Trick for computing log(1+x)

Charles Karney showed me a clever trick for computing log(1+x). It only takes a couple lines of code, but those lines are not obvious. If you understand this trick, you’ll understand a good bit of how floating point arithmetic works.

First of all, what’s the big deal? Why not just add 1 to x and take the log? The direct approach will be inaccurate for small x and completely inaccurate for very small x. For more explanation, see Math library functions that seem unnecessary.

Here’s how Charles Kerney implements the code in C++:

template<typename T> static inline T log1p(T x) throw() {
    volatile T y = 1 + x, z = y - 1;
    return z == 0 ? x : x * std::log(y) / z;
}

The code comes from GeographicLib, and is based on Theorem 4 of this paper paper. I’ve read that paper several times, but I either didn’t notice the trick above or forgot it.

The code includes comments explaining the trick. I cut out the comments because I’ll explain the code below.

If y = 1+x and z = y-1, doesn’t z just equal x? No, not necessarily! This is exactly the kind of code that a well-meaning but uninformed programmer would go through and “clean up.” If x isn’t too small, no harm done. But if x is very small, it could ruin the code. It’s also the kind of code an optimizing compiler might rearrange, hence the volatile keyword telling optimizers to not be clever.

Imagine x = 1e-20. Then y = 1+x sets y to exactly 1; a standard floating point number does not have enough precision to distinguish 1 and 1 + 10^-20. Then z = y-1 sets z to 0. While z will not exactly equal x in general, it will be a good approximation for x.

If z is not zero, log(y)/z is a good approximation to log(1 + x)/x because y is close to 1+x and z is close to x. Multiplying log(y)/z by x gives a good approximation to log(1+x).

If z is zero, x is so small that x is an excellent approximation to log(1 + x) because the error is on the order of x2.

16 thoughts on “Trick for computing log(1+x)

  1. Sliderule users knew about this problem; heavy duty sliderules had separate scales for arguments close to 1.

  2. For me the most interesting part of the code is that it uses x * std::log(y) / z instead of just std::log(y) or, what amounts to the same, std::log(1+x). As you say, this is a good approximation to the logairthm of 1+x, but what is amazing is that it’s a better approximation than computing it directly.

  3. I have one comment and one question.

    Instead of convoluted computation of z (with volatiles etc) which you test against zero, you could just check “x < std::numeric_limits::epsilon()”. This will tell you whether 1+x is representable as a floating point of type T.

    My question is this: For the case where x > epsilon (z != 0), why multiply log(y) by x and then divide by z? Why not just return log(y)? I assume this is somehow compensating for the loss of precision when x is small but not too small, but it is not obvious how.

  4. Stupid comment field dropped my < and >

    That should read:

    “you could just check x < std::numeric_limits<T>::epsilon()”

  5. I’d also love to see the explanation of why x * log(1+x) / z is a better approximation of log(1+x) than just log(1+x) itself. The comments in the code don’t even claim that it is (?!) — they say “The multiplication x * (log(y)/z) introduces little additional error”, which makes it sound undesirable.

  6. So consider z is a crude estimate for x. x = z + s where s is small (not necessarily positive) (actually z is small. s is really small)

    in terms of z and s you want

    log(1+z+s) = (z+s)*log(1+z+s)/(z+s) = (z+s) * [log(1+z+s)/(z+s)]

    so you have z+s exactly, and you can do multiplication without too much loss of precision. so you want a good estimate for [log(1+z+s)/(z+s)].

    but your original problem is you can’t calculate log(1+z+s). You can only calculate log(1+z) = log(y). But note that log(1+z+s)/log(1+z) is close to (z+s)/z. So a good estimate for [log(1+z+s)/(z+s)] is [log(1+z)/z] = [log(y)/z].

    Which brings us to the result.

  7. SteveBrooklineMA

    But why not be more direct? Let F(s)=log(1+z+s), and expand at s=0 to get F(s) ~ log(1+z)+ s/(1+z) ~ log(1+z)+s. So can’t we use log(y)+(x-z) instead? Indeed, can’t we just use this last expression without the z==0 test?

  8. IIt might be instructive to work through the algorithm with x = 1.5 * eps
    where eps is the machine epsilon (the smallest positive number for which 1
    + eps is computed exactly).

    x = 1.5 * eps
    y = 1 + x = 1 + 2 * eps (using the round to even rule)
    z = y – 1 = 2 * eps
    log(y) = 2 * eps
    log(y)/z = 1
    return x * log(y)/z = 1.5 * eps

    Goldberg bounds the error in this method (assuming that the error in
    log(x) is bounded).

  9. @SteveBrooklineMA That’s a very interesting suggestion. I can’t see anything wrong with it.

  10. @SteveBrooklineMA @Jitse Niesen yeah, that does sound right. I did a quick simulation and you get about half the error by using (z+s) * [log(1+z+s)/(z+s)]. Though my calculus is failing me at the moment for explaining why.

  11. SteveBrooklineMA

    If A is the output from the original code, B =log(y)+(x-z) and C=log(y)+(x-z)/y, I haven’t been able to get either (A-B)/A or (A-C)/A to be more than something on the order of 1e-16. The absolute error between B and C seems even lower. This is just by naively testing with simple code and various x. So if it were up to me, I think I’d go with B.

    In the original code, is x * (std::log(y) / z) better, since (x * std::log(y)) / z may overflow for large x?

  12. But why the volatile T in the second line?

    The only thing I can think of is trying to disable compiler optimization for those variables and trivial calculations (addition, subtraction).

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