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If $f(n)=g(n)$, can we just say that $\mathcal{O}(f(n))=\mathcal{O}(g(n))$? ($f$ and $g$ are two $\log$ functions)

Is it definitely yes? if not please describe why.

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I am not sure that the equation $\mathcal{O}(f(n))=\mathcal{O}(g(n))$ makes any sense. When you write $f(n)=\mathcal{O}(g(n))$, you mean that $f$ is bounded by $g$, but that's a shorthand notation, not an equality. – Yves Daoust May 7 '14 at 21:06
@YvesDaoust: Treating $\mathcal{O}(f(n))$ as a set, we do have $\mathcal{O}(f(n))=\mathcal{O}(g(n))$ as an equality of sets, though. – user2357112 May 7 '14 at 21:14
Then $f(n)=\mathcal{O}(g(n))$ cannot be written. And what set would $\mathcal{O}(f(n))$ deemed to be ? – Yves Daoust May 7 '14 at 21:17
@YvesDaoust: People who don't like writing $f(n)=\mathcal{O}(g(n))$ often define $\mathcal{O}(g(n))$ as the set of functions $f$ for which there is some constant $k$ such that $f(x)<kg(x)$ for all integers $x$, and write $f(n) \in \mathcal{O}(g(n))$. I always use $\in$ instead of $=$ for Big O notation where typographical restrictions don't require $=$, but I don't know how common that is. A quick search shows the notation used at – user2357112 May 7 '14 at 22:23
@user2357112: Interesting. It confirms that either way the question is ill-formulated. – Yves Daoust May 8 '14 at 6:38

Of course we can. If $f(n)=g(n)$, then $f$ and $g$ are the exact same function, and it doesn't matter whether we use the name $f$ or $g$ to refer to it in any context we choose.

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Got it. Thank you. And it doesn't matter whether their logarithmic bases are same, because the base can be omitted under big-O. Right? – user3576199 May 7 '14 at 21:00
@user3576199: If their bases are not the same, then $f(n) \ne g(n)$. – user2357112 May 7 '14 at 21:01
what if I have something like log(n) = klog(n)? k is a constant or something else. – user3576199 May 7 '14 at 21:04
and they have different base but log(n) equal to klog(n). – user3576199 May 7 '14 at 21:05
@user3576199: Then that's an entirely different question from what you asked, and it'd be good to be more careful about that in the future. We still have $\mathcal{O}(f(n)) = \mathcal{O}(g(n))$ from the definition of big-O notation, since it's trivial to find constants for which $af(n)<g(n)$ and $bg(n)<f(n)$. – user2357112 May 7 '14 at 21:06

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