Can both $f(n) \neq O(g(n))$ and $g(n) \neq O(f(n))$ happen at once?

Question

For which case are both $f(n) \ne O(g(n))$ and $g(n) \ne O(f(n))$ true?

I have following answer to this which I could not understand:

Sometimes true: For $f(n) = 1$ and $g(n) = |n \sin n|$ it is true, while for any $f(n) = O(g(n))$, e.g. $f(n) = g(n) = 1$, it is not true.

Please someone help in understanding:

For which case is it sometimes true? An explanation with example will be much appreciated.
What is meaning of "||" in this?

Possible duplicate of Sorting functions by asymptotic growth — David Richerby, Oct 03 '18 at 11:17
Here, || means absolute value. In the case of f(n) = 1 and g(n) = |n sin(n) |, we have neither f(n) = O(g(n)) nor g(n) = O(f(n)). — Rick Decker, Oct 03 '18 at 12:47

score 1 · Answer 1 · answered Oct 04 '18 at 03:54

Consider the following statement about two functions $f,g$:

$f(n) \neq O(g(n))$ and $g(n) \neq O(f(n))$.

The question wants you to choose which of the following holds:

The statement holds for every $f,g$.
The statement holds for no $f,g$.
The statement holds for some $f,g$, and doesn't hold for other $f,g$.

The third option is correct in your case, and this is the meaning of "sometimes true".

Indeed, if $f(n) = g(n) = 1$ then the statement doesn't hold. Conversely, if $f(n) = 1$ and $g(n) = |n\sin n|$ (where $\left|\cdot\right|$ denotes absolute value) then the statement does hold.

Can both $f(n) \neq O(g(n))$ and $g(n) \neq O(f(n))$ happen at once?

1 Answers1