sin(x)/x的数值积分

Question

我想用sin计算python中sin(x)/x的正定积分。N= 256。它似乎不太好用：

 from scipy import integrate

 exact = integrate.quad(lambda x : (np.sin(x))/x, 0, 2*np.pi)[0]
 print("Exact value of integral:", exact)

 # Approx of sin(x)/x by Trapezoidal rule
 x = np.linspace(0, 2*np.pi, 257)
 f = lambda x : (np.sin(x))/x
 approximation = np.trapz(f(x), x)
 print ("Estimated value of trapezoidal O(h^2):", round(approximation, 5), 
   '+', round((2/256)**2, 5))
 print ("real error:", exact - approximation)

 # Approx of sin(x)/x by Simpsons 1/3 rule
 approximation = integrate.simps(f(x), x)
 print("Estimated value of simpsons O(h^4):", round(approximation, 9), 
   '+', round((2/256)**4, 9))
 print ("real error:", exact - approximation)

 plt.figure()
 plt.plot(x, f(x))
 plt.show()

精确的积分计算得很好，但是我得到了一个四边形误差.怎么啦？

Exact value of integral: 1.4181515761326284
Estimated value of trapezoidal O(h^2): nan + 6e-05
real error: nan
Estimated value of simpsons O(h^4): nan + 4e-09
real error: nan

提前感谢！

Answer 1

在您的代码中至少存在一些问题：

1. 您的linspace从0开始，因此，当您计算要集成的函数时，在梯形积分的开头，有：sin(0)/0 = nan。您应该使用数字零而不是确切的零(在下面的示例中，我使用了1e-12)。
2. 当您得到第一个nan时，nan + 1.0 = nan：这意味着在您的代码中，当您对间隔上的积分进行求和时，第一个nan会扰乱您的所有结果。
3. 用于python2，只有：除法2/256是两个整数之间的除法，结果是0。尝试使用2.0/256.0 (谢谢@MaxU指出这一点)。

这是您修改的代码(我在python2中运行它，这是我现在使用的pc中安装的)：

from scipy import integrate
import numpy as np

exact = integrate.quad(lambda x : (np.sin(x))/x, 0, 2*np.pi)[0]
print("Exact value of integral:", exact)

# Approx of sin(x)/x by Trapezoidal rule
x = np.linspace(1e-12, 2*np.pi, 257) # <- 0 has become a numeric 0
f = lambda x : (np.sin(x))/x
approximation = np.trapz(f(x), x)
print ("Estimated value of trapezoidal O(h^2):", round(approximation, 5), 
  '+', round((2.0/256.0)**2, 5))
print ("real error:", exact - approximation)

# Approx of sin(x)/x by Simpsons 1/3 rule
approximation = integrate.simps(f(x), x)
print("Estimated value of simpsons O(h^4):", round(approximation, 9), 
  '+', round((2/256)**4, 9))
print ("real error:", exact - approximation)

它的产出：

('Exact value of integral:', 1.4181515761326284)
('Estimated value of trapezoidal O(h^2):', 1.41816, '+', 6e-05)
('real error:', -7.9895502944626884e-06)
('Estimated value of simpsons O(h^4):', 1.418151576, '+', 0.0)
('real error:', 2.7310242955991271e-10)

Discalimer sin(x)/x -> 1 for x -> 0的极限，但由于sin(1e-12)/1e-13 = 1的浮动四舍五入！

Answer 2

您可以使函数返回1( sin(x)/x在0中的限制)，而不是x == 0的NaN。这样，您就不必为了排除0而欺骗和更改集成的时间间隔。

import numpy as np
import matplotlib.pyplot as plt
from scipy import integrate

exact = integrate.quad(lambda x : (np.sin(x))/x, 0, 2*np.pi)[0]
print("Exact value of integral:", exact)


def f(x):
    out = np.sin(x) / x
    # For x == 0, we get nan. We replace it by the 
    # limit of sin(x)/x in 0
    out[np.isnan(out)] = 1
    return out

# Approx of sin(x)/x by Trapezoidal rule

x = np.linspace(0, 2*np.pi, 257)

approximation = np.trapz(f(x), x)
print ("Estimated value of trapezoidal O(h^2):", round(approximation, 5), 
  '+', round((2/256)**2, 5))
print ("real error:", exact - approximation)
 # Approx of sin(x)/x by Simpsons 1/3 rule
approximation = integrate.simps(f(x), x)
print("Estimated value of simpsons O(h^4):", round(approximation, 9), 
  '+', round((2/256)**4, 9))
print ("real error:", exact - approximation)
plt.figure()
plt.plot(x, f(x))
plt.show()

输出：

Exact value of integral: 1.4181515761326284
Estimated value of trapezoidal O(h^2): 1.41816 + 6e-05
real error: -7.98955129322e-06
Estimated value of simpsons O(h^4): 1.418151576 + 4e-09
real error: 2.72103006793e-10

Answer 3

NaN的意思是“不是一个数字”。在你的例子中，基本上是无限的。当您创建：

x = np.linspace(0, 2*np.pi, 257)

创建一个值为0的数组，然后尝试将其除以x，而不能除以0.

一种解决方案是使用以下方法：

x = np.linspace(0.1, 2*np.pi, 257)

这给了你这个

Exact value of integral: 1.4181515761326284
Estimated value of trapezoidal O(h^2): 1.31822 + 6e-05
real error: 0.099935104987
Estimated value of simpsons O(h^4): 1.318207115 + 4e-09
real error: 0.0999444614012

你越接近于零，近似值就越好！