# Numerical Method: Newton’s Forward and Backward Interpolation in C/C++

#include<stdio.h>
#include<math.h>
int main()
{
float x[10],y[15][15];
int n,i,j;
printf("Enter n : ");
scanf("%d",&n);
printf("X\tY\n");
for(i = 0;i<n;i++){
scanf("%f %f",&x[i],&y[i][0]);
}
//forward difference table
for(j=1;j<n;j++)
for(i=0;i<(n-j);i++)
y[i][j] = y[i+1][j-1] - y[i][j-1];
printf("\n***********Forward Difference Table ***********\n");
//display Forward Difference Table
for(i=0;i<n;i++)
{
printf("\t%.2f",x[i]);
for(j=0;j<(n-i);j++)
printf("\t%.2f",y[i][j]);
printf("\n");
}
//backward difference table
for(j=1;j<n;j++)
//for j = 0 initially input is taken so we start from j=1
for(i=n-1;i>(j-1);i--)
y[i][j] = y[i][j-1] - y[i-1][j-1];
printf("\n***********Backward Difference Table ***********\n");
//display Backward Difference Table
for(i=0;i<n;i++)
{
printf("\t%.2f",x[i]);
for(j=0;j<=i;j++)
printf("\t%.2f",y[i][j]);
printf("\n");
}
return 0;
}

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### 2 Responses

1. its use full
but it will be more use full if u did programing for only difference table

2. Anonymous says:

good
very usefull