Lagrange插值法
Lagrange插值公式：
P ( x ) = ∑ k = 0 n L n , k ( x ) f ( x k ) P(x) = \sum_{k=0}^{n}L_{n,k}(x) f(x_{k}) P(x)=∑k=0nLn,k(x)f(xk)
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L_{n,k}(x) = \prod_{i=0,i\neq k}^{n}\frac{x-x_{i}}{x_{k}-x_{i}}
Ln,k(x)=∏i=0,i̸=knxk−xix−xi ,k = 0,1,2,…,n
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\qquad = \frac{(x-x_{0})(x-x_{1})...(x-x_{k-1})(x-x_{k+1})...(x-x_{n})}{(x_{k}-x_{0})(x_{k}-x_{1})...(x_{k}-x_{k-1})(x_{k}-x_{k+1})...(x_{k}-x_{n})}
=(xk−x0)(xk−x1)...(xk−xk−1)(xk−xk+1)...(xk−xn)(x−x0)(x−x1)...(x−xk−1)(x−xk+1)...(x−xn)
Lagrange插值法的算法已知条件：
（1）
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(x_{0},f(x_{0})),(x_{1},f(x_{1})),...,(x_{n},f(x_{n})).
(x0,f(x0)),(x1,f(x1)),...,(xn,f(xn)).
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（3）所要进行内插的点
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（4）Lagrange插值法的公式。
C语言简单实现如下：
#include<cstdio>
#include<iostream>
using namespace std;
double x[30];//x值
double f[30];//对应的y值
double xa;//待求的y值所给出的x值
double l;//l(k)的值
double ans;//结果
int main()
{
int n;
while(cin>>n&&n)
{
cin>>xa;
for(int i=0;i<=n;i++)
cin>>x[i]>>f[i];
ans=0.0;
for(int k=0;k<=n;k++)
{
l=1.0;
for(int i=0;i<=n;i++)//计算l(k)
{
if(k!=i)
l=l*(xa-x[i])/(x[k]-x[i]);
}
ans=ans+l*f[k];
}
printf("The value of P(%.4lf)=%.4lf\n\n",xa,ans);
}
return 0;
}
/*
三组测试用例：
3
1.5
1.0 0.0
2.0 0.693
3.0 1.099
4.0 1.386
3
1.5
1.0 0.000
1.2 0.182
1.4 0.336
2.0 0.693
9
1.5
1.0 0.000
1.2 0.182
1.7 0.531
2.0 0.693
2.2 0.788
2.7 0.993
3.0 1.099
3.2 1.163
3.7 1.308
4.0 1.386
*/