Let the function f(x) is given by the table



x x0 x1 ... xn

y y0 y1 ... yn



An interpolation polynomial of Lagrange is a polynomial



\(L_n(x)=\sum\limits_{i=0}^n y_i\cdot \frac{(x-x_0)(x-x_1)\dots (x-x_{i-1})(x-x_{i+1})\dots (x-x_n)}{(x_i-x_0)(x_i-x_1)\dots (x_i-x_{i-1})(x_i-x_{i+1})\dots (x_i-x_n)}\)



We declare a function for organizing computations using the Lagrange interpolation polynomial




double lagranz(double X[n], double Y[n], double t){
    double sum,prod;
    sum=0;
    for (int j=0; j<n; j++){
        prod=1;
        for (int i=0; i<n; i++){
            if (i!=j){
                prod=prod*(t-X[i])/(X[j]-X[i]);
            }
        }
        sum=sum+Y[j]*prod;
    }
    return sum;
}


Example: Write a program for calculating the values of the Lagrange interpolation polynomial for a function given by the table




x     2 5  -6   7   4  3 8   9   1  -2
f(x) -1	77 -297 249 33 9 389 573 -3 -21


Solution:




#include <iostream>
using namespace std;
const int n=10;

double lagranz(double X[n], double Y[n], double t);

int main() {
    double X[n]={2,5,-6,7,4,3,8,9,1,-2};
    double Y[n]={-1,77,-297,249,33,9,389,573,-3,-21};
    cout << "Ln(5)=" << lagranz(X,Y,5) << " " << endl;
    return 0;
}

double lagranz(double X[n], double Y[n], double t){
    double sum,prod;
    sum=0;
    for (int j=0; j<n; j++){
        prod=1;
        for (int i=0; i<n; i++){
            if (i!=j){
                prod=prod*(t-X[i])/(X[j]-X[i]);
            }
        }
        sum=sum+Y[j]*prod;
    }
    return sum;
}


Result: Ln(5)=77