Can anyone help me solve this problem using Newton's Method?
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Using Newton's Method solve the equations:
x^3 + y^2 + x - 0.715 = 0
x^2 + y^3 - y - 0.523 = 0
with (x0, y0) = (1, 0).
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Sergey Kasyanov
le 13 Mar 2021
Hello,
try that
i = 1;
V0 = [1;0];%[x0;y0]
V = V0 + 1;
while max(abs(V0 - V)) > 1e-10 && i < 1e2
V = V0;
J = [3*V(1)^2+1 2*V(2)
2*V(1) 3*V(2)^2-1];
F = [V(1)^3+V(2)^2+V(1)-0.715
V(1)^2+V(2)^3-V(2)-0.523];
V0 = V - F./(J*V);
i = i + 1;
end
x = V(1);
y = V(2);
3 commentaires
Sergey Kasyanov
le 15 Mar 2021
There are some solutions of the equations system. You can try to find another solutions by changing start point or by adding and varying coefficient k in equation (0<k<1):
V0 = V - k * F./(J*V);
There are three solution near the 0: [-0.46, 1.13], [0.38;-0.54], [0.5;-0.3].
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