Applied Mathematics, 2010, 3, 555-560
doi:10.4236/am.2010.16073 Published Online December 2010 (http://www.SciRP.org/journal/am)
Copyright © 2010 SciRes. AM
On The Eneström-Kakeya Theore m
Gulshan Singh1, Wali Mohammad Shah2
1Bharat hiar Univers ity, Coimbatore, India
2Department of Mat hem at ic s, Kashmir University, Srinagar, India
E-mail: gulshansingh1@rediffmail.com , wmshah@rediffmail.com
Received August 9, 2010; revised November 17, 2010; accepted November 20, 2010
Abstract
In this paper, we prove some generalizations of results concerning the Eneström-Kakeya theorem. The re-
sults obtained considerably improve the bounds by relaxing the hypothesis in some cases.
Keywords: Polynomial, Zeros, Eneström-Kakeya theorem
1. Introduction and Statement of Results
The following result due to Eneström and Kakeya [1] is
well known in the theory of distribution of the zeros of
polynomials.
Theorem A. If

=0
:= n
j
j
j
Pz az
is a polynomial of
degree n such that
12 10
,
nn n
aa aaao


then

Pz does not vanish in >1z
In the literature, [2-8], there exist extensions and ge-
neralizations of Eneström-Kakeya theorem. Joyal, La-
belle and Rahman [9] extended this theorem to a poly-
nomial whose coefficients are monotonic but not neces-
sarily non negative by proving the following result.
Theorem B. Let

=0
:= n
j
j
j
Pz az
be a polynomial
of degree n such that
12 10
,
nn n
aa aaa


then all the zeros of ()Pz lie in

00
1.
n
n
zaaa
a

Dewan and Bidkham [10] generalized Theorem B and
proved the following:
Theorem C. Let

=0
:= n
j
j
j
Pz az
be a polynomial
of degree n such that for some >0t and 0<,n
11
1110
,
nn
nn
ata tatattaa





then

Pz has all the zeros in the circle

00
21.
n
nn
n
a
t
zaaa
att







By using Schwarz's Lemma, Aziz and Mohammad [11]
generalized Eneström-Kakeya theorem in a different way
and proved the following:
Theorem D. Let

=0
:= n
j
j
j
Pz az
be a polynomial
of degree n with real positive coefficients. If 12
>0tt
can be found such that
121 1220
rr r
attatta

 , for =1,2, ,1rn
11
==0,
n
aa

then all the zeros of
Pz lie in 1
zt
Aziz and Zargar [12] also relaxed the hypothesis of
Eneström-Kakeya theorem in a different way and proved
the following result.
Theorem E. Let

=0
:= n
j
j
j
Pz az
be a polynomial
of degree n such that for some 1
K
,
12 10
>0,
nn n
Kaaaa a


then all the zeros of
Pz lie in
1.zK K

While studying Theorem E, a natural question arises
that what happens if we relax the hypothesis of Theorem
D in a similar way and only assume that
121 1220
rr r
attatta

 , for =2,3,,rn
In this paper, we study such a case and prove a more
general result from which many known results follow on
a fairly uniform procedure. Infact we prove:
Theorem 1. Let

=0
:= n
j
j
Pz z
be a polynomial
of degree n such that =
j
jj
aib
where j
a and j
b,
0,1 ,jn
are real numbers and if 12
>0tt can be
found such that for =2,3, ,rn
G. SINGH ET AL.
Copyright © 2010 SciRes. AM
556

121 1220,
rr r
attat ta



121 1220,
rr r
bttbttb


and for some 1
K
,
0,)( 121
nn attKa
0,)( 121
nn bttKb
then all the zeros of

Pz lie in

12
1zK tt 
,R
where




2
1221 11
1
1
=nn nnn
n
t
RKabttabt ab
t


 

001120 121120 12
1
11
11
nn
abattatt bttbtt
tt
 

2
00
1
.
n
t
ab
t

The following interesting result immediately follows
from Theorem 1, if we assume that all the coefficients of
the polynomial

Pz are real.
Corollary 1. Let

=0
:= n
j
j
j
Pz az
be a polynomial
of degree n with real coefficients. If 12
>0tt can be
found such that

121 1220,
rr r
attat ta

 for =2,3, ,rn
and for some 1K,
12 1
0,
nn
Katta
 
then all the zeros of
Pz lie in

12
1zK tt 
*,R
where

*2
122 10
11
11
11
=nn
nn
n
t
RKattataa
att

 

2
1120120
11
1.
nn
t
atta tta
tt
 
Remark 1. If we assume that all the coefficients of

Pz are real and positive, then for =1
K
, Corollary 1
satisfies the statement of Theorem D and a simple calcu-
lation shows that in this case also all the zeros of
Pz
lie in 1.zt
Next, if in the Theorem 1, we take 2=0t and assume
that coefficients to be real, we get the following:
Corollary 2. Let

=0
:= n
j
j
j
Pz az
be a polynomial
of degree n with real coefficients. If for some >0t and
1
K
,
12
12 10
>,
nn n
nn n
K
atatatata


 
then all the zeros of
Pz lie in

00
1.
nnn
n
aa
t
zK tKa
att

 


Remark 2. If we put 1=t in Corollary 2, we get the
result due to Aziz and Zargar [2] and for 1=t, 1=
K
,
Corollary 2 reduces to Theorem B.
We next prove the following more general result
which is of independent interest.
Theorem 2. Let

=0
:= n
j
j
Pz z
be a polynomial
of
degree n such that jjj iba =
where j
a and j
b,
0,1,2, ,jn
are real numbers. If 0>21 tt can be
found such that for 2,3, ,1rn
1211220,
rr r
attat ta

 
121 1220,
rr r
bttbttb

 
and for some real numbers u and v , 11, vu
12 1
0,
nn
uatta

12 1
0,
nn
vbttb

then all the zeros of )(zP lie in

12 1
1,
nn
n
ua ivb
ztt R

 


where




2
112211
1
1
1
=nn nnn
n
t
Ruavbttabtab
t


 

0011201 2112012
1
11
11
nn
abattatt bttbtt
tt
 

2
00
1
}.
n
t
ab
t

If in Theorem 2, we take

1
12
=n
n
a
uat t
and

1
12
=,
n
n
b
vbt t
so that 1,1,  vu we get the following:
Corolla ry 3 . Let

=0
:= n
j
j
Pz z
be a polynomial
of degree n such that jjj iba =
where j
a and j
b,
0,1,2, ,jn
are real numbers. If 0>21 tt can be
found such that
0)( 221121
 rrrattatta , for r = 2,3,,n
121122
() 0
rr r
attat ta

 , for r = n+1
0)( 221121
rrrbttbttb , for r = 2,3,,n
G. SINGH ET AL.
Copyright © 2010 SciRes. AM
557
121 122
() 0
rr r
bttbt tb

, for r = n +1,
then all the zeros of )(zP lie in

*
112 1
,
n
n
zttR

where
 

*2
121111
1
1
1
=nnn nn
n
t
Rabtabab
t



 
0 0112012112 012
1
11
11
nn
abattatt bttbtt
tt


2
00
11
1.
nn
t
ab
tt

In particular, if

1211220
rr r
attat ta

, for 1, 2,,rn

121 1220
rr r
attat ta


, for 1rn,

121 1220
rr r
bttbttb

, for 1, 2,,rn

121 1220
rr r
bttbttb

, for 1rn,
then

1120 120,atta tt

1120 120bttb tt
and we get in this case all the zeros of

Pz lie in

 

11221 1
1.
nnnn n
nn
zttabtab



Remark 3. A result of Shah and Liman [7, Theorem 1]
is a special case of Corollary 3, if we assume that all the
coefficients of
Pz are real.
The following result also follows from Theorem 2, if
we assume that 0=
2
t and 1.=
1
t]
Corollary 4. Let

=0
:= n
j
j
Pz z
be a polynomial
of degree n such that jjj iba =
where j
a and j
b,
0,1,2, ,jn
are real numbers. If for some 1uand
1v,
10
0
nn
ua aa
,
10
0
nn
vb bb
 ,
then all the zeros of
Pz lie in
1
nn nn
nn
ua ivbua vb
z



Many other known results and generalizations simi-
larly follows from Theorem 2 with suitable substitutions.
We leave this to the readers.
2. Proofs of the Theorems
Proof of Theorem 1. Consider the polynomial

21
=
f
ztztzPz



21
12112 1122
=nn n
nn n nnn
zttzttttz
  


  



2
2121 1201120 12012
ttttzttttztt
 
 (1)


21 1
1212112 1122
=1
nn nn
nn nnnnn
zK ttzKttz ttttz
 
 

 



2
2121 1201120 12012
ttttzttttztt
 
 (2)
 

21 1
1212112 1122
=1 ...
nn nn
nn nnnnn
zKttz Kattaz attattaz

 

 



2 1
21211 2011201 20121 21n
nn
attattazattattzattiKbttb z
 


2
1211 2221211 2011201 2012
(())
n
nn n
bttbttbzbttbttbzbttbtt zbtt


.
This gives
   
11
1212112 1122
1
nnn
nnnnnn
fzzzKttKattazattattaz



 
2 1
2121120112012012121
n
nn
attattazattattzattKbttbz
 
 
2
1211 2221211 20112012012
n
nn n
bttbtt bzbttbtt bzbttbttzbtt

 
.
G. SINGH ET AL.
Copyright © 2010 SciRes. AM
558



1
12121121121122
=1
n
nnn nnnnn
zzKttKattaKbttb attatta

 



121 12221211202121 1201
11
nn nn
bttbtt battattabttbtt b
zz

 



112012112 0120120121
11
nn
attattbttbttattbtt
zz
 
.
For 1
>zt, we have by using hypothesis
   

1
1212112 1
1
n
nnn nn
fzzzKt tKat taKbt tb


 

1211 221211221
1
nnnnn n
attat tabttbt tbt
 
 
 

212112021211201
1
1
n
attat tabttbt tbt
 
 


1120 121120 120120121
11
11
>0
nn
attattbttbttattbtt
tt
 
,
if
  
00
22
12121222 11
1 111
1 111
1>
nnnnnnnnn
ab
tt
z KttKattKbttatbtab
t ttt

 
 

22
1120 121120 1200
111
1
nnn
tt
attatt bttbttab
ttt
 
.
Therefore, for 1
zt,

>0,fz if
 





2
121221 100
11
11
11
1> nn nnnn
n
t
z KttKabttabtabab
tt
 
 


2
1120 121120 1200
11
1
nn
t
attattbttbtta b
tt
 
.
Hence all the zeros of

f
z whose modulus is greater than 1
t lie in the circle
 





2
121221 100
11
11
11
1nnnnnn
n
t
zKttKabttabtabab
tt
 
 


2
1120 121120 1200
11
1
nn
t
attattbttbtta b
tt
 
.
Since all the zeros whose modulus is less than 1
t already lie in this circle, we conclude that all the zeros of
f
z and
therefore

Pz lies in
 





2
121221 100
11
11
11
1nnnnnn
n
t
zKttKabttabtabab
tt
 
 


2
1120 121120 1200
11
1
nn
t
attattbttbtta b
tt
 
.
This completes the proof of the Theorem 1.
Proof of Theorem 2. Consider the polynomial
G. SINGH ET AL.
Copyright © 2010 SciRes. AM
559

21
=
f
ztztzPz



21
12112 1122
=...
nn n
nn n nnn
zttzttttz
  


 



2
2121 120112012012
ttttzttttztt
 




21 2
121121122212 1120
= ...
nn n
nn n nnn
zatt azattatt a zattatt az


 




1
11201 20121211211 22
nn
nnnn n
attattzattibtt bzbttbtt bz

 



2
2121 120112012012
bttbtt bzbttbttzbtt



21 1
12121121122
=1
nn nn
nn nnnnn
zuattzuatt azattattaz
 







211
212 112011201201212121
1nn
nnn
attattazatt att zattivbttzvbttbz

  



2
1211 2221211 2011201 2012
...
n
nn n
bttbtt b zbttbtt bzbttbttzbtt

 



211
12121121122
=nnnn
nnnn nnnnn
zuaivbttzuattazattattaz



 

2 1
21211201120120121 21
(()) n
nn
attat tazattat tzattivbt tbz
 



2
1211 22212112011201 2012
n
nn n
bttbtt b zbttbttbzbttbttzbtt


.
This gives
  
1 1
1212112 1122
1
nnn
nn
nnnnnn
n
ua ivb
fzzzttuatt azattattaz
 


 


 
2 1
21211 2011201 20121 21
n
nn
attattazattattzattvbttb z
 
 
2
121 1222121120112012012
n
nn n
bttbtt bzbttbtt bzbttbttzbtt

 .
 

1
1212112 1
=1
nnn
nnnnn
n
ua ivb
zzttuatt avbtt b


 






1211 22121122112012112012
11
nnnnn nn
attattabttb ttbattattbttbtt
zz
 


012 0121
1
n
attbtt z

For 1
>zt, we have
  

1
12121121
1
nnn nnn nn
n
ua ivb
fzzzttuat tavbt tb


 


 

 

1211 2212112211201 211201 2
11
11
nnnnn nn
atta ttabttbttbattattbttbtt
tt
 


012 0121
1
1
n
attbtt t

.
By using hypothesis, this gives
 





12
121221 100
11
11
1
1
nnn nnn nnnn
n
ua ivbt
fzzzttuavbttabtabab
tt


 


G. SINGH ET AL.
Copyright © 2010 SciRes. AM
560
 


2
112012112 01200
11
1>0,
nn
t
attattbttbtta b
tt
 
if






2
121221 100
11
11
11
1>
nn nn nnnn
nn
ua ivbt
zttuavbttabtabab
tt





 


2
1120 121120 1200
11
1
nn
t
attattbttbttab
tt
 
.
Hence all the zeros of

f
z whose modulus is greater than 1
t lie in the circle






2
121221100
11
11
11
1
nn nn nnnn
nn
ua ivbt
zttuavbttabt abab
tt


 


 


2
1120 121120 1200
11
1
nn
t
attattbttbttab
tt
 
.
Since all the zeros whose modulus is less than 1
t already lie in this circle, we conclude that all the zeros of
f
z and
therefore

Pz lies in

12 1
1,
nn
n
ua ivb
ztt R

 


where





 

2
11221 100112012112012
11
11 1
111
=nn nnnn n
n
t
Ruavbttabt ababattattbttbtt
tt t



2
00
1
n
t
ab
t

.
This proves Theorem 2 completely.
3. Acknowledgements
The authors are grateful to the referee for useful sugges-
tions.
4. References
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