UC知道分段线性插值问题
来源：百度知道 编辑：UC知道 时间：2024/06/14 15:52:45
已知:
x=0.0,0.1,0.195,0.3,0.401,0.5时,
f(x)分别=0.39894,0.39695,0.39142,0.38138,0.36812,0.35206
要用C语言编一个程序，分别用分段线性插值，分段二次插值及拉格郎日插值计算f(x)的近似值.
x=0.0,0.1,0.195,0.3,0.401,0.5时,
f(x)分别=0.39894,0.39695,0.39142,0.38138,0.36812,0.35206
要用C语言编一个程序，分别用分段线性插值，分段二次插值及拉格郎日插值计算f(x)的近似值.
#include<stdio.h>
#include<math.h>
double Lagrange1(double *x, double *y, double xx) //拉格郎日插值
{
int i,j;
double *a,yy=0.000;
a=new double[6];
for(i=0;i< 6;i++)
{
a[i]=y[i];
for(j=0;j< 6;j++)
if(j!=i)
a[i]*=(xx-x[j])/(x[i]-x[j]);
yy+=a[i];
}
delete a;
return yy;
}
double Lagrange2(double *x, double *y, double input) //分段线性插值
{
double output;
int i;
for (i=0;i<5;i++)
{
if (x[i] <= input && x[i+1] >= input)
{
output=y[i] +(y[i+1]-y[i])*(input-x[i])/(x[i+1]-x[i]);
break;
}
}
return output;
}
double Lagrange3(double *x,double *y,double u) //分段二次插值
{
int i,k=0;
double v;
for(i=0;i<6;i++)
{
if(u<x[1])
{
k=0;
v=y[k]*(u-x[k+1])*(u-x[k+2])/((x[k]-x[k+1])*(x[k]-x[k+2]))+y[k+1]*(u-x[k])*(u-x[k+2])/((x[k