numpy中高斯-勒让德正交的不同间隔
问题内容：
我们如何numpy.polynomial.legendre.leggauss在非间隔内使用NumPy包[-1, 1]？
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
问题答案：
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
