# picomath

## Python (2.x and 3.x)

### gamma.py

```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
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
```