How to use in Java big double numbers?

bigdecimal, java, pi

Solution

If you need the precision of `BigDecimal`, you need to use it for all calculations. It is not sufficient to convert the result from `double` to `BigDecimal` at the end, because the precision is gone by then.

You need to convert all your `aX` variables to `BigDecimal`, and replace operators with calls to the corresponding methods of `BigDecimal` class:

BigDecimal pi = BigDecimal.ZERO;
for (int k = 0; k < n; k++) {
    BigDecimal a0 = new BigDecimal(16).pow(k);
    BigDecimal a1 = new BigDecimal(4).divide(new BigDecimal(8*k+1), 20, RoundingMode.HALF_UP);
    BigDecimal a2 = new BigDecimal(2).divide(new BigDecimal(8*k+4), 20, RoundingMode.HALF_UP);
    BigDecimal a3 = new BigDecimal(1).divide(new BigDecimal(8*k+5), 20, RoundingMode.HALF_UP);
    BigDecimal a4 = new BigDecimal(1).divide(new BigDecimal(8*k+6), 20, RoundingMode.HALF_UP);
    BigDecimal a5 = a1.subtract(a2).subtract(a3).subtract(a4);
    BigDecimal a6 = BigDecimal.ONE.divide(a0, 20, RoundingMode.HALF_UP);
    pi.add(a5.multiply(a6));
    System.out.println(pi);
}

Demo on ideone.

Problem

I've found a solution for calculating number of Pi by using BBS algorithm. But I encountered a problem. I'm missing a precision if using a double variable. Is there any suggestion to fix it? Here is my code: ``` public class Pi { public static void main(String[] args) { int n = 5; for (int k = 0; k < n; k++) { int a0 = (int) Math.pow(16, k); double a1 = (double) 4 / (8 * k + 1); double a2 = (double) 2 / (8 * k + 4); double a3 = (double) 1 / (8 * k + 5); double a4 = (double) 1 / (8 * k + 6); double a5 = a1 - a2 - a3 - a4; double a6 = (double) 1 / a0; double elem = a5 * a6; System.out.println(new BigDecimal(elem)); } } } ```

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