 Advances in Pure Mathematics, 2011, 1, 23-27 doi:10.4236/apm.2011.12006 Published Online March 2011 (http://www.SciRP.org/journal/apm) Copyright © 2011 SciRes. APM Inequalities for the Polar Derivative of a Polynomial Gulshan Singh1, Wali Mohammad Shah2, Yash Paul1 1Bharathiar University Coimbatore, Tamil Nadu, India 2Department of Mathematics, Kashmir University, Srinagar, India E-mail: {gulshansingh1, wmshah, yashpaul2011}@rediffmail.com Received February 20, 2011; revised March 1, 2011; accepted March 5, 2011 Abstract If =0:njjjPz az is a polynomial of degree n, having all its zeros in ,zK 1K, then it was pro- vied by Aziz and Rather  that for every real or complex number  with K, =1zMaxD Pz =11znnKMaxP zK. In this paper, we sharpen above result for the poly nomials Pz of degree >3n. Keywords: Polynomial, Inequality, Polar Derivative 1. Introduction Let =0:= njjjPz az be a polynomial of degree n and P' z its derivative, then  =1 =1zzMaxPznMaxP z (1) Inequality (1) is a famous result due to Bernstein and is best possible with equality holding for the polynomial =nPz z, where  is a complex number. If we restricted ourselves to a class of polynomial having no zeros in <1z, then the above inequ ality can be sharpened. In fact, Erdös conjectured and later Lax  proved that if 0Pz in <1z, then  =1 =12zznMaxPzMaxPz (2) On the other hand, it was proved by Turán  that if Pz is a polynomial of degree n having all its zeros in 1z, then  =1 =12zznMaxPzMaxPz (3) The inequalities (2) and (3) are also best possible and become equality for polyno mials which have all zeros on 1z. For the class of polynomials having all the zeros in zK, Malik  (See also Govil ) proved that if Pz is a polynomial of degree n having all zeros lie in zK, then  =1=1 , if 1,1zznMaxPzMaxP zKK (4) where as Govil  showed that  =1=1 , if 11zznnMaxPzMaxP zKK (5) Both the inequalities are best possible, with equality in (4) holding for  =nPzz K and in (5) the equality holds for the p olynomial =nnPzzK. Let DP z denote the polar derivative of the polynomial Pz of degree n with respect to , then  =DP znP zz Pz . The polynomial DP z is of degree at most 1n and it generalizes the ordinary derivative in the sense that  =.lim DP zPz  Aziz and Rather  extended (5) to the polar deri- vative of a polynomial and proved the following: Theorem 1: If the polynomial =0:= njjjPz az has all its zeros in zK, 1K, then for every real or complex number  with K, =1 =11zznnKMaxDP zMaxPzK (6) In this paper, we prove th e following result which is a refinement as well as generalization of Theorem 1. G. SINGH ET AL. Copyright © 2011 SciRes. APM 24 Theorem 2: Let =0:= njjjPzaz, 00naa  be a po- lynomial of degree >3n, having all its zeros in zK, 1K, then for every real or complex number  with K,  1=1=1 =221211111 121 2, if >3.123nnzzzKnnnnKnK aMaxDP zMaxP zMinP zKnnKKnKK nKannnn n       (7) Remark 1: For =1K, Theorem 2 provides a refi- ment of a theorem proved by Shah . Remark 2: For >1K, and for >1y, 111yKyKyy  and 1yKy are both increa- sing functions of y and so the expressions 211 121123nnKnKK nKnnn n       and 11nKKn are always non-negative so that for polynomials of degree >3n, Theorem 2 is an improvement of Theo- rem 1. Dividing both sides of (7) by  and letting , we get the following: Corollary 1: Let =0=njjjPzaz, 00naa  be a po- lynomial of degree >3n, having all its zeros in zK, 1K, then   1=1=1 =221211111 121 2, if >3.123nnzzzKnnnnKanMaxPzMaxP zMinP zKnnKKnKK nKannnn n       (8) 2. Lemmas We need the following lemmas. Lemma 1: Let Pz be a polynomial of degree n, then for 1R.  ==1.nzR zMaxPzRMaxPz The above lemma is a simple consequence of the maximum modulus principle . Lemma 2: If =0:= njjjPz az, 0na, is a poly- nomial of degree n having all its zeros in 1z, then  =1=1 =12zzznMaxPzMaxPzMinP z. This lemma is due to Aziz and Dawood . Lemma 3: If =0:= njjjPz az is a polynomial of degree n having no zeros in 1z, and =1=zmMin Pz, then for 1R and >3n, =1211120,122111 121 0123nnnznnRRRPMPRMax PzmRnnRnRR nRPnnn n      . The above result is a special case of a result due to Dewan, Singh and Mir [4, Theorem 1] with =1K and =1. Remark 3: Here we note that for the proof of this result an additional hypothesis that 00P is required. A simple counter example in this case is =nPz z. G. SINGH ET AL. Copyright © 2011 SciRes. APM 253. Proof of Theorem 2 Since Pz has all its zeros in zK, therefore  =Gz PKz has all its zeros in 1z and hence by applying lemma 2 to the polynomial Gz, we get   =1=1=1 .2zzznMaxGzMaxG zMinG z (9) Let 1=nHz zGz. Then it can be easily verified that  =, for =1.HznGzzGzz (10) The polynomial Hz has all its zeros in 1z and  =HzGz for 1z, therefore, by result of a de Bruijn   for =1Hz Gzz (11) Now for every real or complex nu mber α with K, we have    = KDGznGzzG zGzKGznGzzGzK For this, we get by using (10) and (11)  =1 =1zzKKMaxD G zMaxGzK (12) Using (9) in (12), we get  =1=1 =12zzzKKnMaxDG zMaxGzMinG zK. Replacing Gz by PKz, we have ||=1=1 =12zzzKnKMaxDP KzMaxP KzMinPKzK. This gives   =1 =1 =12zzznKMaxnPKzzKPKzMaxP KzMinP KzKK  . Equivalently  ===.2zKzK zKnKMaxDP zMaxP zMinP zK (13) Since the polynomial P(z) has all its zeros in zK, 1K. If 1=nQzzP z be the reciprocal polynomial of Pz. Then the polyn omial zQK has all its zeros in 1z. Hence applying lemma 3 to the polynomial zQK, 1K, we get 1==1=12211 12122111 121 2123nn nnzK zznnnKK KazzzMaxQMax QMin QKKKKnnKnKK nKannn n          This in particular gives  1=1 ==2211 12112211 121 2123nn nnzzKzKnnnnnKK KaMaxP zMaxP zMinP zKnnKKKnKK nKannn n        G. SINGH ET AL. Copyright © 2011 SciRes. APM 26 which is equivalent to  1==1=2211241111111 1214 1231nnn nnzK zzKnn nnnnnnKKaKKMaxP zMaxP zMinP zKnnKK KKnKK nKKannn nK        (14) Using (14) in (13), we get   1==1=22=1124{ 12111111 1214 1231nnn nnzKz zKnn nnnnnzKnKKnK aKKMaxDP zMaxP zMinP zKKnnKK KKnKK nKKaMinPznnn nK      , if >3.n Equivalently  11==1=221211111 121 2, if >3.123nnnzKz zKnnnnKnKK aMaxDP zMaxP zMinP zKnnKKnKK nKannnn n       (15) Since DP z is a polynomial of degree 1n and 1K, therefore by using Lemma 1, we get 1==1nzK zMaxDP zKMaxDP z (16) Combining (16) and (15) we have   1=1=1 =221211111 121 2||, if >3.123nnzzzKnnnnKnK aMaxDP zMaxPzMinP zKnnKKnKK nKannnn n       This completes the proof of Theorem 2. 4. 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