I1057581

International Journal of Engineering Research and Development e-ISSN: 2278-067X, p-ISSN: 2278-800X, www.ijerd.com Volume 10, Issue 5 (May 2014), PP.75-81

On Some Extensions of Enestrom-Kakeya Theorem M. H. Gulzar Department of Mathematics University of Kashmir, Srinagar 19000 Abstract: In this paper we generalize some recent extensions of the Enestrom-Kakeya Theorem concerning the location of zeros of polynomials.

Mathematics Subject Classification: 30 C 10, 30 C 15 Keywords and Phrases: Coefficient, Polynomial, Zero

I.

INTRODUCTION AND STATEMENT OF RESULTS

The following result known as the Enestrom-Kakeya Theorem [7] is an elegant result in the theory of distribution of the zeros of a polynomial: Theorem A: Let P ( z ) 

n

a j 0

j

z j be a polynomial of degree n such that

a n  a n 1  ......  a1  a 0  0 .

Then all the zeros of P(z) lie in the closed disk

z  1.

In the literature [1-8], there exist several extensions and generalizations of Theorem A. Joyal, Labelle and Rahman [5] gave the following generalization of Theorem A: n

Theorem B: Let

P ( z )   a j z j be a polynomial of degree n such that j 0

a n  a n 1  ......  a1  a 0 .

Then all the zeros of P(z) lie in the closed disk z 

an  a0  a0 an

.

Aziz and Zargar [1] generalized Theorems A and B by proving the following results: Theorem C: Let P ( z ) 

n

a j 0

j

z j be a polynomial of degree n such that for some k  1,

kan  a n 1  ......  a1  a0 .

Then all the zeros of P(z) lie in the closed disk

z  k 1 

kan  a0  a0 an

.

Recently, Zargar [8] gave the following generalizations of the above mentioned results: n

j Theorem D: Let P ( z )   a j z be a polynomial of degree n  2 such that for some k  1, either j 0

kan  a n  2  .......  a 3  a1  0 and a n 1  a n 3  .......  a 2  a 0  0 , if n is odd or

kan  a n  2  .......  a 2  a 0  0 and a n 1  a n 3  .......  a 3  a1  0 , if n is even.

Then all the zeros of P(z) lie in the region

z  z    k 

a n 1 , an 75

On Some Extensions of Enestrom-Kakeya Theorem where

, 

are the roots of the quadratic

z2 

a n 1 z  k 1 0 . an

Theorem E: Let P ( z ) 

n

a j 0

j

z j be a polynomial of degree n  2 such that

either

a n  a n  2  .......  a 2 1  a 2 1  .......  a 3  a1  0 and a n 1  a n 3  .......  a 2  a 2  2  ......  a 2  a 0  0 , n 1 for some integer  ,0    , if n is odd 2 or

a n  a n  2  .......  a 2  a 2  2  ...... a 2  a 0  0 and a n 1  a n 3  .......  a 2 1  a 2 1  a 3  a1  0 , n2 for some integer  ,0    , if n is even.

2 Then all the zeros of P(z) lie in the closed disk

z

a n 1 a  2(a 0  a1  a 2  2a 2 1 )  1  n 1 an an

In this paper, we give generalizations of Theorems D and E and prove the following results: n

Theorem

1:

Let

P ( z )   a j z j be a polynomial of degree n  2

such

that

for

some

j 0

k1 , k 2  1;0   1 ,  2  1 , either k1 a n  a n  2  .......  a 3   1 a1  0 and k 2 a n 1  a n 3  .......  a 2   2 a 0  0 , if n is odd or

k1 a n  a n  2  .......  a 2   1 a 0  0 and k 2 a n 1  a n 3  .......  a 3   2 a1  0 ,

if n is even.

Then for odd n all the zeros of P(z) lie in the region

z  z   

k1 a n  (2k 2  1)a n 1  2( 1  1)a1  2( 2  1)a 0 an

and for even n all the zeros of P(z) lie in the region

z  z    where

, 

k1 a n  (2k 2  1)a n 1  2( 1  1)a 0  2( 2  1)a1 , an

are the roots of the quadratic

a n 1 z  k1  1  0 . an Remark 1: For k1  k , k 2  1,  1  1,  2  1 , Theorem 1 reduces to Theorem D of Zargar. z2 

Taking

an1  2an k1  1 , k 2  1 and noting that the quadratic z 2  2 k1  1z  k1  1  0 has

two equal roots each equal to Corollary 1: Let P ( z ) 

a j 0

either

 k1  1 ,we get the following

n

j

z j be a polynomial of degree n  2 such that for some k1  1;0   1 ,  2  1 ,

k1 a n  a n  2  .......  a 3   1 a1  0 and 76

On Some Extensions of Enestrom-Kakeya Theorem

2an k1  1  an1  an3  ....... a2   2 a0  0 , if n is odd, or

k1 a n  a n  2  .......  a 2   1 a 0  0 and

2an k1  1  an1  an3  ....... a3   2 a1  0 , if n is even.

Then for odd n all the zeros of P(z) lie in the region 1

 k1 a n  2a n k1  1  2( 1  1)a1  2( 2  1)a 0  2 z  k1  1    an   and for even n all the zeros of P(z) lie in the region 1

 k1 a n  2a n k1  1  2( 1  1)a 0  2( 2  1)a1  2 z  k1  1    . an   For k1  k ,  1  1,  2  1 , Cor. 1 reduces to a result of Zargar [8, Cor.1] .

Applying Cor. 1 to a polynomial of even degree, we get the following 2n

Corollary 2: Let

Q ( z )   b j z j be a polynomial of even degree 2n such that for some j 0

k1  1;0   1 ,  2  1 , k1b2 n  b2 n  2  .......  b2   1b0  0 and

2 k1  1b2n  b2n1  b2n3  ....... b3   2 b1  0 .

Then all the zeros of Q(z) lie in 1

 k1b2 n  2b2 n k1  1  2( 1  1)b0  2( 2  1)b1  2 z  k1  1    . b2 n   Taking k1  1,  1  1,  2  1 , Cor. 2 reduces to Enestrom-Kakeya Theorem (see [8]).

Taking k1  2, we get the following result from Cor.1: Corollary 3: Let P ( z ) 

n

a j 0

j

z j be a polynomial of degree n  2 such that for some 0   1 ,  2  1 , either

2a n  a n  2  .......  a 3   1 a1  0 and 2a n  a n 1  a n 3  .......  a 2   2 a 0  0 , if n is odd, or

2a n  a n  2  .......  a 2   1 a 0  0 and

2a n  a n 1  a n 3  .......  a3   2 a1  0 , if n is even.

Then for odd n all the zeros of P(z) lie in the region 1

 2  2( 1  1)a1  2( 2  1)a 0  2 z 1    an   and for even n all the zeros of P(z) lie in the region 1

 2  2( 1  1)a 0  2( 2  1)a1  2 z 1    . an   77

On Some Extensions of Enestrom-Kakeya Theorem For

 1  1,  2  1 , Cor.3 reduces to a result of Zargar

[8,Cor.3].

n

Theorem

2:

Let

P ( z )   a j z j be a polynomial of degree n  2

such

that

for

some

for

some

j 0

k1 , k 2  1;0   1 ,  2  1 , either k1 a n  a n  2  .......  a 2 1  a 2 1  .......  a 3   1 a1  0 and k 2 a n 1  a n 3  .......  a 2  a 2  2  ......  a 2   2 a 0  0 , n 1 for some integer  ,0    , if n is odd 2 or

k1 a n  a n  2  .......  a 2  a 2  2  ...... a 2   1 a 0  0 and

k 2 a n 1  a n 3  .......  a 2 1  a 2 1  a 3   2 a1  0 , n2 for some integer  ,0    , if n is even.

2 Then for odd n all the zeros of P(z) lie in the disk

z

a n 1 (2k1  1)a n  (2k 2  1)a n 1  2a 2  2a 2 1  2 1 a1  2 2 a 0  an an

and for even n all the zeros of P(z) lie in the disk

z

a n 1 (2k1  1)a n  (2k 2  1)a n 1  2a 2  2a 2 1  2 1 a 0  2 2 a1 .  an an

Remark 2: For k1  1, k 2  1,  1  1,  2  1 , Theorem 2 reduces to Theorem E of Zargar.

Applying Theorem 2 to the polynomial P(tz), we get the following result: n

Corollary

4:

Let

P ( z )   a j z j be a polynomial of degree n  2

such

that

j 0

k1 , k 2  1;0   1 ,  2  1 and t>0, either k1 a n t n  a n  2 t n  2  .......  a 2  1t 2  1  a 2  1t 2  1  .......  a 3 t 3   1 a1t  0 and k 2 a n 1t n 1  a n 3 t n 3  .......  a 2  t 2   a 2   2 t 2   2  ......  a 2 t 2   2 a 0  0 , for some integer

 ,0   

n 1 , if n is odd 2

or

k1 a n t n  a n  2 t n  2  .......  a 2  t 2   a 2   2 t 2   2  ......  a 2 t 2   1 a 0  0 and k 2 a n 1t n 1  a n 3 t n 3  .......  a 2 1t 2 1  a 2  1t 2 1  ......  a 3 t 3   2 a1t  0 , for some integer

 ,0   

n2 , if n is even.

2

Then for odd n all the zeros of P(z) lie in the disk

z

a n 1 (2k1  1)a n t n 1  (2k 2  1)a n 1t n 1  2a 2 t 2  2a 2 1t 2  2 1 a1t  2 2 a 0  an t n 1 a n

and for even n all the zeros of P(z) lie in the disk

a n 1 (2k1  1)a n t n 1  (2k 2  1)a n 1t n 1  2a 2 t 2  2a 2 1t 2  2 1 a 0  2 2 a1t z  . an t n 1 a n For k1  1, k 2  1,  1  1,  2  1 , Cor. 4 reduces to a result of Zargar [8, Cor.4].

2. Proofs of Theorems

78

On Some Extensions of Enestrom-Kakeya Theorem

Proof of Theorem 1: Suppose n is odd. Consider the polynomial

F ( z )  (1  z 2 ) P ( z )  (1  z 2 )( a n z n  a n 1 z n 1  ......  a1 z  a 0 )

 a n z n  2  a n 1 z n 1  (a n  a n  2 ) z n  ...... (a 2  a 0 ) z  a 0  a n z n  2  a n 1 z n 1  (k1 a n  a n  2 ) z n  (k1  1)a n z n  (k 2 a n 1  a n 3 ) z n 1  (k 2  1)a n 1 z n 1  ......  (a 2   2  a 2  ) z 2   2  (a 2  1  a 2  1 ) z 2  1  (a 2   a 2  2 ) z 2   ......  (a 3   1 a1 ) z 3  ( 1  1)a1 z 3  (a 2   2 a 0 ) z 2  ( 2  1)a 0 z 2  a1 z  a 0 1 For z  1 , so that  1, j  1,2,......,n , we have , by using the hypothesis, j z

k 2 a n1  a n3 (k 2  1)a n1  z z a a  a a  2  2n  2 2  ...... 1 1 n 3 3 z z

F ( z )  z [ a n z 2  a n1 z  (k1  1)a n  {k1 a n  a n 2  n

 ......

a 2  2  a 2 z



(1   1 )a1 z

n 3

n2 2





a 2 1  a 2 1 z

 2 a0  a 2 z

n2



n  2  1

(1   2 )a 0 z

n2



a1 z

n 1



a0 z

n

}]

 z [ a n z 2  a n 1 z  (k 11)a  {k1 a n  a n  2  k 2 a n 1  a n 3  (k 2  1)a n 1 n

 ...... a 2  2  a 2  a 2 1  a 2 1  a 2  a 2  2  a 2  ...... a3   1 a1  (1   1 )a1  a 2   2 a 0  (1   2 )a 0  a1  a 0 }]  z [ a n z 2  a n 1 z  (k 11)a n  {k1 a n  (2k 2  1)a n 1  2( 1  1)a1 n

if

a n z 2  a n1 z  (k 11)a n  k1 a n  (2k 2  1)a n1  2( 1  1)a1  2( 2  1)a0  2( 2  1)a 0 or

z2 

a n 1 1 z k 11  [k1 a n  (2k 2  1)a n 1  2( 1  1)a1  2( 2  1)a 0 ] an an

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

z2 

a n 1 1 z k 11  [k1 a n  (2k 2  1)a n 1  2( 1  1)a1  2( 2  1)a 0 ] an an

or

z  z    where

, 

1 [k1 a n  (2k 2  1)a n 1  2( 1  1)a1  2( 2  1)a 0 ] an

are the roots of the quadratic

z2 

a n 1 z  k1  1  0 . an

Since the zeros of F(z) with modulus less than or equal to 1 already satisfy the above inequality, it follows that all the zeros of F(z)lie in

z  z   

1 [k1 a n  (2k 2  1)a n 1  2( 1  1)a1  2( 2  1)a 0 ] . an 79

On Some Extensions of Enestrom-Kakeya Theorem Since the zeros of P(z) are also the zeros of F(z), , it follows that all the zeros of P(z)lie in

z  z   

1 [k1 a n  (2k 2  1)a n 1  2( 1  1)a1  2( 2  1)a 0 ] . an

The case of even n follows similarly.

Proof of Theorem 2: Suppose n is odd. Consider the polynomial

F ( z )  (1  z 2 ) P ( z )  (1  z 2 )( a n z n  a n 1 z n 1  ......  a1 z  a 0 )

 a n z n  2  a n 1 z n 1  (a n  a n  2 ) z n  ...... (a 2  a 0 ) z  a 0  a n z n  2  a n 1 z n 1  (k1 a n  a n  2 ) z n  (k1  1)a n z n  (k 2 a n 1  a n 3 ) z n 1  (k 2  1)a n 1 z n 1  ......  (a 2   2  a 2  ) z 2   2  (a 2  1  a 2  1 ) z 2  1  (a 2   a 2  2 ) z 2   ......  (a 3   1 a1 ) z 3  ( 1  1)a1 z 3  (a 2   2 a 0 ) z 2  ( 2  1)a 0 z 2  a1 z  a 0 1 For z  1 , so that j  1, j  1,2,......,n  1 , we have , by using the hypothesis, z

F ( z)  z

n 1

[ a n z  a n 1  {

 ......

a 2  2  a 2 z



( 1  1)a1 z

z

n 1

n2

n  2  1



k1 a n  a n  2 (k1  1)a n k 2 a n 1  a n 3 (k 2  1)a n 1    2 2 z z z z 

a 2 1  a 2 1 z

 2 a0  a2 z

n 1



n  2



z

( 2  1)a 0 z

a 2  2  a 2

n 1



n  2  1

a1 z

n



 ......

 1 a1  a 3 z

a0 z

n 1

n2

}]

[ a n z  a n1  {k1 a n  a n2  (k1  1)a n  k 2 a n1  a n3  (k 2  1)a n1

 ...... a 2  2  a 2  a 2 1  a 2 1  a 2 2  a 2  ......  1 a1  a3  ( 1  1)a1   2 a 0  a 2  ( 2  1)a0  a1  a 0 }] z

n 1

[ a n z  a n1  {(2k1  1)a n  (2k 2  1)a n1  2a 2  2a 2 1

 2 1a1  2 2 a0 }]

0 if

an z  an1  (2k1  1)an  (2k 2  1)an1  2a2  2a2 1  2 1a1  2 2 a0 i.e.

z

a n 1 (2k1  1)a n  (2k 2  1)a n 1  2a 2  2a 2 1  2 1 a1  2 2 a 0  an an

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

z

a n 1 (2k1  1)a n  (2k 2  1)a n 1  2a 2  2a 2 1  2 1 a1  2 2 a 0  . an an

Since the zeros of F(z) with modulus less than or equal to 1 already satisfy the above inequality, it follows that all the zeros of F(z) lie in the disk

z

a n 1 (2k1  1)a n  (2k 2  1)a n 1  2a 2  2a 2 1  2 1 a1  2 2 a 0  . an an

Since the zeros of P(z) are also the zeros of F(z), , it follows that all the zeros of P(z)lie in the disk

80

On Some Extensions of Enestrom-Kakeya Theorem

z

a n 1 (2k1  1)a n  (2k 2  1)a n 1  2a 2  2a 2 1  2 1 a1  2 2 a 0 .  an an

The case of even n follows similarly.

REFERENCES [1] [2] [3] [4] [5] [6] [7] [8]

A.Aziz and B.A.Zargar, Some Extensions of Enestrom-Kakeya Theorem, Glasnik Math. 31(1996), 239-244.

N. K. Govil and Q.I. Rahman, On the Enestrom-Kakeya Theorem, Tohoku Math.J.20(1968), 126-136.

M. H. Gulzar, Some Refinements of Enestrom-Kakeya Theorem, Int. Journal of Mathematical Archive -2(9, 2011),1512-1529.

M. H. Gulzar, Bounds for the Zeros and Extremal Properties of Polynomials, Ph.D thesis, Department of Mathematics, University of Kashmir, Srinagar, 2012.

A. Joyal, G. Labelle and Q. I. Rahman, On the Location of Zeros of Polynomials, Canad. Math. Bull., 10(1967), 53-66.

A. Liman and W. M. Shah, Extensions of Enestrom-Kakeya Theorem, Int. Journal of Modern Mathematical Sciences, 2013, 8(2), 82-89.

M. Marden, Geometry of Polynomials, Math. Surveys, No.3; Amer. Math. Soc. Providence R.I. 1966.

B.A.Zargar, On Enestrom-Kakeya Theorem, Int. Journal of Engineering Science and Research Technology, 3(4), April 2014.

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