From b8a6b4e3f1778b53cd45bf2ebe2ed8cac279c71f Mon Sep 17 00:00:00 2001
From: Yigit Sever
Date: Sun, 8 Nov 2020 22:28:06 +0300
Subject: Answer of the last question

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 main.tex | 1 +
 1 file changed, 1 insertion(+)

diff --git a/main.tex b/main.tex
index cabba40..dcb7980 100644
--- a/main.tex
+++ b/main.tex
@@ -650,6 +650,7 @@ From the lectures we know that $\log{n}$ is $\mathcal{O}(n^{d})$ for all $d$.
 
 If $\log{n}$ is $\mathcal{O}(n^{\frac{1}{2}})$ then $n(\log{n})^{2}$ is $\mathcal{O}(n^{2})$. $H=2$.
 
+For $L$; 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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