拉格朗日插值法

#include"iostream"
#include"math.h"
using namespace std;
int n;
int main()
{
int N;
double x[3]={0.2,0.3,0.4}, y[3]={1.2214,1.3499,1.4918}, Y=0, X1=0.285,b,m;
int i,j,k;
for(k=0;k<3;k++)
{
double p=1;
for(j=0;j<3;j++)
{
if(j==k) continue;
else
{
p=p*(X1-x[j])/(x[k]-x[j]);
}

}
Y=Y+p*y[k];
}
/*X1=0.285时的近似值*/
printf("e(%.3lf)=%.6lf\n",X1,Y);
cout<<"Please Input N value( 2 or 3 )"<<endl<<"N=";
while(cin>>N)
{

if(N==2||N==3)
{
double x[3]={0.2,0.3,0.4}, y[3]={1.2214,1.3499,1.4918}, Y=0, X=0.285;
for(k=0;k<N;k++)
{
double p=1;
for(j=0;j<N;j++)
{

if(j==k) continue;
else
p=p*(X-x[j])/(x[k]-x[j]);
}
Y=Y+p*y[k];
}
printf("L(%d)=%.6lf\n", N-1,Y);
double X2;Y=0;

m=1;
if(N==2) X2=0.3;
else X2=0.4;
for(k=0;k<N;k++)
{
double p=1;
for(j=0;j<N;j++)
{
if(j==k) continue;
else
{
p=p*(X2-x[j])/(x[k]-x[j]);
}

}
Y=Y+p*y[k];
m*=(X1-x[k]);
}
/*N==2或N==3的截断误差*/
printf("R%d(%.6lf)<%.5lf\n",N-1,X1,fabs(Y*m/6));
// cout<<"Please Input N value(2 or 3)"<<endl<<"N=";
}
else
{
cout<<"error!"<<endl;
}
cout<<"Please Input N value( 2 or 3 )"<<endl<<"N=";
}
}

同有兴趣的朋友共勉！

转载于:https://www.cnblogs.com/FCWORLD/archive/2010/12/04/1896071.html

来源：https://blog.csdn.net/weixin_30642305/article/details/99910314