picomath

import math

# Visit http://www.johndcook.com/stand_alone_code.html for the source of this code and more like it.

# Note that the functions Gamma and LogGamma are mutually dependent.

def gamma(x):
    if x <= 0:
        raise ValueError("Invalid input")

    # Split the function domain into three intervals:
    # (0, 0.001), [0.001, 12), and (12, infinity)

    ###########################################################################
    # First interval: (0, 0.001)
    #
    # For small x, 1/Gamma(x) has power series x + gamma x^2  - ...
    # So in this range, 1/Gamma(x) = x + gamma x^2 with error on the order of x^3.
    # The relative error over this interval is less than 6e-7.

    gamma = 0.577215664901532860606512090 # Euler's gamma constant

    if x < 0.001:
        return 1.0/(x*(1.0 + gamma*x))

    ###########################################################################
    # Second interval: [0.001, 12)

    if x < 12.0:
        # The algorithm directly approximates gamma over (1,2) and uses
        # reduction identities to reduce other arguments to this interval.
        
        y = x
        n = 0
        arg_was_less_than_one = (y < 1.0)

        # Add or subtract integers as necessary to bring y into (1,2)
        # Will correct for this below
        if arg_was_less_than_one:
            y += 1.0
        else:
            n = int(math.floor(y)) - 1  # will use n later
            y -= n

        # numerator coefficients for approximation over the interval (1,2)
        p = [
            -1.71618513886549492533811E+0,
             2.47656508055759199108314E+1,
            -3.79804256470945635097577E+2,
             6.29331155312818442661052E+2,
             8.66966202790413211295064E+2,
            -3.14512729688483675254357E+4,
            -3.61444134186911729807069E+4,
             6.64561438202405440627855E+4
        ]

        # denominator coefficients for approximation over the interval (1,2)
        q = [
            -3.08402300119738975254353E+1,
             3.15350626979604161529144E+2,
            -1.01515636749021914166146E+3,
            -3.10777167157231109440444E+3,
             2.25381184209801510330112E+4,
             4.75584627752788110767815E+3,
            -1.34659959864969306392456E+5,
            -1.15132259675553483497211E+5
        ]

        num = 0.0
        den = 1.0

        z = y - 1
        for i in range(8):
            num = (num + p[i])*z
            den = den*z + q[i]
        result = num/den + 1.0

        # Apply correction if argument was not initially in (1,2)
        if arg_was_less_than_one:
            # Use identity gamma(z) = gamma(z+1)/z
            # The variable "result" now holds gamma of the original y + 1
            # Thus we use y-1 to get back the orginal y.
            result /= (y-1.0)
        else:
            # Use the identity gamma(z+n) = z*(z+1)* ... *(z+n-1)*gamma(z)
            for _ in range(n):
                result *= y
                y += 1

        return result

    ###########################################################################
    # Third interval: [12, infinity)

    if x > 171.624:
        # Correct answer too large to display. 
        return 1.0/0 # float infinity

    return math.exp(log_gamma(x))

def log_gamma(x):
    if x <= 0:
        raise ValueError("Invalid input")

    if x < 12.0:
        return math.log(abs(gamma(x)))

    # Abramowitz and Stegun 6.1.41
    # Asymptotic series should be good to at least 11 or 12 figures
    # For error analysis, see Whittiker and Watson
    # A Course in Modern Analysis (1927), page 252

    c = [
         1.0/12.0,
        -1.0/360.0,
         1.0/1260.0,
        -1.0/1680.0,
         1.0/1188.0,
        -691.0/360360.0,
         1.0/156.0,
        -3617.0/122400.0
    ]
    z = 1.0/(x*x)
    sum = c[7]
    for i in range(6, -1, -1):
        sum *= z
        sum += c[i]
    series = sum/x

    halfLogTwoPi = 0.91893853320467274178032973640562
    logGamma = (x - 0.5)*math.log(x) - x + halfLogTwoPi + series
    return logGamma