Perturbation on Polynomials

For univariate polynomials with complex coefficients there are many estimates about the roots of polynomials. Moreover, the result corresponding to the “continuity of the zeroes which respect to the coefficients” is generally obtained as a corollary of Rouché’s theorem and is rarely precise. Here we prove an explicit result for algebraically closed fields with an absolute value, in any characteristic.


Introduction
The first motivation of this paper was the following: try to perturb slightly polynomials with integer coefficients with the hope of finding examples with very small root separation.At the beginning of the work we noticed that it is very difficult to find explicit statements of the theorem of "Continuity of the roots of a polynomial with respects to the coefficients".For this reason we decided to write the section 1 of this paper.Then we studied examples and we had the bad surprise to see that in each case a perturbation of a "good" example leads to a "poor" example.
Preliminaries We begin by a general lemma in algebra.
Lemma 1 Let R be any ring and let P ∈ R[X] be a polynomial of degree n 1.Let us define the k−hyperderivative by the formula and the linearity properties Then, for any a ∈ R, we have the formula Moreover, if R = K is a field and if P has all its roots, say α 1 , . . ., α n , in some extension L of K, then for all a ∈ K and all k ∈ N, we have Proof.The first assertion is well-known (in the case of a field of characteristic zero, this is just Taylor's formula for polynomials).To prove it, by linearity, it's enough to verify that it's true if P(X) = X m for m n.Then, one notices that this case is just Newton's formula for the expansion of (X + a) m .
For a = 0, the second relation is again well-known: this is Vieta's formula.The general case is obtained by the translation X → X + a.
Remark 1 If K is of characteristic zero, then where P (k) (X) is the usual k−derivative of P.

Continuity of the Roots of Polynomials in Terms of the Coefficients
We want to prove the following result which is a generalization of a classical one when K is the field of complex numbers, whose proof is generally obtained as a corollary of Rouché's theorem.
Theorem 1 Let K be an algebraically closed field with an absolute value and let f ∈ K[X] be a fixed polynomial with the decomposition f where α i α j for i j and m 1 , m 2 , . . ., m k are positive integers.
Let n = m 1 +m 2 +• • •+m k be the degree of f .Let 0 < ε < sep(P) a given real number then there exist a real number η > 0, such that if g ∈ K[X] is a polynomial of degree n, satisfying H( f − g) < η (where H() is the height of a polynomial and sep(P) = min i j α i − α j , i, j = 1, . . ., k).Then all the roots of g belong to the union for Moreover if ε is sufficiently small, then for i = 1, . . ., k, there exists exactly m i roots of g which belong to the disk D i .
Proof.Without loss of generality, we may assume that f is monic and that f and f are of the same degree n.Let Let us first notice that, the roots α i , i = 1, . . ., n, and any root β satisfy Thus, assuming also η . This shows that each root of g is "close" to some root of f .For small enough η, more precisely when The roots of g belong to the union of the disks D i defined in the theorem.This proves the first assertion.Now, to prove the second assertion, it is enough to prove that each D i does not contain more than m i roots of g.To simplify the notation, let α be a root of f and let m be its multiplicity.

Suppose that the disks
. But, by the lemma, we also have and the assumption β j − α < ε for j = 1, 2, . . ., m + 1 implies . This gives a contradiction.Thus we have proved the second assertion.

Remark 2 We have proved that
• for the first assertion we can take η < η 0 , defined in (2), • and for the second assertion, η < η where M is equal to min {m i , i = 1, . . ., k}.

Perturbation of the Coefficients of a Polynomial
Example 1 Let P ∈ K[X] be the polynomial , where (F n ) n is the Fibonacci sequence.Denote by H the height of P. We have, The height satisfies H = F n+1 + F n = F n+2 α n+2 / √ 5.And we have , and sep(P) 1/H 2 .
Put P 1 (x) = P(x) + x n+1 .Therefore, For ε sufficiently small such that P 1 (1/a + ε) = 0, we have the approximation: Thus we have two possible roots, say ε 1 and ε 2 : Example 3 Let P(z) = z n − a, where a ∈ N, a 0. The roots of P are: thus sep(P) = 2a 1/n sin(π/n) and H = a.Moreover, For a = 2853 and n = 16, P(z) = z 16 − 2853, η 0 = (2 sin π/16) 16 (2853) −15 , η 0 4.2710 −59 .We notice that in this case, for which the roots of P are very well separated, these roots remain very "stable" under perturbation of the coefficients, for example, the real roots of P are: where the real roots of perturbated polynomial are:

±1.644206499888864834627084840920319068838650249550067267404659984536653
We have the same remark with the other roots.
Remark 3 To keep integer coefficients, instead of the above Q, we could consider the polynomial Q where a is a large positive integer.This value of η 0 , is extraordinarily small and seems very pessimistic.But, indeed, Wilkinson shows that a very small perturbation (η = 10 −9 ) causes a dramatic change of some of the roots for example the perturbated polynomial admits the complex roots 16.57173899 ± 0.8833156071 • I.
Example 6 Let P(x) be the polynomial defined by The height of this polynomial is H = a 2 as soon as a > 2.