pi = 4(1-1/3 + 1/5-1/7 + 1/9-1/11 + 1/13-.........- 1/(2i-1) + 1/(2i+1) )

I am starting this program to approximate pi. Would someone be so kind as to direct me to information concerning this formula?

As I can see it, it contains two AP's as:

pi=4(1+1/5+1/9+1/13+......) - 4(1/3+1/7+1/11+...)

or

pi=4(Sigma(1/(2i-1))) - 4(Sigma(1/(2i+1)))
with i=0 to n

Choose a large value of n to approximate (Larger the n, more better the approximation.

/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/

package piproject;
//import java.math.BigDecimal;
/**
*
* @author Administrator
*/
import java.util.*;

public class ComputePi {
private double j;
public double decimal;
private double stop=100000;
private double index=10000;
private double sign=-1;//must alternate the sign by multiplying by -1??????????
private double term =0;
private double result=0;// Import java.text.DecimalFormat;???
private double x;//index or decimal places??
private double i;
private double PI = 0; // Import java.text.DecimalFormat;???
private double series_neg=(1/(2*i-1));
private double series_pos=(1/2*i+1);
private double series_one[];//store results of series_neg incrementing i and use as element of result_term
private double series_two[];//store results of series_pos incrementing i and use as element of result_term
private double result_term[];//incorporate in the summation from i=10000 to 100000
public static void main (String[] args){

Scanner reader = new Scanner (System.in);
System.out.println("This program approximates PI .");
//not sure if this is required
System.out.print("Please enter number of decimal places required: ");
//not sure how input is involved;to double stop approximation or to double numDecimalPlaces to indicate percision?
int numDecimalPlaces = reader.nextInt();
System.out.println ("Number of decimal places = " + numDecimalPlaces);
}
/////////////////////////////////////////////////////////////////////////////////////////////////
//Problem4.25 Computing pi You can approximate pi by using the following series:
//PI=4(1-1/3+1/5-1/7+1/9-1/11+1/13-...1/(2i-1)+1/(2i+i)
//Write a program that displays thePI value for 9=10000,20000...and 100000.
//////////////////////////////////////////26
///////////////////////////////////////////////////

public void calcPi(){
for(i=index,j=term;i<stop;i+index;){
result_term[term]=((sign)(series_neg[i])+(sign*-1)(series_pos[i]));

sign=sign*-1;
result=result_term[term];
}

PI=4*result;//not sure of summation of i from 10000 to 100000 and formula
}
}

if someone is farmiliar with a program could someone evauate the calcPi().
Thanks

This post has been edited by ceyesuma: 7 Feb, 2008 - 04:04 PM