This is a reprint of the paper The roots of a polynomial

depend continuously on its coefficients by Raúl Naulin & Carlos Pabst published in Rev. Colombiana Mat.

28 (1994), pp. 35–37

THE ROOTS OF A POLYNOMIAL DEPEND

CONTINUOUSLY ON ITS COEFFICIENTS

Raúl Naulin^{∗} Carlos Pabst^{†}

Universidad de Valparaíso, CHILE

Abstract. An elementary proof is given of the continuous dependence of the roots of a polynomial on its coefficients. Keywords and phrases. Lexicographic order, metric space, normed space, symmetric functions, roots of a polynomial, Fundamental Theorem of Algebra, continuous dependence on coefficients. 1991 Mathematics Subject Classification. Primary 11C99. Secondary 00A99.

Definition 1. We say that the complex number u = α + iβ is lexicographically less than the complex number v = γ + iδ if α < γ or δ = γ and β < δ. We denote this by writing u ≺ v. The notation u ⪯ v means that u ≺ v or u = v.

With C_{n} we denote the set of n–tuples of complex numbers lexicographically ordered from less to greater. Thus (x_{1}, x_{2}, …, x_{n}) ∈ C_{n} iff x_{1} ⪯ x_{2} ⪯ … ⪯ x_{n}. With 0 we denote the n–tuple (0, 0, …, 0).

On the set C_{n} we define the metric

d(x, y) = \sqrt{∑_{j=1}^{n} |x_{j} − y_{j}|^{2}}.

Typeset by AMS-TEX

∗ Supported by Proyecto Fondecyt 0028–92, Conicyt, Chile.

† Supported by Proyecto DIUV 09/93, Universidad de Valparaíso, Chile.

The pair (C_{n}, d) is a complete metric space. In fact, C_{n} is a closed subset of the normed complex euclidean space, i.e., of (ℂ^{n}, ∥ · ∥), where the norm of a⃗ ∈ ℂ^{n} is

∥a⃗∥ = \sqrt{∑_{j=1}^{n} |a_{j}|^{2}}.

Now let

P (z) = z^{n} − a_{1}z^{n−1} + … + (−1)^{n}a_{n}

be a polynomial, and consider its coefficients as a vector in ℂ^{n}:

a⃗ = (a_{1}, a_{2}, …, a_{n}).

From the fundamental theorem af algebra [1] we know that p(x) has n roots. We will denote these roots by λ_{i}, i = 1, 2, …, n, and assume that λ_{i} ⪯ λ_{i+1}, so that the vector

λ⃗ = (λ_{1}, …, λ_{n})

is in C_{n}. From the well known formulae of Viète we have the identities

a_{1} = λ_{1} + λ_{2} + ⋯ + λ_{n}
a_{2} = λ_{1}λ_{2} + λ_{1}λ_{3} + ⋯ + λ_{n−1}λ_{n}
a_{n} = λ_{1}λ_{2} ⋯ λ_{n},

by means of which we can define in an obvious manner a continuous map

T : C_{n} → ℂ^{n}; λ⃗ ↦ T (λ⃗) = a⃗

From the fundamental theorem of algebra this function is one to one and onto, i.e., T establishes a biyective correspondence between C_{n} and ℂ^{n}.

Let S denote the inverse mapping of T :

S := T^{−1} : ℂ^{n} → C_{n}

Lemma 1.

d(S(a⃗),0) ≤ 2n max{1, ∥a⃗∥}

Proof: Let S(a⃗) = (λ_{1}, …, λ_{n}). Then

|λ_{1}|^{n} ≤ ∑_{j=i}^{n} |a_{j}|(1 + |λ_{1}|^{n−j}).  (1)

Now, if |λ_{i}| ≤ 1, then

|λ_{i}| ≤ 2\sqrt{n} max\{1, ∥a⃗∥\},  (2)

and if |λ_{i}| ≥ 1, then, dividing (1) by |λ_{i}|^{n−1}, we obtain

|λ_{i}| ≤ ∑_{j=1}^{n} |a_{j}|(1 + |λ_{i}|^{1−j})

≤ 2 ∑_{j=1}^{n} |a_{j}| ≤ 2\sqrt{n} ∥a⃗∥

≤ 2\sqrt{n} max\{1, ∥a⃗∥\}.  (3)

From (2) and (3) the proof follows. □

Theorem 1. The function S : ℂ^{n} → C_{n} is continuous.

Proof: Assume that S is not continuous at a point a⃗. Then there exist δ > 0 and a sequence (a⃗_{n})_{n=1}^{∞} such that lim_{n→∞} a⃗_{n} = a⃗ and

d(S(a⃗_{n}), S(a⃗)) ≥ δ.  (4)

Because of Lemma 1, the sequence (S(a⃗_{n}))_{n=1}^{∞} is bounded, and therefore (passing to a subsequence, if necessary) we can assume that this sequence has a limit:

lim_{n→∞} S(a⃗_{n}) = ξ⃗.

But from the continuity of T we have

T (ξ) = lim_{n→∞} T (S(a⃗_{n})) = lim_{n→∞} a⃗_{n} = a⃗,

and therefore ξ⃗ = S(a⃗). Hence, for n sufficiently large

d(S(a⃗_{n}), S(a⃗)) < δ,

which contradicts (4). □

References

1. S. Lang, Linear Algebra, Addison–Wesley, Reading, Mass., 1966.

(Recibido en enero de 1994)

Raúl Naulin

Instituto de Matemáticas y Física (IMAFI)

Universidad de Valparaiso Casilla 1470, Valparaiso, CHILE

e–mail: rnaulin@udo.uunet.ve

Carlos Pabst

Instituto de Matemáticas y Física (IMAFI)

Universidad de Valparaiso Casilla 1470, Valparaiso, CHILE