About the $O$ notation with a subscript The usual $O$ notation was that $f(x) = O(g(x))$ meant that there exists a positive real number $C$ such that $|f(x)| \le C g(x)$ for sufficiently large $x$. However I came across the notation $O_r(\log x)$, where $r$ was a natural number. Having a subscript, what does it mean? I have tried to find if there is an explanation of this in the Wikipedia page about the Big $O$ notation but there was no such explanation. Would you please help me understand what $O_r(\log x)$ means?
 A: I can make no promises without a source for this notation, but I most frequently see this notation to mean the constant $C$ guaranteed by the big oh is allowed to depend on $r$. It can be useful to keep track of this to ensure you don't accidentally have two constants which depend on each other!
For a concrete example, in a power series expansion about the point $z_0$, we might have (when $|z-z_0| < r$):
$$f(z) = \sum_{k=0}^n a_k (z-z_0)^k + O_{n,r}(|z|^{n+1})$$
This means the constant depends both the degree of the approximation and the desired radius of approximation (which, of course, must be less than the radius of convergence). 
For a more detailed explanation of this notation, see the second PDF at this link. It is a set of notes on asymptotic notation by A.J. Hildebrand, which I have always found extremely well written and useful. 
Edit: 
For convenience, here is an example given in the linked notes. I have copied it almost verbatim.
For any $0 < r < 1$, we have
$$\log (1+z) = O_r(|z|)~~~~~~(|z| < r)$$
Notice, for $|z| < 1$, we have:
$$\log(1+z) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} z^n$$
Thus, we have
$$
|\log(1+z)| \leq 
\sum_{n=1}^\infty \frac{1}{n}|z^n| \leq 
\sum_{n=1}^\infty |z|^n =
\frac{1}{1-|z|}|z|
$$
Now if we restrict to $|z|<r$, we can get a constant which does not depend on $|z|$, and thus is suitable for Big-oh:
$$\frac{1}{1-|z|}|z| \leq (1-r)^{-1} |z|$$
Thus we see that the constant in the Big-oh depends on $r$.

I hope this helps ^_^
