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Using the basic definition of theta notation prove that $\max(f(n), g(n)) = \Theta(f(n) + g(n))$

I came across two answer to this question on this website but the answers weren't clear to me. Would you mind to elaborate how this can be proven? I am first year student of computer sciences. Thank you!


What exactly does $\max(f(n), g(n))$ return?

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It's not true in general without some condition, the most obvious being that both are positive functions. – Thomas Andrews Dec 29 '12 at 19:46
@ThomasAndrews is "both are positive functions" the only condition? Am not able to guess any other condition for which above hold – PardonMeForMySuperPoorMaths May 2 at 20:21
up vote 7 down vote accepted

Note that $f(n) \leq f(n) + g(n)$ and $g(n) \leq f(n) + g(n)$. Hence, $$\max(f(n), g(n)) \in \mathcal{O}(f(n) + g(n))$$ Next note that $f(n) + g(n) \leq 2 \max(f(n),g(n))$. Hence, $$\max(f(n), g(n)) \in \mathcal{\Omega}(f(n) + g(n))$$ Hence, we get that $$\max(f(n), g(n)) \in \mathcal{\Theta}(f(n) + g(n))$$

Note that $$\max(f(n),g(n)) = \begin{cases} f(n) & \text{if } f(n) \geq g(n)\\ g(n) & \text{if } g(n) \geq f(n) \end{cases}$$ For instance, if $f(n) = 10n$ and $g(n) = n^2$, we get that $$\max(f(n),g(n)) = \begin{cases} 10n & \text{if } n \leq 10\\ n^2 & \text{if } n \geq 10 \end{cases}$$

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I think I'm starting to get the hang of it. So first we get the upper bound. However what I do not understand is the third and fourth line, namely, "Next note that f(n)+g(n)≤2max(f(n),g(n)). Hence, max(f(n),g(n))∈Ω(f(n)+g(n))". Why is f(n) + g(n) < 2max(f(n), g(n)) ? Basically after we get the upper and lower bounds, we can get the running time, correct? – Peter Dec 29 '12 at 19:44
@Peter For a fixed $n$, if $f(n) \geq g(n)$, then we have that $f(n) +f(n) \geq f(n) + g(n) \implies f(n) + g(n) \leq 2 f(n)$ and if $g(n) \geq f(n)$, then we have that $g(n) +g(n) \geq f(n) + g(n) \implies f(n) + g(n) \leq 2 g(n)$. Hence, we get that $$f(n) + g(n) \leq 2 \max(f(n) + g(n))$$ The key observation is that $f(n) \leq \max(f(n),g(n))$ and $g(n) \leq \max(f(n),g(n))$ and hence $$f(n) + g(n) \leq 2 \max(f(n) + g(n))$$ – user17762 Dec 29 '12 at 19:49
That makes more sense. However, how did you get to f(n) + f(n) or g(n) + g(n) ? – Peter Dec 29 '12 at 19:54
For instance, if we have $f(n) \geq g(n)$ adding $f(n)$ to both sides, we get that $2f(n) \geq f(n) + g(n)$ and similarly for the other case as well. – user17762 Dec 29 '12 at 19:56
Ok I'm off to try and figure this out to the core. Thanks for your help and time. Answer accepted – Peter Dec 29 '12 at 19:59

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