Question

Use java and make sure to output only last 20 values of the series. Thanks

2) 18 points) Calculate the value of π from the infinite series: 4 44, 44 3 5 7911 Print a table that shows the value of π approximated by computing the first 200,000 terms of this series. How many terms do you have to use before you first get a value that begins with 3.14159? Note: As the output include only last 20 values. Do not paste all 200K results.

Answer 1

pr.java

/*
Calculating PI:

Calculate the value of π from the infinite series:  
π = 4 – 4/3 + 4/5 – 4/7 + 4/9 – 4/11 + …
Use a command-line argument to decide how far into the series to go.
Make sure to use a separate class to perform the calculation.
Print a table that shows the value of π approximated by computing the first
    200,000 terms of this series.
*/
import java.util.Scanner;
import java.math.*;
public class pr {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args)
    {
        // TODO code application logic here
      
        int howFar;
      
        System.out.println("How far into the infinite series would you like to go?");
        Scanner sc = new Scanner(System.in);
        howFar = sc.nextInt();
        double PICalced;
        PICalced = Calculation.calculate(howFar);
        System.out.println("PI is " + PICalced + ".");
      
        PICalced = Calculation.calculate(howFar);
        System.out.println("\n\nWould you like to print a table of the first 200,000 terms? (Type 1 for yes, 2 for no)");
        Scanner sc2 = new Scanner(System.in);
        if(sc2.nextInt() == 1);
            makeTable();
    }
    /**
     *
     */
    public static void makeTable()
    {
        double PICalcedBig = Calculation.calculate(200000);
      
        for(int i = 0; i < 200000; i++)
        {
            System.out.println(Calculation.getAllNums().get(i));
        }
    }
  
}

Calculation.java

//π = 4 – 4/3 + 4/5 – 4/7 + 4/9 – 4/11 + …

import java.math.*;
import java.util.*;

public class Calculation
{
    private static ArrayList<String> numByNum = new ArrayList<String>();;
    /**
     *
     * @param numPlaces
     * @return
     */
    public static double calculate(int numPlaces)
    {
        double PInum = 4.0;
        int denominator = 3;
        boolean operatorSwitch=true; //true is minus; false is plus
      
      
        for (int i = 0; i < numPlaces; i++)
        {
            //PT -- notice that both the if and else code are the same,
            //   other than the + or -. Could you get rid of the if?
            //   hint: (-1)^n is -1 if n is odd, and 1 if n is even
            if(operatorSwitch == true) // then subtract
            {
                PInum = PInum - (4.0/denominator);
                //System.out.println(PInum + " with " + i+1 + " additions.");
                numByNum.add("Number " + i + ": " + PInum);
                denominator += 2;
                operatorSwitch = false;
            }
            else // then add
            {
                PInum = PInum + (4.0/denominator);
                //System.out.println(PInum + " with " + i+1 + " additions. (+)");
                numByNum.add("Number " + i + ": " + PInum);
                denominator += 2;
                operatorSwitch = true;
            }
        }
        return PInum;
    }
    /**
     *
     * @return
     */
    public static ArrayList<String> getAllNums()
    {
        return numByNum;
    }
      
}