Smajl
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 Nov 5 awarded Notable Question Aug 6 awarded Popular Question Dec 19 awarded Constituent Dec 9 awarded Caucus Dec 1 awarded Notable Question Oct 17 awarded Popular Question Sep 28 answered Prove or Disprove? $\log(n^n)\text{ is } \Theta(\log n)$ Sep 28 answered how can i prove that square root of n is space constructible Sep 7 accepted Time complexity, proof $\log(n + c) \in O(\log(n))$ Sep 7 comment Time complexity, proof $\log(n + c) \in O(\log(n))$ But what if $c < 1$? Will it still hold? Sep 7 asked Time complexity, proof $\log(n + c) \in O(\log(n))$ Sep 7 awarded Custodian Sep 7 reviewed Approve Solving $1/n^{\lg (n)}$ Sep 7 accepted Solving $1/n^{\lg (n)}$ Sep 7 asked Solving $1/n^{\lg (n)}$ Aug 1 awarded Popular Question Jul 2 awarded Curious Jun 3 accepted $f(n) \in o(g(n))$ and $g(n) \in o(f(n))$ Jun 3 comment $f(n) \in o(g(n))$ and $g(n) \in o(f(n))$ Ok, right, but small o notation requires the second function to grow asymptotically faster than $f(n)$ - equality is not permitted here! Jun 3 comment $f(n) \in o(g(n))$ and $g(n) \in o(f(n))$ That is right, $f$ is not in $o(f(n))$ because small $o(g(n))$ notation means, that $g(n)$ is bigger (not $\geq$) than $f(n)$