Question

JAVA programming:

Construct a computer program that uses the Secant Method to solve the problem: fox)cos(3.0)- 1.0/20.00 Starting with the initial guesses of x 2 and x- 3, obtain an approximation to x such that fx)<0.000000001. On your output, show the iteration number (n), the approximate value of x on the nth iteration (xn), and the value of fxn Show the values to 9 or 10 digits. -0.5

Answer 1

Here is the code for your question

public class secant_method {
  
public static double Function(double x) {
return Math.cos((x*x)/3.0) - (1.0/20.0);
}
  

public static void main(String[] args) {
  
  
double x1, x0,x2,x3, precision,check;
int iter=0;
precision = 0.000000001;
x0 = 2;
x1 = 3;
if (Function(x0) Function(x1) < 0)
{
do {

//Intermediate value
x2 = (x0 Function(x1) - x1*Function(x0))/(Function(x1)-Function(x0));
  
  
//Checking is x2 a root
check = Function(x0) Function(x2);
  
// update the value of interval
x0 = x1;
x1 = x2;
  
// update number of iteration
iter++;
  
//if check is 0 then we will break the loop because we got
//the root
if (check == 0)
break;
x3 = (x0 Function(x1) - x1*Function(x0))/(Function(x1)-Function(x0));
  
System.out.println("Iteration :" + iter);
System.out.println("Approx value of x :" + x3);
System.out.println("Value of f(x) :"+ Function(x3));
//repeat untill we get to the precision
} while (Math.abs(x3 - x0) >= precision);
  
System.out.println("Root is "+ x1);
  
System.out.println("No of iterations are "+ iter);
}
  
else
System.out.print("Once check the interval");
  

}

}

Screenshots of the work

28 29 // update the value ot interval 31 32 // update number of iteration iter++ 34 35 36 37 38 39 40 41 //if check is θ then we will break the loop because we got //the root if (check0) break; (x8 x3 Function ( x1) -x1*Function (x®) ) / ( Function ( x1)-Function (x8)) * = System.out.println("Iteration" iter) System.out.println("Approx value of x"+ x3) System.out.println ( "Value of f (x) Function (x3)); //repeat untill we get to the precision 43 45 46 47 48 49 } while (Math.abs (x3 - x®) >= precision); System.out.println("Root is "+x1); System.out.println ("No of iterations are " iter); 52 53 54 else System.out.print("Once check the interval"); 56 57 58 59 Problems @ Javadoc Declaration Console X3 <terminated> secantmethod [Java Application] /usr/lib/jvm/java-8-openjdk-amd64/bin/java (25-Sep-2018, 11:49:09 A Iteration :1 Approx value of x 2.133062575993711 Value of f(x) :0.004117891289808814 Iteration :2 Approx value of x 2.1359486160796246 Value of f(x) :1.661899193537708E-5 Iteration :3