如何使用多变量牛顿拉夫森方法进行和总循环的功能

Question

我目前正在尝试通过Python中的MLE方法估算分布的参数。这些是我的loglikelihood函数的导数： loglikelihood partialionhienhiohieny parterial partivative ，

您可以看到它具有相当复杂的公式，我需要得出一个非常复杂的公式根部的根（我想使用牛顿拉夫森方法）这是我写代码的方式：

from math import cos, sin, pi, exp, sqrt, log
from scipy.special import psi
import autograd as ad
from autograd import grad, jacobian
import autograd.numpy as np
import sympy as sy

x = [0.1, 0.2, 1, 1, 1, 1, 1, 2, 3, 6, 7, 11, 12, 18, 18, 18, 18, 18, \
 21, 32, 36, 40, 45, 46, 47, 50, 55, 60, 63, 63, 67, 67, 67, 67, 72, 75, 79, \
 82, 82, 83, 84, 84, 84, 85, 85, 85, 85, 85, 86, 86]

p = [2.5, 0.00003, 0.1, 0.02, 0.1]
p0 = p[0]
p1 = p[1]
p2 = p[2]
p3 = p[3]
p4 = p[4]

def Fs():
n = len(x)
sum1 = 0
sum2 = 0
sum3 = 0
sum4 = 0
sum5 = 0
sum6 = 0
sum7 = 0
sum8 = 0
sum9 = 0
sum10 = 0

for i in range(1,n):
    z = exp(-(p1 / p2) * (exp(p2 * x[i]) - 1))
    a = 1 - z
    b = a ** p0
    c = exp(p2 * x[i])
    sum1 = sum1 + log(a)
    sum2 = sum2 + b * log(a) / log(1 - b)
    sum3 = sum3 + c
    sum4 = sum4 + (c - 1) * z / a
    sum5 = sum5 + (c - 1) * z * (a ** (p0 - 1)) / (1 - b)
    sum6 = sum6 + x[i]
    sum7 = sum7 + c * (1 - p2 * x[i])
    sum8 = sum8 + (p2 * x[i] * c - c + 1) * z / a
    sum9 = sum9 + (p2 * x[i] * c - c + 1) * z * \
    (a ** (p0 - 1)) / (1 - b)
    sum10 = sum10 + log(1 - b)

f1 = n / p0 + p3 * sum1 - (p4 - 1) * sum2
f2 = n / p1 + n / p2 - sum3 / p2 + sum4 * (p0 * p3 - 1) / \
 p2 + sum5 * p0 * (p4 - 1) / p2
f3 = -n * p1 / (p2 ** 2) + sum6 + p1 * sum7 / (p2 ** 2) - \
 p1 * (p3 * p0 - 1) * sum8 / (p2 ** 2) + \
 p0 * p1 * (p4 - 1) * sum9 / (p2 ** 2)
f4 = -n * (psi(p3) + psi(p3 + p4)) + p0 * sum1 
f5 = -n * (psi(p4) + psi(p3 + p4)) + p0 * sum10

return np.matrix([[f1], [f2], [f3], [f4], [f5]])

因此，这里的问题是，我想使用jacobian（）获得NR方法的Jacobian矩阵。但是，我意识到，如果我运行我的FS函数，则输出将不以多个元素的形式数字，因此使Jacobian函数无法执行。我想知道有人可以帮助我将自己的功能作为多位数而不是价值观吗？有什么办法可以代表我的功能中的总和而无需循环？还是有什么建议，我如何解决这个问题？先感谢您。