Some compact generalisations of Enestrom-Kakeya type theorem for polynomials

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*Corresponding author Received March 26, 2012

45

Advances in Inequalities and Applications, 1 (2012), No. 1, 45-52

SOME COMPACT GENERALIZATIONS OF ENESTROM-KAKEYA TYPE

THEOREM FOR POLYNOMIALS

B. A. ZARGER*

P.G. Deptt. of Math. University of Kashmir, Srinagar

Abstract. In this paper we present an interesting generalization of Enestrom-Kakeya Theorem which amoung other things yields a number of already known classical results by putting some restrictive conditions on the coefficients of the polynomials.

keywords: coefficients, zeros, parameter

2010 AMS Subject Classification : 20C10,20C15.

1. INTRODUCTION AND STATEMENT OF RESULTS

The following result due to Enestrom-and Kakeya [7] is well-known in the theory of

distribution of the zeros of polynomials

THEOREM A. If

0 1 1

1 ... )

(z a z a z az a P  _n n  _n n    is a polynomial of degree n such that

Applying this result to P(tz), the following more general result is immediate.

THEOREM B. If

0 1 1

1 ... )

(z a z a z az a P  _n n  _n n    is a polynomial of degree n such that

0 ... _{1 0} 1

1    

  _

a ta a

t a

tn _n n _n

then all the zeros of P(z) lie in z t.

In the literature [1,2,4,5,8] these exist some extensions and generalizations of

Enestrom-Kakeya Theorem, Joyal, Labellel and Rahman[6] extended this theorem to

polynomials whose coefficients were monotonic but not necessarily non-negative. Recently

Aziz and Zarger[3] relaxed the hypothesis in several ways and among other things proved the

following interesting generalization of Theorem A:

THEOREM C. If

0 1 1

1 ... )

(z a z a z az a

P n n

n

n    

 



is a polynomial of degree n such that for some k 1.

0 ... _{1 0} 1   

a _ a a

ka_{n n}

then P(z) has all its zeros in

(2) zk1k

In this paper we start by proving the following result which includes Theorems A,B and

C as special cases.

THEOREM 1. Let

0 1 1

1 ... )

(z a z a z az a

P n n

n

n    

 



be a polynomial of degree n, if for some real number k 1,

(3) Maxz kan an  an an   a a zn a zn M

 

 

 0

1 0 1 2

1 1

1( ) ( ) ... ( )

then all the zeros of P(z) lie in the disk

(4)

n

a M k

z 1 

 (ka_n a_n1)  a_n1a_n2 ... a₁a₀  a₀ kanan1an1an2...a1a0a0 ka_n

We take M ka_nin Theorem 1, it follows that all the zeors of P(z) satisfying the conditions of Theorem C lie in the circle

k k

z 1 ,

which is precisely the conclusion of Theorem C.

The following corollary follows by taking

n n

a a k  1

in Theorem 1.

COROLLARY 1. If





 n

j j jz

a z

P

0 )

( , is a polynomial of degree n, such that

1 0

2 1 1

1 a and Max (a a )z ... a z M

an  n z n  n   n 

then all the zeros of P(z) lie in the disk.

n n

n

a M a

a

z 1 1

1 

  

REMARK 2. Let

0 1 1

1 ... )

(z a z a z az a

P n n

n

n    

 



be a polynomial of degree n satisfying the condition

(05) 0a_n a_n1...a₁a₀0 then M1 Maxz1(an1an2)(an2 an3)z...a0zn (an1an2)(an2an3)...a0

a_n1.

Hence from Corollary 1 it follows that all the zeros of P(z) satisfying (5) lie in the circle.

n n

n n

a a a

a

z 1 1

1 

 _{ }



This result was earlier proved by the authors in [3] Cor. 2.

Theorem 1 follows by taking t 1in the following general result:

THEOREM 2. Let

0 1 1

1 ... )

be a polynomial of degree n. If for some positive real numbers t and k 1 (6) Max H z M

t

z1 ( )  , where





n n n

n a ta a z ta a z ta z

tka z

H 1 ₀

0 1 2

1

1) ( ) ... ( )

( )

(   _  _  _     

then all the zeros of P(z) lie in

(7) n a M k t

z ( 1) 

Since        

 H z H _t

Max M t z 1 ) ( 1 n n n n n n n n t ta t a ta t a ta t a ta a

tka ) ( )1 ( ) 1 ... ( ) 1 .1

(  1  1  2  2  3 2   1 0 1  0

 _{     }

n

kta



Theorem 2 follows by taking t

R1in the following more general result which yields a number of other interesting results for various choices of parameters R and t:

THEOREM 3. Let

0 1 1

1 ... )

(z a z a z az a

P n n

n

n    

 



be a polynomial of degree n. If for some positive real numbers t and k1 (8) Max_zRH(z) M, where

. ... ) ( ) ( )

( 1 1 2 0

n n

n n

n a ta a z ta z

kta z

H   _  _  _  

then all the zeros of P(z) lie in the circle

(9) _n

n a t R kt M if a M k t z _                

 ( 1) 1

and in

(10) (2 1) 1 , 1 t a_n.

R kt M if t R k t z _                        

PROOF OF THEOREM 3. Consider the polynomial

F(z)(tz)P(z)

a_nzn1(ta_na_n1)zn (ta_n1a_n2)zn1...(ta₁a₀)zta₀ then we have



      

z F z z

G( ) n 1 1

a_n(ta_na_n1)z(ta_n1a_n2)z2...(ta₁a₀)znta₀zn1 a_n ta_nzkta_nz(tka_na_n1)z(ta_n1a_n2)z2...ta₀zn1 a_nta_nzkta_nzzH(z)

where

H(z)





(11) G(z)  an t(k1)z1 z H(z) Since

) ( ,

)

(z M for z R and H z

H   is analytical for z R, therefore by Maximum Modulus

Theorem H(z) M for z R. Using this fact in (11), we get

. 1

) 1 ( )

(z a t k z zM

G  n   

> 0, if

R z for z

k t z a M

n

 



 ( 1) 1 .

Thus in z R,

G(z) 0 for zE, where

    

  

  

 ; z t(k 1)z 1

a M z E

n

we show if wE, then w Rif

n

a t R kt

M _

  

 

       

 1

w  t(k1)w1

a M

n

t(k1)w 1 This implies,

1 )

1

( 

    

  



t k W

a M

n

or

R a k t M

a W

n

n _

  

) 1 ( if

an MRt(k1)an R or if,





   



 _{ }

  

 t

R kt a k

Rt R a

M n ( 1) 1 _n 1

Thus if

n

a t R kt

M _

  

 

       

 1

then in z R,

if z

G( ) 0,

1 ) 1 (  

 t k z

z a M

n

This shows that all the zeros of G(z) lie in the region defined by

1 ) 1 (  

 t k z z

a M

n

Replacing z by z 1

and noting that F(z)zn1G(z),it follows that all the zeros of F(z) lie in

n

a M k

t

z ( 1) .

(12) _n n a t R kt M if a M k t z                 

 ( 1) 1 .

We now assume that

n a R k Rt M        

 ( 1) 1

then for z R,

G(z)  a_n (ta_n kta_n)zzH(z)

                    n n a z H k t z

a 1 ( 1) ( )

            n n a M k t z

a ( 1)

                      

 t tk

R k t z an 1 ) 1 ( 1 >0, if ) ( 1 ) 1 2 ( 1 R t R k t z            .

Tthis shows that all the zeros of G(z) lie in the region

          t R k t z 1 ) 1 2 ( 1

Replacing z by z 1

and as before noting that

        z G z z

F( ) n 1 1

it follows that all the zeros of F(z) and hence all the zeros of P(z) lie in the circle

(13) 

         t R k t

z (2 1) 1

From (12) and (13), the desired result follows.

REFERENCES

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