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# Ak03102250234

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1. 1. I nternational Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue.1 On The Enestrom –Kakeya Theorem and Its Generalisations 1, M. H. Gulzar 1, Department of Mathematics University of Kashmir, Srinagar 190006Abstract Many extensions of the Enestrom –Kakeya Theorem are availab le in the literature. In this paper we prove someresults which generalize some known results .Mathematics Subject Classification: 30C10,30C15Key-Words and Phrases : Comp lex nu mber, Polynomial , Zero, Enestrom – Kakeya Theorem1. Introduction And Statement Of Results The Enestrom –Kakeya Theorem (see[7]) is well known in the theory of the distribution of zeros of polynomialsand is often stated as follo ws: nTheorem A: Let P( z )   a j z j be a polynomial of degree n whose coefficients satisfy j 0 an  an1  ......  a1  a0  0 .Then P(z) has all its zeros in the closed unit disk z  1.In the literature there exist several generalizations and extensions of this resu lt. Joyal et al [6] extended it to polynomialswith general monotonic coefficients and proved the following result: nTheorem B : Let P( z )   a j z j be a polynomial of degree n whose coefficients satisfy j 0 an  an1  ......  a1  a0 .Then P(z) has all its zeros in a n  a0  a0 z  . anAziz and zargar [1] generalized the result of Joyal et al [6] as follows: nTheorem C: Let P( z )   a j z j be a polynomial of degree n such that for some k  1 j 0 kan  an1  ......  a1  a0 .Then P(z) has all its zeros in kan  a0  a0 z  k 1  . anFor polynomials ,whose coefficients are not necessarily real, Govil and Rahman [2] proved the following generalization ofTheorem A : nTheorem C: If P( z )   a j z j is a polyno mial of degree n with Re( a j )   j and Im(a j )   j , j=0,1,2,……,n, j 0such that  n   n1  ......  1   0  0 ,where  n  0 , then P(z) has all its zeros in||Issn 2250-3005(online)|| ||January || 2013 Page 225
2. 2. I nternational Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue.1 n 2 z  1 ( )(  j ) . n j 0Gov il and Mc-tume [3] p roved the following generalisations of Theorems B and C: nTheorem D: Let P( z )   a j z j be a polynomial of degree n with Re( a j )   j and Im(a j )   j , j=0,1,2,……,n. j 0If for some k  1, k n   n1  ......  1   0 ,then P(z) has all its zeros in n k n   0   0  2  j j 0 z  k 1  . n nTheorem E: Let P( z )   a j z j be a polynomial of degree n with Re( a j )   j and Im(a j )   j , j=0,1,2,……,n. j 0If for some k  1, k n   n1  ......  1   0 ,then P(z) has all its zeros in n k n   0   0  2  j j 0 z  k 1  . nM. H. Gu lzar [4] proved the follo wing generalizations of Theorems D and E: nTheorem F: Let P( z )   a j z j be a polynomial of degree n with Re( a j )   j and Im(a j )   j , j=0,1,2,……,n. j 0If for so me real nu mber  0,    n   n1  ......  1   0 ,then P(z) has all its zeros in the disk n    n   0   0  2  j  j 0 . z  n n nTheorem G: Let P( z )   a j z j be a polynomial of degree n with Re( a j )   j and Im(a j )   j , j=0,1,2,……,n. j 0If for some real nu mber   0,    n   n1  ......  1   0 ,then P(z) has all its zeros in the disk n    n   0   0  2  j  j 0 z  . n nIn this paper we give generalization of Theorems F and G. In fact, we prove the following : nTheorem 1: Let P( z )   a j z j be a polynomial of degree n with Re( a j )   j and Im(a j )   j , j=0,1,2,……,n. If j 0for some real numbers  ,   0 , 1  k  n, ank  0,||Issn 2250-3005(online)|| ||January || 2013 Page 226
3. 3. I nternational Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue.1    n   n1  ... nk 1   nk   nk 1  ...  1   0 ,and  nk 1   nk , then P(z) has all its zeros in the disk n    n  (  1) n k    1  n k   0   0  2  j  j 0 z  , an anand if  nk   nk 1 , then P(z) has all its zeros in the disk n    n  (1   ) n k  1    n k   0   0  2  j  j 0 z  an anRemark 1: Taking   1 , Theorem 1 reduces to Theorem F.Taking k=n ,and   1 in Theorem 1, we get a result due to M.H. Gulzar [5, Theorem 1] .If a j are real i.e.  j  0 for all j, we get the following result: nCorollary 1: Let P( z )   a j z j be a polynomial of degree n. If fo r some real numbers  ,   0 , j 01  k  n, ank  0,   an  an1  ...  ank 1  ank  ank 1  ...  a1  a0 ,and ank 1  ank , then P(z) has all its zeros in the disk    a n  (  1)a nk    1 a nk  a0  a0 z  ,, an anand if ank  ank 1 , then P(z) has all its zeros in the disk    a n  (1   )a nk  1   a nk  a0  a0 z  an an If we apply Theorem 1 to the polynomial –iP(z) , we easily get the follo wing result: nTheorem 2: Let P( z )   a j z j be a polynomial of degree n with Re( a j )   j and Im(a j )   j , j=0,1, 2,……,n. j 0If for some real nu mbers  ,   0 , 1  k  n, ank  0,    n   n1  ....   nk 1   nk   nk 1  ..  1   0 ,and  nk 1   nk , then P(z) has all its zeros in the dis k n    n  (  1)  n k    1  n k   0   0  2  j  j 0 z  , an anand if  nk   nk 1 , then P(z) has all its zeros in the disk n    n  (1   )  n k  1    n k   0   0  2  j  j 0 z  an anRemark 2: Taking   1 , Theorem 2 reduces to Theorem G.||Issn 2250-3005(online)|| ||January || 2013 Page 227
4. 4. I nternational Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue.1 nTheorem 3: Let P( z )   a j z j be a polynomial of degree n such that for some real numbers  ,   0 , j 01  k  n, ank  0, and ,   an  an1  ...  ank 1   ank  ank 1  ...  a1  a0and  arg a j      , j  0,1,......, n. 2If ank 1  ank (i.e.   1 ) , then P(z) has all its zeros in the disk [   a n (cos   sin  )  a n  k (cos   sin    cos    sin     1) n 1   a 0 (cos   sin   1)  2 sin  a j 1, j  n  k j ] z  an anIf ank  ank 1 (i.e.   1), then P(z) has all its zeros in the disk [   a n (cos   sin  )  a n  k (cos   sin    cos    sin   1   ) n 1   a 0 (sin   cos   1)  2 sin  a j 1, j  n  k j ] z  an anRemark 3: Taking   1 in Theorem 3 , we get the follo wing result: nCorollary 2: Let P( z )   a j z j be a polynomial of degree n. If fo r some real number   0 , j 0   an  an1  ......  a1  a0 ,then P(z) has all its zeros in the disk n 1 [   a n (cos   sin  )  a0 (cos   sin   1)  2 sin   a j ]  j 1, z  . an anRemark 4: Taking   (k  1)an , k  1, we get a result of Shah and Liman [8,Theorem 1].2. Lemmas For the proofs of the above results , we need the following results: nLemma 1: . Let P(z)= a z j 0 j j be a polynomial of degree n with complex coefficients such that  arg a j      , j  0,1,...n, for some real  . Then for some t >0, 2 ta j  a j 1     t a j  a j 1 cos   t a j  a j1 sin  . The proof of lemma 1 follows fro m a lemma due to Govil and Rah man [2]. Lemma 2.If p(z) is regular ,p(0)  0 and p( z )  M in z  1, then the number of zeros of p(z)in z   ,0    1, does not exceed 1 M log (see[9],p 171). 1 p(0)log ||Issn 2250-3005(online)|| ||January || 2013 Page 228
5. 5. I nternational Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue.13. Proofs Of Theorems Proof of Theorem 1: Consider the polynomial F ( z)  (1  z) P( z)  (1  z )(a n z n  a n1 z n1  ......  a1 z  a0 )  a n z n1  (a n  a n1 ) z n  ......  (a nk 1  a nk ) z nk 1  (a nk  a nk 1 ) z nk  (ank 1  ank 2 ) z nk 1  ......  (a1  a0 ) z  a0  ( n  i n ) z n 1  ( n   n 1 ) z n  ......  ( n  k 1   n  k ) z n  k 1  ( n  k   n  k 1 ) z n  k  ( n  k 1   n  k  2 ) z n  k 1  ......  ( 1   0 ) z   0 n  i  (  j   j 1 ) z j  i 0 j 1If  nk 1   nk , then  nk 1   nk , and we have F ( z)  ( n  i n ) z n1  z n  (    n   n1 ) z n  ......  ( nk 1   nk ) z nk 1  ( nk   nk 1 ) z nk  (  1)ank z nk  ( nk 1   nk 2 ) z nk 1  ...... n  (1   0 ) z   0  i  (  j   j 1 ) z j  i 0 . j 1For z  1, F ( z )  an z n1  z n  (    n   n1 ) z n  ( n1   n2 ) z n1  ......  ( nk 1   nk ) z nk 1  ( n k   n k 1 ) z n k  (  1)a n k z n k  ( n k 1   n k 2 ) z n k 1  ...... n  ( 1   0 ) z   0  i  (  j   j 1 ) z j  i 0 j 1 n 1 1  z  a n z    (    n   n 1 )  ( n1   n2 )  ......  ( nk 1   n k ) k 1  z z 1 1 1  ( n k   n k 1 ) k  (  1) n k k  ( n k 1   n  k  2 )  .... z z k 1 0 n 0  ( 1   0 ) n 1  n  i  (  j   j 1 ) n  j i n  1 1 z z j 1 z z  z n a z     n    n   n1   n1   n2  ......   nk 1   nk   n k   n k 1    1  n k   n k 1   n k 2  ......   1   0 n   0    j   j 1   0  j 1  z n a z       n n   n1   n1   n2  ......  nk 1   nk   nn  k   nk 1 n    1  nk   nk 1   nk 2  ......  1   0   0  2  j  j 0 n  n  z  a n z       n  (  1) n k    1  nk   0   0  2  j     j 0 >0||Issn 2250-3005(online)|| ||January || 2013 Page 229
6. 6. I nternational Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue.1if n a n z       n  (  1) nk    1  nk   0   0  2  j j 0This shows that the zeros of F(z) whose modulus is greater than 1 lie in n    n  (  1) n k    1  n k   0   0  2  j  j 0 z  . an anBut the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence all the zeros ofF(z) lie in n    n  (  1) n k    1  n k   0   0  2  j  j 0 z  . an anSince the zeros of P(z) are also the zeros of F(z), it fo llo ws that all the zeros of P(z) lie in n    n  (  1) n k    1  n k   0   0  2  j  j 0 z  . an anIf  nk   nk 1 , then  nk   nk 1 , and we have F ( z)  ( n  i n ) z n1  z n  (    n   n1 ) z n  ......  ( nk 1   nk ) z nk 1  ( n k   n k 1 ) z n k  (1   ) n k z n k 1  ( n k 1   n k 2 ) z n k 1  ...... n  (1   0 ) z   0  i  (  j   j 1 ) z j  i 0 . j 1For z  1, F ( z )  an z n1  z n  (    n   n1 ) z n  ( n1   n2 ) z n1  ......  ( nk 1   nk ) z nk 1  ( n k   n k 1 ) z n k  (1   ) n k z n k 1  ( n k 1   n k 2 ) z n k 1  ...... n  ( 1   0 ) z   0  i  (  j   j 1 ) z j  i 0 j 1 n 1 1  z  a n z    (    n   n1 )  ( n1   n2 )  ......  ( nk 1   nk ) k 1  z z 1 1 1  ( n  k   n  k 1 ) k  (1   ) n  k k 1  ( n  k 1   n  k  2 ) k 1  .... z z z 0 n 0  ( 1   0 ) n 1  n  i  (  j   j 1 ) n  j i n  1 1 z z j 1 z z  z n a z     n    n   n1   n1   n2  ......   nk 1   nk   n k   n k 1  1    n k   n k 1   n k 2  ......   1   0 n   0    j   j 1   0  j 1  z n a z       n n   n1   n1   n2  ......  nk 1   nk   nn  k   nk 1||Issn 2250-3005(online)|| ||January || 2013 Page 230
7. 7. I nternational Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue.1 n  1    nk   nk 1   nk 2  ......  1   0   0  2  j  j 0 n  n  z  a n z       n  (1   ) n k  1    nk   0   0  2  j     j 0 >0if n a n z       n  (1   ) nk  1    nk   0   0  2  j j 0This shows that the zeros of F(z) whose modulus is greater than 1 lie in n    n  (1   ) n k  1    n k   0   0  2  j  j 0 z  . an anBut the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence all the zeros ofF(z) lie in n    n  (1   ) n k  1    n k   0   0  2  j  j 0 z  . an anSince the zeros of P(z) are also the zeros of F(z), it fo llo ws that all the zeros of P(z) lie in n    n  (1   ) n k  1    n k   0   0  2  j  j 0 z  . an anThat proves Theorem 1.Proof of Theorem 3: Consider the polynomial F ( z)  (1  z) P( z)  (1  z )(a n z n  a n1 z n1  ......  a1 z  a0 )  a n z n1  (a n  a n1 ) z n  ......  (a nk 1  a nk ) z nk 1  (a nk  a nk 1 ) z nk  (ank 1  ank 2 ) z nk 1  ......  (a1  a0 ) z  a0 .If ank 1  ank , then ank 1  ank ,   1 and we have, for z  1 , by using Lemma1, F ( z )  an z n1  z n  (   an  an1 ) z n  (an1  an2 ) z n1  ......  (ank 1  ank ) z nk 1  (a n k  a n k 1 ) z n k  (  1)a n k z n k  (a n k 1  a n k 2 ) z n k 1  ......  (a1  a0 ) z  a0 n 1 1  z  a n z    (   a n  a n 1 )  (a n1  a n2 )  ......  (a nk 1  a nk ) k 1  z z 1 1 1  (a n k  a n k 1 ) k  (  1)a n k k  (a n k 1  a n k 2 ) k 1  .... z z z  (a1  a0 ) n 1  0  1 a z zn||Issn 2250-3005(online)|| ||January || 2013 Page 231
8. 8. I nternational Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue.1  z n a z     n   an  an1  an1  an2  ......  ank 1  ank  a n k  a n k 1    1 a n k  a n k 1  a n k 2  ......  a1  a0  a0   z n  a z    {(   a n n  an1 ) cos   (   an  an1 ) sin   ( an1  an2 ) cos   ( an1  an2 ) sin   ......  ( ank 1  ank ) cos   ( ank 1  ank ) sin   ( ank  ank 1 ) cos   ( ank  ank 1 ) sin     1 ank  ( ank 1  ank 2 ) cos   ( ank 1  ank 2 ) sin   ......  ( a1  a0 ) cos   ( a1  a0 ) sin   a0 }   z n a z    {   a n n (cos   sin  )  ank (cos   sin    cos    sin  n 1    1)  a0 (sin   cos   1)  2 sin  a }  j 1, j  n  k j 0if a z      a n n (cos   sin  )  ank (cos   sin    cos    sin  n 1    1)  a0 (sin  cos  1)  2 sin a j 1, j  n  k jThis shows that the zeros of F(z) whose modulus is greater than 1 lie in [   a n (cos   sin  )  a n k (cos   sin    cos    sin     1) n 1   a0 (cos   sin   1)  2 sin  a j 1, j  n  k j ] z  .. an anBut the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence all the zeros ofF(z) lie in [   a n (cos   sin  )  a n k (cos   sin    cos    sin     1) n 1   a0 (cos   sin   1)  2 sin  a j 1, j  n  k j ] z  . an anSince the zeros of P(z) are also the zeros of F(z), it fo llo ws that all the zeros of P(z) lie in [   a n (cos   sin  )  a n k (cos   sin    cos    sin     1) n 1   a0 (cos   sin   1)  2 sin  a j 1, j  n  k j ] z  an anIf ank  ank 1 , then ank  ank 1 ,   1 and we have, for z  1 , by using Lemma 1, F ( z )  a n z n 1  z n  (   a n  a n 1 ) z n  (a n 1  a n  2 ) z n 1  ......  (a n k 1  a n  k ) z n k 1  (a n k  a n k 1 ) z n k  (1   )a n k z n k 1  (a n k 1  a n k 2 ) z n k 1  ......  (a1  a0 ) z  a0||Issn 2250-3005(online)|| ||January || 2013 Page 232
9. 9. I nternational Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue.1 n 1 1  z  a n z    (   a n  a n 1 )  (a n1  a n2 )  ......  (a nk 1  a n k ) k 1  z z 1 1 1  (a n k  a n k 1 ) k  (1   )a n k k 1  (a n k 1  a n k 2 ) k 1  .... z z z  (a1  a0 ) n 1  n  1 a0 z z  z  an z       an  an1  an1  an2  ......  ank 1  ank n  a n k  a n k 1  1   a n k  a n k 1  a n k 2  ......  a1  a0  a0   z n  a z    {(   a n n  an1 ) cos   (   an  an1 ) sin   ( an1  an2 ) cos   ( an1  an2 ) sin   ......  ( ank 1  ank ) cos   ( ank 1  ank ) sin   ( ank  ank 1 ) cos   ( ank  ank 1 ) sin   1   ank  ( ank 1  ank 2 ) cos   ( ank 1  ank 2 ) sin   ......  ( a1  a0 ) cos   ( a1  a0 ) sin   a0 }   z n a z    {   a n n (cos   sin  )  ank (cos   sin    cos    sin  n 1  1   )  a0 (sin   cos   1)  2 sin  a }  j 1, j  n  k j 0if a z      a n n (cos   sin  )  ank (cos   sin    cos    sin  n 1  1   )  a0 (sin   cos   1)  2 sin  a j 1, j  n  k jThis shows that the zeros of F(z) whose modulus is greater than 1 lie in [   a n (cos   sin  )  a n k (cos   sin    cos    sin   1   ) n 1   a0 (sin   cos   1)  2 sin  a j 1, j  n  k j ] z  . an anBut the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence all the zeros ofF(z) and therefore P(z) lie in [   a n (cos   sin  )  a nk (cos   sin    cos    sin   1   ) n 1   a0 (sin   cos   1)  2 sin  a j 1, j  n  k j ] z  an anThat proves Theorem 3.||Issn 2250-3005(online)|| ||January || 2013 Page 233
10. 10. I nternational Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue.1References[1] A.Aziz And B.A.Zargar, So me Extensions Of The Enestrom-Kakeya Theorem , Glasnik Mathematiki, 31(1996) , 239-244.[2] N.K.Gov il And Q.I.Reh man, On The Enestrom-Kakeya Theorem, Tohoku Math. J.,20(1968) , 126-136.[3] N.K.Gov il And G.N.Mc-Tu me, So me Extensions Of The Enestrom- Kakeya Theorem , Int.J.Appl.Math. Vo l.11,No.3,2002, 246-253.[4] M.H.Gu lzar, On The Location Of Zeros Of A Polynomial, Anal. Theory. Appl., Vo l 28, No.3(2012), 242-247.[5] M. H. Gu lzar, On The Zeros Of Polynomials, International Journal Of Modern Engineering Research, Vol.2,Issue 6, Nov-Dec. 2012, 4318- 4322.[6] A. Joyal, G. Labelle, Q.I. Rah man, On The Location Of Zeros Of Po lynomials, Canadian Math. Bull.,10(1967) , 55- 63.[7] M. Marden , Geo metry Of Polynomials, Iind Ed.Math. Surveys, No. 3, A mer. Math. Soc. Providence,R. I,1996.[8] W. M. Shah And A.Liman, On Enestrom-Kakeya Theorem And Related Analytic Functions, Proc. Indian Acad. Sci. (Math. Sci.), 117(2007), 359-370.[9] E C. Titch marsh,The Theory Of Functions,2nd Ed ition,Oxfo rd University Press,London,1939.||Issn 2250-3005(online)|| ||January || 2013 Page 234