# Is $O(10^{-6})$ an acceptable notation in numerical analysis?

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.

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?

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I have heard people use O in such a way in non-mathematical contexts, but only in words. (I have never seen this written as $O(10^{-6})$.) But I assumed that it stems from the english usage ("on the order of hundreds") and not from the mathematical notation. Of course, we cannot draw a precise border between the two. – Srivatsan Sep 21 '11 at 14:54
I don't think I've seen any of the books I have use such flagrant abuse of notation. But let me check... – J. M. Sep 21 '11 at 14:55
usually, for the proportional you write $\propto$ - and mathematically speaking $O(c)$ as a class of functions does not on $c$ if $c>0$. – Ilya Sep 21 '11 at 14:55
Most of the time I see people using big-O to represent computational complexities, like $O(n)$, $O(n \log n)$, so... – Shuhao Cao Sep 21 '11 at 15:01
It's not necessarily an "abuse of notions" (or even an abuse of notation); they've just defined a new notation. (Assuming that it is actually defined somewhere explicitly.) – ShreevatsaR Sep 25 '11 at 18:34

As I said in the post, "when one says $O(10^{-6})$, he/she may actually mean $c\times 10^{-6}$ where $1<c<10$."