1
$\begingroup$

I know that there is a difference between absolute error and relative error, but what about the term actual error?

This came up in the notes for an example problem, but I haven't been able to find any clear answers. My intuition is that actual is synonymous with absolute, but I'm not sure. It's also possible that the author made a mistake and meant to use one of the two (or even a different) term.

Please note that I'm not asking or confused about "actual value", which I know is used in the definition of absolute error.

Here is the passage in context:


Consider the error bound formula:

$E_t \leq\frac{K(b-a)^3}{12n^2}$

Find a value for K. Then find an upper bound for the error when calculating $T_4$. How does this upper bound for the error compare with the actual error between $\int_0^\pi\cos (e^x)dx$ and $T_4$?

$\endgroup$
5
  • 3
    $\begingroup$ Most likely it was meant to be "absolute" . $\endgroup$ Commented Oct 19, 2016 at 3:29
  • 1
    $\begingroup$ Might help if you quoted the full context where that was used, literally. $\endgroup$
    – dxiv
    Commented Oct 19, 2016 at 3:30
  • 1
    $\begingroup$ It depends on the context. Let me offer one I have seen. Some methods , being approximations, have an actual error. This is usually a bound determined by some analysis. This is deemed to be the actual error. Whether you realize this in an implementation is a different matter motivating the absolute error that you mention. $\endgroup$
    – felasfa
    Commented Oct 19, 2016 at 3:39
  • $\begingroup$ @felasfa Your answer seems to be what I was asking about, but I'm still not sure what kind of comparisons I'm doing between the true (symbolic, arbitrary precision) answer, and the estimate (T4). $\endgroup$
    – Brian H.
    Commented Oct 19, 2016 at 3:55
  • $\begingroup$ @ah! now I understand what the author means. The actual error, as said above by others, is really the absolute error. $E_t$ gives you a bound on the absolute error. It is just an estimate. The author use of actual error is in relation to the error estimate. There might be cases where the absolute error is significantly less than the error estimate $E_t$. So "actual" in relation to the estimate error $E_t$. Hope it is clear $\endgroup$
    – felasfa
    Commented Oct 19, 2016 at 4:18

1 Answer 1

1
$\begingroup$

Absolute error is the difference between a measured or computed value and the true value. If our approximate value for $\pi$ is $3.14$ the absolute error is about $0.0016$. Relative error is the absolute error divided by the quantity. In this example the relative error is about $\frac {0.0016}{\pi} \approx 0.0005 = 0.05\%$ Each is important in certain cases. When quoting errors, you need to figure out which is more useful, then make sure the error you quote covers the possibilities.

Added: in your example, $K$ depends on a derivative of the function. You are expected to find the value of $K$ that corresponds to the greatest derivative of $\cos (e^x)$ on the interval $[0,\pi]$, then plug it in to get an error bound. After that, compute $T_4$ and a more accurate value for the integral. The actual error is the difference between these. You may find that the error bound is rather loose because the relevant derivative becomes large over a small part of the interval.

$\endgroup$
4
  • $\begingroup$ Yours is a correct answer, but to a different question than asked here. The OP hasn't clarified, yet, the context where actual error was sighted. In your example, it would not be unheard of that someone defined the actual error to be $\;-0.0016\;$ and the absolute error as $\;0.0016\;$. $\endgroup$
    – dxiv
    Commented Oct 19, 2016 at 3:44
  • $\begingroup$ @dxiv: the question has been updated since my answer. $\endgroup$ Commented Oct 19, 2016 at 3:54
  • $\begingroup$ Thanks for the response Ross, but as I stated in my question, I understand what the difference is between absolute and relative error. What I'm not sure is which one is meant by "actual error". I added some context to my question. I'm not quoting an error to use as support for something, instead, I'm using the result of this calculation in a later problem, and the choice of which one I use will produce different results in a later step. $\endgroup$
    – Brian H.
    Commented Oct 19, 2016 at 3:58
  • $\begingroup$ The actual error here would be the absolute error because you compute $T_4$ and the correct answer. Depending on the use of the error, you may want to turn that into a relative error if that is more useful. $\endgroup$ Commented Oct 19, 2016 at 4:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .