Smajl
Reputation
304
Next privilege 500 Rep.
Access review queues
 Dec19 awarded Constituent Dec9 awarded Caucus Dec1 awarded Notable Question Oct17 awarded Popular Question Sep28 answered Prove or Disprove? $\log(n^n)\text{ is } \Theta(\log n)$ Sep28 answered how can i prove that square root of n is space constructible Sep7 accepted Time complexity, proof $\log(n + c) \in O(\log(n))$ Sep7 comment Time complexity, proof $\log(n + c) \in O(\log(n))$ But what if $c < 1$? Will it still hold? Sep7 asked Time complexity, proof $\log(n + c) \in O(\log(n))$ Sep7 awarded Custodian Sep7 reviewed Approve Solving $1/n^{\lg (n)}$ Sep7 accepted Solving $1/n^{\lg (n)}$ Sep7 asked Solving $1/n^{\lg (n)}$ Aug1 awarded Popular Question Jul2 awarded Curious Jun3 accepted $f(n) \in o(g(n))$ and $g(n) \in o(f(n))$ Jun3 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! Jun3 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 $f(n)$ Jun3 comment $f(n) \in o(g(n))$ and $g(n) \in o(f(n))$ But the second definition is basically big O notation, right? Small o notation is stronger... Jun3 asked $f(n) \in o(g(n))$ and $g(n) \in o(f(n))$