Applied Mathematics E-Notes, 14(2014), 209-215 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
On Annular Bound For The Zeros Of A Polynomial
Younseok Chooy
Received 23 August 2014
Abstract
In this paper we present some results on the annular bound for the zeros of a
polynomial based on the identities related to the generalized Fibonacci sequence
with arbitrary initial condition. Several recently reported results in the same
direction are special cases of our results.
1
Introduction
Several attempts have been made to obtain an explicit annular bound containing all
the zeros of a polynomial based on the identities related to the Fibonacci sequence
fFn g1
1) or generalized Fibonacci
n=0 (F0 = 0; F1 = 1; and Fn+1 = Fn + Fn 1 ; n
(a;b;c;d) 1
gn=0 de…ned by
sequence fFn
(
(a;b;c;d)
(a;b;c;d)
aFn 1
+ cFn 2
; if n is even,
Fn(a;b;c;d) =
(n 2)
(a;b;c;d)
(a;b;c;d)
bFn 1
+ dFn 2
; if n is odd,
(a;b;c;d)
(a;b;c;d)
where F0
= 0, F1
Based on the identity
= 1; and a; b; c; d > 0.
n
X
2n
k k
3 Fk C(n; k) = F4n ;
(1)
k=1
where C(n; k) =
n!
(n k)!k! ,
Díaz-Barrero [1] proved the following theorem:
THEOREM A. A complex polynomial P (z) =
in the annulus C = fz : r1 jzj r2 g, where
3
min
r1 =
21 k n
2n Fk C(n; k) jd0 j
F4n
jdk j
1
k
Pn
k=0
dk z k (dk 6= 0) has all its zeros
2
max
and r2 =
31 k n
F4n
jdn k j
n
2 Fk C(n; k) jdn j
1
k
:
Later Bidkham and Shashahani [3] derived the identity
n
X
(a2 + 1)n
k
(a;a;1;1)
(a3 + 2a)k Fk
(a;a;1;1)
C(n; k) = F4n
;
(2)
k=1
Mathematics Subject Classi…cations: 30C10, 30C15.
of Electronic and Electrical Engineering, Hongik University, Sejong 339–701, Korea
y Department
209
210
On Annular Bound for the Zeros of a Polynomial
and extended Theorem A as follows:
THEOREM B. All the zeros of a complex polynomial P (z) =
are contained in the annulus C = fz : r1 jzj r2 g, where
r1 = min
1 k n
and
r2 = max
1 k n
(
(
(a2 + 1)n
k
(a;a;1;1)
(a3 + 2a)k Fk
(a;a;1;1)
F4n
Pn
k=0
C(n; k) jd0 j
jdk j
) k1
(a;a;1;1)
jdn k j
k (a3 + 2a)k F (a;a;1;1) C(n; k) jdn j
F4n
(a2 + 1)n
k
dk z k (dk 6= 0)
;
) k1
:
Recently Rather and Mattoo [5] proved the identity
n
X
(abc + c2 )n
k
(ab + 2c)k a
(k)
k
(a;b;c;c)
(ab)[ 2 ] Fk
(a;b;c;c)
C(n; k) = F4n
;
(3)
k=1
2[ k2 ], and then extended Theorem A and Theorem B as follows:
where (k) = k
THEOREM C. All the zeros of a complex polynomial P (z) =
lie in the annulus C = fz : r1 jzj r2 g, where
r1 = min
1 k n
(
(abc + c2 )n
k
(ab + 2c)k a
(k)
k
(a;b;c;c)
(ab)[ 2 ] Fk
(a;b;c;c)
F4n
Pn
k=0
dk z k (dk 6= 0)
C(n; k) jd0 j
jdk j
) k1
;
and
r2 = max
1 k n
(
(a;b;c;c)
F4n
(abc + c2 )n
jdn k j
k
(a;b;c;c)
k (ab + 2c)k a (k) (ab)[ 2 ] F
C(n; k) jdn j
k
) k1
:
In this paper we present further results in the same direction. Two theorems on the
annular bound for the zeros of a polynomial are given respectively based on the iden(a;b;c;d) 1
(a;b;c;c) 1
gn=0 and fFn
gn=0
tities related to the generalized Fibonacci sequences fFn
with arbitrary initial conditions. The second one includes Theorem C as a special case.
2
Main Results
Before presenting our main results, we state some preliminary results.
LEMMA 1. Let r and s (r 6= s) be nonzero roots of x2
following three statements are equivalent:
(i) Bj = b0
r j+1 sj+1
r s
+ (b1
ab0 )
r j sj
r s
for j
0:
ax
b = 0. Then the
Y. Choo
211
(ii) Bj = aBj
+ bBj
1
(iii) b0 xj+1 + (b1
2;
j
2 with B0 = b0 , B1 = b1 .
ab0 )xj = xBj + bBj
1
for j
1 and x = r; s where B0 = b0 .
PROOF. (i))(ii) follows from the fact that B0 = b0 , B1 = b1 and, for j
rj+2
=
(rj+2
=
0:
sj+2
a(rj+1
sj+2
sj+1 )
(r + s)(rj+1
b(rj
0
sj )
sj+1 ) + rs(rj
sj )
To prove (ii))(iii), we proceed by induction as in [4]. If (ii) holds, (iii) is true for j = 1
since
b0 x2 + (b1
ab0 )x
= b0 (ax + b) + (b1
ab0 )x
= xb1 + bb0
= xB1 + bB0 :
If (iii) holds for j = m, then, for j = m + 1
b0 xm+2 + (b1
ab0 )xm+1
= x[b0 xm+1 + (b1
= x2 Bj + xbBj
=
ab0 )xm ]
1
(ax + b)Bj + xbBj
= x(aBj + bBj
1
1 + bBj
= xBj+1 + bBj ;
hence (iii) follows. Now suppose (iii) holds. Then
b0 rj+1 + (b1
ab0 )rj = rBj + bBj
1;
b0 sj+1 + (b1
ab0 )sj = sBj + bBj
1;
and so
B j = b0
rj+1
r
sj+1
s
+ (b1
ab0 )
rj
r
sj
s
for j
1:
Since B0 = b0 , (i) also holds, and the proof is completed.
REMARK 1. Although the closed-form expression for Bj in (i) satisfying the recurrence relation (ii) can also be computed by using the generating function, Lemma
1 provides another simple way to obtain the formula for Bj .
Lemma 2 and Lemma 3 below are slight generalizations of Theorem 1 and Theorem
2 in [2]. The proof of Lemma 3 is similar to that of [2, Theorem 2] and is omitted.
212
On Annular Bound for the Zeros of a Polynomial
LEMMA 2. Let r and s (r 6= s) be nonzero roots of x2 ax
1
n
n
two sequences fAn g1
n=0 and fBn gn=0 by An = n (cr + ds ), where
r n sn
Bn = r s . Then for j 2 and l 0
n
X
C(n; k)(bBj
1)
n k
(Bj )k
jn+l
b = 0. De…ne
2 R, and
n ; c; d
Ak+l = Ajn+l :
(4)
k+l
k=1
PROOF. Using the equivalence (i) and (iii) in Lemma 1 for b0 = 0 and b1 = 1, we
have
Ajn+l
= crjn+l + dsjn+l
jn+l
= crl (rj )n + dsl (sj )n
= crl (bBj 1 + rBj )n + dsl (bBj 1 + sBj )n
n
X
=
C(n; k)(bBj 1 )n k (Bj )k (crk+l + dsk+l )
=
k=1
n
X
C(n; k)(bBj
k=1
Ak+l
n k
(Bj )k
:
1)
k+l
LEMMA 3. With
Pnthe same notation as in Lemma 2, all the zeros of a complex
polynomial P (z) = k=0 dk z k (dk 6= 0) are contained in the annulus C = fz : r1
jzj r2 g where
)1
(
C(n; k)(bBj 1 )n k (Bj )k jn+l Ak+l jd0 j k
r1 = min
;
1 k n
jdk j
k+l Ajn+l
and
r2 = max
1 k n
k+l Ajn+l
C(n; k)(bBj 1 )n k (Bj )k jn+l Ak+l
jdn k j
jdn j
1
k
:
(a;b;c;d)
g1
Now consider the generalized Fibonacci sequence fFn
n=0 de…ned in Section
(a;b;c;d)
(a;b;c;d)
1 with initial condition F0
= f0 , F1
= f1 . It is easily seen that
(a;b;c;d)
2
(a;b;c;d)
4
Fn(a;b;c;d) = (ab + c + d)Fn
(a;b;c;d)
Let Gn
(a;b;c;d)
= F2n
(a;b;c;d)
; n
0: Then G0
G(a;b;c;d)
= (ab + c +
n
cdFn
for n
(a;b;c;d)
= f0 , G1
(a;b;c;d)
d)Gn 1
(a;b;c;d)
cdGn 2
4:
= af1 + cf0 and
for n
2:
Hence, from Lemma 1, we have
G(a;b;c;d)
n
n+1
(a;b;c;d)
n+1
= G0
(a;b;c;d)
+fG1
= c1
n
+ d1
n;
(a;b;c;d)
(ab + c + d)G0
n
g
n
Y. Choo
where
213
and
are roots of the equation x2
c1 =
(
ab
d)f0 + af1
(ab + c + d)x + cd = 0; and
; d1 =
(
(a;b;c;d)
^ (a;b;c;d)
On the other hand, let G
= F2n+1 for n
n
1 (a;b;c;d)
^ (a;b;c;d)
G
= (Gn+1
n
a
where
ab
d)f0 + af1
:
0: Then
cG(a;b;c;d)
) = c2
n
n
+ d2
n;
c)
d1 (
c)
and d2 =
:
a
a
Hence, from Lemma 2 and Lemma 3, we obtain the following theorem.
Pn
THEOREM 1. Consider a complex polynomial P (z) = k=0 dk z k (dk 6= 0), and
n
n
, n 0, where and are roots of x2 (ab + c + d)x + cd = 0. Then,
let Bn =
for j 2 and l 0, all the zeros of P (z) lie in the annulus C = fz : r1 jzj r2 g or
C^ = fz : r^1 jzj r^2 g where
8
91
< C(n; k)( cdBj 1 )n k (Bj )k F (a;b;c;d) jd j = k
2(k+l)
0
r1 = min
;
(a;b;c;d)
1 k n:
jdk j ;
F
c2 =
c1 (
2(jn+l)
8
<
9 k1
jdn k j =
r2 = max
;
n k (B )k F (a;b;c;d) jdn j ;
1 k n : C(n; k)( cdB
j 1)
j
2(k+l)
8
91
< C(n; k)( cdBj 1 )n k (Bj )k F (a;b;c;d) jd j = k
2(k+l)+1
0
r^1 = min
;
(a;b;c;d)
1 k n:
jdk j ;
F
(a;b;c;d)
F2(jn+l)
2(jn+l)+1
and
r^2 = max
8
<
(a;b;c;d)
F2(jn+l)+1
1 k n : C(n; k)(
cdBj
n k (B )k F (a;b;c;d)
1)
j
2(k+l)+1
9 k1
jdn k j =
:
jdn j ;
Next we consider the case where c = d. To this end we …rst p
…nd the formulae
for
p
(a;b;c;c)
(a;b;c;c)
(a;b;c;c)
^ (a;b;c;c)
= F2n+1
in terms of r =
= F2n
and G
, s =
. It is
Gn
n
p
easily seen that r and s are roots of the equation x2
ab c = 0: Now we have
G(a;b;c;c)
n
(a;b;c;c)
n+1
n+1
= G0
(a;b;c;c)
+fG1
=
(a;b;c;c)
(ab + 2c)G0
1 h (a;b;c;c) r2n+2
p
G0
r
ab
(a;b;c;c)
+fG1
n
n
g
s2n+2
s
(a;b;c;c)
(ab + 2c)G0
g
r2n
r
s2n i
:
s
214
On Annular Bound for the Zeros of a Polynomial
On the other hand
^ (a;b;c;c)
G
n
1 (a;b;c;c)
(G
cG(a;b;c;c)
)
n
a n+1
1 h (a;b;c;c) n (r2 c)r2n+3
G
a 0
r(r2 s2 )
=
=
(s2 c)s2n+3 o
s(r2 s2 )
n (r2 c)r2n+1
(s2 c)s2n+1 oi
(a;b;c;c)
(ab + 2c)G0
g
r(r2 s2 )
s(r2 s2 )
2n+3
2n+3
r
s
r s
r2n+1 r2n+1 i
(a;b;c;c)
(ab + 2c)G0
g
:
r s
(a;b;c;c)
+fG1
1 h (a;b;c;c)
G
a 0
=
(a;b;c;c)
+fG1
Consequently Fn (a; b; c; c) can be expressed as
Fn(a;b;c;c)
=
h
n+2
(a;b;c;c) r
G0
(n)
r
sn+2
s
n
r
sn i
(a;b;c;c)
(ab + 2c)G0
g
r s
1
p
(n)
a
( ab)1
(a;b;c;c)
+fG1
=
where
a
p
(n) (
1
ab)1
(n)
(c3 rn + d3 sn );
(s2 ab c)f0 + af1
c)f0 + af1
; d3 =
:
r s
r s
From Lemma 2, we obtain the identity
p
n
(k+l)
X
( ab) (jn+l) (a;b;c;c)
(a;b;c;c)
n k
ka
p
C(n; k)(cBj 1 )
(Bj )
F
= Fjn+l ;
(jn+l) ( ab) (k+l) k+l
a
k=1
c3 =
where Bn =
r n sn
r s :
(r2
ab
(5)
Finally, from Lemma 3, we obtain the following theorem.
Pn
THEOREM 2. Consider a complex polynomial P (z)
= k=0 dk z k (dk 6= 0), and
p
n
n
abx c = 0. Then, for j 2
let Bn = r r ss , n 0, where r and s are roots of x2
and l 0, all the zeros of P (z) lie in the annulus C = fz : r3 jzj r4 g, where
r3 = min
1 k n
r4 = max
1 k n
(
(
n k
(Bj )k a (k+l) (
1)
C(n; k)(cBj
a
a
C(n; k)(cBj
p
(jn+l) (
(jn+l)
n
1)
ab)
p
( ab)
p
ab)
(jn+l)
(a;b;c;c)
Fk+l
(k+l) F (a;b;c;c)
jn+l
(k+l)
(a;b;c;c)
Fjn+l
p
k (B )k a (k+l) ( ab)
j
jd0 j
jdk
) k1
jdn k j
(jn+l) F (a;b;c;c) jdn j
k+l
REMARK 2. Since fBn g1
n=0 satis…es the recurrence relation
p
Bn+1 = abBn + cBn 1 for j 1
;
) k1
:
Y. Choo
215
with B0 = 0, B1 = 1, we have
B3 = ab + c and B4 =
p
ab(ab + 2c):
Then, setting j = 4, l = 0, (5) reduces to (3), and so Theorem C is also a special case
of Theorem 2.
References
[1] J. L. Díaz-Barrero, An annulus for the zeros of polynomials, J. Math. Anal. Appl.,
273(2002), 349–352.
[2] J. L. Díaz-Barrero and J. J. Egozcue, Bounds for the moduli of zeros, Appl. Math.
Lett., 17(2004), 993–996.
[3] M. Bidkham and E. Shashahani, An annulus for the zeros of polynomials, Appl.
Math. Lett., 24(2011), 122–125.
[4] D. Redmond, Problem 132, Missori J. Math. Sci., 11(1999), 197.
[5] N. A. Rather and S. G. Mattoo, On annulus containing all the zeros of a polynomial,
Appl. Math. E-Notes, 13(2013), 155–159.
[6] O. Yayenie, A note on generalized Fibonacci sequences, Appl. Math. Comput.,
217(2011), 5603–5611.