# Lagrange’s Interpolation

What is Interpolation?
Interpolation is a method of finding new data points within the range of a discrete set of known data points (Source Wiki). In other words interpolation is the technique to estimate the value of a mathematical function, for any intermediate value of the independent variable.
For example, in the given table we’re given 4 set of discrete data points, for an unknown function f(x) :

How to find?
Here we can apply the Lagrange’s interpolation formula to get our solution.
The Lagrange’s Interpolation formula:
If, y = f(x) takes the values y0, y1, … , yn corresponding to x = x0, x1 , … , xn then,

This method is preferred over its counterparts like Newton’s method because it is applicable even for unequally spaced values of x.
We can use interpolation techniques to find an intermediate data point say at x = 3.

Advantages of Lagrange Interpolation:

• This formula is used to find the value of the function even when the arguments are not equally spaced.
• This formula is used to find the value of independent variable x corresponding to a given value of a function.

Disadvantages of Lagrange Interpolation:

• A change of degree in Lagrangian polynomial involves a completely new computation of all the terms.
• For a polynomial of high degree, the formula involves a large number of multiplications which make the process quite slow.
• In the Lagrange Interpolation, the degree of polynomial is chosen at the outset. So it is difficult to find the degree of approximating polynomial which is suitable for given set of tabulated points.

## C++

 // C++ program for implementation of Lagrange's Interpolation #include using namespace std;   // To represent a data point corresponding to x and y = f(x) struct Data {     int x, y; };   // function to interpolate the given data points using Lagrange's formula // xi corresponds to the new data point whose value is to be obtained // n represents the number of known data points double interpolate(Data f[], int xi, int n) {     double result = 0; // Initialize result       for (int i=0; i

## Java

 // Java program for implementation  // of Lagrange's Interpolation   import java.util.*;   class GFG {   // To represent a data point  // corresponding to x and y = f(x) static class Data {     int x, y;       public Data(int x, int y)      {         super();         this.x = x;         this.y = y;     }       };   // function to interpolate the given // data points using Lagrange's formula // xi corresponds to the new data point // whose value is to be obtained n  // represents the number of known data points static double interpolate(Data f[], int xi, int n) {     double result = 0; // Initialize result       for (int i = 0; i < n; i++)     {         // Compute individual terms of above formula         double term = f[i].y;         for (int j = 0; j < n; j++)         {             if (j != i)                 term = term*(xi - f[j].x) / (f[i].x - f[j].x);         }           // Add current term to result         result += term;     }       return result; }   // Driver code public static void main(String[] args) {     // creating an array of 4 known data points     Data f[] = {new Data(0, 2), new Data(1, 3),                  new Data(2, 12), new Data(5, 147)};       // Using the interpolate function to obtain      // a data point corresponding to x=3     System.out.print("Value of f(3) is : " +                      (int)interpolate(f, 3, 4)); } }   // This code is contributed by 29AjayKumar

## Python3

 # Python3 program for implementation # of Lagrange's Interpolation   # To represent a data point corresponding to x and y = f(x) class Data:     def __init__(self, x, y):         self.x = x         self.y = y   # function to interpolate the given data points # using Lagrange's formula # xi -> corresponds to the new data point # whose value is to be obtained # n -> represents the number of known data points def interpolate(f: list, xi: int, n: int) -> float:       # Initialize result     result = 0.0     for i in range(n):           # Compute individual terms of above formula         term = f[i].y         for j in range(n):             if j != i:                 term = term * (xi - f[j].x) / (f[i].x - f[j].x)           # Add current term to result         result += term       return result   # Driver Code if __name__ == "__main__":       # creating an array of 4 known data points     f = [Data(0, 2), Data(1, 3), Data(2, 12), Data(5, 147)]       # Using the interpolate function to obtain a data point     # corresponding to x=3     print("Value of f(3) is :", interpolate(f, 3, 4))   # This code is contributed by # sanjeev2552

## C#

 // C# program for implementation  // of Lagrange's Interpolation using System;   class GFG {   // To represent a data point  // corresponding to x and y = f(x) class Data {     public int x, y;     public Data(int x, int y)      {         this.x = x;         this.y = y;     } };   // function to interpolate the given // data points using Lagrange's formula // xi corresponds to the new data point // whose value is to be obtained n  // represents the number of known data points static double interpolate(Data []f,                            int xi, int n) {     double result = 0; // Initialize result       for (int i = 0; i < n; i++)     {         // Compute individual terms         // of above formula         double term = f[i].y;         for (int j = 0; j < n; j++)         {             if (j != i)                 term = term * (xi - f[j].x) /                            (f[i].x - f[j].x);         }           // Add current term to result         result += term;     }     return result; }   // Driver code public static void Main(String[] args) {     // creating an array of 4 known data points     Data []f = {new Data(0, 2),                  new Data(1, 3),                  new Data(2, 12),                  new Data(5, 147)};       // Using the interpolate function to obtain      // a data point corresponding to x=3     Console.Write("Value of f(3) is : " +                    (int)interpolate(f, 3, 4)); } }   // This code is contributed by PrinciRaj1992

## Javascript

 

Output:

Value of f(3) is : 35`

Complexity:
The time complexity of the above solution is O(n2) and auxiliary space is O(1).

References:

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