一、簡介

高斯求積公式是變步長數值積分的一種，基本形式是計算[-1,1]上的定積分。理論證明對於 n個節點的上述求積公式，最高有 2n - 1 次的代數精度，高斯公式就是使得上述公式具有 2n - 1次代數精度的積分公式。

詳解

二、實現

# -*- coding: utf-8 -*-
"""
Created on Mon Dec 19 13:50:15 2016

    Gauss型求積公式

@author: Administrator
"""

from numpy import *
from scipy import integrate

import matplotlib.pyplot as
 
 plt

#(-1,1)
def f1(x):
    return x**2 / (1 - x**2)**(1/2)

#(0,pi/2]
def f2(x):
    return sin(x) / x

def f2_symbol(x):
    from sympy import sin
    return sin(x) / x

def gauss_legendre(f,a,b,n):
    from sympy import Symbol
    from sympy import diff
    from sympy import solve
    from sympy import
 
 Eq
    from math import factorial
    x = Symbol('x')
    l = diff((x**2 - 1)**(n+1),x,n+1)
    l = l  / (2**(n+1) * factorial(n+1))
    root = solve(Eq(l,0))
    A = []
    result = 0
    l_derivative = diff(l,x,1)
    for i in range(0,n+1):
        l_derivative_value = (l_derivative).evalf(subs={x:root[i]})
        A.append(2
 
 / ((1 - root[i]**2) *  l_derivative_value**2))
        result += (b -a) / 2 * A[i] * f((a+b)/2 + (b-a)/2*root[i])
    return result

if __name__ == '__main__':
    n = 4 # 1 2 4
    print 'f1:'
    print 'Truth-value:',integrate.quad(f1,-1,1)[0]
    print 'Estimated-value:',float(gauss_legendre(f1,-1,1,n))
    print
    print 'f2:'
    print 'Truth-value',integrate.quad(f2,0,pi/2)[0]
    print 'Estimated-value:',float(gauss_legendre(f2_symbol,0,pi/2,n))
    zero = 1e-10
    x1 = linspace(-1,1,500)
    x1[0] += zero
    x1[-1] -= zero
    y1 = f1(x1)
    x2 = linspace(0,pi/2,500)
    x2[0] += zero
    y2 = f2(x2)
    fig = plt.figure(figsize=(8,6))
    ax1 = fig.add_subplot(211)
    ax2 = fig.add_subplot(212)
    ax1.set_xlabel('x')
    ax1.set_ylabel('y')
    ax2.set_xlabel('x')
    ax2.set_ylabel('y')
    ax1.plot(x1,y1,color='b',linestyle='-',label='f1(x)')
    ax1.legend(loc='upper left')
    ax2.plot(x2,y2,color='r',linestyle='-',label='f2(x)')
    ax2.legend(loc='lower left')
    fig.show()
    fig.savefig('a.png')