# 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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its use full

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

good

very usefull