| | | Exercice 3 : Eléments de solutions |
Posons : F(x) = SUM_i >= 0 F_i x^i = F_0 + F_1 + SUM_i >= 2 F_i x^i F(x) = F_0 + F_1 + SUM_i >= 2 (F_i-2 + F_i-1) x^i F(x) = F_0 + F_1 + SUM_i >= 2 F_i-2 x^i + SUM_i >= 2F_i-1 x^i F(x) = F_0 + F_1 + x^2SUM_i >= 2 F_i-2 x^i-2 + x SUM_i >= 2F_i-1 x^i-1 F(x) = F_0 + F_1 + x^2SUM_i >= 0 F_i x^i + x SUM_i >= 1F_i x^i F(x) = F_0 + F_1 + x^2 F(x) + x (SUM_i >= 0F_i x^i) - x F_0 F(x) = F_0 + F_1 + x^2 F(x) + x F(x) - x F_0 (1-x^2-x) F(x) = F_0 + F_1 - xF_0 F(x) = (F_0 + F_1 - xF_0)/(1-x^2-x)
Soit dans le cas qui nous concerne : F(x) = (1)/(1-x^2-x)= (1)/((x-x_1)(x-x_2)) Car 1-x2-x est un polynôme ayant deux racines réelles : x1 = (1+sqrt(5))/(2), x2 = (1-sqrt(5))/(2)
On veut écrire F(x) sous la forme d'une somme de fraction simple : F(x) = (a)/(x-x_1) + (b)/(x-x_2) T_a(x) = (x-x_1) F(x) = (1)/(x-x_2) a=T_a(x_1) (1)/(x_1-x_2)= (1)/(sqrt(5))= (sqrt(5))/(5) T_b(x) = (x-x_2) F(x) = (1)/(x-x_1) b= T_b(x_2) = (1)/(x_2-x_1)= - (1)/(sqrt(5))= - (sqrt(5))/(5) Donc F(x)= (sqrt(5))/(5) (1)/((x-x_1)) - (sqrt(5))/(5) (1)/((x-x_2)) F(x)= -(sqrt(5))/(5) (1)/((x_1-x)) + (sqrt(5))/(5) (1)/((x_2-x)) F(x)= -(sqrt(5))/(5x_1) (1)/((1-(x)/(x_1))) + (sqrt(5))/(5x_2) (1)/(1-(x)/(x_2))
Or (1)/(x1) = (2)/(1+sqrt(5))= (2(1-sqrt(5)))/((1+sqrt(5))(1-sqrt(5)))
(1)/(x1) = - (1-sqrt(5))/(2) = -x2
De même : (1)/(x2) = -x1.
F(x) = (sqrt(5)x_2)/(5) (1)/(1-(-x_2) x) - (sqrt(5)x_1)/(5) (1)/(1-(-x_1) x)
Or
(1)/(1-alphax) = SUM_n >= 0 alpha^n x^n
Donc
F_n = (sqrt(5)x_2)/(5) (-x_2)^n - (sqrt(5)x_1)/(5) (-x_1)^n
| | | Exercice 3 : Eléments de solutions |