Can anyone help me solve this problem using Newton's Method?

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Muzammil Arif
Muzammil Arif 2021-3-13
评论： Sergey Kasyanov 2021-3-15
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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回答（1 个）

Sergey Kasyanov
Sergey Kasyanov 2021-3-13
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);
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Sergey Kasyanov
Sergey Kasyanov 2021-3-15
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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