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# Lagrange’s Interpolation

• Difficulty Level : Medium
• Last Updated : 20 Jul, 2021

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.

## C++

 // C++ program for implementation of Lagrange's Interpolation#includeusing 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 pointsdouble 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 pointsstatic 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 codepublic 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 pointsdef 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 Codeif __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 Interpolationusing 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 pointsstatic 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 codepublic 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: