因此，当我尝试运行/构建代码时，它会卡在该阶段，没有显示错误，但代码没有发生任何事情，因此我尝试调试它，并尝试让它绘制这样的图表，

这是我试图运行的代码。当我试图构建它时，没有输出。更新后，我得到了一个ValueError：一个数学域错误，我想知道如何在我的代码中修复它，以便它可以输出上面的图形。

Traceback (most recent call last):

  File "file_location", line 19, in g
    return (f(a + (2 * j - 2) * h) + 4 * f(a + (2 * j - 1) * h) + f(a + (2 * j) * h))
  File "file_location", line 9, in f
    return math.acos(arg_cos / (1 + 2 * arg_cos))

ValueError: math domain error

代码

import math
import numpy as np
import time
import multiprocessing as mp
import matplotlib.pyplot as plt

def f(x):
    return math.acos(math.cos(x) / (1 + 2 * math.cos(x)))

b = math.pi / 2
a = 0
exact = math.pi ** 2 / 16
h = (b - a)

def g(j):
    global a
    global h
    return (f(a + (2 * j - 2) * h) + 4 * f(a + (2 * j - 1) * h) + f(a + (2 * j) * h))

if __name__ == '__main__':
    mp.freeze_support()  #needed for Windows

    pool = mp.Pool(processes=mp.cpu_count())

    serial_times = []
    parallel_times = []
    intervals = []

    for i in range(1, 10):
        n = 10 ** i
        h = (b - a) / n
        N = int(n / 2)
        intervals.append(N)
    
        start_time = time.time()
        p = pool.map(g, np.linspace(1, N, N, dtype=np.int32))
        P = (h / 3) * np.sum(p, dtype=np.float32)
        elapsedTime = time.time() - start_time
        parallel_times.append(elapsedTime)

        start_time = time.time()
        S = 0
        for j in range(1, N+1):
            S += f(a + (2 * j - 2) * h) + 4 * f(a + (2 * j - 1) * h) + f(a + (2 * j) * h)
        S *= h / 3
        elapsedTime = time.time() - start_time
        serial_times.append(elapsedTime)

    #Plotting
    plt.plot(intervals, serial_times, label='Serial Computation')
    plt.plot(intervals, parallel_times, label='Parallel Computation')
    plt.xlabel('Interval Count (N)')
    plt.ylabel('Runtime (s)')
    plt.xscale('log')
    plt.yscale('log')
    plt.title('Simpson\'s Rule Computation Time vs Interval Count')
    plt.legend()
    plt.show()