fixed dependencies
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94
vendor/github.com/nuknal/goNum/InterpNewtonForward.go
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94
vendor/github.com/nuknal/goNum/InterpNewtonForward.go
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// InterpNewtonForward
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/*
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------------------------------------------------------
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作者 : Black Ghost
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日期 : 2018-12-6
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版本 : 0.0.0
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------------------------------------------------------
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计算x点n次Newton前向插值结果,拟合n+1个等距数据点
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Newton前向等距节点插值,满阶插值,即阶数为给定点数-1
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理论:
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(-1)^y0 (-1)^2y0
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f(x) = f(x0) + --------(x-x0)/h + ---------(x-x0)(x-x1) +
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h 2!h^2
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... +
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(-1)^ny0
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----------(x-x0)(x-x1)...(x-x_(n-1))
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n!h^n
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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 107-110.
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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等距分布;第二列yi
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xq 插值点, xq!=xi
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输出 :
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sol xq点插值结果
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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
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import "math"
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//k阶差分
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func difff_InterpNewtonForward(A Matrix, k int) float64 {
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sol := A.GetFromMatrix(k, 1) //yk
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for s := 1; s <= k; s++ {
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sol += math.Pow(-1.0, float64(s)) * float64(Cnm(k, s)) * A.GetFromMatrix(k-s, 1)
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}
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return sol
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}
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// InterpNewtonForward 计算x点n次Newton前向插值结果,拟合n+1个等距数据点
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func InterpNewtonForward(A Matrix, xq float64) (float64, bool) {
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/*
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计算x点n次Newton前向插值结果,拟合n+1个等距数据点
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输入 :
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A 数据点矩阵,(n+1)x2,第一列xi等距分布;第二列yi
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xq 插值点, xq!=xi
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输出 :
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sol xq点插值结果
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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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//判断xq是否等于xi
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for i := 0; i < A.Rows; i++ {
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if math.Abs(xq-A.GetFromMatrix(i, 0)) < 1e-3 {
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return A.GetFromMatrix(i, 1), true
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}
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}
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//判断xi是否等距节点
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for i := 0; i < A.Rows; i++ {
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x0 := A.GetFromMatrix(0, 0)
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if math.Abs(A.GetFromMatrix(i, 0)-float64(i)*x0) < 1e-3 {
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panic("Error in goNum.InterpNewtonForward: xi is not in equidistance")
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}
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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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h := A.GetFromMatrix(n, 0) - A.GetFromMatrix(n-1, 0)
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BA := ZeroMatrix(n+1, 1)
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//计算
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BA.SetMatrix(0, 0, A.GetFromMatrix(0, 1)) //f(x0)
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sol = BA.GetFromMatrix(0, 0)
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for k := 1; k < n+1; k++ {
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//求差分
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BA.SetMatrix(k, 0, difff_InterpNewtonForward(A, k))
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//乘系数1/(k!h^k)
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BA.Data[k] = BA.Data[k] / (float64(Factorial(k)) * math.Pow(h, float64(k)))
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//求乘积
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for j := 0; j < k; j++ {
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BA.Data[k] = BA.Data[k] * (xq - A.GetFromMatrix(j, 0))
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}
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//累加
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sol += BA.Data[k]
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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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