- #1

ggeo1

- 63

- 0

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 :

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 :

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;
}
```