Scipy optimize minimize scipy.optimize.minimize和Minuit返回初始猜测值

我在使用scipy.minimize.optimize时遇到问题。这是我的密码

from time import process_time 
import numpy as np
from scipy.optimize import minimize
class NMin(object):
    def __init__(self, error):
        self.error=error

    def func(self, N):
        i = np.arange(1, N+1)
        f = np.abs(np.sum(4/(N*(1+((i - 0.5)/N)**2))) - np.pi)-self.error
        return(f)

    def nMin(self):
        x0 = 1
        nMin = minimize(self.func, x0)
        return(nMin.x)


def main():
    t1_start = process_time()
    error=10**(-6)
    nMin = NMin(error).nMin()
    print("the minimum value of N is: " + str(nMin))
    t1_stop = process_time() 
    print("Elapsed time during the whole program in seconds:", 
                                         t1_stop-t1_start)

main ()  

我试图最小化函数

func（x）

以找到N最小值，但是

NMin（错误）。NMin（）

似乎返回

x0=1

而不是N最小值。这是我的输出

the minimum value of N is: [1.]
Elapsed time during the whole program in seconds: 0.015625

---

我真的被这个问题困扰，因为我似乎找不到问题所在，也不明白为什么

scipy.optimize

不起作用。

scipy.optimize.minimize

主要用于连续可微函数。在func中使用arange使其成为一个离散问题。由于这些不连续性，这会导致梯度出现大幅跳跃（见下图）

我添加了一些调试打印：

from time import process_time
import numpy as np
from scipy.optimize import minimize
class NMin(object):
    def __init__(self, error):
        self.error=error

    def func(self, N):
        print("func called N = {}".format(N))
        i = np.arange(1, N+1)
        print("i = {}".format(i))
        f = np.abs(np.sum(4/(N*(1+((i - 0.5)/N)**2))) - np.pi)-self.error
        print("f = {}".format(f))
        return(f)

    def nMin(self):
        x0 = 1
        nMin = minimize(self.func, x0)
        return(nMin.x)


def main():
    t1_start = process_time()
    error=10**(-6)
    nMin = NMin(error).nMin()
    print("the minimum value of N is: " + str(nMin))
    t1_stop = process_time()
    print("Elapsed time during the whole program in seconds:",
                                         t1_stop-t1_start)

main()

这导致：

func called N = [1.]
i = [1.]
f = 0.05840634641020706
func called N = [1.00000001]
i = [1. 2.]
f = 1.289175555623012

也许您希望使用更适合离散问题的不同解算器，或者更改目标以满足基于梯度的优化的连续性前提