Question

enter image description here当我尝试找到非线性方程的根时有一些例外我认为def f（x）的问题：先感谢您。 a = 3

def jacobian(f, x):
     h = 1.0e-4
     n = len(x)
     Jac = zeros([n,n])
     f0 = f(x)
     for i in arange(0, n, 1):
              tt = x[i]
              x[i] = tt + h
              f1= f(x)
              x[i] = tt
              Jac [:, i] = (f1 - f0)/h
     return Jac, f0


def newton(f, x, tol=1.0e-9):
     iterMax = 50
     for i in range(iterMax):
              Jac, fO = jacobian(f, x)
              if sqrt(dot(fO, fO) / len(x)) < tol:
                       return x, i
              dx = linalg.solve(Jac, fO)
              x = x - dx
     print ("Too many iterations for the Newton method")
n = 2


def f(x):
     f = zeros([n])
     for i in arange(0,n):
        f [1] = 3*x[n-1]^2 - x*[n-1] + x[n]*2 - 1  # my two equations
        f[n-1] = x[n] - math.tan(x[n-1])
     return f
x0 = zeros([n])
x, iter = newton(f, x0)

Answer 1

根据图像索引超出范围。我认为您的问题出在评论#my two equations附近的两行中。当您执行x[n]时，数组超出范围，我假设您的数组大小为2。别忘了数组的大小从0开始，所以您访问的数组越界。

def f(x):
     f = zeros([n])
     for i in arange(0,n):
        f [1] = 3*x[n-1]^2 - x*[n-1] + x[n]*2 - 1  # my two equations - error is here when accessing x[n]
        f[n-1] = x[n] - math.tan(x[n-1]) # also here, accessing x[n] again
     return f

牛顿法求解非线性方程组的问题

1 个答案: