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95 lines
2.5 KiB
Go
95 lines
2.5 KiB
Go
// InterpLagrange_test
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/*
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------------------------------------------------------
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作者 : Black Ghost
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日期 : 2018-12-3
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版本 : 0.0.0
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------------------------------------------------------
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求解n次拉格朗日Lagrange插值法拟合n+1个数据点
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满阶插值,即阶数为给定点数-1
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内插/外插
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理论:
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n omega0n+1(xq)
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Ln(xq) = Sum(-----------------------)
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k=0 (xq-xk)*omega1n+1(xk)
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n
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omega0n+1(xq) = Prod(xq-xi)
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i=0
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n
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omega1n+1(xk) = Prod (xk-xi)
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i=0,i!=k
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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 94-100.
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------------------------------------------------------
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输入 :
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A 数据点矩阵,(n+1)x2,第一列xi;第二列yi
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xq 插值点
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n 最大插值阶数 1 <= ... <= n
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输出 :
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sol 插值结果
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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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*/
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package goNum_test
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import (
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"testing"
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"github.com/chfenger/goNum"
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)
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// InterpLagrange 求解n次拉格朗日Lagrange插值法拟合n+1个数据点
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func InterpLagrange(A goNum.Matrix, xq float64) (float64, bool) {
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/*
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求解n次拉格朗日Lagrange插值法拟合n+1个数据点
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输入 :
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A 数据点矩阵,(n+1)x2,第一列xi;第二列yi
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xq 插值点
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n 最大插值阶数 1 <= ... <= n
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输出 :
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sol 插值结果
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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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var sol float64
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var err bool = false
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n := A.Rows - 1
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//计算系数矩阵
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for k := 0; k <= n; k++ {
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//1. 计算分子
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var temp0 float64 = 1.0
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for i := 0; i <= n; i++ {
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temp0 = temp0 * (xq - A.GetFromMatrix(i, 0))
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}
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temp0 = temp0 / (xq - A.GetFromMatrix(k, 0))
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//2. 计算分母
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var temp1 float64 = 1.0
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for i := 0; i <= n; i++ {
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if i != k {
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temp1 = temp1 * (A.GetFromMatrix(k, 0) - A.GetFromMatrix(i, 0))
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}
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}
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//3. 求和
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sol += temp0 * A.GetFromMatrix(k, 1) / temp1
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}
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err = true
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return sol, err
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}
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func BenchmarkInterpLagrange(b *testing.B) {
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A22 := goNum.NewMatrix(3, 2, []float64{1.0, 0.367879441,
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2.0, 0.135335283,
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3.0, 0.049787068})
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for i := 0; i < b.N; i++ {
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InterpLagrange(A22, 2.1)
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}
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}
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