρ5 =pp+ 1|ap − ap−1|M4.Thus the proof of Theorem 3 is completed. Proof of Theorem 4. It is clear that limj→∞ aj = 0. Consider(32)F (z) = (z − 1)f(z)= − a0 + (a0 − a1)z + (a1 − a2)z2 + . . .+ (ap−1 − ap)zp + . . .= φ(z) + ψ(z),whereφ(z) = − a0 +p−1∑j=1(aj−1 − aj)zjandψ(z) = (ap−1 − ap)zp +∞∑j=p+1(aj−1 − aj)zjNow for |z| = ρ6 (ρ6 < 1, by hypothesis for ap−1ap < 2 + 1p),|ψ(z)| ≥ |ap−1 − ap|ρp6 − ρp+16{ ∞∑j=p+1|aj−1 − aj |ρj−(p+1)6}≥ (ap−1 − ap)ρp6 − ρp+16{ ∞∑j=p+1|aj−1 − aj |}= (ap−1 − ap)ρp6 − ρp+16 ap=(pap)p(ap−1 − app+ 1)p+1> |a0|+ ρ6(a0 − ap−1) (by (23)).Thus for |z| = ρ6,(33) |ψ(z)| > |a0|+ ρ6(a0 − ap−1).108 R. Lal, S. Kumar and S. HansNow for |z| = ρ6 (ρ6 < 1)|φ(z)| ≤ |a0|+p−1∑j=1|aj−1 − aj |ρj6≤ |a0|+ ρ6p−1∑j=1|aj−1 − aj |= |a0|+ ρ6(a0 − ap−1).Now the remaining proof of the Theorem 4 follows on the same lines ofTheorem 3.Acknowledgment. The authors are grateful to the referee for valuablesuggestions.References[1] Gardner, R. B., Govil, N. K., Some generalizations of the Enestro¨m–Kakeya theorem,Acta Math. Hungar. 74 (1–2) (1997), 125–134.[2] Govil, N. K., Rahman, Q. I., On the Enestro¨m–Kakeya theorem, Toˆhoku Math. J. 20(1968), 126–136.[3] Jain, V. K., On the zeros of a polynomial, Proc. Indian Acad. Sci. Math. Sci. 119 (1)(2009), 37—43.[4] Joyal, A., Labelle, G. and Rahman, Q. I., On the location of zeros of polynomials,Canadian Math. Bull. 10 (1967), 55–63.[5] Marden, M., The Geometry of Polynomials, Math. Surveys No. 3, Amer. Math. Soc.,Providence, RI, 1966.[6] Pellet, M.A., Sur un mode de se´paration des racines des e´quations et la formule deLagrange, Bull. Sci. Math. 5 (1881), 393–395.Roshan Lal Susheel KumarDepartment of Mathematics Department of MathematicsGovernment Degree College Faculty of Natural SciencesChaubattakhal (Pauri) Jamia Millia Islamia (Central University)Uttrakhand 246 162 New Delhi - 110 025India Indiae-mail: rlkeshtwal@yahoo.co.in e-mail: ahlawat skumar@yahoo.co.inSunil HansDepartment of MathematicsFaculty of Natural SciencesJamia Millia Islamia (Central University)New Delhi - 110 025Indiae-mail: sunil.hans82@yahoo.comReceived February 26, 2010

On the zeros of polynomials and analytic functions

https://journals.umcs.pl/a/article/download/3941/2851

On the zeros of polynomials and analytic functions

Abstract

Similar works

Full text

Available Versions

University of Maria Curie-Skłodowska (UMCS): Scientific e-Journals / Uniwersytet Marii Curie-Skłodowskiej: e-czasopisma naukowe