Question

write a computer program ( C++ ) that gives roots of an equation by newton raphson's methode.

Answer 1

The coding is done based on your requirement and is given below with sample output :

code:

#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double f(double x); //declare the function for the given equation
double f(double x) //define the function here, ie give the equation
{
double a=pow(x,3.0)-x-11.0; //write the equation whose roots are to be determined
return a;
}
double fprime(double x);
double fprime(double x)
{
double b=3*pow(x,2.0)-1.0; //write the first derivative of the equation
return b;
}
int main()
{
double x,x1,e,fx,fx1;
cout.precision(4); //set the precision
cout.setf(ios::fixed);   
cout<<"Enter the initial guess\n"; //take an intial guess
cin>>x1;
cout<<"Enter desired accuracy\n"; //take the desired accuracy
cin>>e;
fx=f(x);   
fx1=fprime(x);
cout <<"x{i}"<<" "<<"x{i+1}"<<" "<<"|x{i+1}-x{i}|"<<endl;   
  
do   
{
x=x1; /*make x equal to the last calculated value of x1*/
fx=f(x); //simplifying f(x)to fx
fx1=fprime(x); //simplifying fprime(x) to fx1
x1=x-(fx/fx1); /*calculate x{1} from x, fx and fx1*/
cout<<x<<" "<<x1<<" "<<abs(x1-x)<<endl;   
}while (fabs(x1-x)>=e); /*if |x{i+1}-x{i}| remains greater than the desired accuracy, continue the loop*/
cout<<"The root of the equation is "<<x1<<endl;
return 0;
}

Sample Output:

gcc version 4.6.3 Enter the initial guess 2 Enter desired accuracy 0001 xi x(i+1) 2.0000 2.4545 2.3764 2.3737 The root of the equation is 2.3736 2.4545 2.3764 2.3737 2.3736 0.4545 0.8781 0.0028 0.000e

Hope this Helps, if you have any doubs please comment i will get back to you and please thumbs up, thank you.