From f904af1768be984c7d6507bfe3690ad8e226ded8 Mon Sep 17 00:00:00 2001
From: Yigit Sever
Date: Mon, 9 Nov 2020 03:04:58 +0300
Subject: final changes

---
 main.tex | 10 ++++++----
 1 file changed, 6 insertions(+), 4 deletions(-)

(limited to 'main.tex')

diff --git a/main.tex b/main.tex
index dcb7980..a67b175 100644
--- a/main.tex
+++ b/main.tex
@@ -131,7 +131,7 @@ $S'$ could not have occurred since men propose in accordance to their preference
 For a proposer agnostic formation, arrange the preference lists of colleges and students as follows;
 
 \begin{gather*}%
-    \label{eq:student_prefs}
+    \label{eq:college_prefs}
     C_{1} \rightarrow \left\{ S_{1}, S_{n}, S_{n-1}, \dots, S_{2} \right\} \\
     C_{2} \rightarrow \left\{ S_{2}, S_{n}, S_{n-1}, \dots, S_{3} \right\} \\
     \dots  \\
@@ -517,6 +517,8 @@ For $c > 1$ and $n_{0} > 1$;
 
 For $f(n) \ge 1$. The conjecture is true for asymptotically positive function $f(n)$ but does not hold for an $f(n) < 1$.
 
+\pagebreak
+
 \begin{info}[]
     $f(n) + o(f(n)) = \Theta(f(n))$
 \end{info}
@@ -629,7 +631,7 @@ The infinite sum approaches 1 as $\lceil{\log n\rceil} \rightarrow \infty$, givi
     f(n) = 2 \cdot n \\
 \end{equation*}
 
-Finally, $H = L = 1$ since $f(n)$ is $\Theta(n)$. Borrowing the Big-$\Theta$ notation from Equation~\ref{eq:a_big_theta} for $c_1 = 1$ and $c_2 = 2$;
+(1,2) Finally, $H = L = 1$ since $f(n)$ is $\Theta(n)$. Borrowing the Big-$\Theta$ notation from Equation~\ref{eq:a_big_theta} for $c_1 = 1$ and $c_2 = 2$;
 
 \begin{equation*}
     n \le f(n) \le 2n
@@ -646,11 +648,11 @@ Finally, $H = L = 1$ since $f(n)$ is $\Theta(n)$. Borrowing the Big-$\Theta$ not
     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 $d$.
+(1) 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$.
+(2) For $L$; we know that $f(n) = n(\log{n})^{2} > n^{1}$ but $n(\log{n})^{2} < n^{2}$ hence $L = 1$.
 
 % 1}}} %
 
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