Program For Newton’s Forward Interpolation

Here we will construct a C program for Newton’s Forward interpolation formula to calculate y(0.5), Correct to 3 decimal places from the following table:

x	0	1	2	3
y	1	0	1	10

Also, To compute the value, we need to construct forward difference table and thereafter, to implement Newton’s Forward Interpolation by generating the formula.

Newton’s Forward Interpolation

Algorithm:

Step 1: Start the program
Step 2: Read n (No. of arguments)
Step 3: For i = 0 to n − 1
             Read x i &y i [0]
             End i
Step 4: Construct the Forward Difference Table
             For j = 1 to n − 1
             For i = 0 to n − 1 − j
             y i [j] = y[i + 1][j − 1] − y[i][j − 1]
             End i
             End j
Step 5: Print the Forward Difference Table
             For i = 0 to n − 1
             For j = 0 to n − 1 − i
             Print y i [j]
             End j
             Next Line
             End i
Step 6: Read a (Point of Interpolation)
Step 7: Assign h = x[1] − x[0] (Step Length)
Step 8: Assign u = (a − x[0])/h
Step 9: Assign sum = y 0 [0] & p = 1.0
Step 10: For j = 1 to n − 1
               p = p ∗ (u − j + 1)/j
               sum = sum + p ∗ y 0 [j]
               End j
Step 11: Display a & sum
Step 12: Stop

Newton’s Backward Interpolation 

The working formula for Newton’s Backward Interpolation is 

To Compute the value, we need to construct a backward difference table and thereafter, to implement Newton’s backward interpolation by generating the formula.

Algorithm:

Step 1: Start the program
Step 2: Read n (No. of arguments)
Step 3: For i = 0 to n − 1
             Read x i &y i [0]
             End i
Step 4: Construct the Backward Difference Table
             For j = 1 to n − 1
             For i = j to n − 1
             y i [j] = y[i][j − 1] − y[i − 1][j − 1]
             End i
             End j
Step 5: Print the Backward Difference Table
             For i = 0 to n − 1
             For j = 0 to i
             Print y i [j]
             End j
             End i
Step 6: Read a (Point of Interpolation)
Step 7: Assign h = x[1] − x[0] (Step Length)
Step 8: Assign u = (a − x[n − 1])/h
Step 9: Assign sum = y n − 1 [0] & p = 1.0
Step 10: For j = 1 to n − 1
               p = p ∗ (u + j − 1)/j
               sum = sum + p ∗ y n − 1 [j]
               End j
Step 11: Display a & sum
Step 12: Stop

Example:

C

// C program to demonstrate
// both Forward and Backward
// Newton's Interpolation
#include <stdio.h>
void forward(float x[4], float y[4][4], int n);
void BackWard(float x[4], float y[4][4], int n);
int main()
{
    int i, j;
    int n = 4; // number of arguments
    float x[4] = { 0, 1, 2, 3 };
    float y[4][4] = {
        { 1, 0, 0, 0 },
        { 0, 0, 0, 0 },
        { 1, 0, 0, 0 },
        { 10, 0, 0, 0 },
    };

    forward(x, y, n);
    BackWard(x, y, n);

    return 0;
}
void forward(float x[4], float y[4][4], int n)
{
    int i, j;
    float a = 0.5; // interpolation point
    float h, u, sum, p;
    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 The forward difference table is:\n");
    for (i = 0; i < n; i++) {
        printf("\n");
        for (j = 0; j < n - i; j++) {
            printf("%f\t", y[i][j]);
        }
    }

    p = 1.0;
    sum = y[0][0];
    h = x[1] - x[0];
    u = (a - x[0]) / h;
    for (j = 1; j < n; j++) {
        p = p * (u - j + 1) / j;
        sum = sum + p * y[0][j];
    }
    printf("\nThe value of y at x=%0.1f is %0.3f", a, sum);
}
void BackWard(float x[4], float y[4][4], int n)
{
    int i, j;
    float a = 0.5; // interpolation point
    float h, u, sum, p;
    for (j = 1; j < n; j++) {
        for (i = j; i < n; i++) {
            y[i][j] = y[i][j - 1] - y[i - 1][j - 1];
        }
    }
    printf("\nThe backward difference table is:\n");
    for (i = 0; i < n; i++) {
        printf("\n");
        for (j = 0; j <= i; j++) {
            printf("%f\t", y[i][j]);
        }
    }

    p = 1.0;
    sum = y[n - 1][0];
    h = x[1] - x[0];
    u = (a - x[n - 1]) / h;
    for (j = 1; j < n; j++) {
        p = p * (u + j - 1) / j;
        sum = sum + p * y[n - 1][j];
    }

    printf("\nThe value of y at x=%0.1f is %0.3f", a, sum);
}

Output

 The forward difference table is:

1.000000    -1.000000    2.000000    6.000000    
0.000000    1.000000    8.000000    
1.000000    9.000000    
10.000000    
The value of y at x=0.5 is 0.625
The backward difference table is:

1.000000    
0.000000    -1.000000    
1.000000    1.000000    2.000000    
10.000000    9.000000    8.000000    6.000000    
The value of y at x=0.5 is 0.625

Time Complexity: O(N*N)

Space Complexity: O(N*N)