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authorYigit Sever2020-11-08 22:28:06 +0300
committerYigit Sever2020-11-08 22:28:06 +0300
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Answer of the last question
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@@ -650,6 +650,7 @@ From the lectures we know that $\log{n}$ is $\mathcal{O}(n^{d})$ for all $d$.
650 650
651If $\log{n}$ is $\mathcal{O}(n^{\frac{1}{2}})$ then $n(\log{n})^{2}$ is $\mathcal{O}(n^{2})$. $H=2$. 651If $\log{n}$ is $\mathcal{O}(n^{\frac{1}{2}})$ then $n(\log{n})^{2}$ is $\mathcal{O}(n^{2})$. $H=2$.
652 652
653For $L$; we know that $f(n) = n(\log{n})^{2} > n^{1}$ but $n(\log{n})^{2} < n^{2}$ hence L = 1$.
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654% 1}}} % 655% 1}}} %
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