fsolve求方程的解
如果要求解方程：
\(\left \{ \begin{array}{ccc}
f1(u1,u2,u3)=0\\
f2(u1,u2,u3)=0\\
f3(u1,u2,u3)=0
\end{array}\right.\)
那么func这么定义：
def func(x):
u1,u2,u3=x
return [f1(u1,u2,u3),f2(u1,u2,u3),f3(u1,u2,u3)]
代码案例1：
import scipy.optimize as opt
import numpy as np
def f(x):
x0,x1,x2=x
return np.array([5*x1+3,4*x0*x0-2*np.sin(x1*x2),x1*x2-1.5])
result=opt.fsolve(f,[1,1,1])
print(result)
print(f(result))
代码案例2：
如果给了Jacobian矩阵，那么迭代速度更快
Jacobian矩阵的定义是：
\(\left [ \begin{array}{ccc}
\frac{\partial f1}{\partial u1}&\frac{\partial f1}{\partial u2}&\frac{\partial f1}{\partial u2}\\ \\
\frac{\partial f2}{\partial u1}&\frac{\partial f2}{\partial u2}&\frac{\partial f2}{\partial u2}\\ \\
\frac{\partial f3}{\partial u1}&\frac{\partial f3}{\partial u2}&\frac{\partial f3}{\partial u2}
\end{array} \right ]\)
import scipy.optimize as opt
import numpy as np
def obj_func(x):
x0,x1,x2=x
return [5*x1+3,4*x0*x0-2*np.sin(x1*x2),x1*x2-1.5]
def jacobian(x):
x0, x1, x2 = x
return [
[0,5,0],
[8*x0,-2*x2*np.cos(x1*x2),-2*x1*np.cos(x1*x2)],
[0,x2,x1]
]
result=opt.fsolve(obj_func,[1,1,1],fprime=jacobian)
print(result)
print(obj_func(result))
