EXTENSION AND GENERALISATION OF AN INEQUALITY OF PAUL TURAN'S THEOREM ON POLYNOMIAL

. In this paper, certain new results concerning the maximum modulus of the polar derivative of a polynomial with restricted zeros are obtained. These estimates strengthen some well known inequalities for polynomial due to Turån, Dubinin and others.


INTRODUCTION
In scientific disciplines like physics, engineering, computer science, biology, physical chemistry, economics, and other applied areas, experimental observations and investigations when translated into mathematical language are called mathematical models.The solution of these models could lead to problems of estimating how large or small the maximum modulus of the derivative of an algebraic polynomial can be in terms of the maximum modulus of that polynomial.Bounds for such type of problems are of some practical importance.Since, there are no closed formulae for precise evaluation of these bounds and whatever is available in literature is in the form of approximations.However for practical purposes, nobody ever needs exacts bounds and mathematicians must only indicate methods for obtaining approximate bounds.These approximate bounds, when computed efficiently, are quite satisfactory for the needs of investigators and scientists.Therefore there is always a desire to look for better and improved bounds than those available in literature.It is this aspiration of obtaining more refined and revamped bounds that has inspired our work in this article.In this paper, we have generalized and refined some well known results concerning the polynomials due to Turån [16], Dubinin [6] and others.Let begin with the polynomial of the form of degree n > 1 and let P' (z) be the derivative of P (z).Then concerning the lower bound for the maximum of P'(z)l in terms of maximum of for class of polynomials P G {Pn not vanishing outside unit disc, Turån [16] showed that In this paper, we obtain certain refinements and generalizations of inequalities (1.1), (1.2), (1.3) and (1.4).We first prove the following result.The above lemma is due to Laguerre (see [10]).The following lemma is due to Frappier, Rahman and Ruscheweyh [7].We show that for G C with < l, f (z) 4-/\mz n 0 in Izl < 1.This is trivially true if m 0. Henceforth we suppose that m 0, so that all the zeros of f (z) lie in Izl > 1.By the maximum modulus theorem (3.9) mlz n l < for Izl < 1.
In view of this inequality, choosing argument of in the left hand side of inequality inequality (1.1) holds for those polynomials P G {Pn which have all their zeros on Izl = 1.As an extension of (1.1), Govil [8] proved that if P G (Pn and P (z) has all its zeros in Izl k, k 1, then (1.2) max Izl=l In literature, there exist several generalizations and extensions of (1.1) and (1.2) (see [1, 2, 4, 5, 13-15]).Dubinin [6] refined inequality (1.1) by proving that if all the zeros of P e {Pn lie in 1, then (1.3) 2n + The polar derivative DOP(z) of P e {Pn with respect to the point a G C is defined by DuP(z) nP(z) + (a -The polynomial DnP(z) is of degree at most n -I and it generalizes the ordinary derivative P'(z) of P (z) in the sense that DuP(z) lim uniformly for Izl R, R > O. A. Aziz [Il, Aziz and Rather ( [4, 5]) obtained several sharp estimates for maximum modulus of DuP(z) on Izl = I and among other things they extended inequality (1.2) to the polar derivative of a polynomial by showing that if P e (Pn has all its zeros in Izl k, k 1

Theorem 2 . 1 . 2 .
If P e [Pn and P (z) has all its zeros in 121 k, k 1, then for every G C, with k max (2.1) 21=1 where +(k) (k-1)2 and q/'(k) 1 or I-according as n > 2 or n = 2. Remark 2.1.Since all the zeros of p (z) lie in zl k, k 2 1, it follows that laol k n an l.In view of this, inequalities (2.1) refines inequality (1.4).If we divide the two sides of inequality (2.1) by la and let al 00, we get the following result.Corollary 2.1.If P 'J)n and P (z) has all its zeros in Izl k, k or n = 2.The result is best possible and equality in inequality (2.2) holds for P (z) = z n + k Remark 2.2.As before, it can be easily seen that inequality (2.2) refines inequality (1.2).Further for k: = I, inequality (2.2) reduces to inequality (1.3).Next, we present the following result which is generalisation of Theorem 2.1 and in particular, includes refinement of inequality (1.2) as a special case.If all the zeros of P e 'J)n lie in Iz k, k 2 1, then for every a C with al 2 k, Of I < 1, 1according as n > 2 or n = 2.If we divide both sides of inequality (2.3) by al and let 00, we get the following result.Corollary 2.2.If all the zeros minlzl=kor Iaccording as n > 2 or n = 2. Remark 2.3.For I = O, Corollary 2.2 reduces to Corollary 2.1 and for k = I, inequality (2.4) refines inequality (1.3).
need the following lemmas for the proof of our theorems.The first lemma is due to Dubinin Lemma 3.1.If P (Pn and P (z) has all its zeros in zl I, then max is special case of a result due to Aziz and Rather [3, 4].Lemma 3.2.If P (Pn and P (z) has its all zeros in zl I, then for zl = I where Q(z) = z n p (I/Z).Lemma 3.3.If all the zeros of P G {Pn lie in a circular region C and w is any zero of DuP(z), the polar derivative of P(z), then at most one of the points w and a may lie outside C.

Lemma 3. 4 .( 2 IP
If P(z) is a polynomial of degree at most n lemma is the famous result of P. D. Lax [9]. 1, then for R 2 1 if n 2 2, if n = 1.Lemma 3.5.If P G {Pn does not vanish in Izl < 1, then max Il) (z) l, for Izl We also need the following lemma.i0 ) -P(e t0 ) 1 which gives with the help of (3.2) of Lernrna 3.4 and Lernrna 3.5 for n > leads to (3.4).Similarly we can prove inequality (3.5) by using inequality (3.3) of Lemma 3.4.This proves Lemma 3.6.Finally we require the following lemma.3.7.Since all the zeros of P G [Pn lie in Izl k, k 1, therefore, all the zeros of g(z) P(kz) lie in Izl 1 and hence all the zeros of f (z) z n TFj z n P(k/Z) lie in Izl 1.Moreover, m = milllzl=k milllzl-l so that rnlznl for Izl = 1.

1 . 1 ( 4 . 7 )
If(kz) + n mz n I+ /\mz n l -Again since all the zeros of f (z) = P(kz) lic in Izl 1, thcrcforc, using Lemma 3.1, wc have for zl = I.Replacing f (z) by P(kz), wc obtain max P (z) is a polynomial of dcgrcc at most n -I, using inequality (3.2) of Lemma 3.4, wc havc for n > 2 -(IV 1 -R TE 3 )Inao .Using this inequality and inequality (3.7) of Lemma 3.7 with I -0 and R k > I in (4is equivalent to the inequality (2.1) for n > 2. For n = 2, the result follows on similar lines by using inequality (3.3) of Lemma 3.4 and inequality (3.8) Lemma 3.7 in the inequality (4.6).This cornpletes the proof of Theorem 2.Proof of Theorem 2.2.By hypothesis P e has all zeros in Izl k, k 2 1.If P (z) has a zero on Izl = k, then m = minlzl=k = 0 and result follows from Theorem 2.1.Henceforth, we suppose that P (z) has all its zeros in Izl < k, k 2 1, so that m > O. Now if f (z) = P(kz), then f e {Pn and f (z) has all zeros in zl < I and m = minlzl=k = milllzl=l This implies for Izl n n By the Rouche's Theorem, we conclude that for every e C with < I, the polynomial g(z) = f (z)has all zeros in Izl < I. Applying inequality (4.3) to the polynomial g(z), it follows for Izl = I and lal Since all the zeros of g(z) lie in Izl < I, using Lemma 3.1, we obtain for Izl = I and lal 2k Using the fact that the function S(x) x > 0, is non-decreasing function of x and lc n an -2 Ic n lanl we get for every e C with < I and z Replacing g(z) by f (z) in (4.7), we get for Izl = I and al 2k Do/kf(z) - DaP(z) is a polynomial of degree at most n -1, applying inequality (3.2) of Lemma 3.4 and inequality (3.7) of Lemma 3.7 with R = k 2 1, we obtain for lal 2 k, 0 s I < 1 and Izl = 1 we get for That proves the inequality (2.3) for n > 2. For the case n = 2, the result follows on similar lines by using inequality (3.3) of Lemma 3.4 and inequality (3.8) of Lemma 3.7 in the inequality (4.10).This completes the proof of Theorem 2.2.