 Applied Mathematics, 2011, 2, 141-144 doi:10.4236/am.2011.21016 Published Online January 2011 (http://www.SciRP.org/journal/am) Copyright © 2011 SciRes. AM Integral Mean Estimates for Polynomials Whose Zeros are within a Circle Yash Paul1, Wali Mohammad Shah2, Gulshan Singh1 1Bharathiar University Coimbatore, Tamil Nadu, India 2Department of Mat hem at i cs, K ash mi r Uni versity Srinagar , In di a E-mail: yashpaul2011@rediffmail.com, wmshah@rediffmail.com, gulshan si ng h 1@rediffmail.com Received December 28, 2010, revised January 12, 2011, accepted January 18, 2011 Abstract Let Pz be a polynomial of degree n having all its zeros in 1zK, then for each >0r, >1p, >1q with 11=1pq, Aziz and Ahemad (1996) proved that 11122 200 01.rprqrrprqriiinPedKed Ped   In this paper, we extend the above inequality to the class of polynomials =:= ,nnnjnnjjmPzaza z1mn, having all its zeros in 1zK, and obtain a generalization as well as refinement of the above result. Keywords: Derivative of a Polynomial, Integral Mean Estimates, Complex Domain Inequalities 1. Introduction and Statement of Results Let Pz be a polynomial of degree n and Pz be its derivative. If Pz has all its zeros in 1z, then it was shown by Turan  that  =1=1 .2zznMaxPzMaxP z (1) Inequality (1) is best possible with equality for =nPz z, where =. As an extension of (1) Malik  proved that if Pz has all its zeros in zK, where 1K, then  =1=1 .1zznMaxPzMaxP zK (2) Malik  obtained a generalization of (1) in the sense that the right-hand side of (1) is replaced by a factor involving the integral mean of Pz on =1z. In fact he proved the following theorem. Theorem A. If Pz has all its zeros in 1z, then for each >0r 1122=1001.rrrriiznPed edMaxPz (3) The result is sharp and equality in (3) holds for  =1nPz z. If we let r in (3), we get (1). As a generalization of Theorem A, Aziz and Shah  proved the following: Theorem B. If =:= nnnjnnjjmPzaza z, 1mn is a polynomial of degree n having all its zeros in the disk zK, 1K, then for each >0r, 12012=101.rrirrmi znPedKed MaxPz (4) Aziz and Ahemad  generalized (3) in the sense that =1zMaxP z on =1z on the right-hand side of (3) is replaced by a factor involving the integral mean of Pz on =1z and proved the following: Theorem C. If Pz is a polynomial of degree n having all its zeros in 1zK, then for >0r, >1p, >1q with 11=1pq, Y. PAUL ET AL. Copyright © 2011 SciRes. AM 1421201122001.rriprqrprqriinPedKedP ed (5) If we let r and p (so that 1q) in (5), we get (2). In this paper, we extend Theorem B to the class of polynomi als =:= ,nnnjnnjjmPzazaz1,mn having all the zeros in 1zK, and thereby obtain a more general result by proving the following. Theorem 1. If =:= nnnjnnjjmPzaza z, 1mn is a polynomial of degree n having all its zeros in the disk zK, 1K, then for each >0r, >1s, >1t with 11=1st, 11 12122 2100 01,sr srmmrtrrtrnnmiiimnnmna KmaKnPeded Pedna Kma   (6) If we take m = 1 in Theorem 1, we get the fol- lowing: Corollary 1. If =0:= njjjPz az is a polynomial of degree n having all its zeros in the disk zK, 1K, then for each >0r, >1s, >1t with 11=1st, 11 1222 2100 011.sr srrtrrtrnniiinnna KanPeded Pedna a   (7) The next result immediately follows from Theorem 1, if we let t so that 1s Corollary 2. If =:= ,nnnjnnjjmPzaza z1mn is a polynomial of degree n having all its zeros in the disk zK, 1K, then for each >0r, 112122=11001.rrmmrrnnmii zmnnmna KmaKnPed edMaxPzna Kma  (8) Also if we let r in the Theorem 1 and note that 12=101=lim 2rrirzPedMax Pz . We get the following: Corollary 3. If =:= nnnjnnjjmPzaza z, 1mn is a polynomial of degree n having all its zeros in the disk zK, 1K, then =11=1211 .1zmnnm zmm mnnmMaxP zna KmaMaxP zna KKmaK (9) For K = 1, Corollary 3 reduces to Inequality (1) (the result of Turan) . 2. Lemmas For the proof of this theorem, we need the following lemmas. The first lemma is due to Qazi . Lemma 1. If 0=:= njjjmPz aaz is a polynomial of degree n having no zeros in the disk 0r and 0<2, 212102121011.rmmnnm imnnmrmmnnmimnnmna KmaKWe dnaKmana KmaKedna Kma (17) Also from (16), we have   2111=.mmnnmmnnmna KmaKnQ zWz nQ zzQzna Kma Therefore,  211=1 .mmnnmmnnmna KmaKnQ zWznQ zzQzna Kma (18) Using (12) and the fact that  =QzPz for =1z, we get from (18)   211=1for =1.mmnnmmnnmna KmaKnP zWzPzzna Kma (19) Y. PAUL ET AL. Copyright © 2011 SciRes. AM 144 Combining (17) and (19), we get  21221001 for >0.rmmr rnnmri iimnnmna KmaKn PedePedrna Kma (20) Now applying Hölder’s inequality for 1>s, 1>t, with 1=11ts to (20), we get  112122 2100 01 for >0.sr smmrtr tnnmrii imnnmna KmaKnPededPedrna Kma  (21) This is equivalent to 11 12122 2100 01 for >0.sr srmmrtrrtrnnmiiimnnmna KmaKnPededPedrna Kma  (22) which proves the desired resu lt. 3. Acknowledgements The authors are grateful to the referee for useful sug- gestions. 4. References  P. Turan, “Über die Ableitung von Polynomen,” Compo-sitio Mathematica, Vol. 7, 1939, pp. 89-95.  M. A. Malik, “On the Derivative of a Polynomial,” Journal of the London Mathematical Society, Vol. 2, No. 1, 1969, pp. 57-60. doi:10.1112/jlms/s2-1.1.57  M. A. 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