some generalizations of enestrom –kakeya theorem · international journal of scientific and...

International Journal of Scientific and Research Publications, Volume 3, Issue 2, February 2013 1 ISSN 2250-3153

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Some Generalizations of Enestrom –Kakeya Theorem

M. H. Gulzar

Department of Mathematics, University of Kashmir, Srinagar 190006

E-mail: [email protected]

Abstract- Many generalizations of the Enestrom –Kakeya Theorem are available in the literature. In this paper we give generalizations

of some recent results on the subject.

Mathematics Subject Classification : 30C10, 30C15

Index Terms- Complex number, Polynomial , Zero, Enestrom –Kakeya Theorem

I. INTRODUCTION AND STATEMENT OF RESULTS

he following result known as the Enestrom –Kakeya Theorem [3] is well-known in the theory of the distribution of zeros of

polynomials:

Theorem A: Let

n

j

j

j zazP0

)(

be a polynomial of degree n whose coefficients satisfy

0...... 011 aaaa nn .

Then P(z) has all its zeros in the closed unit disk 1z

.

In the literature there exist several generalizations and extensions of this result .Recently, Y. Choo [1] proved the following results:

Theorem B: Let

n

j

j

j zazP0

)(

be a polynomial of degree n such that for some real number 1 ,, ,0,1 knank

01111 ...... aaaaaaa knknknnn .

If ,1 knkn aa then P(z) has all its zeros in the disk 1Kz

, where 1K is the greatest positive root of the equation

011

1 kk KK

and

n

kn

a

a

)1(1

, n

knn

a

aaaa 00

1

)1(

.

If ,1 knkn aathen P(z) has all its zeros in the disk 2Kz


02

1

2 kk KK

and

n

kn

a

a

)1(2

, n

knn

a

aaaa 00

2

)1(

.

Theorem C: Let

n

j

j

j zazP0

)(

be a polynomial of degree n with jja )Re( and jja )Im(

, j=0,1,2,……,n and for

some real numbers 1 , ,0,1 knnk

01111 ...... knknknnn

011 ...... nn .

If ,1 knkn

then P(z) has all its zeros in the disk 1Kz , where 1K

is the greatest positive root of the equation

T

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International Journal of Scientific and Research Publications, Volume 3, Issue 2, February 2013 2

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011

1 kk KK

and

n

kn

a

)1(1

,

n

nknn

a

0000

1

)1(

.

If ,1 knkn then P(z) has all its zeros in the disk 2Kz


02

1

2 kk KK

and

n

kn

a

)1(2

,

n

nknn

a

0000

2

)1(

.

Theorem D: Let

n

j

j

j zazP0

)(

be a polynomial of degree n such that for some real number

,

.,......,1,0,2

arg nja j

and for some 1 and 0kna

,


If knkn aa 1 (i.e. 1 ), then P(z) has all its zeros in the disk 1Kz , where 1K

is the greatest positive root of the

equation

011

1 kk KK

and

n

kn

a

a

)1(1

,

n

n

j

jknn

a

aaa

1

0

1

sin2)sin}(cos)1({

.

If 1 knkn aa (i.e. 10 ), then P(z) has all its zeros in the disk 2Kz

, where 2K is the greatest positive root of the

equation

02

1

2 kk KK

and

n

kn

a

a

)1(2

,

n

n

j

jknn

a

aaa

1

0

2

sin2)sin}(cos)1({

.

In this paper we generalize the above results with less restrictive conditions on the coefficients. In fact, we prove the following results:


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Theorem 1 : Let

n

j

j

j zazP0

)(

be a polynomial of degree n such that for some real numbers 10,0

, 1 , ,

,0,1 knank




011

1 kk KK

and

n

kn

a

a

)1(1

,

n

knn

a

aaaaa 000

1

2)()1(2

. .



02

1

2 kk KK

and

n

kn

a

a

)1(2

,

n

knn

a

aaaaa 000

2

2)()1(2

.

Remark 1: Taking ,1,0 Theorem 1 reduces to Theorem B.

Taking 0

, Theorem 1 gives the following result:

Corollory 1: Let

n

j

j

j zazP0

)(

be a polynomial of degree n such that for some real number 10

, 1 , ,

,0,1 knank




011

1 kk KK

and

n

kn

a

a

)1(1

,

n

knn

a

aaaaa 000

1

2)()1(

. .



02

1

2 kk KK

and

n

kn

a

a

)1(2

,


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n

knn

a

aaaaa 000

2

2)()1(

.

Theorem 2: Let

n

j

j

j zazP0

)(


, j=0,1,2,……,n and for

some real numbers 10,0 1 ,

,0,1 knnk

01111 ...... knknknnn

011 ...... nn .

If ,1 knkn



011

1 kk KK

and

n

kn

a

)1(1

,

n

nknn

a

00000

1

2)()1(2

.



02

1

2 kk KK

and

n

kn

a

)1(2

,

n

nknn

a

00000

2

2)()1(2

.

Remark 2: Taking ,1,0 Theorem 2 reduces to Theorem C.

Taking 0


Corollary 2: Let

n

j

j

j zazP0

)(


, j=0,1,2,……,n and for

some real numbers 10

, 1 , ,0,1 knnk

01111 ...... knknknnn

011 ...... nn .

If ,1 knkn



011

1 kk KK

and

n

kn

a

)1(1

,

n

nknn

a

00000

1

2)()1(

.




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02

1

2 kk KK

and

n

kn

a

)1(2

,

n

nknn

a

00000

2

2)()1(

.

Theorem 3: Let

n

j

j

j zazP0

)(


,

.,......,1,0,2

arg nja j


,




equation

011

1 kk KK

and

n

kn

a

a

)1(1

,

n

n

j

jknn

a

aaaaa

1

1

00

1

]sin22)1sin(cos)sin}(cos)1({[

.



equation

02

1

2 kk KK

and

n

kn

a

a

)1(2

,

n

n

j

jknn

a

aaaaa

1

1

00

2

]sin22)1sin(cos)sin}(cos)1({[

Remark 3: Taking ,1,0 Theorem 4 reduces to Theorem D.

Taking 0


Corollory3 : Let

n

j

j

j zazP0

)(


,

.,......,1,0,2

arg nja j


,



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equation

011

1 kk KK

and

n

kn

a

a

)1(1

,

n

n

j

jknn

a

aaaaa

1

1

00

1

]sin22)1sin(cos)sin}(cos)1([{

.



equation

02

1

2 kk KK

and

n

kn

a

a

)1(2

,

.

]sin22)1sin(cos)sin}(cos)1([{1

1

00

2

n

n

j

jknn

a

aaaaa

II. LEMMAS

For the proofs of the above results , we need the following result:

Lemma 1: Let P(z)=

n

j

j

j za0 be a polynomial of degree n with complex coefficients such that

fornja j ,,...1,0,2

arg

some real

.Then for some t>0,

.sincos

111

jaataatata jjjjj

The proof of lemma 1 follows from a lemma due to Govil and Rahman [2].

III. PROOFS OF THE THEOREMS

Proof of Theorem 1: We first prove that 1K>1 and 2K

>1.

Let

11

1

1 )( kk KKKf.

To prove 1K>1, it suffices to prove that

.0)1(1 f

If ,1 knkn aa then one of the following four cases happens:

(a) 011 knknkn aaa

and 1 .

(b) knknkn aaa 011 and 0 .


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(c) knknkn aaa 11 0 and 1 .

(d) knknkn aaa 110 and 10 .

It is easy to see that 01

and 11 1 for the cases (a), (b) and (c) and

11 1 for the case (d). Thus 111 1)1( f

<0 and we have 1K>1.

Again , let

2

1

22 )( kk KKKf .

If ,1 knkn aathen the four possible cases to occur are the following:

(a) 011 knknkn aaa

and 10 .

(b) 11 0 knknkn aaa and 1 .

(c) 110 knknkn aaa and 0 .

(d) 110 knknkn aaa and 1 .

For the first three cases 02

and 22 1 and for the last case

22 1 . Hence 222 1)1( f

<0 and we have 2K>1.

Now, consider the polynomial

)()1()( zPzzF

kn

knkn

kn

knkn

n

nn

n

n

n

n

n

n

zaazaazaaza

azazazaz

)()(......)(

)......)(1(

1

1

11

1

01

1

1

001

1

21 )(......)( azaazaa kn

knkn

If ,1 knkn aa

then ,1 knkn aa

and we have 1

11

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

kn

knkn

n

nn

nkn

kn

n

n zaazaazzazazF

......)()( 1

211

kn

knkn

kn

knkn zaazaa

.)1()( 0001 azazaa

For 1z

,

1

11

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

kn

knkn

n

nn

nkn

kn

n

n zaazaazzazazF

......)()( 1

211

kn

knkn

kn

knkn zaazaa

0001 )1()( azazaa

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

1

nn

n

kn

k

n

knaazazaz

nnn

kknknkknknkknkn

z

a

zzaa

zaa

zaa

zaa

0

1101

121111

1)1(

1)(

......1

)(1

)(1

)(

......)([)1( 1

1

nn

n

kn

k

n

knaazazaz

])1()(

......)()()(

0001

2111

aaaa

aaaaaa knknknknknkn

)()1(2[)1( 00

1 aaaazazaz knn

n

kn

k

n

kn


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]2 0a

0 if

k

knn

n

knkzaaaaa

a

az ]2)()1(2[)1( 000

1

or

.11

1 kk zz

This inequality holds if

.11

1 kkzz

Hence all the zeros of F(z) whose modulus is greater than 1 lie in the disk 1Kz , where 1K


equation

011

1 kk KK.

But the zeros of F(z) whose modulus is less than or equal to 1 are already contained in the disk 1Kz since

.11 KTherefore, all

the zeros of F(z) and hence P(z) lie in 1Kz .

If ,1 knkn aa then

,1 knkn aaand we have

1

11

11 )(......)()1()(

kn

knkn

n

nn

nkn

kn

n

n zaazaazzazazF

......)()( 1

211

kn

knkn

kn

knkn zaazaa

.)1()( 0001 azazaa

For 1z

,

1

11

11 )(......)()1()(

kn

knkn

n

nn

nknn

n zaazaazzzazF

......)()( 1

211

kn

knkn

kn

knkn zaazaa

0001 )1()( azazaa

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

1

nn

n

kn

k

n

knaazazaz

nnn

kknknkknknkknkn

z

a

zzaa

zaa

zaa

zaa

0

1101

121111

1)1(

1)(

......1

)(1

)(1

)(

......)([)1( 1

1

nn

n

kn

k

n

knaazazaz

])1()(

......)()()(

0001

2111

aaaa

aaaaaa knknknknknkn

)()1(2[)1( 00

1aaaazazaz knn

n

kn

k

n

kn

]2 0a

0 if

1

000 ]2)()1(2[)1(

k

knn

n

knkzaaaaa

a

az

or


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.

1

22

kk zz


.

1

22

kkzz

Hence all the zeros of F(z) whose modulus is greater than 1lie the disk 2Kz , where 2K


equation

02

1

2 kk KK.

But those zeros of F(z) whose modulus is less than or equal to 1 are already contained in the disk 2Kz since

.12 KTherefore,

all the zeros of F(z) and hence P(z) lie in 2Kz .

That proves Theorem 1.

Proof of Theorem 2: Consider the polynomial

)()1()( zPzzF

kn

knkn

kn

knkn

n

nn

n

n

n

n

n

n

zaazaazaaza

azazazaz

)()(......)(

)......)(1(

1

1

11

1

01

1

1

001

1

21 )(......)( azaazaa kn

knkn

1

11

1 )(......)(

kn

knkn

n

nn

n

n zzza

0

1

100

01

1

211

)()1(

)(......)()(

iziz

zzz

n

j

j

jj

kn

knkn

kn

knkn

If ,1 knkn then

,1 knkn and we have

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

1

n

nn

nkn

kn

n

n zzzzazF

1

1 )(

kn

knkn z ......)()( 1

211

kn

knkn

kn

knkn zz

.)()1()( 0

1

10001 izizzn

j

j

jj

For 1z

,

111

1 1)(......)([)1()(

kknknnn

nkn

kn

n

n

zzzzazF

]1

)(......)[(])1(

1)(......

1)(

1)(

0

1011

0

1

0

1011211

nnnn

n

nn

nkknknkknkn

zzz

zz

zzz

)()1(2[)1( 00

1

knn

nkn

kn

n

n zzza

]2 000 n

>0

if

.11

1 kk zz

But this inequality holds if


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kkzz 11

1

.

Hence, it follows that all the zeros of F(z) whose modulus is greater than 1 lie in the disk 1Kz , where 1K

is the greatest positive

root of the equation

011

1 kk KK.

As in the proof of Theorem 1, it can be shown that11 K

, so that the zeros of F(z) whose modulus is less than or equal to 1 are

already contained in the disk 1Kz . Therefore, all the zeros of F(z) and hence P(z) lie in 1Kz

.

Now,, suppose that ,1 knkn

then ,1 knkn

and we have

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

11

n

nn

nkn

kn

n

n zzzzazF

zz

zz

kn

knkn

kn

knkn

kn

knkn

)(......)(

)()(

01

1

21

1

1

1

.)()1( 0

1

100 izizn

j

j

jj

For 1z

,

111

11 1)(......)([)1()(

kknknnn

nkn

kn

n

n

zzzzazF

]1

)(......)[(])1(

1)(......

1)(

1)(

0

1011

0

1

0

1011211

nnnn

n

nn

nkknknkknkn

zzz

zz

zzz

)()1(2[)1( 00

11

knn

nkn

kn

n

n zzza

]2 000 n

>0

if

.

1

22

kk zz

But this inequality holds if

1

22

kkzz

.

Hence, it follows that all the zeros of F(z) whose modulus is greater than 1 lie in the disk 2Kz , where 2K



01

1

2 kk KK.

As in the proof of Theorem 1, it can be shown that12 K

.Thus the zeros of F(z) whose modulus is less than or equal to 1 are

already contained in the disk 2Kz . Therefore, all the zeros of F(z) and hence P(z) lie in 2Kz

and the proof of Theorem 2 is

complete.

Proof of Theorem 3: Consider the polynomial

)()1()( zPzzF

kn

knkn

kn

knkn

n

nn

n

n

n

n

n

n

zaazaazaaza

azazazaz

)()(......)(

)......)(1(

1

1

11

1

01

1

1


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001

1

21 )(......)( azaazaa kn

knkn

.

If knkn aa 1 , then knkn aa 1 , 1 and we have,

F(z) 1

11

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

kn

knkn

n

nn

n

kn

n

n zaazaazaza

.)1()(

......)()(

0001

1

211

azazaa

zaazaa kn

knkn

kn

knkn

For 1z

, by using Lemma1,

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

211

1 n

nn

n

nn

nkn

kn

n

n zaazaazzazazF

0001

1

211

1

1

)1()(......

)()()(

azazaa

zaazaazaa kn

knkn

kn

knkn

kn

knkn

1

1

1

1 ......[)1(

k

knkn

nn

nkn

kn

n

n

z

aaaazzaza

])1(

......0

1

0

1

011

nnnk

knkn

z

a

z

a

z

aa

z

aa

knknnn

nkn

kn

n

n aaaazzaza

11

1 ......[)1(

])1(...... 00011 aaaaaa knkn

)sin}(cos)1({[)1(1

knn

nkn

kn

n

n aazzaza

]sin22)1sin(cos1

,0

00

n

knjj

jaaa

0 if

kk zz 11

1

.


kkzz 11

1

and thus all the zeros of F(z )and hence P(z) with modulus greater than 1 lie in the disk 1Kz , where 1K



011

1 kk KK.

It is easy to see that 11 K

and all the zeros of P(z) with modulus less than or equal to 1 are already contained in 1Kz .

Again , If 1 knkn aa, then 1 knkn aa

, 1 and we have,

F(z) ......)()1( 1

11

n

nn

nkn

kn

n

n zaazzaza

1

1 )(

kn

knkn zaa

......)()( 1

211

kn

knkn

kn

knkn zaazaa

.)1()( 0001 azazaa

For 1z

, by using Lemma1,

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

211

11 n

nn

n

nn

nkn

kn

n

n zaazaazzazazF


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0001

1

211

1

1

)1()(......

)()()(

azazaa

zaazaazaa kn

knkn

kn

knkn

kn

knkn

1

1

1

11 ......[)1(

k

knkn

nn

nkn

kn

n

n

z

aaaazzaza

])1(

......0

1

0

1

011

nnnk

knkn

z

a

z

a

z

aa

z

aa

knknnn

nkn

kn

n

n aaaazzaza

11

11 ......[)1(

])1(...... 00011 aaaaaa knkn

)sin}(cos)1({[)1( 11

knn

nkn

kn

n

n aazzaza

]sin22)1sin(cos1

,0

00

n

knjj

jaaa

0 if

1

22

kk zz .


1

22

kkzz

and thus all the zeros of F(z )and hence P(z) with modulus greater than 1 lie in the disk 2Kz , where 2K



02

1

2 kk KK.

It is easy to see that 12 K

and all the zeros of P(z) with modulus less than or equal to 1 are already contained in 2Kz .

That proves Theorem 3.

REFERENCES

[1] Y. Choo, Further Generalizations of Enestrom-Kakeya Theorem, Int. Journal of Math. Analysis. Vol. 5, 2011,no. 20, 983-995.

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

[3] M. Marden , Geometry of Polynomials, IInd Ed.Math. Surveys, No. 3, Amer. Math. Soc. Providence,R. I,1996.

some generalizations of enestrom –kakeya theorem · international journal of scientific and...

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