Question

我想使用scipy在python中使用sin（x）/ x的quadratures来定量积分。 n = 256.它看起来效果不好：

 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()

精确积分计算得很好，但是我得到了一个带有正交误差的错误......出了什么问题？

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

提前致谢！

Answer 1

您的代码中至少存在一些问题：

linspace 从0开始，因此当您评估要集成的函数时，在梯形积分的开头，您有：sin(0)/0 = nan。您应该使用数字零而不是精确零（在下面的示例中我使用1e-12）
当您获得第一个nan，nan + 1.0 = nan时：这意味着在您的代码中，当您在某个时间间隔内总计积分时，第一个nan会弄乱所有你的结果。
仅适用于python 2 ：除法2/256是2个整数之间的除法，结果为0。请尝试使用2.0/256.0（感谢@MaxU指出这一点）。

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

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)

输出：

('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的限制，但由于sin(1e-12)/1e-13 = 1的浮动四舍五入！

Answer 2

对于x == 0，你可以使函数返回1（sin（x）/ x的极限为0）而不是NaN。这样，你就不必作弊和改变您为了排除0而整合的时间间隔。

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表示＆＃34;不是数字＆＃34;。在你的情况下基本无限。创建时：

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

你创建一个值为0的数组，然后你试着除以x，你就不能被0 ...

一种解决方案是使用它：

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

给你这个：

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

越接近零，近似值就越好！

sin（x）/ x的数字积分

3 个答案: