Code Golf: Leibniz formula for Pi

Answer 1

Language: dc, Char count: 35

dc -e '9k0 1[d4r/r2+sar-lad274899>b]dsbxrp'

Answer 2

Oracle SQL 73 chars

select -4*sum(power(-1,level)/(level*2-1)) from dual connect by level<1e6

Answer 3

Using the formula for the error term in an alternating series (and thus the necessary number of iterations to achieve the desired accuracy is not hard coded into the program):

public static void Main(string[] args) {
    double tolerance = 0.000001;
    double piApproximation = LeibnizPi(tolerance);
    Console.WriteLine(piApproximation);
}

private static double LeibnizPi(double tolerance) {
    double quarterPiApproximation = 0;

    int index = 1;
    double term;
    int sign = 1;
    do {
        term = 1.0 / (2 * index - 1);
        quarterPiApproximation += ((double)sign) * term;
        index++;
        sign = -sign;
    } while (term > tolerance);

    return 4 * quarterPiApproximation;
}

Answer 4

J, 14 chars

4*-/%>:+:i.1e6

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Explanation

• 1e6 is number 1 followed by 6 zeroes (1000000).
• i.y generates the first y non negative numbers.
• +: is a function that doubles each element in the list argument.
• >: is a function that increments by one each element in the list argument.

So, the expression >:+:i.1e6 generates the first one million odd numbers:

1 3 5 7 ...

• % is the reciprocal operator (numerator "1" can be omitted).
• -/ does an alternate sum of each element in the list argument.

So, the expression -/%>:+:i.1e6 generates the alternate sum of the reciprocals of the first one million odd numbers:

1 - 1/3 + 1/5 - 1/7 + ...

• 4* is multiplication by four. If you multiply by four the previous sum, you have π.

That's it! J is a powerful language for mathematics.

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Edit: since generating 9! (362880) terms for the alternate sum is sufficient to have 5 decimal digit accuracy, and since the Leibniz formula can be written also this way:

4 - 4/3 + 4/5 - 4/7 + ...

...you can write a shorter, 12 chars version of the program:

-/4%>:+:i.9!

Answer 5

Another C# version:

(60 characters)

4*Enumerable.Range(0, 500000).Sum(x => Math.Pow(-1, x)/(2*x + 1));  // = 3,14159

Answer 6

F# (Interactive Mode) (59 Chars)

{0.0..1E6}|>Seq.fold(fun a x->a+ -1.**x/(2.*x+1.))0.|>(*)4.

(Yields a warning but omits the casts)