Mathematics Scientific Journal Vol. 6, No. 1, S. N. 12, (2010), 59-65
On the Zeros and Coefficients of a Polynomial with Restricted zeros Abdullah Mira,1 , Sajad Amin Babab,2 a
Department of Mathematics, University of Kashmir, Srinagar, 190006, India, b
Govt. Hr. Sec., Kurhama, Ganderbal-191201, Kashmir, India. Received 4 September 2010; Accepted 22 November 2010
Abstract Let p(z) =
n P
aj z j be a polynomial of degree n and let M (p, r) = max|z|=r |p(z)|.
j=0
In this paper we prove some sharp inequalities concerning the coefficients of the polynomial p(z) with restricted zeros. We also establish a sufficient condition for the separation of zeros of p(z). Keywords: Inequalities, Coefficient estimates, Separation of zeros
Mathematics AMS Subject Classification (2000): 30A64, 30C10, 30C15, 41A17
1
Introduction
Let p(z) =
n P
aj z j be a polynomial of degree n. By a simple method Visser [5]
j=0
observed: |a0 | + |an | ≤ M (p, 1).
(1.1)
Corput and Visser [1] generalized inequality (1.1) by proving the following sharp estimate for |aj | + |ak |, 0 < j < k < n, when |k − j| is not too small. 1 Corresponding 2 Email:
Author, Email address: mabdullah
[email protected] [email protected]
60
Abdullah Mir, Sajad Amin Baba
Theorem 1.1. Let aj and ak where 0 < j < k < n, be any two coefficients of the n P polynomial p(z) := av z v such that for no other coefficients at 6= 0 do we have v=0
t ≡ j(mod(k − j)).Then |aj | + |ak | ≤ M (p, 1).
(1.2)
In this case Rahman [4] proved the following more general result: Theorem 1.2. Let p(z) :=
n P
av z v and q(z) :=
v=0
m P
bµ z µ , where m < n. Suppose in
µ=0
addition that |p(z)| < |q(z)| for |z| = 1. If q(z) 6= 0 for |z| < 1, then |a0 | + |an | ≤ |bm |.
(1.3)
Concerning the inequalities involving three coefficients Frappier, Rahman and Ruscheweyh [2] proved the following much stronger result: Theorem 1.3. Let p(z) :=
n P
av z v and suppose that |p(z)| < 1, for |z| = 1, then
v=0
|a0 | + 1/2(|aj | + |ak |) ≤ 1, 1 ≤ j ≤ k, k ≥ n + 1 − j.
(1.4)
Next, concerning the size of M (p, r), the following inequality is well known M (p, r) M (p, R) ≥ , 0 0, then the inequality j=0
k n |an | ≤ |a0 |
(1.7)
is trivial.We first prove the following result,which in particular provides an improvement of inequality (1.7) and yields a number of results as special cases:
61
On the Zeros and Coefficients of a Polynomial with Restricted zeros
Theorem 1.4. Let p(z) :=
n P
aj z j be a polynomial of degree n which does not vanish
j=0
in |z| < k, k ≥ 1, then |an |k n + (1/n)||a1 |k − |an−1 |k n−1 | ≤ |a0 |
(1.8)
|a1 |k + |an−1 |k n−1 ≤ n(|a0 | + |an |k n )
(1.9)
and n
The result is best possible with equality for p(z) = (z + k) . Instead of proving 1.4,we prove the following more general result. Theorem 1.5. Let p(z) :=
n P
aj z j be a polynomial of degree n which does not vanish
j=0
in |z| < k, k ≥ 1, then for s = 1, 2, · · · , n, |an |k n + (1/C(n, s))||as |k s − |an−s |k n−s | ≤ |a0 |
(1.10)
|as |k s + |an−s |k n−s ≤ C(n, s){|a0 | + |an |k n }
(1.11)
and where C(n, s) = n(n − 1) · · · (n − s + 1)/s(s − 1) · · · 2.1.
(1.12) n
The result is best possible and equality holds for the polynomial p(z) = (z + k) . Applying Theorem 1.5 to the polynomial z n p(1/z), we immediately get the following: Corollary 1.6. If p(z) :=
n P
aj z j is a polynomial of degree n having all its zeros in
j=0
|z| < k, then |a0 | + (1/C(n, s))||as |k s − |an−s |k n−s | ≤ |an |k n
(1.13)
|as |k s + |an−s |k n−s |a0 |, then p(z) has at least one zero in |z| < k. From inequality (1.11) and inequality (1.14),we deduce the following:
(1.15)
62
Abdullah Mir, Sajad Amin Baba
Corollary 1.8. Let p(z) be a polynomial of degree n. If for some integer s, 1 < s < n, k s |as | + k n−s |an−s | > C(n, s){|a0 | + k n |an |},
(1.16)
then the circle |z| = k separates the zeros of p(z). We next present the following sharp result when there is a restriction on the zeros of p(z). Theorem 1.9. Let p(z) be a polynomial of degree n such that |p(z)| < 1 for |z| = 1. If the geometric mean of the zeros of p(z) is at least k, (k > 0), then (1 + k n )|an | ≤ 1
(1.17)
and |aj |2 ≤ 1 − (1 + k n )2 |an |2
j = 1, 2, · · · , n − 1.
(1.18)
Finally, concerning the estimates of M (p, r),we prove the following result: Theorem 1.10. Let p(z) be a polynomial of degree n with real coefficients having all zeros with non-positive real part. If for some r, R (0 < r < R < k); k > 0 n−m m k+r r p(r) < p(R) (1.19) k+R R where m is a non-negative integer, then p(z) has atleast (m + 1) zeros in |z| < k. The result is best possible and equality holds for the polynomial p(z) = (z + k)n−m−1 z m+1 .
2
LEMMA
For the proofs of these theorems, we need the following lemma,which is due to Corput and Visser [1]. Lemma 2.1. If p(z) :=
n P
aj z j is a polynomial of degree n such that |p(z)| < 1 for
j=0
|z| = 1, then 2|a0 ||an | +
n X
|aj |2 ≤ 1.
(2.1)
j=0
3
PROOFS OF THEOREMS
Proof of Theorem 2 By hypothesis all the zeros of polynomial p(z) = an z n + an−1 z n−1 + · · · + as z s + · · · + a1 z + a0 lie in |z| ≥ k where k ≥ 1. Therefore,all the zeros of polynomial F (z)
= p(kz) = an k n z n + an−1 ¯ k n−1 + · · · + an−s k n−s z s + · · · + a¯1 kz n−1 + a¯0 z n ,
On the Zeros and Coefficients of a Polynomial with Restricted zeros
63
lie in |z| ≥ 1. If G(z)
= z n F (1/¯ z) = z n p(k/¯ z) = a¯n k n + an−1 k n−1 z n−1 + · · · + as k s z s + · · · + a1 kz + a0
(3.1)
then all the zeros of G(z) lie in |z| ≤ 1 and |F (z)| = |G(z)| for |z| = 1. If G(z) has m, 1 ≤ m ≤ n zeros in |z| < 1 and the remaining n − m zeros on |z| = 1, then F (z) has m, 1 ≤ m ≤ n zeros in |z| > 1 and the remaining n − m zeros on |z| = 1. Thus the function S(z) = G(z)/F (z) is analytic for |z| ≤ 1, and |S(z)| = |G(z)/F (z)| = 1 for |z| = 1. Therefore, by the maximum modulus principle, it follows that |S(z)| ≤ 1 f or
|z| ≤ 1.
(3.2)
Equivalently, |G(z)| ≤ |F (z)| f or |z| ≤ 1. Replacing z by 1/¯ z , we conclude that |F (z)| ≤ |G(z)| f or |z| ≥ 1. This implies |F (z)| < |G(z)| f or |z| > 1,
(3.3)
(3.4)
unless F (z)/G(z) is a constant of absolute value unity.If F (z)/G(z) is constant, then it is obvious that for every complex number β with |β| ≥ 1, the zeros of F (z) + βG(z) lie in |z| ≤ 1. In case F (z)/G(z) is not constant, we have from (3.1) with the help of Rouche’s theorem that for every real or complex number β with |β| ≥ 1, all the zeros of F (z) + βG(z) lie in |z| ≤ 1. Hence by Gauss-Lucas theorem all the zeros of F (s) (z) + βG(s) (z) lie in |z| ≤ 1. That is, all the zeros of n(n − 1) · · · (n − s + 1)z n−s an k n + · · · + s!as k s + β{n(n − 1) · · · (n − s + 1)z n−s a ¯0 + · · · + s!¯ an−s k n−s }
(3.5)
n(n − 1) · · · (n − s + 1){an k n + β¯ a0 }z n−s + · · · + s!(as k s + β¯ an−s k n−s )
(3.6)
or
lie in |z| < 1. This implies C(n, s)|an k n + β¯ a0 | ≥ |as k s + β¯ an−s k n−s |.
(3.7)
Choosing the argument of β such that |as k s + β¯ an−s k n−s | = |as |k s + |β||an−s |k n−s and making |β| → 1, we get from (3.7) |as |k s + |an−s |k n−s ≤ C(n, s){|a0 | + |an |k n }, which proves inequality (1.11).
(3.8)
64
Abdullah Mir, Sajad Amin Baba
Again choosing the argument of β in the left hand side of (3.7) such that |an k n + β¯ a0 | = |β|(|a0 | − |an |k n ), which is possible by (1.7). Making |β| → 1, we get from (3.7) ||as |k s − |an−s |k n−s | ≤ C(n, s){|a0 | − |an |k n }, (3.9) from which inequality (1.10) follows and this completes the proof of Theorem 2. Proof of Theorem 1.9. Since the geometric mean of the moduli of the zeros of the polynomial p(z) is less than k > 0, then by combining (1.1) and (1.7), we get (1.16). Again from inequality (2.1), we have |a0 |2 + 2|a0 ||αn | + |αn |2 + or (|a0 | + |an |)2 +
X n−1 lim |aj |2 ≤ 1, j=1
X n−1 lim |aj |2 ≤ 1, f or all j = 1, 2, · · · , n − 1. j=1
(3.10)
(3.11)
This implies (|a0 | + |an |)2 + |aj |2 ≤ 1, f or all j = 1, 2, · · · , n − 1.
(3.12)
Using (1.7) in (3.4),we get (k n |an | + |an |)2 + |aj |2 ≤ 1,
j = 1, 2, · · · , n − 1,
(3.13)
from which the result follows and this proves Theorem 1.9. Proof of Theorem 1.10. Suppose that p(z) has t zeros in |z| < k where t < m. Let p(z) = (z−z1 )(z−z2 ) · · · (z−zt )(z−zt+1 ) · · · (z−zn ) and assume that |zj | < k; j = 1, 2, · · · , t. Put g(z) = (z − z1 )(z − z2 ) · · · (z − zt ) and h(z) = (z − zt+1 ) · · · (z − zn ). The polynomial p(z), g(z) and h(z) have positive coefficients. Hence for r, R with 0 < r < R ≤ k, we have t r g(r) ≥ g(R) (3.14) R by (1.5), and n−t k+r h(r) ≥ h(R) k+R
(3.15)
by (1.6). On combining (3.14) and (3.15), we get p(r)
= g(r)h(r) n−t t k+r r ≥ g(R)h(R) k+R R n t k+r r k+R = p(R) k+R k+r R n m k+r r k+R ≥ p(R) k+R k+r R
(3.16)
which is clearly a contradiction. This establishes the Theorem 1.10 completely.
On the Zeros and Coefficients of a Polynomial with Restricted zeros
65
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