多项式插值，Lagrange，Neville，Newton

2024-07-15 16:59| 来源: 网络整理| 查看: 265

理论公式

Lagrange插值公式为： $P(x)=\sum_{k=1}^{n+1}f(x_{k})q_{k}(x)$ ，其中 $q_{k}=\frac{\prod_{i=1}^{n+1}(x-x_{i})}{(x-x_{k})\prod_{i=1,i\neq k}^{n+1}(x_{k}-x_{i})}$ 。

$P(x_{k})=f(x_{k})$ 可以从 $q_{k}$ 的结构看出 $q_{k}(x_{k})=1,q_{k}(x_{i,i\neq k})=0$ ,则 $P(x_{k})=f(x_{k})$ .

Neville插值公式为： $P_{1,2,3,...,n,n+1}(x)=\frac{1}{x_{n+1}-x_{1}}\begin{vmatrix} x-x_{1}&P_{1,2,...n}(x) \\ x-x_{n+1}&P_{2,3,...n,n+1}(x) \end{vmatrix}$ ，其中 $P_{i,i+1}(x)=\frac{f(x_{i})-f(x_{i+1})}{x_{i}-x_{i+1}}(x-x_{i})+f(x_{i})$ ,实质就是从两个n-1阶的插值多项式构造出一个n阶的插值多项式。

Newton插值公式为 $Q(x)=f(x_{1})+f(x_{1},x_{2})(x-x_{1})+...+f(x_{1},x_{2},...x_{n+1})\prod_{i=1}^{n}(x-x_{i})$ 其中 $f(x_{1},x_{2},...x_{k+1})$ 为k阶差商， $f(x_{1},x_{2},...x_{k+1})=\frac{f(x_{1},x_{2},...x_{k})-f(x_{2},...x_{k,}x_{k+1})}{x_{1}-x_{k+1}}$

代码部分

例题 x=[-0.89,-0.5,0.04,0.57,0.92]; y=[0.4529,0.7788,0.9984,0.7266,0.429];

clear;clc;close all; % Neville插值公式 x=[-0.89,-0.5,0.04,0.57,0.92]; y=[0.4529,0.7788,0.9984,0.7266,0.429]; n=length(x); x1=linspace(-1,1,20); y1=[]; for l=1:20 h=y; for i=1:n-1 for j=1:n-i h(j)=((x(j+i)-x1(l))*h(j)+(x1(l)-x(j))*h(j+1))/(x(j)-x(i+j)); end end y1=[y1 h(1)]; end plot(x1,y1,'--ks'); hold on; plot(x,y,'sb'); clear;clc;close all; % Lagrange插值公式 x=[-0.89,-0.5,0.04,0.57,0.92]; y=[0.4529,0.7788,0.9984,0.7266,0.429]; n=length(x); x1=linspace(-1,1,20); y1=zeros(1,20); for k=1:20 h=0; for i=1:n za=1; for j=1:n if i~=j za=za*(x1(k)-x(j))/(x(i)-x(j)); end end za=za*y(i); h=h+za; end y1(k)=h; end plot(x1,y1,'--k'); hold on; plot(x,y,'bs'); clear;clc;close all; pause(1) % Newton插值公式 x=[-0.89,-0.5,0.04,0.57,0.92]; y=[0.4529,0.7788,0.9984,0.7266,0.429]; n=length(x); x1=linspace(-1,1,20); h=y y1=[] for i=1:n-1 for j=1:n-i h(j)=(h(j)-h(j+1))/(x(j)-x(i+j)); end y1=[y1 h(1)]; end y2=zeros(1,5); for k=1:20 ko=y(1); for i=1:4 ji=1; for j=1:i ji=ji*(x1(k)-x(j)) end ko=ko+ji*y1(i) end y2(k)=ko end plot(x1,y2,'--ks'); hold on; plot(x,y,'sb');

此代码是直接计算，也可以通过设置函数变量来进行计算，这里不再展示

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