Engineering Mathematics VII Newton’s Divided Difference Interpolation Formula

Interpolation is an estimation of a value within two known values in a sequence of values.

Newton’s divided difference interpolation formula is a interpolation technique used when the interval difference is not same for all sequence of values.

Suppose f(x₀), f(x₁), f(x₂)………f(x_n) be the (n+1) values of the function y=f(x) corresponding to the arguments x=x₀, x₁, x₂…x_n, where interval differences are not same
Then the first divided difference is given by

The second divided difference is given by

and so on…
Divided differences are symmetric with respect to the arguments i.e independent of the order of arguments.
so,
f[x₀, x₁]=f[x₁, x₀]
f[x₀, x₁, x₂]=f[x₂, x₁, x₀]=f[x₁, x₂, x₀]
By using first divided difference, second divided difference as so on .A table is formed which is called the divided difference table.

Divided difference table:

NEWTON’S DIVIDED DIFFERENCE INTERPOLATION FORMULA

$f(x)=f(x_0)+f[x_0, x_1]+(x-x_0)(x-x_1)f[x_0, x_1, x_2]+..........................+(x-x_0)(x-x_1)...(x-x_k)f[x_0, x_1, x_2...x_k]$

Examples:

Input : Value at 7
       
Output :
      
      Value at 7 is 13.47

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Below is the implementation for Newton’s divided difference interpolation method.

C++
Java
Python3
C#
PHP

// CPP program for implementing 
// Newton divided difference formula 
#include <bits/stdc++.h> 
using namespace std; 
 
// Function to find the product term 
float proterm(int i, float value, float x[]) 
{ 
    float pro = 1; 
    for (int j = 0; j < i; j++) { 
        pro = pro * (value - x[j]); 
    } 
    return pro; 
} 
 
// Function for calculating 
// divided difference table 
void dividedDiffTable(float x[], float y[][10], int n) 
{ 
    for (int i = 1; i < n; i++) { 
        for (int j = 0; j < n - i; j++) { 
            y[j][i] = (y[j][i - 1] - y[j + 1] 
                         [i - 1]) / (x[j] - x[i + j]); 
        } 
    } 
} 
 
// Function for applying Newton's 
// divided difference formula 
float applyFormula(float value, float x[], 
                   float y[][10], int n) 
{ 
    float sum = y[0][0]; 
 
    for (int i = 1; i < n; i++) { 
      sum = sum + (proterm(i, value, x) * y[0][i]); 
    } 
    return sum; 
} 
 
// Function for displaying  
// divided difference table 
void printDiffTable(float y[][10],int n) 
{ 
    for (int i = 0; i < n; i++) { 
        for (int j = 0; j < n - i; j++) { 
            cout << setprecision(4) <<  
                                 y[i][j] << "\t "; 
        } 
        cout << "\n"; 
    } 
} 
 
// Driver Function 
int main() 
{ 
    // number of inputs given 
    int n = 4; 
    float value, sum, y[10][10]; 
    float x[] = { 5, 6, 9, 11 }; 
 
    // y[][] is used for divided difference 
    // table where y[][0] is used for input 
    y[0][0] = 12; 
    y[1][0] = 13; 
    y[2][0] = 14; 
    y[3][0] = 16; 
 
    // calculating divided difference table 
    dividedDiffTable(x, y, n); 
 
    // displaying divided difference table 
    printDiffTable(y,n); 
 
    // value to be interpolated 
    value = 7; 
 
    // printing the value 
    cout << "\nValue at " << value << " is "
               << applyFormula(value, x, y, n) << endl; 
    return 0; 

Engineering Mathematics VII Newton’s Divided Difference Interpolation Formula

Newton’s Divided Difference Interpolation Formula

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

ASSIGNMENT : Engineering Mathematics VII Newton’s Divided Difference Interpolation Formula Assignment MARKS : 10 DURATION : 1 week, 3 days

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Engineering Mathematics VII Newton’s Divided Difference Interpolation Formula

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

ASSIGNMENT : Engineering Mathematics VII Newton’s Divided Difference Interpolation Formula Assignment MARKS : 10 DURATION : 1 week, 3 days

Search

Girls in STEM

About FAWE Elearning

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