数值分析(2)-多项式插值: 拉格朗日插值法

整理一下数值分析的笔记~
目录：

1. 误差

2. 多项式插值与样条插值(THIS)

3. 函数逼近

4. 数值积分与数值微分

5. 线性方程组的直接解法

6. 线性方程组的迭代解法

7. 非线性方程求根

8. 特征值和特征向量的计算

9. 常微分方程初值问题的数值解

1. 拉格朗日基函数

定义. 设lk(x)l_k(x)lk(x)是n次多项式，在插值节点x0,x1,...,xnx_0,x_1,...,x_nx0,x1,...,xn上满足：

lk(xj)={1,j=k0,j≠kl_k(x_j)=\begin{cases} 1,&j=k\\ 0,&j \neq k \end{cases} lk(xj)={1,0,j=kj̸=k

则称lk(x)l_k(x)lk(x)为节点x0,x1,...,xnx_0,x_1,...,x_nx0,x1,...,xn上的拉格朗日插值基函数。

n=1 →\rarr→ 线性插值

问题定义：给定区间[xk,xk+1][x_k,x_{k+1}][xk,xk+1]及端点函数值：yk=f(xk),yk+1=f(xk+1)y_k=f(x_k),y_{k+1}=f(x_{k+1})yk=f(xk),yk+1=f(xk+1),求线性插值多项式L1(x)L_1(x)L1(x),使其满足L1(xk)=yk,L1(xk+1)=yk+1L_1(x_k)=y_k,L_1(x_{k+1})=y_{k+1}L1(xk)=yk,L1(xk+1)=yk+1。

不难看出几何上就是通过两点的直线，两点式有：L1(x)=xk+1−xxk+1−xkyk+x−xkxk+1−xkyk+1L_1(x)=\frac{x_{k+1}-x}{x_{k+1}-x_k}y_k+\frac{x-x_k}{x_{k+1}-x_k}y_{k+1}L1(x)=xk+1−xkxk+1−xyk+xk+1−xkx−xkyk+1

即L1(x)L_1(x)L1(x)是两个线性函数lk(x)=xk+1−xxk+1−xk,lk+1(x)=x−xkxk+1−xkl_k(x)=\frac{x_{k+1}-x}{x_{k+1}-x_k},l_{k+1}(x)=\frac{x-x_k}{x_{k+1}-x_k}lk(x)=xk+1−xkxk+1−x,lk+1(x)=xk+1−xkx−xk

的线性组合，系数分别是yk和yk+1y_k和y_{k+1}yk和yk+1,即：

L1(x)=yklk(x)+yk+1lk+1(x)L_1(x)=y_kl_k(x)+y_{k+1}l_{k+1}(x) L1(x)=yklk(x)+yk+1lk+1(x)

显然：

lk(xk)=1,lk(xk+1)=0lk+1(xk)=0,lk+1(xk+1)=1l_k(x_k)=1,l_k(x_{k+1})=0\\ l_{k+1}(x_k)=0,l_{k+1}(x_{k+1})=1 lk(xk)=1,lk(xk+1)=0lk+1(xk)=0,lk+1(xk+1)=1

称lk(x)和lk+1(x)l_k(x)和l_{k+1}(x)lk(x)和lk+1(x)为线性插值基函数。

n=2 →\rarr→ 抛物插值

同理，假定插值节点为xk−1,xk,xk+1x_{k-1},x_k,x_{k+1}xk−1,xk,xk+1,求二次插值多项式L2(x)L_2(x)L2(x)使其满足L2(xj)=yj,(j=k−1,k,k+1)L_2(x_j)=y_j,(j=k-1,k,k+1)L2(xj)=yj,(j=k−1,k,k+1),几何上其为通过三点(xk−1,yk−1),(xk,yk),(xk+1,yk+1)(x_{k-1},y_{k-1}),(x_k,y_k),(x_{k+1},y_{k+1})(xk−1,yk−1),(xk,yk),(xk+1,yk+1)的抛物线，用基函数的方法求L2(x)L_2(x)L2(x)的表达式，此时基函数lk−1(x),lk(x),lk+1(x)l_{k-1}(x),l_k(x),l_{k+1}(x)lk−1(x),lk(x),lk+1(x)是二次函数，且在节点上满足条件：

lk−1(xk−1)=1,lk−1(xk)=0,lk−1(xk+1)=0lk(xk−1)=0,lk(xk)=1,lk(xk+1)=0lk+1(xk−1)=0,lk+1(xk)=0,lk+1(xk+1)=1l_{k-1}(x_{k-1})=1,l_{k-1}(x_{k})=0,l_{k-1}(x_{k+1})=0\\ l_k(x_{k-1})=0,l_k(x_k)=1,l_k(x_{k+1})=0\\ l_{k+1}(x_{k-1})=0,l_{k+1}(x_k)=0,l_{k+1}(x_{k+1})=1 lk−1(xk−1)=1,lk−1(xk)=0,lk−1(xk+1)=0lk(xk−1)=0,lk(xk)=1,lk(xk+1)=0lk+1(xk−1)=0,lk+1(xk)=0,lk+1(xk+1)=1

以lkl_{k}lk为例，由上面的式子可以知道它有两个零点xk−1,xk+1x_{k-1},x_{k+1}xk−1,xk+1,有lk(x)=A(x−xk−1)(x−xk+1)l_k(x)=A(x-x_{k-1})(x-x_{k+1})lk(x)=A(x−xk−1)(x−xk+1),其中A为待定系数，根据lk(xk)=1l_{k}(x_{k})=1lk(xk)=1定出：

A=1(xk−xk−1)(xk−xk+1)A=\frac{1}{(x_k-x_{k-1})(x_k-x_{k+1})} A=(xk−xk−1)(xk−xk+1)1

于是有：

lk(x)=(x−xk−1)(x−xk+1)(xk−xk−1)(xk−xk+1)l_k(x)=\frac{(x-x_{k-1})(x-x_{k+1})}{(x_k-x_{k-1})(x_k-x_{k+1})} lk(x)=(xk−xk−1)(xk−xk+1)(x−xk−1)(x−xk+1)

同理：

lk−1(x)=(x−xk)(x−xk+1)(xk−1−xk)(xk−1−xk+1)lk+1(x)=(x−xk−1)(x−xk)(xk+1−xk−1)(xk+1−xk)l_{k-1}(x)=\frac{(x-x_{k})(x-x_{k+1})}{(x_{k-1}-x_k)(x_{k-1}-x_{k+1})}\\ l_{k+1}(x)=\frac{(x-x_{k-1})(x-x_{k})}{(x_{k+1}-x_{k-1})(x_{k+1}-x_{k})} lk−1(x)=(xk−1−xk)(xk−1−xk+1)(x−xk)(x−xk+1)lk+1(x)=(xk+1−xk−1)(xk+1−xk)(x−xk−1)(x−xk)

可得二次插值多项式：

L2(x)=yk−1lk−1(x)+yklk(x)+yk+1lk+1(x)L_2(x)=y_{k-1}l_{k-1}(x)+y_{k}l_{k}(x)+y_{k+1}l_{k+1}(x) L2(x)=yk−1lk−1(x)+yklk(x)+yk+1lk+1(x)

2. 拉格朗日插值多项式

推广至一般情形，构造通过n+1个节点的n次插值多项式Ln(x)L_n(x)Ln(x).根据插值定义有：Ln(xj)=yj,(j=0,1,...,n)L_n(x_j)=y_j,(j=0,1,...,n)Ln(xj)=yj,(j=0,1,...,n)。为此先定义n次插值基函数：

定义1 若n次多项式Lj(x)(j=0,1,2,...,n)L_j(x)(j=0,1,2,...,n)Lj(x)(j=0,1,2,...,n)在n+1个节点x0<x1<...<xnx_0<x_1<...<x_nx0<x1<...<xn上满足条件：

lk(xj)={1,j=k0,j≠k,(j,k=0,1,...,n)l_k(x_j)=\begin{cases} 1,&j=k\\ 0,&j \neq k \end{cases},(j,k=0,1,...,n) lk(xj)={1,0,j=kj̸=k,(j,k=0,1,...,n)

就称这n+1个n次多项式l0(x)，l1(x),...,ln(x)l_0(x)，l_1(x),...,l_n(x)l0(x)，l1(x),...,ln(x)为节点x0,x1,...,xnx_0,x_1,...,x_nx0,x1,...,xn上的n次插值基函数，类似地：

lk(x)=(x−x0)...(x−xk−1)(x−xk+1)...(x−xn)(xk−x0)...(xk−xk−1)(xk−xk+1)...(xk−xn)l_k(x)=\frac{(x-x_0)...(x-x_{k-1})(x-x_{k+1})...(x-x_n)}{(x_k-x_0)...(x_k-x_{k-1})(x_k-x_{k+1})...(x_k-x_n)} lk(x)=(xk−x0)...(xk−xk−1)(xk−xk+1)...(xk−xn)(x−x0)...(x−xk−1)(x−xk+1)...(x−xn)

由此可得插值多项式：

Ln(x)=∑k=0nyklk(x)L_n(x)=\sum^n_{k=0}y_kl_k(x) Ln(x)=k=0∑nyklk(x)

称为拉格朗日插值多项式。为了方便表示，引入记号：

ωn+1=(x−x0)(x−x1)...(x−xn)且有ωn+1′(xk)=(xk−x0)...(xk−xk−1)(xk−xk+1)...(xk−xn)\omega_{n+1}=(x-x_0)(x-x_1)...(x-x_n)\\ 且有\omega'_{n+1}(x_k)=(x_k-x_0)...(x_k-x_{k-1})(x_k-x_{k+1})...(x_k-x_n) ωn+1=(x−x0)(x−x1)...(x−xn)且有ωn+1′(xk)=(xk−x0)...(xk−xk−1)(xk−xk+1)...(xk−xn)

则拉格朗日插值多项式可以写为：

Ln(x)=∑k=0nωn+1(x)(x−xk)ωn+1′(xk)L_n(x)=\sum^n_{k=0}\frac{\omega_{n+1}(x)}{(x-x_k)\omega'_{n+1}(x_k)} Ln(x)=k=0∑n(x−xk)ωn+1′(xk)ωn+1(x)

关于插值多项式存在惟一性有一个定理：

定理1 在次数不超过n的多项式集合 HnH_nHn中，满足Ln(xj)=yj,(j=0,1,...,n)L_n(x_j)=y_j,(j=0,1,...,n)Ln(xj)=yj,(j=0,1,...,n)的插值多项式Ln(x)∈HnL_n(x)\in H_nLn(x)∈Hn存在且惟一。

惟一性证明：假定另有P(x)∈Hn使得P(xj)=f(xj),i=0,1,...,nP(x) \in H_n使得P(x_j)=f(x_j),i=0,1,...,nP(x)∈Hn使得P(xj)=f(xj),i=0,1,...,n成立，于是有Ln(xi)−P(xi)=0对i=0,1,2...,nL_n(x_i)-P(x_i)=0对i=0,1,2...,nLn(xi)−P(xi)=0对i=0,1,2...,n成立，表明多项式Ln(xi)−P(xi)∈HnL_n(x_i)-P(x_i) \in H_nLn(xi)−P(xi)∈Hn有n+1个零点，这与n次多项式只有n个零点的基本定理矛盾。

eg,已知f(144)=12,f(169)=13,f(225)=15,作f(x)二次Lagrange插值多项式并求f(175)1的近似值。

解：x0=144,x1=169,x2=225,y0=12,y1=13,y2=15x_0=144,x_1=169,x_2=225,y_0=12,y_1=13,y_2=15x0=144,x1=169,x2=225,y0=12,y1=13,y2=15，则f(x)的二次Lagrange插值基函数为：

l0(x)=(x−x1)(x−x2)(x0−x1)(x0−x2)=(x−169)(x−225)(144−169)(144−225)l1(x)=(x−x0)(x−x2)(x1−x0)(x1−x2)=(x−144)(x−225)(169−144))(169−225)l2(x)=(x−x0)(x−x1)(x2−x0)(x2−x1)=(x−144)(x−169)(225−144)(225−169)l_0(x)=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}=\frac{(x-169)(x-225)}{(144-169)(144-225)}\\ l_1(x)=\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}=\frac{(x-144)(x-225)}{(169-144))(169-225)}\\ l_2(x)=\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}=\frac{(x-144)(x-169)}{(225-144)(225-169)} l0(x)=(x0−x1)(x0−x2)(x−x1)(x−x2)=(144−169)(144−225)(x−169)(x−225)l1(x)=(x1−x0)(x1−x2)(x−x0)(x−x2)=(169−144))(169−225)(x−144)(x−225)l2(x)=(x2−x0)(x2−x1)(x−x0)(x−x1)=(225−144)(225−169)(x−144)(x−169)

因此f(x)的二次Lagrange插值多项式为：

L2(x)=y0l0(x)+y1l1(x)+y2l2(x)L_2(x)=y_0l_0(x)+y_1l_1(x)+y_2l_2(x) L2(x)=y0l0(x)+y1l1(x)+y2l2(x)

知f(175)≈12l0175+13l1(175)+15l2(175)=13.23015873f(175) \approx 12l_0{175}+13l_1(175)+15l_2(175)=13.23015873f(175)≈12l0175+13l1(175)+15l2(175)=13.23015873

3. 插值余项和误差估计

在[a,b]上用Ln(x)L_n(x)Ln(x)近似f(x)，则其截断误差Rn(x)=f(x)−Ln(x)R_n(x)=f(x)-L_n(x)Rn(x)=f(x)−Ln(x),也称为插值多项式的余项，若f(x)的n阶导数连续，n+1阶导数存在，则:

Rn(x)=f(x)−Ln(x)=f(n+1)(ξ)(n+1)!ωn+1(x),其中ξ∈(a,b)且依赖于xR_n(x)=f(x)-L_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\omega_{n+1}(x),其中\xi \in (a,b)且依赖于x Rn(x)=f(x)−Ln(x)=(n+1)!f(n+1)(ξ)ωn+1(x),其中ξ∈(a,b)且依赖于x

通常情况下ξ\xiξ在(a,b)上具体位置未知，若可以求出maxa<x<b∣f(n+1)(x)∣=Mn+1max_{a<x<b}|f^{(n+1)}(x)|=M_{n+1}maxa<x<b∣f(n+1)(x)∣=Mn+1，则Rn(x)≤Mn+1(n+1)!∣ωn+1(x)∣R_n(x) \leq \frac{M_{n+1}}{(n+1)!}|\omega_{n+1}(x)|Rn(x)≤(n+1)!Mn+1∣ωn+1(x)∣。

Lagrange插值多项式的缺点：

插值基函数计算复杂
当增加一个新点时需重新计算
插值多项式从n次增加到n+1次需重新计算
插值曲线在节点处有尖点，不光滑，节点处不可导
高次插值的精度不一定高

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