3. Newton rapshon method
//equation x^3-x-1
#include<bits/stdc++.h>
using namespace std;
#define maxval 1000000
double f(double x)
{
return x*x*x - x - 1;
}
double df(double x)
{
return 3*x*x-1;
}
void rapshon(double a)
{
double erro=f(a)/df(a);
while(abs(erro)>.001)
{
erro=f(a)/df(a);
a=a-erro;
}
cout<<"Root : "<<a<<endl;
}
int main()
{
cout<<"Enter value : ";
double x,y;
cin>>x;
rapshon(x);
return 0;
}
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