数值积分 Python实现

原理：

利用复化梯形公式，复化Simpson公式，计算积分。

步骤：

import math

"""测试函数"""
def f(x,i):
    if i == 1:
        return (4 - (math.sin(x)) ** 2) ** 0.5
    if i == 2:
        if x == 0:
            return 1
        else:
            return math.sin(x) / x
    if i == 3:
        return (math.exp(x)) / (4 + x ** 2)
    if i == 4:
        return math.log(1+x,math.e) / (1 + x ** 2)

"""打印显示函数"""     
def p(i,n):
    return "第" + str(i) + "题,n=" + str(n) + "时的积分值为："

"""复化Simpson函数"""
def Simpson(a, b, n, i):
    h = (b - a) / (2 * n)
    F0 = f(a,i) + f(b,i)
    F1 = 0
    F2 = 0
    for j in range(1,2 * n):
        x = a + (j * h)
        if j % 2 == 0:
            F2 = F2 + f(x,i)
        else:
            F1 = F1 + f(x,i)
    SN = (h * (F0 + 2 * F2 + 4 * F1)) / 3
    print("复化Simpson函数" + p(i,n) + str("%-10.7f"%(SN)))
    return SN

def T(a, b, n, i):
    h = (b - a) / n
    F0 = f(a,i) + f(b,i)
    F = 0
    for j in range(1,n):
        x = a + (j * h)
        F = F + f(x,i)
    SN = (h * (F0 + 2 * F)) / 2
    print("复化梯形函数" + p(i,n) + str("%-10.7f"%(SN)))
    return SN

def SimpsonTimes(x):
    n = 1
    y = Simpson(0, math.pi/4, n, 1)
    while(abs(y - 1.5343916) > x):
        n = n + 1
        y = Simpson(0, math.pi/4, n, 1)
    else:
        return n

def Times(x):
    n = 1
    y = T(0, math.pi/4, n, 1)
    while(abs(y - 1.5343916) > x):
        n = n + 1
        y = T(0, math.pi/4, n, 1)
    else:
        return n


"""
    测试部分    
"""
Simpson(0, math.pi/4, 10, 1)
Simpson(0, 1, 10, 2)
Simpson(0, 1, 10, 3)
Simpson(0, 1, 10, 4)
Simpson(0, math.pi/4, 20, 1)
Simpson(0, 1, 20, 2)
Simpson(0, 1, 20, 3)
Simpson(0, 1, 20, 4)

T(0, math.pi/4, 10, 1)
T(0, 1, 10, 2)
T(0, 1, 10, 3)
T(0, 1, 10, 4)
T(0, math.pi/4, 20, 1)
T(0, 1, 20, 2)
T(0, 1, 20, 3)
T(0, 1, 20, 4)

print("复化梯形函数求解第一问，精度为0.00001时需要" + str(Times(0.00001)) + "个步数")
print("复化Simpson函数求解第一问，精度为0.00001时需要" + str(SimpsonTimes(0.00001)) + "个步数")
print("复化梯形函数求解第一问，精度为0.000001时需要" + str(Times(0.000001)) + "个步数")
print("复化Simpson函数求解第一问，精度为0.000001时需要" + str(SimpsonTimes(0.000001)) + "个步数")