Lagrange Approximation for y(x)=cos(π x) Using 5 Points

[ggeo1]

Hello, i have done the following code in c++ which gives me a result but :

1) I compute the product "prod*=(y-x)/(x[k]-x)" which has variable "y" inside ,but the result i take is just a number.(i want the result to be for example in this form "p(x)=1.02*x^3+2*x^2..."
I can't understand how the program manipulates the variable y which does hasn't a value!

2) I don't know if i have it right..

The exercise is :
Find Lagrange's polynomial approximation for y(x)=cos(π x), x ∈−1,1 using 5 points
(x = -1, -0.5, 0, 0.5, and 1).


My code :

using namespace std;

const double pi=3.1415;

// my function
double f(double x){

return (cos(pi*x));
}

//function to compute lagrange polynomial
double lagrange_polynomial(int N){
//N = degree of polynomial

double *x;//hold the values
x=new double [N+1];

int i,k;
double y;

//these points are given
x[0]=-1.0;
x[1]=-0.5;
x[2]=0;
x[3]=0.5;
x[4]=1.0;

//computations for finding lagrange polynomial
double sum=0;
for (k=0;k<=N;k++){
    double prod=1.0;
        for (i=0;i<=N;i++){
            if (i==k) prod=1.0;
            else if (i!=k){
            prod*=(y-x[i])/(x[k]-x[i]);

                      }
              }
sum+=prod*f(x[k]);
}
return sum;
}

int main()
{

    double deg,result;

    cout <<"Give the degree of the polynomial :"<<endl;
    cin >>deg;

    result=lagrange_polynomial(deg);

    cout <<"The Lagrange approximation is :"<<endl;
    cout <<"p(x) = "<<result;

    return 0;
}

 

[Future Bruno]

The answer to the first question is that in your code, you are not manipulating the variable y. You are simply calculating the product of a series of terms, each of which involves the variable y. The result of the product will be a single number, but it is still a function of y. To get the polynomial form of the result, you need to expand the product out into the sum of its factors, and then reorganize the terms into powers of y.