## Sunday, January 22, 2012

### 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;}

#### 1 comment:

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