Question

Translate the following code to java with comments

x0 = 1  # The initial guess
f(x) = x^2 - 2  # The function whose root we are trying to find
fprime(x) = 2 * x  # The derivative of the function
tolerance = 10^(-7)  # 7 digit accuracy is desired
epsilon = 10^(-14)  # Do not divide by a number smaller than this
maxIterations = 20  # Do not allow the iterations to continue indefinitely
solutionFound = false  # Have not converged to a solution yet

for i = 1:maxIterations
  y = f(x0)
  yprime = fprime(x0)
 
  if abs(yprime) < epsilon  # Stop if the denominator is too small
    break
  end
 
  global x1 = x0 - y/yprime  # Do Newton's computation
 
  if abs(x1 - x0) <= tolerance  # Stop when the result is within the desired tolerance
    global solutionFound = true
    break
  end
 
  global x0 = x1  # Update x0 to start the process again
end

if solutionFound
  println("Solution: ", x1)  # x1 is a solution within tolerance and maximum number of iterations
else
  println("Did not converge")  # Newton's method did not converge
end

Answer 1

`Hey,

Note: Brother if you have any queries related the answer please do comment. I would be very happy to resolve all your queries.

public class HelloWorld{
public static double f(double x)
{
return x*x-2.0;
}
public static double fprime(double x)
{
return x*2.0;
}
  
public static void main(String []args){

double x0 = 1; //The initial guess

double tolerance = 1e-7; // 7 digit accuracy is desired
double epsilon = 1e-14; // Do not divide by a number smaller than this
int maxIterations = 20; // Do not allow the iterations to continue indefinitely
boolean solutionFound = false; // Have not converged to a solution yet
double x1=x0;
for(int i=1;i<=maxIterations;i++)
{
double y=f(x0);
double yprime = fprime(x0);
if(Math.abs(yprime) < epsilon)
{
break;
}
x1 = x0 - y/yprime;
if(Math.abs(x1 - x0) <= tolerance) // Stop when the result is within the desired tolerance
{
solutionFound = true;
break;
}
x0 = x1;

}
if(solutionFound)
{
System.out.println("Solution: "+x1); // x1 is a solution within tolerance and
  
}
else
System.out.println("Did not converge"); // Newton's method did not converge

}
}

Kindly revert for any queries

Thanks.