LAGRANGIAN INTERPOLATION

           

            The lagrangian interpolation or lagrangian polynomials are perhaps the simplest way to exhibit the existence of a polynomial for interpolation with unequally spaced data. Data where the x-value is not equi-spaced often occurs as the result of experimental observation or when historical data are examined.

          The lagrangian polynomial equation is made up of  ‘n’ terms, each of which is a ‘n-1’ – ary degree where ‘n’ is the number of set of data. It, in fact passes through each point used in its construction.

SOURCE CODE

#include<iostream>

#include<cmath>

#include<iomanip>

using namespace std;

int main()

{

    double n,xx,mul;int i,j;

    int x[50],y[50];

    double sum=0;

    cout<<"input the value of n ";

    cin>>n;

    cout<<"enter the value of x and y:";

    for(i=0;i<n;i++)

    {

        cin>>x[i];

        cin>>y[i];

    }

cout<<"input point at which value is unknown:";

cin>>xx;

for(i=0;i<n;i++)

{

    mul=1;

    for(j=0;j<n;j++)

    {

        if(i!=j)

            mul=mul*(xx-x[i])/(x[i]-x[j]);

    }

sum=sum+mul*y[i];

}

cout<<"result="<<sum;

}

It can also be implemented as following:

#include<iostream>

using namespace std;

int main(){

    float *x, *y, result = 0, a;

    int size;

    cout<<"Enter the number of data : ";

    cin>>size;

    x = new float[size];

    y = new float[size];

    cout<<"Enter the data (x y):"<<endl;

    for (int i = 0; i < size; i++)

        cin>>x[i]>>y[i];

    cout<<"Enter the desired x : ";

    cin>>a;

    for (int i = 0; i < size; i++){

        float product = 1;

        for (int j = 0; j < size; j++){

            if(j != i)

                product *= ((a - x[j])/(x[i] - x[j]));

        }

        result += product*y[i];

    }

    cout<<"y("<<a<<") = "<<result<<endl;

    return 0;

}