# how can be prove that $\max(f(n),g(n)) = \Theta(f(n)+g(n))$

how can be prove that $\max(f(n),g(n)) = \Theta(f(n)+g(n))$

though the big O case is simple since $\max(f(n),g(n)) \leq f(n)+g(n)$

edit : where $f(n)$ and $g(n)$ are asymptotically nonnegative functions.

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It is false by the common mathematical definition: $f(n) = n, g(n) = -n$. What definition of BigOh are you using? (I know you have tagged it algorithms, so I believe I can guess, but just making sure...) –  Aryabhata Dec 22 '10 at 1:49
@Jonas Yes.they are non negative. –  Bunny Rabbit Dec 22 '10 at 1:53

## 1 Answer

Hint...: $2 + 100 \le 100 + 100$

For the sake of completeness:

Use the fact that: $\max(f,g) \le f + g \le 2\max(f,g)$ when $f,g$ are non-negative

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i don't know whatever the constant n0 we choose , the $f(n0)+g(n0) < = max(f(n0),g(n0))$ is not going to hold, since they are non negative. –  Bunny Rabbit Dec 22 '10 at 2:13
@Bunny: How is that relevant to the hint above? –  Aryabhata Dec 22 '10 at 2:18
only if the above condition holds for some constant n0 then we can prove the $\omega$ case i.e the lower bound. –  Bunny Rabbit Dec 22 '10 at 2:19
Okay got it , i had actually not noticed that there can be two different constant for lower bound and the upper bound respectively , Thanks,that you didn't give away the whole solution , the hint made this lazy fella work a little , thanks a lot. –  Bunny Rabbit Dec 22 '10 at 2:30
@Bunny: Glad it helped. If the constant is same for lower and upper bound, then they both have to be equal! So you can rule that out :-) –  Aryabhata Dec 22 '10 at 2:33