I have a question about rounding error. I have created a function to approximate exp(x) by summing the terms of its taylor series until the sum stops changing. (That is, 1 + x + x^2/2! ...).

This function works well for seemingly any positive value of x that i have tried, but when i try for instance -30, it does not give me anything even close to what exp(-30) gives me.

I am aware of how to solve this problem, but I'm more curious as to why this happens. Does it have something to do with the alternating + and - signs when summing the series? Thank you.


Yes, that's exactly why: it's due to floating-point roundoff error, due to the alternating signs.

Suppose you have $x=10^{100}$ and $y=10^{100}-1$, and you ask your computer to subtract $x-y$. We know what the correct answer should be: it should be $1$. However, when $x$ and $y$ are represented as floating-point numbers, you don't get $1$ as the answer. Floating-point representation is not exact for large numbers, so $10^{100}$ will get rounded off to the nearest number that can be represented in the floating-point representation. As a result, $x$ won't be exactly $10^{100}$; it'll be a bit larger or a bit smaller -- and while the relative error will be very small ($< 0.1\%$), the absolute error might be very large ($x$ might be $10^{100}+\delta$ where $\delta$ is in the billions or much larger). The same is true for $y$. Thus, when you subtract $x-y$, the result won't be exactly $1$; it could be off by an absolute error of billions or much more. As a result, computing $x-y$ in floating-point arithmetic will give you answer that is wildly wrong.

In contrast, if you compute $x+y$ where $x=10^{100}$ and $y=10^{100}-1$, the answer will be reasonable. It won't be exactly correct, but it will be very close to correct.

To put it another way: suppose you compute $x+y$ and $x-y$ where $x \approx y$ and $x,y$ are very large. Then $x+y$ will be reasonably accurate, but $x-y$ will be wildly inaccurate. In particular, suppose that the closest representable floating-point number to $x$ is $\tilde{x}$, and the closest representable floating-point number to $y$ is $\tilde{y}$. We know that floating-point arithmetic has small relative error, so $\tilde{x} = x \times (1+\epsilon_x)$ where $\epsilon_x$ is small (say, $|\epsilon_x| \le 2^{-20}$), and $\tilde{y} = y \times (1+\epsilon_y)$ where $\epsilon_y$ is small. Now

$$\tilde{x} - \tilde{y} = x - y + x \epsilon_x - y \epsilon_y \approx x-y + x \times (\epsilon_x - \epsilon_y),$$

so the absolute error could be as large as $x \times (\epsilon_x- \epsilon-y)$. We know that $\epsilon_x - \epsilon_y$ will be no bigger than $2^{-19}$ or so, but if $x \approx 10^{100}$, their product will be enormous: the absolute error could be as large as $10^{100} \times 2^{-19}$, which is enormous. Since the correct answer is $x-y \approx 0$, the relative error will also be enormous.

In contrast,

$$\tilde{x} + \tilde{y} = x+y x \epsilon_x + y \epsilon_y \approx x+y + x \times (\epsilon_x + \epsilon_y),$$

and here $x \times (\epsilon_x + \epsilon_y)$ might be as large as $10^{100} \times 2^{-19}$ or so, which is enormous. So, the absolute error in this case is also large. However, the relative error is small: $10^{100} \times 2^{-19}$ is quite small compared to the correct answer, $x+y \approx 2 \times 10^{100}$. In particular, the relative error will be on the order of $2^{-19}$ when computing $x+y$ using floating-point math.

So, that's why absolute signs cause problems but the same sign doesn't cause problems.

For further explanation, you might enjoy reading the following article, especially Section 1.4:

What Every Computer Scientist Should Know About Floating-Point Arithmetic. David Goldberg. ACM Computing Surveys, 23:1, March 1991.

Separately, one other thing to know about Taylor series approximations is to have a feeling for the range where they are likely to be accurate. In practice, we always truncate the Taylor series after a finite number of terms -- the full Taylor series is infinite, so we can't sum the entire series; instead, we sum just the first $n$ terms, for some finite $n$.

If you truncate the Taylor series for some function $f(x)$ after $n$ terms, you get an approximation to $f(x)$. How good an approximation is this? There are various bounds on the error of this approximation. One bound is

$$\text{error} \le {|x-a|^{n+1} \over (n+1)!} \times \max \{f^{(n+1)}(v) : a \le v \le x\},$$

where we're using the first $n$ terms of the Taylor series approximation around $a$, and where we evaluate this at $x$ for some $x\ge a$. So, if you can upper-bound the value of the $n+1$-th derivative of $f(x)$, you can get a bound on how good an approximation the truncated Taylor series will be. You can read more about this at https://en.wikipedia.org/wiki/Taylor_series#Approximation_and_convergence and http://norsemathology.org/wiki/index.php?title=Taylor_Series_as_Approximations and https://en.wikipedia.org/wiki/Taylor%27s_theorem#Example.

In general, when you use a Taylor series approximation centered around $a$, you usually want to choose $a$ to be near $x$: the closer it is to $x$, the more accurate the approximation will be. Also, the more terms in your sum, the more accurate the approximation will be.

Once you have an upper-bound for the quality of the Taylor series approximation, you can now figure out how many terms you need to sum. Note that this analysis assumes infinite-precision numbers and ignores the issues due to floating-point round-off. So, to get an accurate answer, you need to ensure two things: (a) pick a Taylor series approximation with enough accuracy that it would give a good answer if you used infinite-precision numbers, and (b) you need to account for round-off error, to ensure that when you do the calculation with floating-point numbers you get a result that is close enough to what you would have gotten with infinite-precision math.

  • $\begingroup$ Thank you for your response. I think I understand what you are saying here, but I am still confused as to how the alternating signs cause this error whereas having non alternating signs doesn't. Could you please elaborate on that a little bit? $\endgroup$
    – myriagon
    Oct 10 '16 at 23:34
  • 1
    $\begingroup$ @myriagon, done. If that's still not enough, read the paper I linked to. $\endgroup$
    – D.W.
    Oct 11 '16 at 0:06

The Taylor series for the exponential function converges (mathematically) for any input value to the correct result. The problem is that with this particular series, the terms that you add will be huge compared to the final result. Rounding errors for addition are roughly proportional to the magnitude of each the result, so adding the Taylor series even for slightly negative arguments like -2 or -3 is less precise than you would want in this case, and for an argument like -30 you will be adding huge values and the rounding error will be larger than the actual result.

So you want to avoid the Taylor series as a method to evaluate a function if the individual terms are large compared to the result (exp (+100) for example is no problem there; the individual terms will be huge, but so is the result; the rounding error will be large, but not large compared to the result). In the case of exp (), trivially exp (-x) = 1.0 / exp (x), so the problem is easily avoided.

You will also notice that when the Taylor series is useful, the first terms or middle terms are usually large compared to the last terms that you add (because the Taylor series converges). Since each addition gives rounding errors proportional to the result of that addition, you should add the values starting with the smallest ones and add the largest ones last. Or not add the terms at all, but use the Horner scheme which does this automatically.


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