I have a simple question that is really hard to Google (besides the canonical What Every Computer Scientist Should Know About Floating-Point Arithmetic paper).

When should functions such as log1p or expm1 be used instead of log and exp? When should they not be used? How do different implementations of those functions differ in terms of their usage?

  • 2
    $\begingroup$ Welcome to Scicomp.SE! That's a very reasonable question, but would be easier to answer if you explained a bit which log1p you're referring to (especially how it's implemented, so we don't have to guess). $\endgroup$ – Christian Clason Sep 2 '15 at 14:53
  • 4
    $\begingroup$ For real-valued arguments, log1p$(x)$ and expm1$(x)$ should be used when $x$ is small, e.g., when $1 + x = 1$ in floating point accuracy. See, e.g., docs.scipy.org/doc/numpy/reference/generated/numpy.expm1.html and docs.scipy.org/doc/numpy/reference/generated/numpy.log1p.html. $\endgroup$ – GoHokies Sep 2 '15 at 14:58
  • $\begingroup$ @ChristianClason thanks, I am referring mostly to C++ std or R, but as you ask I start to think that learning about differences in the implementations would be also very interesting. $\endgroup$ – Tim Sep 2 '15 at 15:04
  • $\begingroup$ Here's an example: scicomp.stackexchange.com/questions/8371/… $\endgroup$ – Juan M. Bello-Rivas Sep 2 '15 at 16:01
  • 1
    $\begingroup$ @user2186862 "when $x$ is small" is correct, but not only "when $1+x=1$ in floating point accuracy" (which happens for $x\approx 10^{-16}$ in the usual double precision arithmetic). The documentation pages you linked show that they are useful already for $x\approx 10^{-10}$, for instance. $\endgroup$ – Federico Poloni Sep 3 '15 at 8:25

We all know that \begin{equation} \exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac12 x^2 + \dots \end{equation} implies that for $|x| \ll 1$, we have $\exp(x) \approx 1 + x$. This means that if we have to evaluate in floating point $\exp(x) -1$, for $|x| \ll 1$ catastrophic cancellation can occur.

This can be easily demonstrated in python:

>>> from math import (exp, expm1)

>>> x = 1e-8
>>> exp(x) - 1
>>> expm1(x)

>>> x = 1e-22
>>> exp(x) - 1
>>> expm1(x)

Exact values are \begin{align} \exp(10^{-8}) -1 &= 0.000000010000000050000000166666667083333334166666668 \dots \\ \exp(10^{-22})-1 &= 0.000000000000000000000100000000000000000000005000000 \dots \end{align}

In general an "accurate" implementation of exp and expm1 should be correct to no more than 1ULP (i.e. one unit of the last place). However, since attaining this accuracy results in "slow" code, sometimes a fast, less accurate implementation is available. For example in CUDA we have expf and expm1f, where f stands for fast. According to the CUDA C programming guide, app. D the expf has an error of 2ULP.

If you do not care about errors in the order of few ULPS, usually different implementations of the exponential function are equivalent, but beware that bugs may be hidden somewhere... (Remember the Pentium FDIV bug?)

So it is pretty clear that expm1 should be used to compute $\exp(x)-1$ for small $x$. Using it for general $x$ is not harmful, since expm1 is expected to be accurate over its full range:

>>> exp(200)-1 == exp(200) == expm1(200)

(In the above example $1$ is well below 1ULP of $\exp(200)$, so all three expression return exactly the same floating point number.)

A similar discussion holds for the inverse functions log and log1p since $\log(1+x) \approx x$ for $|x| \ll 1$.

  • 1
    $\begingroup$ This answer was already contained in the comments to the OP question. However I felt useful to give a longer (although basic) account just for clarity, in the hope it will useful to some readers. $\endgroup$ – Stefano M Sep 3 '15 at 9:44
  • $\begingroup$ OK, but then one can simply conclude "so I can always use expm1 instead of exp"... $\endgroup$ – Tim Sep 3 '15 at 10:03
  • 1
    $\begingroup$ @tim your conclusion is wrong: you can always use expm1(x) instead of exp(x)-1. Of course exp(x) == exp(x) - 1 does not hold in general. $\endgroup$ – Stefano M Sep 3 '15 at 10:13
  • $\begingroup$ OK, that is clear. And are there any clear cut criteria for $x \ll 1$? $\endgroup$ – Tim Sep 3 '15 at 13:13
  • 1
    $\begingroup$ @Tim there is no clear cut threshold, and the answer depends on the precision of the floating point implementation (and the problem being solved). While expm1(x) should be accurate to 1ULP over the entire range $0\leq x \leq 1$, exp(x) - 1 progressively looses precision from few ULP's when $x\approx 1$ to a complete breakdown when $x < \epsilon$, where $\epsilon$ is machine-epsilon. $\endgroup$ – Stefano M Sep 3 '15 at 20:17

To expand on the difference between log and log1p it might help to recall the graph if the logarithm:


If your data contains zeroes, then you probably don't want to use log since it's undefined at zero. And as $x$ approaches $0$, the value of $\ln(x)$ approaches $- \infty$. So if your $x$ values are close to $0$, then the value of $\ln(x)$ is potentially a large negative number. For example $\ln(\frac{1}{e}) = -1$ and $\ln(\frac{1}{e^{10}}) = -10$ and so on. This can be useful, but it can also distort your data towards large negative numbers, especially if your dataset also contains numbers much larger than zero.

On the other hand, as $x$ approaches $0$, the value of $\ln(x+1)$ approaches $0$ from the positive direction. For example $\ln(1+\frac{1}{e}) \sim 0.31$ and $\ln(1+\frac{1}{e^{10}}) \sim 0.000045$. So log1p produces only positive values and removes the 'danger' of large negative numbers. This generally insures a more homogeneous distribution when a dataset contains numbers close to zero.

In short, if the dataset is all greater than $1$, then log is usually fine. But, if the dataset has numbers between $0$ and $1$, then log1p is usually better.


Your Answer

