见鬼吧，拉格朗日插值法

一、理论证明

　　笔者看过比较多博文后发现一个比较的推导的博文：http://www.cnblogs.com/ECJTUACM-873284962/p/6833391.html

　　因而此处就不过多的证明了

　　主要两个公式:

 

二、实现(c++)

#include 
     
      #include 
      
       using namespace std;inline double coefficient_l(double* x, int k, double num,int n)//x是插值的x向量，k为第几个l，n是插值节点个数{    double res = 1.0;    for (int i = 0; i < n; i++)    {        if (i != k)        {            res *= (num - x[i]) / (x[k] - x[i]);        }    }    return res;}inline double lagrange(double* x, double* y, double num, int n)/*    x，y都是插值的节点*/{    double res = 0.0;    for (int i = 0; i < n; i++)    {        res += y[i] * coefficient_l(x,i,num,n);    }    return res;}int main(int argc, char const *argv[]){    int n = 0;    cout<<"请输入插值结点个数n:";    cin>>n;    double* x = new double[n];    double* y = new double[n];    cout<<"请输入n个x:";    for(int i=0;i
       
        >x[i];    }    cout<<"请输入n+1个y:";    for(int i=0;i
        
         >y[i];    }    cout<<"请输入要计算的num:";    double num = 0;    cin>>num;    double res = lagrange(x,y,num,n);    cout<<"res:"<
         
          <