numpy中高斯-勒让德正交的不同间隔

Question

我们如何numpy.polynomial.legendre.leggauss在非间隔内使用NumPy包[-1, 1]？

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下面的示例比较scipy.integrate.quad了整个时间间隔内的Gauss-
Legendre方法[-1, 1]。

import numpy as np
from scipy import integrate

# Define function and interval
a = -1.
b =  1.
f = lambda x: np.cos(x)

# Gauss-Legendre (default interval is [-1, 1])
deg = 6
x, w = np.polynomial.legendre.leggauss(deg)
gauss = sum(w * f(x))

# For comparison
quad, quad_err = integrate.quad(f, a, b)

print 'The QUADPACK solution: {0:.12} with error: {1:.12}'.format(quad, quad_err)
print 'Gauss-Legendre solution: {0:.12}'.format(gauss)
print 'Difference between QUADPACK and Gauss-Legendre: ', abs(gauss - quad)

输出：

The QUADPACK solution: 1.68294196962 with error: 1.86844092378e-14
Gauss-Legendre solution: 1.68294196961
Difference between QUADPACK and Gauss-Legendre:  1.51301193796e-12

Answer 1

要更改间隔，请使用以下方法将x值从[-1，1]转换为[a，b]：

t = 0.5*(x + 1)*(b - a) + a

然后将正交公式缩放为（b-a）/ 2：

gauss = sum(w * f(t)) * 0.5*(b - a)

这是您的示例的修改版本：

import numpy as np
from scipy import integrate

# Define function and interval
a = 0.0
b = np.pi/2
f = lambda x: np.cos(x)

# Gauss-Legendre (default interval is [-1, 1])
deg = 6
x, w = np.polynomial.legendre.leggauss(deg)
# Translate x values from the interval [-1, 1] to [a, b]
t = 0.5*(x + 1)*(b - a) + a
gauss = sum(w * f(t)) * 0.5*(b - a)

# For comparison
quad, quad_err = integrate.quad(f, a, b)

print 'The QUADPACK solution: {0:.12} with error: {1:.12}'.format(quad, quad_err)
print 'Gauss-Legendre solution: {0:.12}'.format(gauss)
print 'Difference between QUADPACK and Gauss-Legendre: ', abs(gauss - quad)

它打印：

QUADPACK解决方案：1.0，错误：1.11022302463e-14
高斯-勒根德雷解决方案：1.0
QUADPACK和Gauss-Legendre之间的差异：4.62963001269e-14