PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 93 (107) (2013), 65–67 DOI: 10.2298/PIM1307065B

GENERALIZATION OF THE GRACE–HEAWOOD THEOREM

Radoš Bakić

Communicated by Žarko Mijajlović

Abstract. Using the theorem of Walsh we give a generalization of the theorems of Grace.

1. Introduction

All the polynomials in this paper are complex. The following theorem is well known and is due to Walsh [1].

Theorem 1.1. If all the zeros of a polynomial f(z) = a_{0} + a_{1}z + ⋯+ z^{n} lie in the circle |z| ⩽ r, then all the zeros of the polynomial F (z) = f(z) + c, lie in the circle |z| ⩽ r + \sqrt[n]{|c|}.

Due to translation, previous theorem can be slightly generalized:

Theorem 1.2. If all the zeros of a polynomial f(z) = a_{0} + a_{1}z + ⋯+ z^{n} lie in the circle |z − d| ⩽ r, then all the zeros of the polynomial F (z) = f(z) + c, lie in the circle |z − d| ⩽ r + \sqrt[n]{|c|}.

For polynomials f(z) = a_{0} + a_{1}z + ⋯+ z^{n} and g(z) = b_{0} + b_{1}z + ⋯+ z^{n} let

us define

A(f, g) = a_{0} − \frac{a_{1}b_{n−1}}{\binom{n}{1}} + \frac{a_{2}b_{n−2}}{\binom{n}{2}} − ⋯+ (−1)^{n−1} \frac{a_{n−1} b_{1}}{\binom{n}{n−1}} + (−1)^{n}b_{0}

The polynomials f(z) and g(z) are called apolar if A(f, g) = 0. The following theorem, due to Grace, is the basic one for us.

Theorem 1.3. If f(z) and g(z) are apolar and if one of them has all its zeros in a circular region C, then at least one zero of the other one is in C.

2010 Mathematics Subject Classification: Primary 26C10.

Key words and phrases: Zeros of polynomials.

2. Main result

A question arises what happens when f(z) and g(z) are not apolar. We prove the following simple result.

Theorem 2.1. For polynomials f(z) = a_{0} + a_{1}z + ⋯+ z^{n} and g(z) = b_{0} +

b_{1}z + ⋯+ z^{n} the following holds: if all the zeros of f(z) are contained in some circular region of radius r, then a zero of g(z) is contained in concentric circular region of radius r + \sqrt[n]{|A(f, g)|}.

Proof. Polynomials f(z) − A(f, g) and g(z) are apolar. If C is a circular region of a radius r containing all zeros of f(z), then the concentric circular region of a radius r + \sqrt[n]{|A(f, g)|} contains all the zeros of the polynomial f(z) − A(f, g) (Theorem 1.2). Then, applying Theorem 1.3, we obtain the desired result. □

Now we want to prove a theorem that is, in a sense, dual to Theorem 1.1.

Lemma 2.1. Let f(z) = a_{0} + a_{1}z + ⋯+ z^{n}, be a polynomial having a zero in a circular region |z| ⩽ r. Then f(z) + c has a zero in the circular region

|z| ⩽ r + \sqrt[n]{|c|}.

Proof. Let b be a zero of f(z), |b| ⩽ r. It is easy to see that 0 = A(f(z), (z − b)^{n}) = A(f(z) + c, (z − b)^{n} − c). Hence, polynomials f(z) + c and (z − b)^{n} − c are apolar. Since all zeros of (z − b)^{n} are in the circle |z| ⩽ r, by Theorem 1.1. it follows that all the zeros of (z − b)^{n} − c are in the circle |z| ⩽ r + \sqrt[n]{|c|}. Finally, because of apolarity of f(z) + c and (z − b)^{n} − c, we conclude that f(z) + c, has a zero in the circle |z| ⩽ r + \sqrt[n]{|c|}. □

Similarly as before, due to translation, we can slightly reformulate the statement of the previous lemma.

Lemma 2.2. Let f(z) = a_{0} + a_{1}z + ⋯+ z^{n}, be a polynomial having a zero in a circular region |z − d| ⩽ r. Then f(z) + c has a zero in the circular region

|z − d| ⩽ r + \sqrt[n]{|c|}.

Our main result is a generalization of the following theorem due to Grace and Heawood [2].

Theorem 2.2. If z_{1} and z_{2} are zeros of the polynomial f(z), deg(f) = n > 1, then its derivative has a zero in the circular region

|z − \frac{1}{2}(z_{1} + z_{2})| ⩽ \frac{1}{2}|z_{1} − z_{2}| cot( \frac{π}{n}) .

Now we are going to prove the following theorem.

Theorem 2.3. Let f(z) = a_{0} + a_{1}z + ⋯+ z^{n} be a polynomial. Then for any distinct z_{1} and z_{2}, the circular region

|z − \frac{1}{2}(z_{1} + z_{2})| ⩽ \frac{1}{2}|z_{1} − z_{2}| cot( \frac{π}{n}) + \sqrt[n−1]{\frac{|a|}{n − 1}} , where a = −\frac{f(z_{2}) − f(z_{1})}{z_{2} − z_{1}}

contains a zero of the derivative f′(z).

Proof. Choose some distinct z_{1} and z_{2}, and consider the polynomial

F (z) = f(z) + az + b, where a = −\frac{f(z_{2}) − f(z_{1})}{z_{2} − z_{1}} , b = \frac{z_{1}f(z_{2}) − z_{2}f(z_{1})}{z_{2} − z_{1}} .

Then we have F (z_{1}) = F (z_{2}) = 0, and F′(z) = f′(z) + a. By Theorem 2.2, the circular region

|z − \frac{1}{2}(z_{1} + z_{2})| ⩽ \frac{1}{2}|z_{1} − z_{2}| cot( \frac{π}{n})

contains a zero of F′(z). Then, by Lemma 2.2 applied on F′(z) − a = f′(z) and r = \frac{1}{2}|z_{1} − z_{2}| cot( \frac{π}{n} ), we conclude that a zero of f′(z) is contained in the circular region

|z − \frac{1}{2}(z_{1} + z_{2})| ⩽ \frac{1}{2}|z_{1} − z_{2}| cot( \frac{π}{n}) + \sqrt[n−1]{\frac{|a|}{n − 1}} ,

and the theorem is proved. □

References

1. J. L. Walsh, On the location of the roots of certain types of polynomials, Trans. Amer. Math. Soc. 24 (1922), 163–180.

2. M. Marden Geometry of Polynomials, American Mathematical Society, 1966.

Zavod za unapređivanje (Received 06 02 2013) obrazovanja i vaspitanja 11000 Beograd Serbia

    bakicr@gmail.com