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diff --git a/main.tex b/main.tex
index fe20d1f..cabba40 100644
--- a/main.tex
+++ b/main.tex
@@ -639,6 +639,17 @@ Finally, $H = L = 1$ since $f(n)$ is $\Theta(n)$. Borrowing the Big-$\Theta$ not
     $f(n) = n(\log n)^{2}$
 \end{info}
 
+\begin{align*}
+    n^{2} \quad v.s. \quad n(\log{n})^{2} \\
+    n^{1} \quad v.s. \quad (\log{n})^{2} \\
+    \sqrt{n} \quad v.s. \quad \sqrt{(\log{n})^{2}} \\
+    n^{\frac{1}{2}} \quad v.s. \quad \log{n}
+\end{align*}
+
+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 $\mathcal{O}(n^{2})$. $H=2$.
+
 
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