J. Comp. & Math. Sci. Vol.2 (1), 99-102 (2011)
Cauchy’s Bound for Zero’s of Polynomials M. S. PUKHTA and A. A. BHAT* Division of Agri. Engineering S. K. University of Agricultural Sciences and Technology of Kashmir, Srinagar, India - 191121. email: mspukhta_67@yahoo.co.in *Department of Mathematics, Islamia College Srinagar, Kashmir ABSTRACT A well known result of Cauchy1 about the moduli of the zeros of n −1
polynomial states that if
p (z ) = z n + ∑ ak z k the all the k =0
zeros of the polynomial lie in the circle
z ≤ 1 + A , where
A = max a j . 0 ≤ j ≤ n −1
In this paper we improve Cauchy’s result and proved that if n
p( z ) = z n + ∑ ai z i i =0
the zeros of
, ai ∈ R be a real polynomial, then all
p(z ) lie in the ring shaped region given by
R2 ≤ z ≤ R1 , Here
R1 = 1 + max{Aij } , where Aij =
i , j = 0,1,..., n . R2 =
j >1
{
ai a j −1 − a j ai −1 ai + a j
an =1 , a−1 = 0 and
− R12 b ( M 2 − a0 ) + R14 b
2
(M
2
− a0
)
2
+ 4 a0 R12 M 23
}
,
1 2
2 M 22 n −1
where
M 2 = R1n [ R1 +1 + 2 ∑ ak ] k =0
Keywords and phrases : Polynomials, Zero’s, refinements and Cauchy’s bound.
1. INTRODUCTION AND STATEMENT OF RESULTS n −1
For the polynomial
p ( z ) = z n + ∑ ak z k k =0
the result of Cauchy was improved by Joyal,
Labelle and Rahman2. They obtained a better region and proved the following Theorem 1.1. If p (z ) = z n +
n −1
∑a z k =0
all its zeros lie in the circle
Journal of Computer and Mathematical Sciences Vol. 2, Issue 1, 28 February, 2011 Pages (1-169)
k
k
then
M. S. Pukhta, et al., J. Comp. & Math. Sci. Vol.2 (1), 99-102 (2011)
100
{
1 z ≤ 1 + an −1 + (1 − an −1 2
)
}
1 2
+ 4B
2
R2 =
{
− R12 b ( M 2 − a0 ) + R14 b
2
(M
2
− a0
)
2
+ 4 a0 R12 M 23
}
1 2
2 M 22
where n −1
where B = max a j
M 2 = R1n [ R1 + 1 + 2 ∑ ak ]
0 ≤ j ≤ n −1
k =0
3
Ezra Zaheb extended the result of Cauchy in providing two circular bounds for the zero’s of a polynomial. In fact he proved Theorem 1.2. If n
p( z ) = z n + ∑ ai z i i =0
, ai ∈ R be a real
polynomial.Then all the zeros of p( z ) lie in the circle z < 1 + max Aij where
{ }
Aij =
b = a1 − a0
1
ai a j −1 − a j ai −1 ai + a j
, i , j = 0,1,..., n ,
j > i , an =1 , a−1 = 0 In this paper we improve up on Theorems 1.1 and 1.2 by obtaining a ring shaped region. More precisely we prove Theorem. Let p ( z ) = z + n
(1.2)
Remark . As remarked earlier our theorem clearly improves upon the result obtained by Cauchy’s theorem , Theorem 1.1 and Theorem 1.2. We illustrate this by means of the following example: Example .
Consider the polynomial
p( z ) = z 3 + 3 z 2 + 2 z +1
then according to Cauchy’s theorem p( z )
z ≤ 4 , by Theorem
has all its zero’s in
1.1, p( z ) has all its zero’s in
z ≤ 3.225
and by Theorem 1.2, p( z ) has all its zero’s in
z ≤ 2.75 , where as by our theorem all
zero’s of the polynomial p( z ) lie in the ring shaped region 0.00058 ≤ z < 2.75
n
∑a z ,a ∈ R i
i =0
i
i
2. LEMMA
be a real polynomial, then all the zeros of p(z ) lie in the ring shaped region given by
For the proof of the theorem , we require the following lemmas:
R2 ≤ z ≤ R1 ,
Lemma 2.1 . If p ( z ) is analytic in
{ }
where a < 1 , p (0) = b,
Here R1 = 1 + max Aij ,
'
Aij =
ai + a j
an =1 , a−1 = 0 and
p (z ) ≤ 1
on
z = 1 then for z ≤ 1
where
ai a j −1 − a j ai −1
z ≤1, p( 0) = a
, i , j = 0,1,..., n . j >1 (1.1)
p (z) ≤
(1 − a ) z a (1 − a ) z
+ b z + a (1 − a )
2 2
(
+ b z + 1− 1− a
)
The above lemma is a generalization of Schwart’z lemma and is due to Govil,
Journal of Computer and Mathematical Sciences Vol. 2, Issue 1, 28 February, 2011 Pages (1-169)
M. S. Pukhta, et al., J. Comp. & Math. Sci. Vol.2 (1), 99-102 (2011)
Rehman and Schmeissear4 . From Lemma 2.1 one can easily obtain the following:
≤ R1
If p ( z )
an = 1
Lemma 2.2 .
p (0) = 0 ,
pz ≤R
is
analytic in
p (0) = b '
and
p z ≤ M for z = R than for z ≤ R p (z ) ≤
M z
M z + R2 b
R2
M + z b
3. PROOF OF THEOREM
= R1
n +1
+ R1
n +1
= a0 +
∑ (a k =1
k
− ak −1 ) z − z
∑a n −1
n
∑a k =1
n +1
, an = 1
= a0 + P( z ) , say
(3.1)
where
R1 = 1 + max{Aij } and Aij is defined
P(z ) = z
n +1
+
n
∑a k =1
k
− ak −1 z
k
,
R1 = 1 + max{Aij } ≥ 1 .
n +1
n
n
n −1
∑a k =0
k
(3.2)
P / (0) = a1 − a0 = b , hence by lemma 2.2
we have
P (z ) ≤
M 2 z M 2 z + R1 b 2
R1
(3.3)
M2 + b z
2
for z ≤ R1 . Combining (3.1) and (3.3) we get for
F ( z ) ≥ a0 −
=
M ( R1 ) = max P ( z ) n
+ ∑ ak − ak −1 R1
k
k =1
+ R1
n
1
(M
2
2
R
{z
M 2 z M 2 z + R1 b 2
R1
2
M
, an = 1
∑a k =1
k
− ak −1
,
2
+ z b +R
1
2
)
b z
(M
2
− a
0
)−
2
a R M 0
1
2
}
if
an = 1
z <
{
− R1 b (M 2 − a0 ) + R1 b 2
n
2
−1
M2 + b z
2
> 0
z = R1
≤ R1
+ R1 + R1
k
,
k −1
n −1 n ≤ R1 R1 +1+ 2∑ ak k =0 =M 2.
an = 1
n +1
k =1
z ≤ R1
in ( 1.1 ) . Clearly
≤ R1
∑a
for z = R1 . Further since P (0 ) = 0 ,
z = R1
Hence
n
P( z ) ≤ M 2
Let M (R1 ) = max P (z ) ,
and
+ R1
k
n
Clearly from ( 3.1 )
F ( z ) = (1− z ) p ( z ) k
n k =1
+ R1
Consider
n
n
101
4
2
(M
2M2
− a0 ) + 4 a0 R1 M 2 2
2
2
= R2 from this theorem follows.
Journal of Computer and Mathematical Sciences Vol. 2, Issue 1, 28 February, 2011 Pages (1-169)
2
3
}
1
2
M. S. Pukhta, et al., J. Comp. & Math. Sci. Vol.2 (1), 99-102 (2011)
102 REFERECES
1. L. Cauchy, Exercise de mathematique in Oeuvres, (2) Vol. 9, pp 122 (1829). 2. A. Joyal, G Labelle and Q. I. Rahman, On the location of Zeroâ&#x20AC;&#x2122;s of polynomials, Canad. Math. Bull., Vol. 10, pp. 53-63 (1967).
3. Ezra Zeheb, On the largest modulus of polynomial zeroâ&#x20AC;&#x2122;s, IEEE Transctions on circuits and systems, Vol. 38, March, pp. 333-337 (1991). 4. N. K. Govil, Q. I. Rehman and G. Schmeisser, On the derivative of a polynomial, Illinois Jour. of Mathematics, Vol 23, pp. 319-330 (1979).
Journal of Computer and Mathematical Sciences Vol. 2, Issue 1, 28 February, 2011 Pages (1-169)