146 lines
3.6 KiB
Go
146 lines
3.6 KiB
Go
// IntegralNewtonCotes
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/*
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------------------------------------------------------
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作者 : Black Ghost
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日期 : 2018-12-11
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版本 : 0.0.0
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------------------------------------------------------
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1-8级Newton-Cotes求积分公式
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理论：
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对于积分
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b n
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|f(x)dx ~= Sum Ak*f(xk)
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a k=0
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(n)
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Ak = (b-a)C
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k
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(n) (-1)^(n-k) n
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C = ------------ |t(t-1)(t-2)...(t-(k-1))(t-(k+1))...(t-n)dt
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k k!(n-k)!n 0
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特别的，n=1为梯形公式；
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n=2为Simpson（辛浦生）公式；
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n=4为Cotes（科特斯）公式
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参考 李信真, 车刚明, 欧阳洁, 等. 计算方法. 西北工业大学
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出版社, 2000, pp 145-153.
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------------------------------------------------------
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输入 :
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fun 被积分函数
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a, b 积分范围
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n Newton-Cotes公式级数
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输出 :
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sol 解
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err 解出标志：false-未解出或达到步数上限；
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true-全部解出
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------------------------------------------------------
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注意 ：
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由于误差得不到有效控制，稳定性无法保证，故而并不是n值越
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大越好，实际应用中很少使用n值较大的Newton-Cotes公式
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------------------------------------------------------
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*/
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package goNum
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// IntegralNewtonCotes 1-8级Newton-Cotes求积分公式
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func IntegralNewtonCotes(fun func(float64) float64, a, b float64, n int) (float64, bool) {
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/*
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1-8级Newton-Cotes求积分公式
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输入 :
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fun 被积分函数
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a, b 积分范围
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n Newton-Cotes公式级数
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输出 :
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sol 解
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err 解出标志：false-未解出或达到步数上限；
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true-全部解出
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*/
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//判断n
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if (n < 1) || (n > 8) {
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panic("Error in goNum.IntegralNewtonCotes: n is not correct")
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}
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//判断a, b
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if a == b {
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return 0.0, true
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}
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var sol float64
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var err bool = false
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//计算xi
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xi := ZeroMatrix(n+1, 1)
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for i := 0; i < n+1; i++ {
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xi.Data[i] = a + (b-a)*float64(i)/float64(n)
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}
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//系数切片
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coeff := [][]float64{
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{1.0, 1.0},
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{1.0, 4.0, 1.0},
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{1.0, 3.0, 3.0, 1.0},
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{7.0, 32.0, 12.0, 32.0, 7.0},
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{19.0, 75.0, 50.0, 50.0, 75.0, 19.0},
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{41.0, 216.0, 27.0, 272.0, 27.0, 216.0, 41.0},
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{751.0, 3577.0, 1323.0, 2989.0, 2989.0, 1323.0, 3577.0, 751.0},
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{989.0, 5888.0, -928.0, 10496.0, -4540.0, 10496.0, -928.0, 5888.0, 989.0},
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}
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//计算积分值
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switch n {
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case 1:
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for i := 0; i < n+1; i++ {
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sol += coeff[0][i] * fun(xi.Data[i])
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}
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sol = sol * (b - a) / 2.0
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return sol, true
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case 2:
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for i := 0; i < n+1; i++ {
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sol += coeff[1][i] * fun(xi.Data[i])
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}
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sol = sol * (b - a) / 6.0
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return sol, true
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case 3:
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for i := 0; i < n+1; i++ {
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sol += coeff[2][i] * fun(xi.Data[i])
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}
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sol = sol * (b - a) / 8.0
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return sol, true
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case 4:
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for i := 0; i < n+1; i++ {
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sol += coeff[3][i] * fun(xi.Data[i])
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}
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sol = sol * (b - a) / 90.0
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return sol, true
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case 5:
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for i := 0; i < n+1; i++ {
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sol += coeff[4][i] * fun(xi.Data[i])
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}
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sol = sol * (b - a) / 288.0
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return sol, true
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case 6:
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for i := 0; i < n+1; i++ {
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sol += coeff[5][i] * fun(xi.Data[i])
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}
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sol = sol * (b - a) / 840.0
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return sol, true
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case 7:
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for i := 0; i < n+1; i++ {
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sol += coeff[6][i] * fun(xi.Data[i])
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}
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sol = sol * (b - a) / 17280.0
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return sol, true
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case 8:
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for i := 0; i < n+1; i++ {
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sol += coeff[7][i] * fun(xi.Data[i])
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}
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sol = sol * (b - a) / 28350.0
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return sol, true
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default:
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return 0.0, err
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}
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return sol, err
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}
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