NEWTON'S BACKWARD INTERPOLATION FORMULA IN C++ & WITH EXPLANATION

Backward Interpolation.Interpolation is the technique of estimating the value of a function for any intermediate value of the independent variable, while the process of computing the value of the function outside the given range is called extrapolation.

newtons backword interpolation formula





// CPP Program to interpolate using 
// newton backward interpolation 
#include <bits/stdc++.h>

using namespace std; 

// Calculation of u mentioned in formula 
float backward(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[] = { 1900,1910,1920,1930,1940,1950}; 
     
 //input the values in tale
 float y[n][n]; 
 y[0][0] = 65; 
 y[1][0] = 76; 
 y[2][0] = 91; 
 y[3][0] = 97; 
 y[4][0] = 110; 

 // code for to display 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 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; 
 } 


 float value = 1945; //which we have to interpolate


 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 + (backward(u, i) * y[n - 1][i]) / 
         fact(i); 
 } 

 cout << "\n Value at " << value << " is "
  << sum << endl; 
 return 0; 
} 

OUTPUT 

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this progaran is running through the codechef ide

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