In this paper we obtain certain results for the polar derivative of a polynomial \(p(z) = c_nz^n +\sum_{j=\mu}^n c_{n-j}z^{n-j}\), \(1\leq\mu\leq n\), having all its zeros on \(|z| = k\), \(k\leq 1\), which generalizes the results due to Dewan and Mir, Dewan and Hans. We also obtain certain new inequalities concerning the maximum modulus of a polynomial with restricted zeros.  [Editor's note: There are flaws in the paper, see  M. A. Qazi, Remarks on some recent results about polynomials with restricted zeros, Ann. Univ. Mariae Curie-Skłodowska Sect. A 67 (2), (2013), 59-64 

Ahuja, Arty

Dewan, K. K.

Hans, Sunil

English

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

doi: 10.2478/v10062-011-0001-xANNALESU 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 1–9ARTY AHUJA, K. K. DEWAN and SUNIL HANSInequalities concerningpolar derivative of polynomialsAbstract. In this paper we obtain certain results for the polar derivative ofa polynomial p(z) = cnzn +∑nj=µ cn−jzn−j , 1 ≤ µ ≤ n, having all its zeroson |z| = k, k ≤ 1, which generalizes the results due to Dewan and Mir, Dewanand Hans. We also obtain certain new inequalities concerning the maximummodulus of a polynomial with restricted zeros.1. Introduction and statement of results. Let p(z) be a polynomialof degree n and p′(z) its derivative, then according to Bernstein’s inequality(for reference see [1]), we havemax|z|=1|p′(z)| ≤ nmax|z|=1|p(z)|.(1.1)The result is sharp and equality holds in (1.1) for p(z) = λzn, where |λ| = 1.For the class of polynomials not vanishing in |z| < k, k ≥ 1, Malik [8]provedmax|z|=1|p′(z)| ≤ n1 + kmax|z|=1|p(z)|.(1.2)The result is sharp and the extremal polynomial is p(z) = (z + k)n.While trying to obtain inequality analogous to (1.2) for polynomials notvanishing in |z| < k, k ≤ 1, Govil [5] proved that if p(z) has all its zeros on2000 Mathematics Subject Classification. 30A10, 30C10, 30C15.Key words and phrases. Polynomials, maximum modulus, inequalities in the complexdomain, polar derivative.2 A. Ahuja, K. K. Dewan and S. Hans|z| = k, k ≤ 1, thenmax|z|=1|p′(z)| ≤ nkn−1 + knmax|z|=1|p(z)|.(1.3)While seeking for a better bound in the inequality (1.3), Dewan and Mir[4] proved the following result.Theorem A. If p(z) =∑nj=0 cjzj is a polynomial of degree n having all itszeros on |z| = k, k ≤ 1, thenmax|z|=1|p′(z)| ≤ nkn(n|cn|k2 + |cn−1|n|cn|(1 + k2) + 2|cn−1|)max|z|=1|p(z)|.(1.4)Dewan and Hans [3] generalized the above result to the class of polyno-mials of the type p(z) = cnzn +∑nj=µ cn−jzn−j , 1 ≤ µ ≤ n and proved thefollowing result.Theorem B. If p(z) = cnzn +∑nj=µ cn−jzn−j, 1 ≤ µ < n, is a polynomialof degree n having all its zeros on |z| = k, k ≤ 1, then(1.5)max|z|=1|p′(z)|≤ nkn−µ+1(n|cn|k2µ + µ|cn−µ|kµ−1µ|cn−µ|(1 + kµ−1) + n|cn|kµ−1(1 + kµ+1))max|z|=1|p(z)|.Let α be a complex number. If p(z) is a polynomial of degree n, thenpolar derivative of p(z) with respect to the point α, denoted by Dαp(z), isdefined byDαp(z) = np(z) + (α− z)p′(z).(1.6)Clearly Dαp(z) is a polynomial of degree at most n − 1 and it generalizesthe ordinary derivative in the sense thatlimα→∞[Dαp(z)α]= p′(z) .(1.7)In this paper, we first prove the following result which extends Theorem Aand Theorem B to the polar derivative of a polynomial having all its zeroson |z| = k, k ≤ 1.Theorem 1. If p(z) = cnzn +∑nj=µ cn−jzn−j, 1 ≤ µ < n, is a polynomialof degree n having all its zeros on |z| = k, k ≤ 1, then for every real orcomplex number α with |α| ≥ k, we have(1.8)max|z|=1|Dαp(z)|≤ n(|α|+kµ)kn−µ+1(n|cn|k2µ+µ|cn−µ|kµ−1µ|cn−µ|(1 +kµ−1)+n|cn|kµ−1(1 + kµ+1))max|z|=1|p(z)|.Instead of proving Theorem 1 we prove the following theorem which givesa better bound than the above theorem. Briefly, we prove:Inequalities concerning polar derivative of polynomials 3Theorem 2. If p(z) = cnzn +∑nj=µ cn−jzn−j, 1 ≤ µ < n, is a polynomialof degree n having all its zeros on |z| = k, k ≤ 1, then for every real orcomplex number α with |α| ≥ k, we have(1.9)max|z|=1|Dαp(z)|≤ n(|α|+Sµ)kn−µ+1(n|cn|k2µ+µ|cn−µ|kµ−1µ|cn−µ|(1 + kµ−1)+n|cn|kµ−1(1+kµ+1))max|z|=1|p(z)|,whereSµ =n|cn|k2µ + µ|cn−µ|kµ−1n|cn|kµ−1 + µ|cn−µ| .(1.10)To prove that the bound obtained in the above theorem is better thanthe bound obtained in Theorem 1, we show thatSµ ≤ kµorn|cn|k2µ + µ|cn−µ|kµ−1µ|cn−µ|+ n|cn|kµ−1 ≤ kµwhich is equivalent ton|cn|k2µ + µ|cn−µ|kµ−1 ≤ µ|cn−µ|kµ + n|cn|k2µ−1,which impliesn|cn|(k2µ − k2µ−1) ≤ µ|cn−µ| (kµ − kµ−1)ornµ∣∣∣∣ cncn−µ∣∣∣∣ ≥ 1kµ ,which is always true (see Lemma 6).Remark 1. Dividing both sides of inequalities (1.8) and (1.9) by |α| andletting |α| → ∞, we get Theorem B due to Dewan and Hans [3].If we choose µ = 1 in Theorem 2 , we have the following result.Corollary 1. If p(z) =∑nj=0 cjzj is a polynomial of degree n having allits zeros on |z| = k, k ≤ 1, then for every real or complex number α with|α| ≥ k, we have(1.11)max|z|=1|Dαp(z)|≤ n(|α|+ S1)kn(n|cn|k2 + |cn−1|2|cn−1|+ n|cn|(1 + k2))max|z|=1|p(z)|,whereS1 =(n|cn|k2 + |cn−1|n|cn|+ |cn−1|).(1.12)4 A. Ahuja, K. K. Dewan and S. HansRemark 2. Dividing both sides of (1.11) by |α| and letting |α| → ∞, weobtain Theorem A due to Dewan and Mir [4].We next prove the following interesting results for the maximum modulusof polynomials.Theorem 3. If p(z) =∑nj=0 cjzj is a polynomial of degree n having allits zeros on |z| = k, k ≤ 1, then for every real or complex number α with|α| ≥ k and 0 ≤ r ≤ k ≤ R, we have(1.13)max|z|=R|Dαp(z)| ≤ nRn−1(|α|+RS′1)kn(n|cn|k2 +R|cn−1|2R|cn−1|+ n|cn|(R2 + k2))×(Rn + kRn−1rn + krn−1)max|z|=r|p(z)|,whereS′1 =1Rn|cn|k2 +R|cn−1|nR|cn|+ |cn−1| .(1.14)Dividing both sides of (1.13) by |α| and letting |α| → ∞, we obtain thefollowing result.Corollary 2. If p(z) =∑nj=0 cjzj is a polynomial of degree n having all itszeros on |z| = k, k ≤ 1, then for 0 ≤ r ≤ k ≤ R, we have(1.15)max|z|=R|p′(z)| ≤ nRn−1kn(n|cn|k2 +R|cn−1|2R|cn−1|+ n|cn|(R2 + k2))×(Rn + kRn−1rn + krn−1)max|z|=r|p(z)|.By involving the coefficients c0 and c1 of p(z) =∑nj=0 cjzj , we prove thefollowing generalization of Theorem 3.Theorem 4. If p(z) =∑nj=0 cjzj is a polynomial of degree n having allits zeros on |z| = k, k ≤ 1, then for every real or complex number α with|α| ≥ k and 0 ≤ r ≤ k ≤ R, we have(1.16)max|z|=R|Dαp(z)| ≤nRn−1(|α|+RS′1)kn(n|cn|k2 +R|cn−1|2R|cn−1|+ n|cn|(R2 + k2))×(2k2Rn|c1|+Rn−1(R2 + k2)n|c0|2k2rn|c1|+ rn−1(r2 + k2)n|c0|)max|z|=r|p(z)|,where S′1 is the same as defined in Theorem 3.On dividing both sides of (1.16) by |α| and letting |α| → ∞, we get thefollowing result.Inequalities concerning polar derivative of polynomials 5Corollary 3. If p(z) =∑nj=0 cjzj is a polynomial of degree n having all itszeros on |z| = k, k ≤ 1, then for 0 ≤ r ≤ k ≤ R, we have(1.17)max|z|=R|p′(z)| ≤nRn−1kn(n|cn|k2 +R|cn−1|2R|cn−1|+ n|cn|(R2 + k2))×(2k2Rn|c1|+Rn−1(R2 + k2)n|c0|2k2rn|c1|+ rn−1(r2 + k2)n|c0|)max|z|=r|p(z)|.2. Lemmas. We need the following lemmas for the proof of these theorems.Lemma 1. If p(z) is a polynomial of degree n, then for |z| = 1|p′(z)|+ |q′(z)| ≤ nmax|z|=1|p(z)|,(2.1)where here and throughout this paper q(z) = znp(1z¯).This is a special case of a result due to Govil and Rahman [6].Lemma 2. Let p(z) = cnzn +∑nν=µ cn−νzn−ν , 1 ≤ µ < n, be a polynomialof degree n having no zero in the disk |z| < k, k ≤ 1. Then for |z| = 1kn−µ+1 max|z|=1|p′(z)| ≤ max|z|=1|q′(z)|.(2.2)The above lemma is due to Dewan and Hans [3].Lemma 3. Let p(z) = c0 +∑nυ=µ cυzυ, 1 ≤ µ ≤ n, be a polynomial ofdegree n having no zero in the disk |z| < k, k ≥ 1. Then for |z| = 1kµ|p′(z)| ≤ |q′(z)|.(2.3)The above lemma is due to Chan and Malik [2].Lemma 4. Let p(z) = cnzn +∑nν=µ cn−νzn−ν , 1 ≤ µ ≤ n, be a polynomialof degree n having all its zeros on |z| = k, k ≤ 1. Then for |z| = 1kµ|p′(z)| ≥ |q′(z)|.(2.4)Proof of Lemma 4. If p(z) has all its zeros on |z| = k, k ≤ 1, then q(z)has all its zeros on |z| = 1k , 1k ≥ 1. Now applying Lemma 3 to the polynomialq(z), the result follows. Lemma 5. Let p(z) = c0 +∑nυ=µ cυzυ, 1 ≤ µ ≤ n, be a polynomial ofdegree n having no zero in the disk |z| < k, k ≥ 1. Then for |z| = 1,kµ+1{µ|cµ|kµ−1 + n|c0|µ|cµ|kµ+1 + n|c0|} ∣∣p′ (z)∣∣ ≤ ∣∣q′ (z)∣∣ ,(2.5)andµn∣∣∣∣cµc0∣∣∣∣ kµ ≤ 1.(2.6)The above lemma was given by Qazi [10, Remark 1].6 A. Ahuja, K. K. Dewan and S. HansLemma 6. Let p(z) = cnzn +∑nν=µ cn−νzn−ν , 1 ≤ µ ≤ n, be a polynomialof degree n having all its zeros on |z| = k, k ≤ 1. Then for |z| = 1,kµ−1{µ|cn−µ|+ n|cn|kµ+1µ|cn−µ|+ n|cn|kµ−1} ∣∣p′ (z)∣∣ ≥ ∣∣q′ (z)∣∣(2.7)andµn∣∣∣∣cn−µcn∣∣∣∣ ≤ kµ.(2.8)Proof of Lemma 6. Since p(z) has all its zeros on |z| = k, k ≤ 1, thenq(z) has all its zeros on |z| = 1k , 1k ≥ 1. Now applying Lemma 5 to thepolynomial q(z), Lemma 6 follows. Lemma 7. If p(z) =∑nυ=0 cυzυ be a polynomial of degree n, having all itszeros in the disk |z| ≥ k, k > 0, then for r ≤ k and R ≥ kM(p, r)rn + krn−1≥ M(p,R)Rn + kRn−1.(2.9)The above lemma is due to Jain [7].Lemma 8. If p(z) =∑nυ=0 cυzυ be a polynomial of degree n, having all itszeros in the disk |z| ≥ k, k > 0, then for r ≤ k and R ≥ k(2.10)M(p, r)2k2rn|c1|+rn−1(r2+k2)n|c0| ≥M(p,R)2k2Rn|c1|+Rn−1(R2+k2)n|c0| .The above lemma is due to Mir [9].3. Proofs of the theorems.Proof of Theorem 1. The proof of this theorem follows on the same linesas that of Theorem 2, but instead of using Lemma 6, we use Lemma 4. Weomit the details. Proof of Theorem 2. Since q(z) = znp(1z¯), then it can be easily verifiedthat|q′(z)| = |np(z)− zp′(z)| for |z| = 1 .Now for every real or complex number α, we haveDαp(z) = np(z) + (α− z)p′(z).This implies with the help of Lemma 6 that(3.1)|Dαp(z)| ≤ |αp′(z)|+ |np(z)− zp′(z)|= |α||p′(z)|+ |q′(z)|≤ (|α|+ Sµ)|p′(z)|.Inequalities concerning polar derivative of polynomials 7Let z0 be a point on |z| = 1, such that |q′(z0)| = max|z|=1 |q′(z)|, then byLemma 1, we get|p′(z0)|+max|z|=1 |q′(z)| ≤ nmax|z|=1|p(z)|,(3.2)which on using Lemma 6, gives1kµ−1(µ|cn−µ|+ n|cn|kµ−1n|cn|kµ+1 + µ|cn−µ|)|q′(z0)|+max|z|=1 |q′(z)| ≤ nmax|z|=1|p(z)|or(µ|cn−µ|(1 + kµ−1) + n|cn|kµ−1(1 + kµ+1)n|cn|k2µ + µ|cn−µ|kµ−1)max|z|=1|q′(z)| ≤ nmax|z|=1|p(z)|.The above inequality when combined with Lemma 2, gives(3.3)kn−µ+1(µ|cn−µ|(1 + kµ−1) + n|cn|kµ−1(1 + kµ+1)n|cn|k2µ + µ|cn−µ|kµ−1)max|z|=1|p′(z)|≤ nmax|z|=1|p(z)|.On combining the inequalities (3.1) and (3.3), we get the desired result. Proof of Theorem 3. Let 0 ≤ r ≤ k ≤ R. Since p(z) has all its zero on|z| = k, k ≤ 1, then the polynomial p(Rz) has all its zeros on |z| = kR , kR ≤ 1,therefore, applying Corollary 1 to the polynomial p(Rz) with |α| ≥ k, wegetmax|z|=1|D αRp(Rz)|≤n( |α|R + S′1)knRn nRn|cn| k2R2 +Rn−1|cn−1|2Rn−1|cn−1|+ nRn|cn|(1 + k2R2)max|z|=1|p(Rz)|ormax|z|=1∣∣∣np(Rz) + (αR− z)Rp′(Rz)∣∣∣≤n( |α|R + S′1)knRn nRn|cn| k2R2 +Rn−1|cn−1|2Rn−1|cn−1|+ nRn|cn|(1 + k2R2)max|z|=R|p(z)|which is equivalent tomax|z|=R|Dαp(z)|≤ nRn−1(|α|+RS′1)kn(nRn−2|cn|k2 +Rn−1|cn−1|2Rn−1|cn−1|+ nRn−2|cn|(R2 + k2))max|z|=R|p(z)|.8 A. Ahuja, K. K. Dewan and S. HansFor 0 ≤ r ≤ k ≤ R, the above inequality in conjunction with Lemma 7,yieldsmax|z|=R|Dαp(z)|≤ nRn−1(|α|+RS′1)kn(n|cn|k2 +R|cn−1|2R|cn−1|+ n|cn|(R2 + k2))×(2Rn + kRn−1rn + krn−1)max|z|=r|p(z)|,which completes the proof of Theorem 3. Proof of Theorem 4. The proof follows on the same lines as that of The-orem 3, but instead of using Lemma 7 we use Lemma 8. Remark 3. For µ = n, Theorems 1 and 2 hold if the polynomial satisfiesthe condition |c0| ≤ k|cn|.References[1] Bernstein, S., Lecons sur les proprie´te´s extre´males et la meilleure approximation des-fonctions analytiques d’une variable re´ele, Gauthier Villars, Paris, 1926 (French).[2] Chan, T. N., Malik, M. A., On Erdo¨s–Lax theorem, Proc. Indian Acad. Sci. 92 (3)(1983), 191–193.[3] Dewan, K. K., Hans, S., On maximum modulus for the derivative of a polynomial,Ann. Univ. Mariae Curie-Skłodowska Sect. A 63 (2009), 55–62.[4] Dewan, K. K., Mir, A., Note on a theorem of S. Bernstein, Southeast Asian Bulletinof Math. 31 (2007), 691–695.[5] Govil, N. K., On the theorem of S. Bernstein, J. Math. Phys. Sci. 14 (1980), 183–187.[6] Govil, N. K., Rahman, Q. I., Functions of exponential type not vanishing in a halfplane and related polynomials, Trans. Amer. Math. Soc. 137 (1969), 501–517.[7] Jain, V. K., On polynomials having zeros in closed exterior or interior of a circle,Indian J. Pure Appl. Math. 30 (1999), 153–159.[8] Malik, M. A., On the derivative of a polynomial, J. London Math. Soc. 1 (1969),57–60.[9] Mir, A., On extremal properties and location of zeros of polynomials, Ph.D. Thesissubmitted to Jamia Millia Islamia, New Delhi, 2002.[10] Qazi, M. A., On the maximum modulus of polynomials, Proc. Amer. Math. Soc. 115(1992), 337–343.Arty Ahuja K. K. DewanDepartment of Mathematics Department of MathematicsFaculty of Natural Sciences Faculty of Natural SciencesJamia Millia Islamia Jamia Millia Islamia(Central University) (Central University)New Delhi–110025 New Delhi–110025India Indiae-mail: aarty ahuja@yahoo.com e-mail: kkdewan123@yahoo.co.inInequalities concerning polar derivative of polynomials 9Sunil HansDepartment of MathematicsFaculty of Natural SciencesJamia Millia Islamia (Central University)New Delhi - 110 025Indiae-mail: sunil.hans82@yahoo.comReceived July 2, 2010

Inequalities concerning polar derivative of polynomials

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Inequalities concerning polar derivative of polynomials

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