Perturbation on Polynomials

Transcription

1 Perturbation on Polynoials Isaila Diouf 1, Babacar Diakhaté 1 & Abdoul O Watt 2 1 Départeent Maths-Infos, Université Cheikh Anta Diop, Dakar, Senegal Journal of Matheatics Research; Vol 5, No 3; 2013 ISSN E-ISSN Published by Canadian Center of Science and Education 2 Départeent Mathéatiques, Ecole polytechnique, Thiès, Senegal Correspondence: Isaila Diouf, Départeent Maths-infos, FST, Université Cheikh Anta Diop (UCAD), Dakar, BP 5005, Senegal Tel: E-ail: isailadiouf@ucadedusn Received: June 12, 2013 Accepted: July 23, 2013 Online Published: August 8, 2013 doi:105539/jrv5n3p51 URL: Abstract For univariate polynoials with coplex coefficients there are any estiates about the roots of polynoials Moreover, the result corresponding to the continuity of the zeroes which respect to the coefficients is generally obtained as a corollary of Rouché s theore and is rarely precise Here we prove an explicit result for algebraically closed fields with an absolute value, in any characteristic Keywords: polynoials, roots, perturbation, separation of roots, Rouché s theore 1 Introduction The first otivation of this paper was the following: try to perturb slightly polynoials with integer coefficients with the hope of finding exaples with very sall root separation At the beginning of the work we noticed that it is very difficult to find explicit stateents of the theore of Continuity of the roots of a polynoial with respects to the coefficients For this reason we decided to write the section 1 of this paper Then we studied exaples and we had the bad surprise to see that in each case a perturbation of a good exaple leads to a poor exaple Preliinaries We begin by a general lea in algebra Lea 1 Let R be any ring and let P R[X] be a polynoial of degree n 1 Let us define the k hyperderivative by the forula ( ) ( ) (X ) <k> = X k, where = 0ifk >, k k and the linearity properties (S + T) <k> = S <k> + T <k>, (λs ) <k> = λs <k>, when λ R So that P <k> R[X] for all k Then, for any a R, we have the forula P(x + a) = P(X) + ap <1> (X) + a 2 P <2> (X) + + a n P <n> (X) Moreover, if R = K is a field and if P has all its roots, say α 1,,α n, in soe extension L of K, then for all a K and all k N, we have P <k> (a) = ( 1) n k (α j1 a) (α jn k a) 1 j 1 < < j n k n Proof The first assertion is well-known (in the case of a field of characteristic zero, this is just Taylor s forula for polynoials) To prove it, by linearity, it s enough to verify that it s true if P(X) = X for n Then, one notices that this case is just Newton s forula for the expansion of (X + a) For a = 0, the second relation is again well-known: this is Vieta s forula The general case is obtained by the translation X X + a Reark 1 If K is of characteristic zero, then P <k> (X) = P(k) (X), k! where P (k) (X) is the usual k derivative of P 51

2 2 Continuity of the Roots of Polynoials in Ters of the Coefficients We want to prove the following result which is a generalization of a classical one when K is the field of coplex nubers, whose proof is generally obtained as a corollary of Rouché s theore Theore 1 Let K be an algebraically closed field with an absolute value and let f K[X] be a fixed polynoial with the decoposition f (x) = a(x α 1 ) 1 (x α 2 ) 2 (x α k ) k, a 0, where α i α j for i j and 1, 2,, k are positive integers Let n = k be the degree of f Let 0 <ε<sep(p) a given real nuber then there exist a real nuber η>0, such that if g K[X] is a polynoial of degree n, satisfying H( f g) <η(where H() is the height of a polynoial and sep(p) = in αi i j α j, i, j = 1,,k) Then all the roots of g belong to the union for 1 i k of the disks D i = {z K; z α i <ε} Moreover if ε is sufficiently sall, then for i = 1,,k, there exists exactly i roots of g which belong to the disk D i Proof Without loss of generality, we ay assue that f is onic and that f and f are of the sae degree n Let H = H( f ) Let us first notice that, the roots α i, i = 1,,n, and any root β satisfy Put r = f g If we assue η<1 Since g(β) = 0, we have k i=0 α i < and β < H + 2 (1) β α i i = f (β) = r(β) η(1 + (H + 2) + + (H + 2) n ) = η [(H + 2)n+1 1] Thus, assuing also η [(H + 2)n+1 1] < 1, we get ( in { β α i } η [(H + ) 1/n 2)n+1 1] i This shows that each root of g is close to soe root of f For sall enough η, ore precisely when η<in { 1, (sep( f )) n} (H + 2) n+1 1 =: η 0 (2) The roots of g belong to the union of the disks D i defined in the theore This proves the first assertion Now, to prove the second assertion, it is enough to prove that each D i does not contain ore than i roots of g To siplify the notation, let α be a root of f and let be its ultiplicity Suppose that the disks D i = {z K; z α i < sep( f )} contains + 1 roots of g, say β 1,, β +1 First notice that g <> (α) μ 2, where μ = in { f < 1 > (α 1,, f < k > (α } k, we add the condition η<2 n H μ () n But, by the lea, we also have 1 g <> (α) = ± (β j1 α) (β jn α), 1< j 1 < < j n <n and the assuption β j α <εfor j = 1, 2,, + 1 iplies g <> (α) ( ) n ε (2H + 3) n so that, if ε is sall enough, we obtain g <> (α) μ This gives a contradiction Thus we have proved the second 2 assertion 52

3 Reark 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 1 = 2 n H μ () n M 1 where M is equal to in { i, i = 1,,k} Notice that the arguent works if η<in {η 0,η 1 } 3 Perturbation of the Coefficients of a Polynoial Exaple 1 Let P K[X] be the polynoial P(X) = (ax b)(x 2 X 1), then α = (1 + 5)/2 is a root of P Let a = F n and b = F n+1, where (F n ) n is the Fibonacci sequence Denote by H the height of P We have, P(x) = F n x 3 (F n + F n+1 )x 2 + (F n+1 F n )x + F n+1 The height satisfies H = F n+1 + F n = F n+2 α n+2 / 5 And we have F n+1 = αn+1 β n+1 F n α n β n where β = Thus, b/a = F n+1 /F n = α + O(α 2n ) = α + O(1/H 2 ), and sep(p) 1/H 2 Put P = P 1 If α + ε is a root of P where ε is sall, then P(α + ε) = P(α + ε) 1 = εp (α) + ε2 2 P (α) + ε3 3! P (α) 1 = 0 We have P (α) a 1, P (b/a) a 1 and P (b/a) 2a(2b/a 1) 2 5a Hence a 2 5ε 2 + ε a = 0 ε ±1/ a Therefore, sep( P) 1/ H Conclusion: When perturbating P by η = 1, the roots of perturbated polynoial deviate considerably Exaple 2 Let us consider the following polynoial P(x) = (ax 1)(x n ax + 1) Denote H the height of P Then H = a 2 (a 1) Then one root, say α, of the second factor of this polynoial is in the neighborhood of 1/a Indeed, P(1/a) = 0 and aα 1 = α n a n Let ε sufficiently sall such that P(1/a + ε) = 0 We have P(1/a + ε) = εa(1/a + ε)n εa Thus ε a (n+1) Therefore, Put P 1 (x) = P(x) + x n+1 Therefore, sep(p) a (n+1) P 1 (1/a + ε) = (1/a + ε) n+1 + εa ( (1/a + ε) n εa ) For ε sufficiently sall such that P 1 (1/a + ε) = 0, we have the approxiation: Thus we have two possible roots, say ε 1 and ε 2 : a (n+1) + εa (n 1) ε 2 a 2 = 0 ε a (n 1) 2 2 and ε a (n 1) 2 2 Hence sep(p 1 ) a (n 1)

4 Exaple 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, z k = a 1/n exp(i2kπ/n), k = 0, 1,,n 1, η<η 0 = in { 1, sep(p) n} (H + 2) n+1 1 in {1, a(2 sin π/n)n } a n For a = 2853 and n = 16, P(z) = z , η 0 = (2 sin π/16) 16 (2853) 15, η We notice that in this case, for which the roots of P are very well separated, these roots reain very stable under perturbation of the coefficients, for exaple, the real roots of P are: ± where the real roots of perturbated polynoial are: ± We have the sae reark with the other roots Exaple 4 Let P(x) = (x 1)(x 2)(x 3), and Q(x) = 0009x 3 + P(x) We have sep(p) = 1 The roots of Q are: , ± i and sep(q) For ε 1/2, D 2 and D 3 (see notation in the theore) do not contain any root of Q Wehave η 0 = (H + 2) 4 1 < Now, by taking η = (< η 0 ), Q(x) = ηx 3 + P(x) And then, the roots of Q are , , and , which are rather close to the roots of the initial polynoial P, and each disk D i contain exactly one root of Q Reark 3 To keep integer coefficients, instead of the above Q, we could consider the polynoial Q 1 (x) = x 3 +ap(x) where a is a large positive integer Exaple 5 The polynoial P(x) = 20 i=1 (x i) of Wilkinson satisfies H = and sep(p) = 1 By the theore, η<η 0 = (H + 2) This value of η 0, is extraordinarily sall and sees very pessiistic But, indeed, Wilkinson shows that a very sall perturbation (η = 10 9 ) causes a draatic change of soe of the roots for exaple the perturbated polynoial adits the coplex roots ± I Exaple 6 Let P(x) be the polynoial defined by P(x) = x n (ax 1) 2 The height of this polynoial is H = a 2 as soon as a > 2 Moreover, P(1/a) = a n We consider a 1 Let ε sufficiently sall such that P(1/a + ε) = 0 Then P(1/a + ε) = (1/a + ε)n ((1/a + ε)a 1) 2 = (1/a + ε) n ε 2 a 2 Hence, Then fro a n ε 2 a 2 0 we obtain P(1/a + ε) a n ε 2 a 2 ε a (n+2)/2, 54

5 thus we obtain sep(p) 2a (n+2)/2 Put P 1 (x) = P(x) + x 2 WehaveP 1 (x) = x n (ax 1) 2 + x 2 Let ε sufficiently sall such that P(1/a + ε) = 0 Then Since a 1 and ε 1, we have P 1 (1/a + ε) = (1/a + ε) n ε 2 a 2 + (1/a + ε) 2 = 0 P 1 (1/a + ε) ε 2 a 2 + a 2 Thus ε ±a 2 Hence sep(p 1 ) 2a 2 If we consider the polynoial P 2 (x) = P(x) + x n, the height of P 2 is a 2 for a 1 Let ε sufficiently sall such that P(1/a + ε) = 0 We have, 2a n ε 2 a 2 0 Thus ε 2 1/2 a (n+2)/2, hence sep(p 2 ) 2 2a (n+2)/2 References Henrici, P (1974) Applied and coputional coplex analysis (Vol 1, Pure and applied atheatics) New York: John Wiley & Sons Inc Mignotte, M, Cerlienco, L, & Piras, F (1987) Coputing the easure of a polynoial Journal of Sybolic Coputation, 4(1), Marden, M (1989) Geoetry of polynoials (Matheatical surveys and onographs) United States of Aerica: Aerican Matheatical Society Ostrowski, A (1966) Solutions of equations and systes of equations New York: Acad Press Wilkinson, J H (1965) The algebraic eigenvalue proble (Monographs on nuerical analysis) London: Clarendon Press Copyrights Copyright for this article is retained by the author(s), with first publication rights granted to the journal This is an open-access article distributed under the ters and conditions of the Creative Coons Attribution license ( 55