Culegere de probleme de Analiz˘a numeric˘a

10.2. Sisteme neliniare 163
Solut¸ie. F(x,y) = 0
G(x,y) = 0
x = xn +hn
F,g ∈ C 1
y = yn +kn
F(xn +hn,yn +kn) = 0
G(xn +hn,yn +kn) = 0
Utilizând formula lui Taylor se obt¸ine
F(xn,yn)+hnF ′ x (xn,yn)+knF ′ (xn,yn) = 0
G(xn,yn)+hnG ′ x (xn,yn)+knG ′ (xn,yn) = 0
Dacă jacobianul
J(xn,yn) =
obt¸inem
F′ x (xn,yn) F ′ y (xn,yn)
G ′ x(xn,yn) G ′ y(xn,yn)
= 0
1
hn = −
J(xn,yn) F(xn,yn) F ′ y (xn,yn)
G(xn,yn) G ′ y (xn,yn)
1
kn = −
J(xn,yn) F′ x (xn,yn) F(xn,yn)
G ′
x(xn,yn) G(xn,yn)
x0 = 1.2, y0 = 1.7
F(x0,y0) = −0.434
G(x0,y0) = 0.1956
J(x,y) =
6x2
−2y
=
8.64 −3.40
4.91 5.40 = 57.91
y 3 3xy 2 −1
h0 = 0.6349
k0 = −0.0390