generating pi to nth digit java

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无人共我
无人共我 2020-12-16 20:51

I wanted to know how I can generate pi to the nth digit. I have a couple of basic ideas.

  1. Use Math.PI and increase the precision (if that\'s possib
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  • 2020-12-16 20:56

    You can use this code

    import java.math.BigDecimal;
    import java.math.RoundingMode;
    
    public final class Pi {
    
    private static final BigDecimal TWO = new BigDecimal("2");
    private static final BigDecimal FOUR = new BigDecimal("4");
    private static final BigDecimal FIVE = new BigDecimal("5");
    private static final BigDecimal TWO_THIRTY_NINE = new BigDecimal("239");
    
    private Pi() {}
    
    public static BigDecimal pi(int numDigits) {
    
      int calcDigits = numDigits + 10;
    
      return FOUR.multiply((FOUR.multiply(arccot(FIVE, calcDigits)))
        .subtract(arccot(TWO_THIRTY_NINE, calcDigits)))
        .setScale(numDigits, RoundingMode.DOWN);
    }
    
     private static BigDecimal arccot(BigDecimal x, int numDigits) {
    
    BigDecimal unity = BigDecimal.ONE.setScale(numDigits,
      RoundingMode.DOWN);
    BigDecimal sum = unity.divide(x, RoundingMode.DOWN);
    BigDecimal xpower = new BigDecimal(sum.toString());
    BigDecimal term = null;
    
    boolean add = false;
    
    for (BigDecimal n = new BigDecimal("3"); term == null ||
      term.compareTo(BigDecimal.ZERO) != 0; n = n.add(TWO)) {
    
      xpower = xpower.divide(x.pow(2), RoundingMode.DOWN);
      term = xpower.divide(n, RoundingMode.DOWN);
      sum = add ? sum.add(term) : sum.subtract(term);
      add = ! add;
    }
    return sum;
    }
    }
    

    resource

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  • 2020-12-16 21:12
    • Math.PI is of type double. That means about 15 decimal digits of precision, and that is all the data you have; nothing will magically make additional digits of PI appear.
    • BigDecimal has arbitrary precision. setScale() allows you to create BigDecimal objects with as much precision as you want and most of the arithmetic methods will automatically increase precision as required, but of course the more precision, the slower all calculations will be.
    • The most difficult part of implementing Ramanujan's formula will ironically be the sqrt(2) in the constant factor, because there is not built-in sqrt() for BigDecimal, so you'll have to write your own.
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  • 2020-12-16 21:20

    You need to use MathContext to increase the precision of the BigDecimal

    e.g.

    MathContext mc = new MathContext(1000);
    BigDecimal TWO = new BigDecimal(2, mc);
    

    It's important that ALL the BigDecimals you use in your calculations use that MathContext. Heron's method should give you 1000 digits precision with only 10 iterations and a million digits with 20 iterations so it's certainly good enough. Also, create all the constant BigDecimals like e.g. 26390 only once at the start of your program.

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