AK03102250234

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 190006

Abstract Many extensions of the Enestrom –Kakeya Theorem are availab le in the literature. In this paper we prove some results which generalize some known results .

Mathematics Subject Classification:

30C10,30C15

Key-Words and Phrases : Comp lex nu mber, Polynomial , Zero, Enestrom – Kakeya Theorem 1. Introduction And Statement Of Results The Enestrom –Kakeya Theorem (see[7]) is well known in the theory of the distribution of zeros of polynomials and is often stated as follo ws: n

Theorem 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 polynomials with general monotonic coefficients and proved the following result: n

Theorem 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

z 

a n  a0  a0 an

.

Aziz and zargar [1] generalized the result of Joyal et al [6] as follows: n

Theorem 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

z  k 1 

kan  a0  a0 an

.

For polynomials ,whose coefficients are not necessarily real, Govil and Rahman [2] proved the following generalization of Theorem A : n

Theorem 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 0

such that where

 n   n1  ......  1   0  0 ,

 n  0 , then P(z) has all its zeros in

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z  1 (

n

2

n

)(  j ) . j 0

Gov il and Mc-tume [3] p roved the following generalisations of Theorems B and C: n

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.

Theorem D: Let

j 0

If 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

n

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.

Theorem E: Let

j 0

If 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

.

M. H. Gu lzar [4] proved the follo wing generalizations of Theorems D and E: n

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.

Theorem F: Let

j 0

  0,    n   n1  ......  1   0 ,

If for so me real nu mber

then P(z) has all its zeros in the disk n

z

  n

   n   0   0  2  j j 0

n

.

n

Theorem 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 0

  0,    n   n1  ......  1   0 ,

If for some real nu mber

then P(z) has all its zeros in the disk n

z

  n

   n   0   0  2  j j 0

n

.

In this paper we give generalization of Theorems F and G. In fact, we prove the following : n

Theorem 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 0

for some real numbers

 ,   0 , 1  k  n, ank  0,

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and

 nk 1

   n   n1  ... nk 1   nk   nk 1  ...  1   0 ,   nk , then P(z) has all its zeros in the disk n

z and if

 an



   n  (  1) n k    1  n k   0   0  2  j j 0

an

,

 nk   nk 1 , then P(z) has all its zeros in the disk n

z

 an



   n  (1   ) n k  1    n k   0   0  2  j j 0

an

Remark 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: n

P( z )   a j z j be a polynomial of degree n. If fo r some real numbers  ,   0 ,

Corollary 1: Let

j 0

1  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

z and if

 an



  a n  (  1)a nk    1 a nk  a0  a0 an

,,

ank  ank 1 , then P(z) has all its zeros in the disk

z

 an



  a n  (1   )a nk  1   a nk  a0  a0 an

If we apply Theorem 1 to the polynomial –iP(z) , we easily get the follo wing result: n

Theorem 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 0

 ,   0 , 1  k  n, ank  0,    n   n1  ....   nk 1   nk   nk 1  ..  1   0 ,   nk , then P(z) has all its zeros in the dis k

If for some real nu mbers

and

 nk 1

n

z and if

 an



   n  (  1)  n k    1  n k   0   0  2  j j 0

an

,

 nk   nk 1 , then P(z) has all its zeros in the disk n

z

 an

Remark 2: Taking



   n  (1   )  n k  1    n k   0   0  2  j j 0

an

  1 , Theorem 2 reduces to Theorem G.

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P( z )   a j z j be a polynomial of degree n such that for some real numbers  ,   0 ,

Theorem 3: Let

j 0

1  k  n, ank  0, and ,

  an  an1  ...  ank 1   ank  ank 1  ...  a1  a0 and

arg a j      If

 2

, j  0,1,......, n.

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)

z If

 a 0 (cos   sin   1)  2 sin 





an

n 1

a

j 1, j  n  k

j

]

an

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   )

z

 a 0 (sin   cos   1)  2 sin 





an

n 1

a

j 1, j  n  k

j

]

an

Remark 3: Taking

  1 in Theorem 3 , we get the follo wing result: n

Corollary 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

z

 an



[   a n (cos   sin  )  a0 (cos   sin   1)  2 sin   a j ] j 1,

an

.

  (k  1)an , k  1, we get a result of Shah and Liman [8,Theorem 1].

Remark 4: Taking

2. Lemmas For the proofs of the above results , we need the following results: n

Lemma 1: .

Let P(z)=

a z j

j 0

arg a j     

ta j  a j 1

j

be a polynomial of degree n with complex coefficients such that



, j  0,1,...n, for some real  . Then for some t >0, 2  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 1 log

1



log

z  1, then

the

number

of

zeros

of

p(z)in

z   ,0    1, does

not

exceed

M (see[9],p 171). p(0)

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3. 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 1

If

 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 1

For

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

1 1 n  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 )  .... k 1 z z n   1 1  ( 1   0 ) n 1  n0  i  (  j   j 1 ) n  j i n0  z z z z j 1  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  



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if n

a n z       n  (  1) nk    1  nk   0   0  2  j j 0

This shows that the zeros of F(z) whose modulus is greater than 1 lie in n

z

 an



   n  (  1) n k    1  n k   0   0  2  j j 0

an

.

But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence all the zeros of F(z) lie in n

z

 an



   n  (  1) n k    1  n k   0   0  2  j j 0

an

.

Since 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

z If

 an



   n  (  1) n k    1  n k   0   0  2  j j 0

an

.

 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 1

For

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

1 1 n  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 n   1 1  ( 1   0 ) n 1  n0  i  (  j   j 1 ) n  j i n0  z z z z j 1  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

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n

  n1   n1   n2  ......  nk 1   nk   nn  k   nk 1 ||January || 2013

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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  



>0 if n

a n z       n  (1   ) nk  1    nk   0   0  2  j j 0

This shows that the zeros of F(z) whose modulus is greater than 1 lie in n

z

 an



   n  (1   ) n k  1    n k   0   0  2  j j 0

an

.

But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence all the zeros of F(z) lie in n

z

 an



   n  (1   ) n k  1    n k   0   0  2  j j 0

an

.

Since 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

z

 an



   n  (1   ) n k  1    n k   0   0  2  j j 0

an

.

That 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

1 1 n  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 a 1  (a1  a0 ) n 1  0n  z z ||Issn 2250-3005(online)||

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 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 

   1)  a0 (sin   cos   1)  2 sin 

n 1

a } 

j 1, j  n  k

j

0 if

a z      a n

n

(cos   sin  )  ank (cos   sin    cos    sin 

   1)  a0 (sin  cos  1)  2 sin

n 1

a

j 1, j  n  k

j

This 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)

z

 an

 a0 (cos   sin   1)  2 sin  

n 1

a

j 1, j  n  k

j

] ..

an

But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence all the zeros of F(z) lie in

[   a n (cos   sin  )  a n k (cos   sin    cos    sin     1)

z

 an

 a0 (cos   sin   1)  2 sin  

n 1

a

j 1, j  n  k

j

] .

an

Since 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)

z If

 an

 a0 (cos   sin   1)  2 sin  

n 1

a

j 1, j  n  k

j

]

an

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)||

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1 1 n  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 a0 1  (a1  a0 ) n 1  n  z z n  z  an z       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 

 1   )  a0 (sin   cos   1)  2 sin 

n 1

a } 

j 1, j  n  k

j

0 if

a z      a n

n

(cos   sin  )  ank (cos   sin    cos    sin 

 1   )  a0 (sin   cos   1)  2 sin 

n 1

a

j 1, j  n  k

j

This 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   )

 z  an

 a0 (sin   cos   1)  2 sin 

n 1

a

j 1, j  n  k

j

]

an

.

But the zeros of F(z) whose modulus is less than or equal to 1 already satisfy the above inequality. Hence all the zeros of F(z) and therefore P(z) lie in

[   a n (cos   sin  )  a nk (cos   sin    cos    sin   1   )

 z  an

 a0 (sin   cos   1)  2 sin 

n 1

a

j 1, j  n  k

j

]

an

That proves Theorem 3.

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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.

M.H.Gu lzar, On The Location Of Zeros Of A Polynomial, Anal. Theory. Appl., Vo l 28, No.3(2012), 242-247.

M. H. Gu lzar, On The Zeros Of Polynomials, International Journal Of Modern Engineering Research, Vol.2,Issue 6, Nov-Dec. 2012, 4318- 4322.

A. Joyal, G. Labelle, Q.I. Rah man, On The Location Of Zeros Of Po lynomials, Canadian Math. Bull.,10(1967) , 5563.

M. Marden , Geo metry Of Polynomials, Iind Ed.Math. Surveys, No. 3, A mer. Math. Soc. Providence,R. I,1996.

W. M. Shah And A.Liman, On Enestrom-Kakeya Theorem And Related Analytic Functions, Proc. Indian Acad. Sci. (Math. Sci.), 117(2007), 359-370.

E C. Titch marsh,The Theory Of Functions,2nd Ed ition,Oxfo rd University Press,London,1939.

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