On annulus containing all the zeros of a polynomial

ON ANNULUS CONTAINING ALL THE ZEROS OF A POLYNOMIAL

arXiv:1312.0135v1 [math.CV] 30 Nov 2013

N. A. RATHER AND SUHAIL GULZAR

Department of Mathematics, University of Kashmir Hazratbal Srinagar 190006, India emails: [email protected], [email protected], Abstract. In this paper, we obtain an annulus containing all the zeros of the polynomial involving binomial coefficients and generalized Fibonacci numbers. Our result generalize some of the recently obtained results in this direction.

1. Introduction and Statements 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 coefficients, then all zeros of P (z) lie in the disk |z| ≤ r where r is the unique positive root of the real-coefficient polynomial Q(x) = xn − |an−1 |xn−1 − |an−2 |xn−2 − · · · − |a1 |x − |a0 |.

Recently D´ıaz-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 coefficients and Fibonacci numbers. In fact he has proved the following result. P Theorem B. Let P (z) = nj=0 aj z j be a non-constant complex polynomial. Then all its zeros lie in the annulus C = {z ∈ C : r1 ≤ |z| ≤ r2 } where ( (
) k1 ) k1 2n Fk nk a0 3 2 F4n an−k
r1 = min , r2 = max . 2 1≤k≤n F4n ak 3 1≤k≤n 2n Fk nk an

Here Fj are Fibonacci’s numbers, that is, F0 = 0, F1 = 1 and for j ≥ 2, Fj = Fj−1 + Fj−2 .

More recently, M. 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 j Theorem C. Let P (z) = j=0 aj z be a non-constant complex polynomial of degree n and
(t3 + 2t)k (t2 + 1)n Ft,k nk λk = (t2 + 1)k Ft,4n 2010 Mathematics Subject Classification. primary: 30C10, 30C15. Key words and phrases. Polynomials; Location of zeros of polynomials. 1

N. A. RATHER AND SUHAIL GULZAR

2

for any real positive number t. Then all the zeros of P (z) lie in the annulus R = {z ∈ C : s1 ≤ |z| ≤ s2 } where  1   k1  a0 1 an−k k s1 = min λk , s2 = max . 1≤k≤n 1≤k≤n λk an ak

In this paper, we determine in the complex plane an annulus containing all the zeros of a polynomial involving binomial coefficients and generalized Fibonacci numbers (see [5]) defined recursively by (a,b,c)

(a,b,c)

F0

= 0, F1 = 1, ( (a,b,c) (a,b,c) a Fn−1 + c Fn−2 if n is even, = (a,b,c) (a,b,c) b Fn−1 + c Fn−2 if n is odd,

Fn(a,b,c)

(1.1)

(n ≥ 2)

where a, b, c are any three positive real numbers. Our result include Theorems B, C as special cases. More precisely, we prove the following result. Pn Theorem 1.1. 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 = {z ∈ C : r1 ≤ |z| ≤ r2 } where ( ) k1 k (u,v,w) n
(uvw + w2 )n uξ(k) (uv)⌊ 2 ⌋ Fk a uv + 2w 0 k min , r1 = 2 (u,v,w) uvw + w 1≤k≤n ak F4n abc + c2 r2 = max ab + 2c 1≤k≤n

(

(a,b,c)

F4n

(a,b,c)

k

(abc + c2 )n aξ(k) (ab)⌊ 2 ⌋ Fk

k

k − 2⌊ k2 ⌋

a, b, c, u, v, w are any positive real numbers, ξ(k) := as in (1.1).

) 1 an−k k ,
n an (a,b,c)

and Fm

is defined

Remark 1.2. By taking a, b, c and u, v, w suitably, we shall obtain Theorems B, C. For example, if we take a = b = u = v = t and c = w = 1, we obtain Theorem C. Example 1.3. We consider the polynomial P (z) = z 3 + 0.1z 2 + 0.3z + 0.7, which is the only example considered by D´ıaz-Barrero [1] and by using Theorem B, the annulus containing all the zeros of P (z) comes out to be 0.58 < |z| < 1.23. We improved the upper bound of this annulus by taking a = 1/2, b = 1 and c = 3/8 in Theorem 1.1 and obtained the disk, |z| < 1.185, which contains all the zeros of polynomial P (z). We can similarly improve the lower bound by choosing u, v, w suitably. 2. Lemma To prove the above theorem, we need the following lemma. (a,b,c)

Lemma 2.1. If Fk (2.1)

n X

n−k

(ab + c)

is defined as in (1.1), then k ξ(k)

(ab + 2c) a

⌊k ⌋ n−k 2

(ab)

c

(a,b,c) Fk

k=1

where ξ(k) = k − 2⌊ k2 ⌋. (a,b,c)

Proof. For Fk

, we have [5] (a,b,c)

Fk

=

a1−ξ(k) k

(ab)⌊ 2 ⌋



αk − β k α−β



  n (a,b,c) = F4n k

ANNULUS FOR THE ZEROS OF A POLYNOMIAL

3

√ √ ab+ (ab)2 +4abc ab− (ab)2 +4abc where α = , β = and ξ(k) = k − 2⌊ k2 ⌋. 2 2 Consider, n   X  n−k  k k n (a,b,c) (ab)3 + 2(ab)2 c aξ(k) (ab)⌊ 2 ⌋ Fk (abc)n−k (ab)2 + abc k k=1 !k  !n−k  n   3 2 X X X n αk − β k j 3−j j 2−j n−k n−k α β α β a = (−1) (αβ) α−β k j=0 j=0 k=1 ( n   !k ) !n−k 3 2 X n X X aαn j 3−j j 3−j n−k = α β α β (−1) α−β k j=0 j=0 k=1 ( n   !k ) !n−k 3 2 X n X X aβ n − αj β 3−j α1+j β 2−j (−1)n−k α−β k j=0 j=0 k=1 )n )n ( 3 ( 3 2 2 n n X X X X aβ aα − αj β 3−j α1+j β 2−j αj β 3−j − αj β 3−j − = α − β j=0 α − β j=0 j=0 j=0  n 3 n  α (α ) − β n (β 3 )n (a,b,c) =a = (ab)2n F4n . α−β Equivalently, we have n   X k n (a,b,c) (a,b,c) (ab + c)n−k (ab + 2c)k aξ(k) (ab)⌊ 2 ⌋ cn−k Fk = F4n . k k=1



3. Proof of Theorem Proof of Theorem 1.1. We first show that all the zeros of P (z) lie in ( ) 1 (a,b,c) an−k k (ab + c)k ck F4n (3.1) |z| ≤ r2 = max k (a,b,c) n
a 1≤k≤n n (ab + c)n (ab + 2c)k aξ(k) (ab)⌊ 2 ⌋ cn Fk k where a, b, c are any three positive real numbers. From (3.1), it follows that k (a,b,c) n
n k ξ(k) an−k (ab)⌊ 2 ⌋ cn Fk k ≤ r2k (ab + c) (ab + 2c) a , k = 1, 2, 3, · · · , n an (a,b,c) (ab + c)k ck F 4n

or

(3.2)

k (a,b,c) n n X X an−k 1 (ab + c)n (ab + 2c)k aξ(k) (ab)⌊ 2 ⌋ cn Fk ≤ an r k (a,b,c) (ab + c)k ck F4n 2 k=1 k=1

n k




.

Now, for |z| > r2 , we have

|P (z)| =|an z n + an−1 z n−1 + · · · + a1 z + a0 | ) ( n X 1 a n−k ≥|an ||z|n 1 − an |z|k k=1 ( ) n X an−k 1 n 1− >|an ||z| an r k . 2 k=1

Using (2.1) and (3.2), we have for |z| > r2 , |P (z)| > 0. Consequently all the zeros of P (z) lie in |z| ≤ r2 and this proves the second part of theorem.

N. A. RATHER AND SUHAIL GULZAR

4

To prove the first 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−1 z + 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) ak (uv + w)k wk F4n |z| ≤ max
k a0 (au,v,w) n 1≤k≤n (uv + w)n (uv + 2w)k uξ(k) (uv)⌊ 2 ⌋ wn F k

k

1

=

min

1≤k≤n

(

(u,v,w)

(uv + w)k wk F4n

k

(u,v,w)

(uv + w)n (uv + 2w)k aξ(k) (ab)⌊ 2 ⌋ wn Fk

a0 ak
n k

)1/k

1 = . r1 Now replacing z by 1/z and observing that all the zeros of P (z) lie in ( ) k1 (u,v,w) a0 (uv + w)k wk F4n |z| ≥ r1 = min .
k (u,v,w) n ak 1≤k≤n (uv + w)n (uv + 2w)k uξ(k) (uv)⌊ 2 ⌋ wn Fk k This completes the proof of theorem 1.1.



Acknowledgement The second author is supported by Council of Scientific and Industrial Research, New Delhi, under grant F.No. 09/251(0047)/2012-EMR-I. References [1] J. L. D´ıaz-Barrero, An annulus for the zeros of polynomials, J. Math. Anal. Appl., 273 (2002) 349-352. [2] M. Bidkham, E. Shashahani, An annulus for the zeros of polynomials, Appl. Math. 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 scientific Publishing Co., Singapore, (1994). [5] Omer Yayenie, A note on generalized Fibonacci sequences, Appl. Math. Comput., 208 (2009) 180-185. Department of Mathematics, University of kashmir, Srinagar, Hazratbal 190006, India