Unit 3.2 Backword Differentiation

#include <stdio.h>
#include <math.h>
int main()
{
    int i,n,j,fact=1;
    float x[20],y[20][20],h,term=1,u,sum,xval;
    printf("Enter the value of n:");
    scanf("%d",&n);
    printf("\nEnter the elements of x: ");
    for(i=0 ; i<n ; i++){
        scanf("%f",&x[i]);
    }
    printf("Enter the elements of y: ");
    for(i=0 ; i<n ; i++){
        scanf("%f",&y[i][0]);
    }
    h=x[1]-x[0];
    printf("\nEnter x for which y is to be calculated:");
    scanf("%f",&xval);
    u = (xval - x[n-1] )/h;
    printf("\nh=%f u= %f",h,u);
    sum = y[n-1][0];

    for(j=1 ; j<n ; j++){
        for(i=0 ; i< (n-j) ; i++){
            // when j =2 then i =0 ,1 & when j =3 then i =0)
            y[i][j]=y[i+1][j-1] - y[i][j-1];
        }
    }
    printf("The table is :\n");
    for(i=0; i<n ; i++){
        printf("\t %.2f ",x[i]);
        for(j=0 ; j<(n-i)+1 ; j++){
            printf("\t%.2f,",y[i][j]);
        }
        printf("\n");
    }

    for(j=1 ; j<n ; j++){
        for(i=1 ; i<=j ; i++){
            fact = fact * i;
        }
        term = term * (u-(j-1));
        printf("\nTerm=%f",term);
        printf("\nFactorial=%d",fact);
        sum = sum + ((y[0][j] * term) / fact);
        fact = 1;
    }
    printf("\n\nfor x = %f the ultimate answer is sum =  %f",xval,sum);

    return 0;
}

/*
Enter the value of n:5

Enter the elements of x: 1891
1901
1911
1921
1931
Enter the elements of y: 12
15
20
27
39

Enter x for which y is to be calculated:1925

h=10.000000 u= -0.600000The table is :
         1891.00        12.00,  3.00,   2.00,   0.00,   3.00,   0.00,
         1901.00        15.00,  5.00,   2.00,   3.00,   0.00,
         1911.00        20.00,  7.00,   5.00,   0.00,
         1921.00        27.00,  12.00,  -0.00,
         1931.00        39.00,  0.00,

Term=-0.600000
Factorial=1
Term=0.960000
Factorial=2
Term=-2.496000
Factorial=6
Term=8.985600
Factorial=24

for x = 1925.000000 the ultimate answer is sum =  39.283199
*/