write a program to interpolate using Newton's Backward Difference interpolation in c++.

write a program to interpolate using Newton's Backward Difference interpolation in c++.

Source Code:





/********************************************

    program: solution of non-linear equation

                    bisection method

    language: C

    Author : Tanveer Khan

    Rajasthan University, Jaipur

************************************************/

#include<stdio.h>
#include<math.h>
int main()

{
    float x[10],y[15][15];
    int n,i,j;
//no. of items
    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;
}




OUTPUT


Enter n : 4
X       Y
10      1.1
20      2.0
30      4.4
40      7.9

***********Forward Difference Table ***********
        10.00   1.10    0.90    1.50    -0.40
        20.00   2.00    2.40    1.10
        30.00   4.40    3.50
        40.00   7.90

***********Backward Difference Table ***********
        10.00   1.10
        20.00   2.00    0.90
        30.00   4.40    2.40    1.50
        40.00   7.90    3.50    1.10    -0.40