Applied Mathematics E-Notes, 13(2013), 155-159 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
On Annulus Containing All The Zeros Of A
Polynomial
Nisar Ahmad Rathery, Suhail Gulzar Mattooz
Received 21 May 2013
Abstract
In this paper, we obtain an annulus containing all the zeros of the polynomial
involving binomial coe¢ cients and generalized Fibonacci numbers. Our result
generalize some of the recently obtained results in this direction.
1
Introduction
Gauss and Cauchy were the earliest contributors in the theory of the location of zeros
of a polynomial, since then this subject has been studied by many people (for example,
see [3, 4]). There is always a need for better and better results in this subject because
of its application in many areas, including signal processing, communication theory
and control theory.
A classical result due to Cauchy (see [3, p.122]) on the distribution of zeros of a
polynomial may be stated as follows:
THEOREM A. If P (z) = z n + an 1 z n 1 + an 2 z n 2 + + a0 is a polynomial with
complex coe¢ cients, then all zeros of P (z) lie in the disk jzj r where r is the unique
positive root of the real-coe¢ cient polynomial
Q(x) = xn
jan
1 jx
n 1
jan
2 jx
n 2
ja1 jx
ja0 j:
Recently Díaz-Barrero [1] improved this estimate by identifying an annulus containing
all the zeros of a polynomial, where the inner and outer radii are expressed in terms
of binomial coe¢ cients and Fibonacci numbers. In fact he has proved the following
result.
Pn
THEOREM B. Let P (z) = j=0 aj z j be a non-constant complex polynomial. Then
all its zeros lie in the annulus C = fz 2 C : r1 jzj r2 g where
3
r1 = min
21 k n
(
2n Fk nk
F4n
a0
ak
) k1
2
and r2 = max
31 k n
(
F4n
2n Fk nk
an k
an
) k1
Mathematics Subject Classi…cations: 30C10, 30C15.
of Mathematics, University of Kashmir, Hazratbal, Srinagar 190003, India
z Department of Mathematics, University of Kashmir, Hazratbal, Srinagar 190003, India
y Department
155
:
156
On Annulus Containing all the Zeros of a Polynomial
Here Fj are Fibonacci’s numbers, that is, F0 = 0; F1 = 1 and for j
Fj 1 + Fj 2 :
2; Fj =
More recently, Bidkham et. al [2] considered t-Fibonacci numbers, namely Ft;n =
tFt;n 1 + Ft;n 2 for n
2 with initial condition Ft;0 = 0; Ft;1 = 1 where t is any
positive real number and obtained the following generalization of Theorem B.
Pn
THEOREM C. Let P (z) =
degree n and
k
j=0
=
aj z j be a non-constant complex polynomial of
(t3 + 2t)k (t2 + 1)n Ft;k
(t2 + 1)k Ft;4n
n
k
for any real positive number t: Then all the zeros of P (z) lie in the annulus R = fz 2
C : s1 jzj s2 g where
s1 = min
a0
ak
k
1 k n
1
k
1
and s2 = max
1 k n
k
1
k
an k
an
:
In this paper, we determine in the complex plane an annulus containing all the zeros
of a polynomial involving binomial coe¢ cients and generalized Fibonacci numbers (see
[5]) de…ned recursively by
(a;b;c)
(a;b;c)
F0
= 0; F1
=1
and
Fn(a;b;c)
=
(
(a;b;c)
(a;b;c)
aFn 1 + cFn 2 ; if n is even
(n
(a;b;c)
(a;b;c)
bFn 1 + cFn 2 ; if n is odd
2)
(1)
where a; b; c are any three positive real numbers. Our result includes Theorem B and
Theorem C as special cases. More precisely, we prove the following result.
2
Main Result
The main result is
Pn
THEOREM 1. Let P (z) = j=0 aj z j be a non-constant complex polynomial of
degree n: Then all its zeros lie in the annulus C = fz 2 C : r1 jzj r2 g where
uv + 2w
min
r1 =
uvw + w2 1 k n
abc + c2
r2 =
max
ab + 2c 1 k n
(
(
(uvw + w2 )n u
(k)
k
(u;v;w) n
k
(uv)b 2 c Fk
(u;v;w)
F4n
(a;b;c)
F4n
(abc + c2 )n a
(a;b;c) n
(k) (ab)b k
2 cF
k
k
a; b; c; u; v; w are any positive real numbers, (k) := k
in (1).
) k1
a0
ak
an k
an
) k1
(a;b;c)
2b k2 c and Fm
;
;
is de…ned as
N. A. Rather and S. Gulzar
157
REMARK 1. By taking a; b; c and u; v; w suitably in Theorem 1, we shall obtain
Theorems B and C. For example, if we take a = b = u = v = t and c = w = 1; in
Theorem 1 we obtain Theorem C.
EXAMPLE 1. We consider the polynomial P (z) = z 3 + 0:1z 2 + 0:3z + 0:7; which is
the only example considered by Díaz-Barrero [1] and by using Theorem B, the annulus
containing all the zeros of P (z) comes out to be 0:58 < jzj < 1:23. We improved the
upper bound of this annulus by taking a = 1=2; b = 1 and c = 3=8 in Theorem 1
and obtained the disk, jzj < 1:185; which contains all the zeros of polynomial P (z):
Similarly, we can improve the lower bound by choosing u; v; w suitably.
3
Lemma
To prove Theorem 1, we need the following lemma.
(a;b;c)
LEMMA 1. If Fk
n
X
is de…ned as in (1), then
(ab + c)n
k
(ab + 2c)k a
(k)
k
(ab)b 2 c cn
k
Fk
k=1
(a;b;c)
PROOF. For Fk
; we have [5]
(a;b;c)
Fk
=
= F4n
(2)
2b k2 c:
where (k) = k
where
n
k
p
ab+
(ab)2 +4abc
;
2
n
X
n
(abc)n
k
k
ab
=
(ab)2 + abc
=
p
n k
a1
(k)
k
k
k
(ab)b 2 c
(ab)2 +4abc
2
!
and (k) = k
k
(ab)3 + 2(ab)2 c a
(k)
2b k2 c: We consider that
k
(a;b;c)
(ab)b 2 c Fk
k=1
!n k
!k
n
2
3
k
X
X
X
n
n k
n k
j 2 j
j 3 j
=
( 1)
( )
a
k
j=0
j=0
k=1
( n
!
!k )
n
k
2
3
X n
X
X
a n
j 3 j
j 3 j
=
( 1)n k
k
j=0
j=0
k=1
( n
!n k
!k )
2
3
n
X n
X
X
a
n k
1+j 2 j
j 3 j
( 1)
k
j=0
j=0
k=1
0
0
1n
3
2
3
2
a n @X j 3 j X j 3 j A
a n @X j 3 j X 1+j
=
j=0
n
=a
(
3 n
)
j=0
n
(
3 n
)
j=0
(a;b;c)
= (ab)2n F4n
:
j=0
k
!
1n
2 jA
158
On Annulus Containing all the Zeros of a Polynomial
Equivalently, we have
n
X
n
(ab + c)n
k
k
(ab + 2c)k a
(k)
k
(ab)b 2 c cn
k
(a;b;c)
Fk
(a;b;c)
= F4n
:
k=1
4
Proof of Theorem
We …rst show that all the zeros of P (z) lie in
(
(a;b;c)
(ab + c)k ck F4n
jzj r2 = max
k
(a;b;c)
1 k n
(ab + c)n (ab + 2c)k a (k) (ab)b 2 c cn Fk
n
k
an k
an
) k1
(3)
where a; b; c are any three positive real numbers. From (3), it follows that
an k
an
or
r2k
(ab + c)n (ab + 2c)k a
(k)
(a;b;c) n
k
k
(ab)b 2 c cn Fk
;
(a;b;c)
(ab + c)k ck F4n
n
X
an k 1
an r2k
k=1
Now, for jzj > r2 ; we have
n
X
(ab + c)n (ab + 2c)k a
(a;b;c) n
k
k
(ab)b 2 c cn Fk
(a;b;c)
(ab + c)k ck F4n
k=1
jP (z)j = jan z n + an
(
jan jjzjn
> jan jjzjn
(k)
k = 1; 2; 3; :::; n
1
(
1
1z
n 1
n
X
k=1
n
X
k=1
+
an k
an
an k
an
:
(4)
+ a1 z + a0 j
)
1
jzjk
)
1
:
r2k
Using (2) and (4), we have for jzj > r2 ; jP (z)j > 0: Consequently all the zeros of P (z)
lie in jzj r2 and this proves the second part of theorem.
To prove the …rst part of the theorem, we will use second part. If a0 = 0; then
r1 = 0 and there is nothing to prove. Let a0 6= 0; consider the polynomial
Q(z) = z n P (1=z) = a0 + a1 z n
1
+
+ an
1z
+ an :
By second part of the theorem for any three positive real numbers u; v; w, if Q(z) = 0;
then
)1=k
(
(u;v;w)
(uv + w)k wk F4n
ak
jzj
max
k
(au;v;w) n
1 k n
a0
(uv + w)n (uv + 2w)k u (k) (uv)b 2 c wn Fk
k
1
=
(
)1=k
(u;v;w)
(uv + w)k wk F4n
a0
min
k
ak
(u;v;w) n
1 k n
(uv + w)n (uv + 2w)k a (k) (ab)b 2 c wn Fk
k
1
= :
r1
N. A. Rather and S. Gulzar
159
Now replacing z by 1=z and observing that all the zeros of P (z) lie in
jzj
r1 = min
1 k n
(
(u;v;w)
(uv + w)k wk F4n
(uv + w)n (uv + 2w)k u
(u;v;w) n
(k) (uv)b k
2 c wn F
k
k
a0
ak
) k1
:
This completes the proof of theorem 1.
Acknowledgement. The authors are highly grateful to the referee for his valuable
suggestions and comments. The second author is supported by Council of Scienti…c
and Industrial Research, New Delhi, under grant F.No. 09/251(0047)/2012-EMR-I.
References
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273(2002), 349–352.
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Lett., 24(2011), 122–125.
[3] M. Marden, Geometry of Polynomials, Math. Surveys No. 3, Amer. Math. Soc.
Providence R. I. 1966.
[4] G. V. Milovanovic, D. S. Mitrinovic and Th. M. Rassias, Topics in Polynomials:
Extremal Properties, Inequalities, Zeros, World scienti…c Publishing Co., Singapore,
(1994).
[5] O. Yayenie, A note on generalized Fibonacci sequences, Appl. Math. Comput.,
208(2009), 180–185.