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Further Generalizations of Enestrom-Kakeya Theorem
1. International Journal of Engineering Inventions
e-ISSN: 2278-7461, p-ISBN: 2319-6491
Volume 2, Issue 3 (February 2013) PP: 49-58
www.ijeijournal.com P a g e | 49
Further Generalizations of Enestrom-Kakeya Theorem
M. H. Gulzar
Department of Mathematics, University of Kashmir, Srinagar 190006
Abstract:- Many generalizations of the Enestrom –Kakeya Theorem are available in the literature. In this paper
we prove some results which further generalize some known results.
Mathematics Subject Classification: 30C10,30C15
Keywords and phrases:- Complex number, Polynomial , Zero, Enestrom – Kakeya Theorem.
I. INTRODUCTION AND STATEMENT OF RESULTS
The Enestrom –Kakeya Theorem (see[6]) is well known in the theory of the distribution of zeros of
polynomials and is often stated as follows:
Theorem A: Let
n
j
j
j zazP
0
)( 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. Joyal et al [5] extended
it to polynomials with general monotonic coefficients and proved the following result:
Theorem B: Let
n
j
j
j zazP
0
)( be a polynomial of degree n whose coefficients satisfy
011 ...... aaaa nn .
Then P(z) has all its zeros in
n
n
a
aaa
z
00
.
Aziz and zargar [1] generalized the result of Joyal et al [6] as follows:
Theorem C: Let
n
j
j
j zazP
0
)( be a polynomial of degree n such that for some 1k
011 ...... aaaka nn .
Then P(z) has all its zeros in
n
n
a
aaka
kz
00
1
.
For polynomials ,whose coefficients are not necessarily real, Govil and Rahman [2] proved the following
generalization of Theorem A:
Theorem C: If
n
j
j
j zazP
0
)( is a polynomial of degree n with jja )Re( and jja )Im( ,
j=0,1,2,……,n, such that
0...... 011 nn ,
where 0n , then P(z) has all its zeros in
))(
2
(1
0
n
j
j
n
z
.
Govil and Mc-tume [3] proved the following generalisations of Theorems B and C:
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Theorem D: Let
n
j
j
j zazP
0
)( be a polynomial of degree n with jja )Re( and jja )Im( ,
j=0,1,2,……,n. If for some 1k ,
011 ...... nnk ,
then P(z) has all its zeros in
n
n
j
jnk
kz
0
00 2
1 .
Theorem E: Let
n
j
j
j zazP
0
)( be a polynomial of degree n with jja )Re( and jja )Im( ,
j=0,1,2,……,n. If for some 1k ,
011 ...... nnk ,
then P(z) has all its zeros in
n
n
j
jnk
kz
0
00 2
1 .
M. H. Gulzar [4] proved the following generalizations of Theorems D and E:
Theorem F: Let
n
j
j
j zazP
0
)( be a polynomial of degree n with jja )Re( and jja )Im( ,
j=0,1,2,……,n. If for some real numbers 0 , ,10
011 ...... nn ,
then P(z) has all its zeros in the disk
n
n
j
jn
n
z
0
000 2)(2
.
Theorem G: Let
n
j
j
j zazP
0
)( be a polynomial of degree n with jja )Re( and jja )Im( ,
j=0,1,2,……,n. If for some real number 0 ,
011 ...... nn ,
then P(z) has all its zeros in the disk
n
n
j
jn
n
z
0
000 2)(2
.
In this paper we give generalization of Theorems F and G. In fact, we prove the following:
Theorem 1: Let
n
j
j
j zazP
0
)( be a polynomial of degree n with jja )Re( and jja )Im( ,
j=0,1,2,……,n. If for some real numbers , 0 , ,10,0,1 knank
01111 ...... knknknnn ,
and ,1 knkn then P(z) has all its zeros in the disk
n
n
j
jknknn
n aa
z
0
000 2(21)1(
,
and if ,1 knkn then P(z) has all its zeros in the disk
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n
n
j
jknknn
n aa
z
0
000 2(21)1(
Remark 1: Taking 1 , Theorem 1 reduces to Theorem F.
If ja are real i.e. 0j for all j, we get the following result:
Corollary 1: Let
n
j
j
j zazP
0
)( be a polynomial of degree n. If for some real numbers , 0 ,
,0,1 knank ,10
01111 ...... aaaaaaa knknknnn ,
and ,1 knkn aa then P(z) has all its zeros in the disk
n
knknn
n a
aaaaaa
a
z
)(21)1( 000
,,
and if ,1 knkn aa then P(z) has all its zeros in the disk
n
knknn
n a
aaaaaa
a
z
)(21)1( 000
If we apply Theorem 1 to the polynomial –iP(z) , we easily get the following result:
Theorem 2: Let
n
j
j
j zazP
0
)( be a polynomial of degree n with jja )Re( and jja )Im( ,
j=0,1, 2,……,n. If for some real numbers , 0 , ,0,1 knank ,10
01111 ...... knknknnn ,
and ,1 knkn then P(z) has all its zeros in the disk
n
n
j
jknknn
n aa
z
0
000 2)(21)1(
,
and if ,1 knkn then P(z) has all its zeros in the disk
n
n
j
jknknn
n aa
z
0
000 2)(21)1(
Remark 2: Taking 1 , Theorem 2 reduces to Theorem G.
Theorem 3: Let
n
j
j
j zazP
0
)( be a polynomial of degree n such that for some real numbers , 0 ,
,10,,0,1 knank ,
01111 ...... aaaaaaa knknknnn
and
.,......,1,0,
2
arg nja j
If knkn aa 1 (i.e. 1 ) , then P(z) has all its zeros in the disk
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n
n
knjj
j
knn
n a
aaa
aa
a
z
1
,1
00 ]sin22)1cos(sin
)1sincossin(cos)sin(cos[
If 1 knkn aa (i.e. 1 ), then P(z) has all its zeros in the disk
n
n
knjj
j
knn
n a
aaa
aa
a
z
1
,1
00 ]sin22)1cos(sin
)1sincossin(cos)sin(cos[
Remark 3: Taking 1 in Theorem 3 , we get the following result:
Corollary 2: Let
n
j
j
j zazP
0
)( be a polynomial of degree n. If for some real numbers 0 , ,10
011 ...... aaaa nn ,
then P(z) has all its zeros in the disk
n
n
j
jn
n a
aaaa
a
z
1
,1
00 ]sin22)1cos(sin)sin(cos[
.
Remark 4: Taking 1,)1( kak n in Cor.2, we get the following result:
Corollary 3: Let
n
j
j
j zazP
0
)( be a polynomial of degree n. If for some real numbers 0 , ,10
011 ...... aaaak nn ,
then P(z) has all its zeros in the disk
n
n
j
jn
n a
aaaak
a
z
1
,1
00 ]sin22)1cos(sin)sin(cos[
Taking 1,)1( kak n , and 1 in Cor. 3, we get a result of Shah and Liman [7,Theorem 1].
II. LEMMAS
For the proofs of the above results , we need the following results:
Lemma 1: . Let P(z)=
n
j
j
j za
0
be a polynomial of degree n with complex coefficients such that
fornjaj ,,...1,0,
2
arg
some real . Then for some t>0,
.sincos 111
j
aataatata jjjjj
The proof of lemma 1 follows from a lemma due to Govil and Rahman [2].
Lemma 2.If p(z) is regular ,p(0) 0 and Mzp )( in ,1z then the number of zeros of p(z)in
,10, z does not exceed
)0(
log
1
log
1
p
M
(see[8],p171).
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III. PROOFS OF THEOREMS
Proof of Theorem 1: 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
nn zzzi
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
11
1
)(......)()()(
kn
knkn
n
nn
nn
nn zzzzizF
......)()1()( 1
211
kn
knkn
kn
kn
kn
knkn zzaz
.)()1()( 0
1
10001 izizz
n
j
j
jj
For 1z ,
1
1
1
211
1
)(......)()()(
kn
knkn
n
nn
n
nn
nn
n zzzzzazF
0
1
10001
1
211
)()1()(
......)()1()(
izizz
zzaz
n
j
j
jj
kn
knkn
kn
kn
kn
knkn
11211
1
)(......
1
)()( kknknnnnnn
n
zz
zaz
n
n
j
jnjjnnn
knknkknkknkn
z
i
z
i
zzz
kzz
0
1
1
0
1
0
101
211
1
)()1(
1
)(
....
1
1
)(
1
)1(
1
)(
knknnnnnn
n
zaz 1211 ......
0
1
100
01211
1
......1
n
j
jj
knknknknkn
11211 ...... knnknknnnnnn
n
kn
zaz
n
j
j
knknkn
0
0
00121
2
)1(......1
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n
j
j
knknnn
n
zaz
0
000
2
2)(1)1({[
>0
if
n
j
jknknnn za
0
000 22)(1)1(
This shows that the zeros of F(z) whose modulus is greater than 1 lie in
n
n
j
jknknn
n aa
z
0
000 2)(21)1(
.
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
n
j
jknknn
n aa
z
0
000 2)(21)1(
.
Since the zeros of P(z) are also the zeros of F(z), it follows that all the zeros of P(z) lie in
n
n
j
jknknn
n aa
z
0
000 2)(21)1(
.
If ,1 knkn then ,1 knkn and we have
1
11
1
)(......)()()(
kn
knkn
n
nn
nn
nn zzzzizF
.)()1()(
......)()1()(
0
1
10001
1
21
1
1
izizz
zzz
n
j
j
jj
kn
knkn
kn
kn
kn
knkn
For 1z ,
1
1
1
211
1
)(......)()()(
kn
knkn
n
nn
n
nn
nn
n zzzzzazF
0
1
10001
1
21
1
1
)()1()(
......)()1()(
izizz
zzz
n
j
j
jj
kn
knkn
kn
kn
kn
knkn
11211
1
)(......
1
)()( kknknnnnnn
n
zz
zaz
n
n
j
jnjjnnn
kknknkknkknkn
z
i
z
i
zzz
zzz
0
1
1
0
1
0
101
12111
1
)()1(
1
)(
....
1
)(
1
)1(
1
)(
knknnnnnn
n
zaz 1211 ......
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0
1
100
01211
1
......1
n
j
jj
knknknknkn
11211 ...... knnknknnnnnn
n
kn
zaz
n
j
j
knknkn
0
0
00121
2
)1(......1
n
j
j
knknnn
n
za
z
0
0
00
}22
)(1)1({
>0
if
n
j
jknknnn za
0
000 22)(1)1(
This shows that the zeros of F(z) whose modulus is greater than 1 lie in
n
n
j
jknknn
n aa
z
0
000 2)(21)1(
.
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
n
j
jknknn
n aa
z
0
000 2)(21)1(
.
Since the zeros of P(z) are also the zeros of F(z), it follows that all the zeros of P(z) lie in
n
n
j
jknknn
n aa
z
0
000 2)(21)1(
.
That proves Theorem 1.
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
001
1
21 )(......)( azaazaa kn
knkn
.
If knkn aa 1 , then knkn aa 1 , 1 and we have, for 1z , by using Lemma1,
1
1
1
211
1
)(......)()()(
kn
knkn
n
nn
n
nn
nn
n zaazaazaazzazF
0001
1
211
)1()(
......)()1()(
azazaa
zaazazaa kn
knkn
kn
kn
kn
knkn
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11211
1
)(......
1
)()( kknknnnnnn
n
z
aa
z
aaaazaz
nnn
kknknkknkknkn
z
a
z
a
z
aa
z
aa
z
a
z
aa
0
1
0
101
1211
)1(
1
)(
....
1
)(
1
)1(
1
)(
knknnnnnn
n
aaaaaazaz 1211 ......
00
01211
1
......1
aa
aaaaaaa knknknknkn
sin)(cos){( 11 nnnnn
n
aaaazaz
cos)(......sin)(cos)( 12121 knknnnnn aaaaaa
( sin)(cos)(sin) 111 knknknknknkn aaaaaa
sin)(cos)(1 2121 knknknknkn aaaaa
})1(sin)(cos)(...... 000101 aaaaaa
sincossin(cos)sin(cos{ knnn
n
aazaz
}sin22)1cos(sin)1
1
,1
00
n
knjj
jaaa
0
if
sincossin(cos)sin(cos knnn aaza
1
,1
00 sin22)1cos(sin)1
n
knjj
jaaa
This shows that the zeros of F(z) whose modulus is greater than 1 lie in
n
n
knjj
j
knn
n a
aaa
aa
a
z
1
,1
00 ]sin22)1cos(sin
)1sincossin(cos)sin(cos[
. .
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
n
knjj
j
knn
n a
aaa
aa
a
z
1
,1
00 ]sin22)1cos(sin
)1sincossin(cos)sin(cos[
.
Since the zeros of P(z) are also the zeros of F(z), it follows that all the zeros of P(z) lie in
n
n
knjj
j
knn
n a
aaa
aa
a
z
1
,1
00 ]sin22)1cos(sin
)1sincossin(cos)sin(cos[
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If 1 knkn aa , then 1 knkn aa , 1 and we have, for 1z , by using Lemma 1,
1
1
1
211
1
)(......)()()(
kn
knkn
n
nn
n
nn
nn
n zaazaazaazzazF
0001
1
21
1
1
)1()(
......)()1()(
azazaa
zaazazaa kn
knkn
kn
kn
kn
knkn
11211
1
)(......
1
)()( kknknnnnnn
n
z
aa
z
aaaazaz
nnn
kknknkknkknkn
z
a
z
a
z
aa
z
aa
z
a
z
aa
0
1
0
101
12111
)1(
1
)(
....
1
)(
1
)1(
1
)(
knknnnnnn
n
aaaaaazaz 1211 ......
00
01211
1
......1
aa
aaaaaaa knknknknkn
sin)(cos){( 11 nnnnn
n
aaaazaz
cos)(......sin)(cos)( 12121 knknnnnn aaaaaa
( sin)(cos)(sin) 111 knknknknknkn aaaaaa
sin)(cos)(1 2121 knknknknkn aaaaa
})1(sin)(cos)(...... 000101 aaaaaa
sincossin(cos)sin(cos{ knnn
n
aazaz
}sin22)1cos(sin)1
1
,1
00
n
knjj
jaaa
0
if
sincossin(cos)sin(cos knnn aaza
1
,1
00 sin22)1cos(sin)1
n
knjj
jaaa
This shows that the zeros of F(z) whose modulus is greater than 1 lie in
n
n
knjj
j
knn
n a
aaa
aa
a
z
1
,1
00 ]sin22)1cos(sin
)1sincossin(cos)sin(cos[
.
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
n
n
knjj
j
knn
n a
aaa
aa
a
z
1
,1
00 ]sin22)1cos(sin
)1sincossin(cos)sin(cos[
That proves Theorem 3.
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Vol.11,No.3,2002, 246-253.
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5) A. Joyal, G. Labelle, Q.I. Rahman, On the location of zeros of polynomials, Canadian Math.
Bull.,10(1967) , 55-63.
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I,1996.
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Indian Acad. Sci. (Math. Sci.), 117(2007), 359-370.
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