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Newton Forward And Backward Interpolation
• Difficulty Level : Medium
• Last Updated : 28 Jan, 2019

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.

Forward Differences : The differences y1 – y0, y2 – y1, y3 – y2, ……, yn – yn–1 when denoted by dy0, dy1, dy2, ……, dyn–1 are respectively, called the first forward differences. Thus the first forward differences are :

NEWTON’S GREGORY FORWARD INTERPOLATION FORMULA :

This formula is particularly useful for interpolating the values of f(x) near the beginning of the set of values given. h is called the interval of difference and u = ( x – a ) / h, Here a is first term.

Example :

Input : Value of Sin 52

Output :

Value at Sin 52 is 0.788003


Below is the implementation of newton forward interpolation method.

## C++

 // CPP Program to interpolate using // newton forward interpolation#include using namespace std;  // calculating u mentioned in the formulafloat 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 number nint 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 = 4;    float x[] = { 45, 50, 55, 60 };          // y[][] is used for difference table    // with y[][0] used for input    float y[n][n];    y[0][0] = 0.7071;    y[1][0] = 0.7660;    y[2][0] = 0.8192;    y[3][0] = 0.8660;      // Calculating the forward difference    // table    for (int i = 1; i < n; i++) {        for (int j = 0; j < n - i; j++)            y[j][i] = y[j + 1][i - 1] - y[j][i - 1];    }      // Displaying the forward difference table    for (int i = 0; i < n; i++) {        cout << setw(4) << x[i]              << "\t";        for (int j = 0; j < n - i; j++)            cout << setw(4) << y[i][j]                  << "\t";        cout << endl;    }      // Value to interpolate at    float value = 52;      // initializing u and sum    float sum = y[0][0];    float u = (value - x[0]) / (x[1] - x[0]);    for (int i = 1; i < n; i++) {        sum = sum + (u_cal(u, i) * y[0][i]) /                                 fact(i);    }      cout << "\n Value at " << value << " is "          << sum << endl;    return 0;}

## Java

 // Java Program to interpolate using // newton forward interpolation  class GFG{// calculating u mentioned in the formulastatic double u_cal(double u, int n){    double temp = u;    for (int i = 1; i < n; i++)        temp = temp * (u - i);    return temp;}  // calculating factorial of given number nstatic int fact(int n){    int f = 1;    for (int i = 2; i <= n; i++)        f *= i;    return f;}  public static void main(String[] args){    // Number of values given    int n = 4;    double x[] = { 45, 50, 55, 60 };          // y[][] is used for difference table    // with y[][0] used for input    double y[][]=new double[n][n];    y[0][0] = 0.7071;    y[1][0] = 0.7660;    y[2][0] = 0.8192;    y[3][0] = 0.8660;      // Calculating the forward difference    // table    for (int i = 1; i < n; i++) {        for (int j = 0; j < n - i; j++)            y[j][i] = y[j + 1][i - 1] - y[j][i - 1];    }      // Displaying the forward difference table    for (int i = 0; i < n; i++) {        System.out.print(x[i]+"\t");        for (int j = 0; j < n - i; j++)            System.out.print(y[i][j]+"\t");        System.out.println();    }      // Value to interpolate at    double value = 52;      // initializing u and sum    double sum = y[0][0];    double u = (value - x[0]) / (x[1] - x[0]);    for (int i = 1; i < n; i++) {        sum = sum + (u_cal(u, i) * y[0][i]) /                                fact(i);    }      System.out.println("\n Value at "+value+" is "+String.format("%.6g%n",sum));}}// This code is contributed by mits

## Python3

 # Python3 Program to interpolate using # newton forward interpolation  # calculating u mentioned in the formuladef u_cal(u, n):      temp = u;    for i in range(1, n):        temp = temp * (u - i);    return temp;  # calculating factorial of given number ndef fact(n):    f = 1;    for i in range(2, n + 1):        f *= i;    return f;  # Driver Code  # Number of values givenn = 4;x = [ 45, 50, 55, 60 ];      # y[][] is used for difference table# with y[][0] used for inputy = [[0 for i in range(n)]        for j in range(n)];y[0][0] = 0.7071;y[1][0] = 0.7660;y[2][0] = 0.8192;y[3][0] = 0.8660;  # Calculating the forward difference# tablefor i in range(1, n):    for j in range(n - i):        y[j][i] = y[j + 1][i - 1] - y[j][i - 1];  # Displaying the forward difference tablefor i in range(n):    print(x[i], end = "\t");    for j in range(n - i):        print(y[i][j], end = "\t");    print("");  # Value to interpolate atvalue = 52;  # initializing u and sumsum = y[0][0];u = (value - x[0]) / (x[1] - x[0]);for i in range(1,n):    sum = sum + (u_cal(u, i) * y[0][i]) / fact(i);  print("\nValue at", value,       "is", round(sum, 6));  # This code is contributed by mits

## C#

 // C# Program to interpolate using // newton forward interpolationusing System;  class GFG{// calculating u mentioned in the formulastatic double u_cal(double u, int n){    double temp = u;    for (int i = 1; i < n; i++)        temp = temp * (u - i);    return temp;}  // calculating factorial of given number nstatic int fact(int n){    int f = 1;    for (int i = 2; i <= n; i++)        f *= i;    return f;}  // Driver codepublic static void Main(){    // Number of values given    int n = 4;    double[] x = { 45, 50, 55, 60 };          // y[,] is used for difference table    // with y[,0] used for input    double[,] y=new double[n,n];    y[0,0] = 0.7071;    y[1,0] = 0.7660;    y[2,0] = 0.8192;    y[3,0] = 0.8660;      // Calculating the forward difference    // table    for (int i = 1; i < n; i++) {        for (int j = 0; j < n - i; j++)            y[j,i] = y[j + 1,i - 1] - y[j,i - 1];    }      // Displaying the forward difference table    for (int i = 0; i < n; i++) {        Console.Write(x[i]+"\t");        for (int j = 0; j < n - i; j++)            Console.Write(y[i,j]+"\t");        Console.WriteLine();    }      // Value to interpolate at    double value = 52;      // initializing u and sum    double sum = y[0,0];    double u = (value - x[0]) / (x[1] - x[0]);    for (int i = 1; i < n; i++) {        sum = sum + (u_cal(u, i) * y[0,i]) /                                fact(i);    }      Console.WriteLine("\n Value at "+value+" is "+Math.Round(sum,6));}}// This code is contributed by mits

## PHP

 

Output:
  45    0.7071    0.0589    -0.00569999    -0.000699997
50    0.766    0.0532    -0.00639999
55    0.8192    0.0468
60    0.866

Value at 52 is 0.788003


Backward Differences : The differences y1 – y0, y2 – y1, ……, yn – yn–1 when denoted by dy1, dy2, ……, dyn, respectively, are called first backward difference. Thus the first backward differences are :

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.

## C++

 // CPP Program to interpolate using// newton backward interpolation#include using namespace std;  // Calculation of u mentioned in formulafloat 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 nint 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;}

## Java

 // Java Program to interpolate using// newton backward interpolationclass GFG{      // Calculation of u mentioned in formulastatic double u_cal(double u, int n){    double temp = u;    for (int i = 1; i < n; i++)        temp = temp * (u + i);    return temp;}  // Calculating factorial of given nstatic int fact(int n){    int f = 1;    for (int i = 2; i <= n; i++)        f *= i;    return f;}  // Driver codepublic static void main(String[] args){    // number of values given    int n = 5;    double x[] = { 1891, 1901, 1911,                 1921, 1931 };                      // y[][] is used for difference     // table and y[][0] used for input    double[][] y = new double[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++)            System.out.print(y[i][j] + "\t");        System.out.println("");;    }      // Value to interpolate at    double value = 1925;      // Initializing u and sum    double sum = y[n - 1][0];    double 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);    }    System.out.println("\n Value at " + value +                     " is " + String.format("%.6g%n",sum));}}  // This code is contributed by mits

## C#

 // C# Program to interpolate using// newton backward interpolationusing System;  class GFG{      // Calculation of u mentioned in formulastatic double u_cal(double u, int n){    double temp = u;    for (int i = 1; i < n; i++)        temp = temp * (u + i);    return temp;}  // Calculating factorial of given nstatic int fact(int n){    int f = 1;    for (int i = 2; i <= n; i++)        f *= i;    return f;}  // Driver codestatic void Main(){    // number of values given    int n = 5;    double[] x = { 1891, 1901, 1911,                 1921, 1931 };                      // y[][] is used for difference     // table and y[][0] used for input    double[,] y = new double[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++)            Console.Write(y[i,j]+"\t");        Console.WriteLine("");;    }      // Value to interpolate at    double value = 1925;      // Initializing u and sum    double sum = y[n - 1,0];    double 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);    }      Console.WriteLine("\n Value at "+value+" is "+Math.Round(sum,4));}}  // This code is contributed by mits

## PHP

 = $i; $j--)        $y[$j][$i] = $y[$j][$i - 1] -                     $y[$j - 1][$i - 1];}  // Displaying the backward difference tablefor ($i = 0; $i < $n; $i++){ for ($j = 0; $j <= $i; $j++) print($y[$i][$j] . "\t");    print("\n");}  // Value to interpolate at$value = 1925;  // Initializing u and sum$sum = $y[$n - 1][0];$u = ($value - $x[$n - 1]) / ($x[1] - $x[0]);for ($i = 1; $i < $n; $i++) {    $sum = $sum + (u_cal($u, $i) *            $y[$n - 1][$i]) / fact($i);}  print("\n Value at " . $value .  " is " . round($sum, 4));  // This code is contributed by chandan_jnu?>

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

Value at 1925 is 96.8368
`

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