Hi all

http://i42.photobucket.com/albums/e316/K007/1c.gif

I'm probably missing something, it's not obvious that g(n) = O(f(n)) for all n (I don't see how 2^1000 boundaries of g(n) make any different here) ?

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- Mar 9th 2011, 06:01 PMNivgBasic growth of functions...Asymptotic notations question
Hi all

http://i42.photobucket.com/albums/e316/K007/1c.gif

I'm probably missing something, it's not obvious that g(n) = O(f(n)) for all n (I don't see how 2^1000 boundaries of g(n) make any different here) ? - Mar 9th 2011, 11:10 PMCaptainBlack
Yes, $\displaystyle g(n)=O(f(n))$, but is that what the question is asking for? The wording seems pretty obscure.

Could it want something like: $\displaystyle f(n)=O(n^3\log(n))$ and $\displaystyle g(n)=O(n^3)$ and so $\displaystyle g(n)=O(n^3\log(n))$.

Also you probably are expected to justify $\displaystyle f(n)=O(n^3\log(n))$

CB - Mar 10th 2011, 11:14 AMNivg
ok thanks didn't think it through.