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I want to find out how to check, if the following relationships are true or false.

f(n) = nlog(n!); g(n) = nlog(2n^3n); Check, if f(n) = O(g(n)) and/or f(n) = Ω(g(n)) true/false;

f(n) = 3n^2; g(n) = 9^(base3-logn); Check, if f(n) = O(g(n)) and/or f(n) = Ω(g(n)) true/false;

I am very new to this topic, so i would be very grateful, if someone could explain it to me. If something is not understandable, don't hesitate and ask me.

Thank you in advance.

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  • $\begingroup$ Have you tried anything? $\endgroup$
    – zkutch
    Jul 10, 2022 at 15:06
  • $\begingroup$ Yes, but I am absolutely unsure about the result. Second example: I simplified g(n) and got n^2. From this I conclude that g(n) is an upper bound for f(n) and O(g(n)) is valid. Since, to my knowledge, prefactors are negligible in the dominant term, I would say at this point that g(n) is also a lower bound for f(n). But it is exactly at this point that I have the big uncertainty. $\endgroup$
    – roeas
    Jul 10, 2022 at 15:37
  • $\begingroup$ in the first example is it (2n)^(3n) or 2n^3n ? $\endgroup$
    – Rinkesh P
    Jul 10, 2022 at 16:24
  • $\begingroup$ It is nlog(2n)^(3n) $\endgroup$
    – roeas
    Jul 10, 2022 at 16:57
  • $\begingroup$ cs.stackexchange.com/q/824/755 $\endgroup$
    – D.W.
    Jul 11, 2022 at 0:30

1 Answer 1

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In second example if you have $f(n)=3n^2$ and $g(n)=n^2$, then, of course $f(n)=\Theta(g(n))$.

For first enough to note, that $n! < (2n)^{3n}$.

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  • $\begingroup$ I understand that (2n)^3n is the dominant term. Is that why we just need to concentrate on (2n)^3n and n!? If that is the case, then only O(g(n)) is true, isn't it? $\endgroup$
    – roeas
    Jul 10, 2022 at 19:56
  • $\begingroup$ Yes. Written inequality implies $n \log n! < n \log (2n)^{3n}$ for logarithm with base $>1$. $\endgroup$
    – zkutch
    Jul 10, 2022 at 23:18

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