The document presents several theorems that generalize the Enestrom-Kakeya theorem, which concerns the distribution of zeros of polynomials.
Theorem 1 generalizes previous results by Govil and Mc-tume by allowing the coefficients of a polynomial to satisfy certain monotonicity conditions with respect to both their real and imaginary parts, as well as introducing parameters λ and ρ.
Theorem 2 is obtained by applying Theorem 1 to the polynomial -iP(z), and concerns the monotonicity of the imaginary coefficients.
Theorem 3 introduces an additional parameter β to allow the arguments of the coefficients to be bounded. It provides disks within which all the zeros must lie under the given conditions.
1. 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 an1 ...... 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 an1 ...... a1 a0 .
Then P(z) has all its zeros in
a n a0 a0
z .
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 an1 ...... a1 a0 .
Then P(z) has all its zeros in
kan a0 a0
z k 1 .
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
n n1 ...... 1 0 0 ,
where n 0 , then P(z) has all its zeros in
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n
2
z 1 ( )( j ) .
n j 0
Gov il and Mc-tume [3] p roved the following generalisations of Theorems B and C:
n
Theorem D: 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
If for some k 1,
k n n1 ...... 1 0 ,
then P(z) has all its zeros in
n
k n 0 0 2 j
j 0
z k 1 .
n
n
Theorem E: 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
If for some k 1,
k n n1 ...... 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
Theorem F: 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
If for so me real nu mber 0,
n n1 ...... 1 0 ,
then P(z) has all its zeros in the disk
n
n 0 0 2 j
j 0 .
z
n 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
If for some real nu mber 0,
n n1 ...... 1 0 ,
then P(z) has all its zeros in the disk
n
n 0 0 2 j
j 0
z .
n 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, ank 0,
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3. I nternational Journal Of Computational Engineering Research (ijceronline.com) Vol. 3 Issue.1
n n1 ... nk 1 nk nk 1 ... 1 0 ,
and nk 1 nk , then P(z) has all its zeros in the disk
n
n ( 1) n k 1 n k 0 0 2 j
j 0
z ,
an an
and if nk nk 1 , then P(z) has all its zeros in the disk
n
n (1 ) n k 1 n k 0 0 2 j
j 0
z
an 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
Corollary 1: Let P( z ) a j z j be a polynomial of degree n. If fo r some real numbers , 0 ,
j 0
1 k n, ank 0,
an an1 ... ank 1 ank ank 1 ... a1 a0 ,
and ank 1 ank , then P(z) has all its zeros in the disk
a n ( 1)a nk 1 a nk a0 a0
z ,,
an an
and if ank ank 1 , then P(z) has all its zeros in the disk
a n (1 )a nk 1 a nk a0 a0
z
an 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
If for some real nu mbers , 0 , 1 k n, ank 0,
n n1 .... nk 1 nk nk 1 .. 1 0 ,
and nk 1 nk , then P(z) has all its zeros in the dis k
n
n ( 1) n k 1 n k 0 0 2 j
j 0
z ,
an an
and if nk nk 1 , then P(z) has all its zeros in the disk
n
n (1 ) n k 1 n k 0 0 2 j
j 0
z
an an
Remark 2: Taking 1 , Theorem 2 reduces to Theorem G.
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n
Theorem 3: Let P( z ) a j z j be a polynomial of degree n such that for some real numbers , 0 ,
j 0
1 k n, ank 0, and ,
an an1 ... ank 1 ank ank 1 ... a1 a0
and
arg a j , j 0,1,......, n.
2
If ank 1 ank (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)
n 1
a 0 (cos sin 1) 2 sin a
j 1, j n k
j ]
z
an an
If ank ank 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 )
n 1
a 0 (sin cos 1) 2 sin a
j 1, j n k
j ]
z
an 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 an1 ...... a1 a0 ,
then P(z) has all its zeros in the disk
n 1
[ a n (cos sin ) a0 (cos sin 1) 2 sin a j ]
j 1,
z .
an an
Remark 4: Taking (k 1)an , k 1, we get a result of Shah and Liman [8,Theorem 1].
2. Lemmas
For the proofs of the above results , we need the following results:
n
Lemma 1: . Let P(z)= a z
j 0
j
j
be a polynomial of degree n with complex coefficients such that
arg a j , j 0,1,...n, for some real . Then for some t >0,
2
ta j a j 1
t a j a j 1 cos t a j a j1 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 z 1, then the number of zeros of p(z)in z ,0 1, does not exceed
1 M
log (see[9],p 171).
1 p(0)
log
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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 n1 z n1 ...... a1 z a0 )
a n z n1 (a n a n1 ) z n ...... (a nk 1 a nk ) z nk 1 (a nk a nk 1 ) z nk
(ank 1 ank 2 ) z nk 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 nk 1 nk , then nk 1 nk , and we have
F ( z) ( n i n ) z n1 z n ( n n1 ) z n ...... ( nk 1 nk ) z nk 1
( nk nk 1 ) z nk ( 1)ank z nk ( nk 1 nk 2 ) z nk 1 ......
n
(1 0 ) z 0 i ( j j 1 ) z j i 0 .
j 1
For z 1,
F ( z ) an z n1 z n ( n n1 ) z n ( n1 n2 ) z n1 ...... ( nk 1 nk ) z nk 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
n 1 1
z a n z ( n n 1 ) ( n1 n2 ) ...... ( nk 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 ) ....
z z k 1
0 n
0
( 1 0 ) n 1 n i ( j j 1 ) n j i n
1 1
z z j 1 z z
z
n
a z
n n n1 n1 n2 ...... nk 1 nk
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 n1 n1 n2 ...... nk 1 nk nn k nk 1
n
1 nk nk 1 nk 2 ...... 1 0 0 2 j
j 0
n n
z a n z n ( 1) n k 1 nk 0 0 2 j
j 0
>0
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if
n
a n z n ( 1) nk 1 nk 0 0 2 j
j 0
This shows that the zeros of F(z) whose modulus is greater than 1 lie in
n
n ( 1) n k 1 n k 0 0 2 j
j 0
z .
an 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
n ( 1) n k 1 n k 0 0 2 j
j 0
z .
an 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
n ( 1) n k 1 n k 0 0 2 j
j 0
z .
an an
If nk nk 1 , then nk nk 1 , and we have
F ( z) ( n i n ) z n1 z n ( n n1 ) z n ...... ( nk 1 nk ) z nk 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 n1 z n ( n n1 ) z n ( n1 n2 ) z n1 ...... ( nk 1 nk ) z nk 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
n 1 1
z a n z ( n n1 ) ( n1 n2 ) ...... ( nk 1 nk ) 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
0 n
0
( 1 0 ) n 1 n i ( j j 1 ) n j i n
1 1
z z j 1 z z
z
n
a z
n n n1 n1 n2 ...... nk 1 nk
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 n1 n1 n2 ...... nk 1 nk nn k nk 1
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n
1 nk nk 1 nk 2 ...... 1 0 0 2 j
j 0
n n
z a n z n (1 ) n k 1 nk 0 0 2 j
j 0
>0
if
n
a n z n (1 ) nk 1 nk 0 0 2 j
j 0
This shows that the zeros of F(z) whose modulus is greater than 1 lie in
n
n (1 ) n k 1 n k 0 0 2 j
j 0
z .
an 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
n (1 ) n k 1 n k 0 0 2 j
j 0
z .
an 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
n (1 ) n k 1 n k 0 0 2 j
j 0
z .
an 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 n1 z n1 ...... a1 z a0 )
a n z n1 (a n a n1 ) z n ...... (a nk 1 a nk ) z nk 1 (a nk a nk 1 ) z nk
(ank 1 ank 2 ) z nk 1 ...... (a1 a0 ) z a0 .
If ank 1 ank , then ank 1 ank , 1 and we have, for z 1 , by using Lemma1,
F ( z ) an z n1 z n ( an an1 ) z n (an1 an2 ) z n1 ...... (ank 1 ank ) z nk 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
n 1 1
z a n z ( a n a n 1 ) (a n1 a n2 ) ...... (a nk 1 a nk ) 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
(a1 a0 ) n 1 0
1 a
z zn
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z
n
a z
n an an1 an1 an2 ...... ank 1 ank
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 an1 ) cos ( an an1 ) sin
( an1 an2 ) cos ( an1 an2 ) sin ...... ( ank 1 ank ) cos
( ank 1 ank ) sin ( ank ank 1 ) cos ( ank ank 1 ) sin
1 ank ( ank 1 ank 2 ) cos ( ank 1 ank 2 ) sin
...... ( a1 a0 ) cos ( a1 a0 ) sin a0 }
z
n
a z { a
n n (cos sin ) ank (cos sin cos sin
n 1
1) a0 (sin cos 1) 2 sin a }
j 1, j n k
j
0
if
a z a
n n (cos sin ) ank (cos sin cos sin
n 1
1) a0 (sin cos 1) 2 sin 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)
n 1
a0 (cos sin 1) 2 sin a
j 1, j n k
j ]
z ..
an 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)
n 1
a0 (cos sin 1) 2 sin a
j 1, j n k
j ]
z .
an 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)
n 1
a0 (cos sin 1) 2 sin a
j 1, j n k
j ]
z
an an
If ank ank 1 , then ank ank 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
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n 1 1
z a n z ( a n a n 1 ) (a n1 a n2 ) ...... (a nk 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
(a1 a0 ) n 1 n
1 a0
z z
z an z an an1 an1 an2 ...... ank 1 ank
n
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 an1 ) cos ( an an1 ) sin
( an1 an2 ) cos ( an1 an2 ) sin ...... ( ank 1 ank ) cos
( ank 1 ank ) sin ( ank ank 1 ) cos ( ank ank 1 ) sin
1 ank ( ank 1 ank 2 ) cos ( ank 1 ank 2 ) sin
...... ( a1 a0 ) cos ( a1 a0 ) sin a0 }
z
n
a z { a
n n (cos sin ) ank (cos sin cos sin
n 1
1 ) a0 (sin cos 1) 2 sin a }
j 1, j n k
j
0
if
a z a
n n (cos sin ) ank (cos sin cos sin
n 1
1 ) a0 (sin cos 1) 2 sin 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 )
n 1
a0 (sin cos 1) 2 sin a
j 1, j n k
j ]
z .
an 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 nk (cos sin cos sin 1 )
n 1
a0 (sin cos 1) 2 sin a
j 1, j n k
j ]
z
an an
That proves Theorem 3.
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References
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239-244.
[2] N.K.Gov il And Q.I.Reh man, On The Enestrom-Kakeya Theorem, Tohoku Math. J.,20(1968) , 126-136.
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Vo l.11,No.3,2002, 246-253.
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[6] A. Joyal, G. Labelle, Q.I. Rah man, On The Location Of Zeros Of Po lynomials, Canadian Math. Bull.,10(1967) , 55-
63.
[7] M. Marden , Geo metry Of Polynomials, Iind Ed.Math. Surveys, No. 3, A mer. Math. Soc. Providence,R. I,1996.
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(Math. Sci.), 117(2007), 359-370.
[9] E C. Titch marsh,The Theory Of Functions,2nd Ed ition,Oxfo rd University Press,London,1939.
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