find the value of pi till 50 digits

问题:

I want to calculate the value of PI till 50 digits.

How to do this in java for 50 decimal places?

回答1:

You cant do that with default data types, as you need for 50 digits: 50 / log(2) * log(10) = 166 bits. Here BigDecimal is one type you could use instead. But you should have in mind, that 22/7 is just an approximation of pi, and to get it right for 50 digits you need much better formula (e.g. Monte-Carlo method, taylor series, ...).

回答2:

You are using a double variable and instead should use something that has a greater precision. Look into the BigDecimal class.

回答3:

public class PiReCalc {
  public static final int N = 1000; // # of terms
   public static void main(String[] args) {
  BigDecimal sum = new BigDecimal(0);      // final sum
  BigDecimal term = new BigDecimal(0);           // term without sign
  BigDecimal sign = new BigDecimal(1.0);     // sign on each term

  BigDecimal one = new BigDecimal(1.0);
  BigDecimal two = new BigDecimal(2.0);

  for (int k = 0; k < N; k++) {
     BigDecimal count = new BigDecimal(k); 
     //term = 1.0/(2.0*k + 1.0);
     BigDecimal temp1 = two.multiply(count);
     BigDecimal temp2 = temp1.add(one);
     term = one.divide(temp2,50,BigDecimal.ROUND_FLOOR);

     //sum = sum + sign*term;
     BigDecimal temp3 = sign.multiply(term);
     sum = sum.add(temp3);

     sign = sign.negate();
  }
  BigDecimal pi = new BigDecimal(0);
  BigDecimal four = new BigDecimal(4);
  pi = sum.multiply(four);

  System.out.println("Calculated pi (approx., " + N + " terms and 50 Decimal Places): " + pi);
  System.out.println("Actual pi: " + Math.PI);
   }
}

The output is

Calculated pi (approx., 1000 terms and 50 Decimal Places): 3.14059265383979292596359650286939597045138933077984
Actual pi: 3.141592653589793

回答4:

Here is the break through paper of Bailey, Borwein and Plouffe: http://oldweb.cecm.sfu.ca/projects/pihex/p123.pdf

In the meantime, even faster formulas (following the same principles) were found: http://en.wikipedia.org/wiki/Bellard%27s_formula

回答5:

This is a quick and dirty implementation of Bellard's formula bigPi(200,2000) is good for over 500 decimal places in 75ms.

public static BigDecimal bigPi(int max,int digits) {
    BigDecimal num2power6 = new BigDecimal(64);
    BigDecimal sum = new BigDecimal(0);
    for(int i = 0; i < max; i++ ) {
        BigDecimal tmp;
        BigDecimal term ; 
        BigDecimal divisor;
        term = new BigDecimal(-32); 
        divisor = new BigDecimal(4*i+1); 
        tmp =  term.divide(divisor, digits, BigDecimal.ROUND_FLOOR);
        term = new BigDecimal(-1); 
        divisor = new BigDecimal(4*i+3); 
        tmp = tmp.add(term.divide(divisor, digits, BigDecimal.ROUND_FLOOR));
        term = new BigDecimal(256); 
        divisor = new BigDecimal(10*i+1); 
        tmp = tmp.add(term.divide(divisor, digits, BigDecimal.ROUND_FLOOR));
        term = new BigDecimal(-64); 
        divisor = new BigDecimal(10*i+3); 
        tmp = tmp.add(term.divide(divisor, digits, BigDecimal.ROUND_FLOOR));
        term = new BigDecimal(-4); 
        divisor = new BigDecimal(10*i+5); 
        tmp = tmp.add(term.divide(divisor, digits, BigDecimal.ROUND_FLOOR));
        term = new BigDecimal(-4); 
        divisor = new BigDecimal(10*i+7); 
        tmp = tmp.add(term.divide(divisor, digits, BigDecimal.ROUND_FLOOR));
        term = new BigDecimal(1); 
        divisor = new BigDecimal(10*i+9); 
        tmp = tmp.add(term.divide(divisor, digits, BigDecimal.ROUND_FLOOR));
        int s = ((1-((i&1)<<1)));
        divisor = new BigDecimal(2); 
        divisor = divisor.pow(10*i).multiply(new BigDecimal(s));
        sum = sum.add(tmp.divide(divisor, digits, BigDecimal.ROUND_FLOOR));
    }
    sum = sum.divide(num2power6,digits, BigDecimal.ROUND_FLOOR);
    return sum;

}