Implimentation of Newton’s Forward Interpolation Formula.
#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<process.h>
#include<string.h>
void main()
{
int n,x,p;
int i,j;
float ax[10];
float ay[10];
float y=0;
float h;
float d;
float diff[20][20];
float y1,y2,y3,y4;
clrscr();
printf("\n Enter the number of term -");
scanf("%d",&n);
printf("\n\n Enter the value in the form of x -");
for(i=0;i<n;i++)
{
printf("\n\n Enter the value of x%d-",i+1);
scanf("%f",&ax[i]);
}
printf("\n\n Enter the value in the form of y -");
for(i=0;i<n;i++)
{
printf("\n\n Enter the value in the form y%d-",i+1);
scanf("%f",&ay[i]);
}
printf("\n\n Enter the value of x for -");
printf("\n which you want the value of y -");
h=ax[1]-ax[0];
for(i=0;i<n-1;i++)
{
diff[i][2]=ay[i+1]-ay[i];
}
for(j=2;j<=4;j++)
{
for(i=0;i<n-j;i++)
{
diff[i][j]=diff[i+1][j-1];
}
i=0;
do
{
i++;
}while(ax[i]<x);
i--;
p=(x-ax[i])/h;
y1=p*diff[i-1][1];
y2=p*(p+1)*diff[i-1][2]/2;
y3=(p+1)*p*(p-1)*diff[i-2][3]/6;
y4=(p+2)*(p+1)*p*(p-1)*diff[i-3][4]/24;
y=ay[i]+y1+y2+y3+y4;
printf("\n whenn=%6.4f,y=%6.8f",x,y);
printf("\n\n\n Press Enter to Exit");
}
getch();
}
