Python实现数值积分方式
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")
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