MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following question has been on math.SE for several days. Without having a satisfying answer, I'd like to ask the experts here.

In mathematics, the big $O$ notation is used to describe the limiting behavior of a function. It is abuse of notation to say $$ f(x)=O(g(x)). $$ But this is understandable. However, in the class of numerical analysis, I found that the teacher used the big $O$ notation as the following:

If $\kappa = O(10^{-6})$, $\epsilon_{machine} = O(10^{-16})$, then we can only expect $O(10^{-10})$ accuracy.

I am surprised that they regard $O(10^{-6})$, $O(10^{-16})$, $O(10^{-10})$ as different things. Since according to the definition, they are nothing but $O(1)$.

I guess this is another kind of abuse of notions: when one says $O(10^{-6})$, s/he actually means $c\times 10^{-6}$ where $1<c<10$. But I don't know if this is a "standard" use in numerical analysis and I have never seen this in any paper or articles before.

So here are my questions:

Is there anyone who has seen this kind of usage before? Can anyone come up with the references (books, paper, etc.) that have the similar usage I mentioned above of the big $O$ notation?

[UPDATE:] Before this seemingly stupid question being closed (probably? I don't know), I would like to add my motivation here. I HAVE asked such question in class when the professor wrote down such notation, "what's your underlying limit for the big-O notation, sir?" Unfortunately, I got unsatisfying response: "there is no limit behavior for the notation", and he didn't elaborate any more. This is an applied math course, and it is really frustrating that one cannot understand things rigorously.

Anyway, I can get enough information from the comments in case people don't bother to answer such question.

share|cite|improve this question

closed as not a real question by Igor Rivin, Noah Snyder, Dan Petersen, José Figueroa-O'Farrill, quid Sep 25 '11 at 15:38

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Go to office hours and ask the teacher. It's a waste of time asking other people what someone else meant. I'd expect they actually meant something non-rigorous. – Noah Snyder Sep 25 '11 at 14:36
Just to add: during the OPERA neutrino presentation you could've seen that they use it in a similar manner, stating that cosmic ray influence is $O(10^{-4}$. Analysts may look at this as something strange, but people in applied mathematics got used to that symbol overload, I guess. – Harun Šiljak Sep 25 '11 at 14:48
I have seen in Computer Science literature this notation can be used informally. This is half-way between asymptotic formula (in which one has no information at all about the constant multiple) and a concrete bound (giving complete information about constant factors). Something like $A = O(10^{10})$ can be used to mean that $A$ is either a small factor larger or a small factor smaller than $10^{10}$. The small constant is deliberately left vague. – David Harris Sep 25 '11 at 15:15
This is a bit off-topic, but according to various authors big Oh is not (only) for limiting behavior yet for the inequality on the full domain (that is explictly or implictly specified). – user9072 Sep 25 '11 at 15:45
The formalisation you suggest has a problem: it would be meaningless to say that the machine epsilon is c*10^(-16) for some 0 < c < 10 as in this case it could still be arbitrily small. Perhps 1 instead of 0. – user9072 Sep 25 '11 at 16:30
up vote 6 down vote accepted

I have never seen this in civilized literature. Your professor is probably a visigoth.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.