EWTON’S GREGORY BACKWARD INTERPOLATION FORMULA

NEWTON’S GREGORY BACKWARD INTERPOLATION FORMULA :

This formula is useful when the value of f(x) is required near the end of the table. h is called the interval of difference and u = ( x – an ) / h, Here an is last term.

Example :

Input : Population in 1925

Output :

Value in 1925 is 96.8368 

Below is the implementation of newton backward interpolation method.

// CPP Program to interpolate using 

// newton backward interpolation 

#include <bits/stdc++.h> 

using namespace std; 

// Calculation of u mentioned in formula 

float u_cal(float u, int n) 

{ 

float temp = u; 

for (int i = 1; i < n; i++) 

temp = temp * (u + i); 

return temp; 

} 

// Calculating factorial of given n 

int fact(int n) 

{ 

int f = 1; 

for (int i = 2; i <= n; i++) 

f *= i; 

return f; 

} 

int main() 

{ 

// number of values given 

int n = 5; 

float x[] = { 1891, 1901, 1911, 

1921, 1931 }; 

// y[][] is used for difference 

// table and y[][0] used for input 

float y[n][n]; 

y[0][0] = 46; 

y[1][0] = 66; 

y[2][0] = 81; 

y[3][0] = 93; 

y[4][0] = 101; 

// Calculating the backward difference table 

for (int i = 1; i < n; i++) { 

for (int j = n - 1; j >= i; j--) 

y[j][i] = y[j][i - 1] - y[j - 1][i - 1]; 

} 

// Displaying the backward difference table 

for (int i = 0; i < n; i++) { 

for (int j = 0; j <= i; j++) 

cout << setw(4) << y[i][j] 

<< "\t"; 

cout << endl; 

} 

// Value to interpolate at 

float value = 1925; 

// Initializing u and sum 

float sum = y[n - 1][0]; 

float u = (value - x[n - 1]) / (x[1] - x[0]); 

for (int i = 1; i < n; i++) { 

sum = sum + (u_cal(u, i) * y[n - 1][i]) / 

fact(i); 

} 

cout << "\n Value at " << value << " is "

<< sum << endl; 

return 0; 

} 

Output:

  46    
  66      20    
  81      15      -5    
  93      12      -3       2    
 101       8      -4      -1      -3    

 Value at 1925 is 96.8368