Stand-alone Python code for log(1+x)
If p is very small, directly computing log(1+p) can be inaccurate.
See the C++ version of this page for
details.
NB: Starting with version 2.6, Python has a function math.log1p for this purpose.
import math
# compute log(1+x) without loss of precision for small values of x
def log_one_plus_x(x):
if x <= -1.0:
raise FloatingPointError, "argument must be > -1"
if abs(x) > 1e-4:
# x is large enough that the obvious evaluation is OK
return math.log(1.0 + x)
else:
# Use Taylor approx.
# log(1 + x) = x - x^2/2 with error roughly x^3/3
# Since |x| < 10^-4, |x|^3 < 10^-12,
# and the relative error is less than 10^-8
return (-0.5*x + 1.0)*x
This code is in the public domain. Do whatever you want to with it, no strings attached.
Other versions of this code: C++, C#