On 06/05/15 14:55, Paul Royik wrote:
> For example,
> x^5+y^5=7
> x*sin(y)=1
Newton method is perfectly fine here
var('x,y')
f(x,y) = x^5 + y^5 - 7
g(x,y) = x*sin(y) - 1
F(x,y) = (f, g)
m = F.derivative()
V = VectorSpace(RDF, 2)
v = V((2,2))
for _ in range(10):
fv = F(*v)
v = V(m(*v).solve_right(-fv) + v)
print v
print F(*v)
I obtain the approximate solution v which is
x = 1.0102139894432696
y = 1.428474139287505
Vincent
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