Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have been asked to show that:

$$ \mathcal{O}(Max\{ f(n), g(n) \}) = \mathcal{O}(f(n) + g(n)) $$

I have seen explanations of similar problems, but this is the first time I have encountered the equals ($=$) sign being used to describe a relationship between two complexity classes (e.g.—$ \mathcal{O}(f) = \mathcal{O}(g) $).

Here is my attempt at the problem:

If $f(n) > 0$ and $g(n) > 0$ for all $n \geq n_{0}$, then

    $f(n) < f(n) + g(n)$, and
    $g(n) < f(n) + g(n)$.

Therefore, by definition, $f(n) \in \mathcal{O}(f(n) + g(n))$ and $g(n) \in \mathcal{O}(f(n) + g(n))$. Given that $Max\{f(n), g(n)\}$ must be either $f(n)$ or $g(n)$, it follows then that:

$$Max\{f(n), g(n)\} \in \mathcal{O}(f(n) + g(n)).$$

However, I am not sure whether showing this is the same as showing: $$ \mathcal{O}(Max\{ f(n), g(n) \}) = \mathcal{O}(f(n) + g(n)). $$

For example, if I know that $n^2+1 \in \mathcal{O}(n^2)$, may I then write that $\mathcal{O}(n^2+1)$ equals $\mathcal{O}(n^2)$? I understand that such a relationship is not necessarily symmetric (as might be suggested by the '$=$' symbol); would this statement simply imply that $\mathcal{O}(n^2+1) \subseteq \mathcal{O}(n^2)$?

I'd greatly appreciate any clarification and feedback on my approach. I may have completely missed the mark here.

share|improve this question
    
The use of an equals sign is generally considered an abuse of notation, it's really a one way equality. So it's better to use $\in$. –  Silynn Jun 22 at 21:14

1 Answer 1

up vote 1 down vote accepted

To show that $f(n) + g(n) \in \mathcal{O}(\text{Max}\{f(n),g(n)\})$, note that $f(n) \le \text{Max}\{f(n),g(n)\}$ and $g(n) \le \text{Max}\{f(n),g(n)\}$. So, what do you know about $f(n)+g(n)$?

After you've shown $\text{Max}\{f(n),g(n)\} \in \mathcal{O}(f(n)+g(n))$ and $f(n)+g(n) \in \mathcal{O}(\text{Max}\{f(n),g(n)\})$, then you know that $\mathcal{O}(\text{Max}\{f(n),g(n)\}) = \mathcal{O}(f(n)+g(n))$.

share|improve this answer
    
Thanks for the clear response! –  Michael Zalla Jun 22 at 21:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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