International Journal of Engineering Inventions e-ISSN: 2278-7461, p-ISBN: 2319-6491 Volume 2, Issue 3 (February 2013) PP: 49-58
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 P ( z )
n
a j 0
j
z j be a polynomial of degree n whose coefficients satisfy
a n a n 1 ...... 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 result. Joyal et al [5] extended it to polynomials with general monotonic coefficients and proved the following result: Theorem B: Let P ( z )
n
a j 0
j
z j be a polynomial of degree n whose coefficients satisfy
a n a n 1 ...... a1 a 0 . Then P(z) has all its zeros in
z
a n a0 a0 an
.
Aziz and zargar [1] generalized the result of Joyal et al [6] as follows: 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 P(z) has all its zeros in
z k 1
kan a0 a0 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 polynomial of degree n with Re( a j ) j and Im( a j ) j , j 0
j=0,1,2,……,n, such that
n n 1 ...... 1 0 0 ,
where
n 0 , then P(z) has all its zeros in 2 n z 1 ( )( j ) . n j 0
Govil and Mc-tume [3] proved the following generalisations of Theorems B and C:
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Further Generalizations of Enestrom-Kakeya Theorem Theorem D: Let P ( z )
n
a
z j be a polynomial of degree n with Re( a j ) j and Im( a j ) j ,
j
j 0
k 1, k n n 1 ...... 1 0 ,
j=0,1,2,……,n. If for some
then P(z) has all its zeros in n
z k 1
Theorem E: Let P ( z )
k n 0 0 2 j j 0
n n
a
z j be a polynomial of degree n with Re( a j ) j and Im( a j ) j ,
j
j 0
.
k 1, k n n 1 ...... 1 0 ,
j=0,1,2,……,n. If for some
then P(z) has all its zeros in n
k n 0 0 2 j
j 0 . n M. H. Gulzar [4] proved the following generalizations of Theorems D and E:
z k 1
Theorem F: Let P ( z )
n
a j 0
j
z j be a polynomial of degree n with Re( a j ) j and Im( a j ) j ,
0 , 0 1, ...... 1 0 ,
j=0,1,2,……,n. If for some real numbers
n n 1
then P(z) has all its zeros in the disk n
z
n
n 2 0 ( 0 0 ) 2 j j 0
n
Theorem G: Let P ( z )
n
a j 0
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 for some real number 0 ,
n n 1 ...... 1 0 , then P(z) has all its zeros in the disk n
z n
n 2 0 ( 0 0 ) 2 j j 0
n
.
In this paper we give generalization of Theorems F and G. In fact, we prove the following: Theorem 1: Let P ( z )
n
a j 0
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 for some real numbers , 0 , 1 k n, a n k 0,0 1,
and
n k 1
n n 1 ... n k 1 n k n k 1 ... 1 0 , n k , then P(z) has all its zeros in the disk n
z and if
an
n k n k 1 ,
n ( 1) n k 1 n k 2 0 ( 0 0 2 j j 0
an
,
then P(z) has all its zeros in the disk
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Further Generalizations of Enestrom-Kakeya Theorem n
z
an
Remark 1: Taking If a j are real i.e.
n (1 ) n k 1 n k 2 0 ( 0 0 2 j j 0
an
1 , Theorem 1 reduces to Theorem F.
j 0 for all j, we get the following result:
Corollary 1: Let P ( z )
n
a j 0
j
z j be a polynomial of degree n. If for some real numbers , 0 ,
1 k n, a n k 0, 0 1,
a n a n 1 ... a n k 1 a n k a n k 1 ... a1 a 0 , and a n k 1 a n k , then P(z) has all its zeros in the disk
z
an
a n ( 1)a n k 1 a n k 2 a 0 (a 0 a 0 ) an
,,
and if a n k a n k 1 , then P(z) has all its zeros in the disk
z
an
a n (1 )a n k 1 a n k 2 a 0 (a 0 a 0 ) an
If we apply Theorem 1 to the polynomial –iP(z) , we easily get the following result: Theorem 2: Let P ( z )
n
a j 0
j
z j be a polynomial of degree n with Re( a j ) j and Im( a j ) j ,
, 0 , 1 k n, a n k 0, 0 1, n n 1 .... n k 1 n k n k 1 .. 1 0 , n k , then P(z) has all its zeros in the disk
j=0,1, 2,……,n. If for some real numbers
and
n k 1
n
z and if
an
n k n k 1 ,
n ( 1) n k 1 n k 2 0 ( 0 0 ) 2 j j 0
,
an then P(z) has all its zeros in the disk n
z
an
Remark 2: Taking
n (1 ) n k 1 n k 2 0 ( 0 0 ) 2 j j 0
an
1 , Theorem 2 reduces to Theorem G.
Theorem 3: Let P ( z )
n
a j 0
j
z j be a polynomial of degree n such that for some real numbers , 0 ,
1 k n, a n k 0, ,0 1, ,
an an1 ... ank 1 ank ank 1 ... a1 a0 and
arga j If
2
, j 0,1,......,n.
ank 1 ank (i.e. 1 ) , then P(z) has all its zeros in the disk
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Further Generalizations of Enestrom-Kakeya Theorem
[ a n (cos sin ) a n k (cos sin cos sin 1) z If
a 0 (sin cos 1) 2 a 0 2 sin
an
n 1
a
j 1, j n k
j
]
an
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 )
z
a 0 (sin cos 1) 2 a 0 2 sin
an
n 1
a
j 1, j n k
j
]
an
Remark 3: Taking
1 in Theorem 3 , we get the following result:
Corollary 2: Let P ( z )
n
a j 0
j
z j be a polynomial of degree n. If for some real numbers 0 , 0 1,
an an1 ...... a1 a0
,
then P(z) has all its zeros in the disk n 1
z
an
[ a n (cos sin ) a 0 (sin cos 1) 2 a 0 2 sin a j ] j 1,
.
an
Remark 4: Taking
(k 1)a n , k 1 in Cor.2, we get the following result:
Corollary 3: Let P ( z )
n
a j 0
j
z j be a polynomial of degree n. If for some real numbers 0 , 0 1,
k an an1 ...... a1 a0 , then P(z) has all its zeros in the disk n 1
z Taking
an
[k a n (cos sin ) a 0 (sin cos 1) 2 a 0 2 sin a j ] j 1,
an
(k 1)a n , k 1 , 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: n
Lemma 1: .
Let P(z)=
a z j 0
arg a j
ta j a j 1
j
j
be a polynomial of degree n with complex coefficients such that
, j 0,1,...n, for some real . Then for some t>0, 2 t a j a j 1 cos t a j a j1 sin .
The proof of lemma 1 follows from a lemma due to Govil and Rahman [2].
p( z) M in z 1, then the number of zeros of p(z)in 1 M (see[8],p171). z ,0 1, does not exceed log 1 p ( 0 ) log
Lemma 2.If p(z) is regular ,p(0) 0 and
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Further Generalizations of Enestrom-Kakeya Theorem III.
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 n k 1 ) z nk (a n k 1 a n k 2 ) z n k 1 ...... (a1 a 0 ) z a 0 ( 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 n
( 1) 0 z 0 i ( j j 1 ) z j i 0 j 1
If
n k 1 n k ,
then
n k 1 n k , and we have
F ( z ) ( n i n ) z n 1 z n ( n n 1 ) z n ...... ( n k 1 n k ) z n k 1
( n k n k 1 ) z n k ( 1)an k z n k ( n k 1 n k 2 ) z n k 1 ...... n
( 1 0 ) z ( 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 ( 1) 0 z 0 i ( j j 1 ) z j i 0 j 1
1 1 n z an 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 ( n k 1 n k 2 ) .... k 1 z z n 1 1 ( 1 0 ) n 1 ( 1) n 01 n0 i ( j j 1 ) n j i n0 z z z z z j 1 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
1 0 0 j j 1 0
z
n
a z n
j 1
n
n1 n1 n2 ...... nk 1 nk nn k nk 1
1 n k n k 1 n k 2 ...... 1 0 (1 ) 0 n
0 2 j
j 0
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Further Generalizations of Enestrom-Kakeya Theorem
z
a
n
n
z [{ n ( 1) n k 1 n k ( 0 0 ) 2 0
n
2 j
j 0
>0 if n
a n z n ( 1) n k 1 n k ( 0 0 ) 2 0 2 j j 0
This shows that the zeros of F(z) whose modulus is greater than 1 lie in n
z
an
n ( 1) n k 1 n k 2 0 ( 0 0 ) 2 j j 0
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
z
an
n ( 1) n k 1 n k 2 0 ( 0 0 ) 2 j j 0
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 n
z If
an
n ( 1) n k 1 n k 2 0 ( 0 0 ) 2 j j 0
an
.
n k n k 1 , then n k n k 1 , and we have F ( z ) ( n i n ) z n 1 z n ( n n 1 ) z n ...... ( n k 1 n k ) z n k 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 ( 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 ( 1) 0 z 0 i ( j j 1 ) z j i 0 j 1
1 1 n z an 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 n 1 1 ( 1 0 ) n 1 ( 1) n 01 n0 i ( j j 1 ) n j i n0 z z z z z j 1 z
n
a z n
n n1 n1 n2 ...... nk 1 nk
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Further Generalizations of Enestrom-Kakeya Theorem
n k n k 1 1 n k n k 1 n k 2 ...... 1 0 n
1 0 0 j j 1 0
z
n
a z n
j 1
n
n1 n1 n2 ...... nk 1 nk nn k nk 1
1 n k n k 1 n k 2 ...... 1 0 (1 ) 0 n
0 2 j
j 0
a n z { n (1 ) n k 1 n k ( 0 0 ) n z 2 2 j } 0 j 0 n
>0 if n
a n z n (1 ) n k 1 n k ( 0 0 ) 2 0 2 j j 0
This shows that the zeros of F(z) whose modulus is greater than 1 lie in n
z
an
n (1 ) n k 1 n k 2 0 ( 0 0 ) 2 j j 0
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
z
an
n (1 ) n k 1 n k 2 0 ( 0 0 ) 2 j j 0
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 n
z
an
n (1 ) n k 1 n k 2 0 ( 0 0 ) 2 j j 0
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 n k 1 ) z nk (a n k 1 a n k 2 ) z n k 1 ...... (a1 a 0 ) z a 0 . 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 a 0 ) z ( 1)a 0 z a 0
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Further Generalizations of Enestrom-Kakeya Theorem
1 1 n z an z ( an an1 ) (an1 an2 ) ...... (ank 1 ank ) 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 a a 1 (a1 a 0 ) n 1 ( 1) n01 0n z z z n z an z 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 a 0 1 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 (1 ) a0 a0 }
z
n
a z { a n
n
(cos sin ) ank (cos sin cos sin
1) a 0 (sin cos 1) 2 a 0 2 sin
n 1
a
j 1, j n k
j
}
0 if
a z a n
n
(cos sin ) ank (cos sin cos sin
1) a 0 (sin cos 1) 2 a 0 2 sin
n 1
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)
z
an
a 0 (sin cos 1) 2 a 0 2 sin
n 1
a
j 1, j n k
j
] ..
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)
z
an
a 0 (sin cos 1) 2 a 0 2 sin
n 1
a
j 1, j n k
j
] .
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
[ a n (cos sin ) a n k (cos sin cos sin 1)
z
an
a 0 (sin cos 1) 2 a 0 2 sin
n 1
a
j 1, j n k
j
]
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Further Generalizations of Enestrom-Kakeya Theorem If
ank ank 1 , then ank ank 1 , 1 and we have, for z 1, by using Lemma 1,
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 1 (a n k 1 a n k 2 ) z n k 1 ...... (a1 a 0 ) z ( 1)a 0 z a 0
1 1 n z an z ( an an1 ) (an1 an2 ) ...... (ank 1 ank ) 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 a a 1 (a1 a 0 ) n 1 ( 1) n01 0n z z z n z an z 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 a 0
1 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 (1 ) a0 a0 }
z
n
a z { a n
n
(cos sin ) ank (cos sin cos sin
1 ) a 0 (sin cos 1) 2 a 0 2 sin
n 1
a
j 1, j n k
j
}
0 if
a z a n
n
(cos sin ) ank (cos sin cos sin
1 ) a 0 (sin cos 1) 2 a 0 2 sin
n 1
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 )
z
an
a 0 (sin cos 1) 2 a 0 2 sin
n 1
a
j 1, j n k
j
]
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 n k (cos sin cos sin 1 )
z
an
a 0 (sin cos 1) 2 a 0 2 sin
n 1
a
j 1, j n k
j
]
an
That proves Theorem 3.
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Further Generalizations of Enestrom-Kakeya Theorem REFERENCES 1) 2) 3) 4) 5) 6) 7) 8)
A.Aziz and B.A.Zargar, Some extensions of the Enestrom-Kakeya Theorem , Glasnik Mathematiki, 31(1996) , 239-244. N.K.Govil and Q.I.Rehman, On the Enestrom-Kakeya Theorem, Tohoku Math. J.,20(1968) , 126-136. N.K.Govil and G.N.Mc-tume, Some extensions of the Enestrom- Kakeya Theorem , Int.J.Appl.Math. Vol.11,No.3,2002, 246-253. M. H. Gulzar, On the Zeros of polynomials, International Journal of Modern Engineering Research, Vol.2,Issue 6, Nov-Dec. 2012, 4318- 4322. A. Joyal, G. Labelle, Q.I. Rahman, On the location of zeros of polynomials, Canadian Math. Bull.,10(1967) , 55-63. M. Marden , Geometry of Polynomials, IInd Ed.Math. Surveys, No. 3, Amer. Math. Soc. Providence,R. I,1996. W. M. Shah and A.Liman, On Enestrom-Kakeya Theorem and related Analytic Functions, Proc. Indian Acad. Sci. (Math. Sci.), 117(2007), 359-370. E C. Titchmarsh,The Theory of Functions,2 nd edition,Oxford University Press,London,1939.
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