Generating Functions edits

src/Advanced/Generating Functions/main.tex

@@ -1,7 +1,7 @@
 % use [nosolutions] flag to hide solutions.
 % use [solutions] flag to show solutions.
 \documentclass[
-	solutions,
+	%solutions,
 	singlenumbering
 ]{../../../lib/tex/handout}
 \usepackage{../../../lib/tex/macros}
@@ -19,4 +19,5 @@
 	\input{parts/01 fibonacci.tex}
 	\input{parts/02 dice.tex}
 	\input{parts/03 coins.tex}
+	\input{parts/04 bonus.tex}
 \end{document}

src/Advanced/Generating Functions/parts/01 fibonacci.tex

@@ -77,7 +77,7 @@ A \textit{rational function} $f$ is a function that can be written as a quotient
 That is, $f(x) = \frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials.
 
 \problem{}
-Solve the equation from \ref<fibo> for $F(x)$, expressing it as a rational function.
+Solve the equation from \ref{fibo} for $F(x)$, expressing it as a rational function.
 
 \begin{solution}
 	\begin{align*}
@@ -99,8 +99,8 @@ Solve the equation from \ref<fibo> for $F(x)$, expressing it as a rational funct
 
 
 \definition{}
-\textit{Partial fraction decomposition} is an algebreic technique that works as follows: \par
+\textit{Partial fraction decomposition} is an algebraic technique that works as follows: \par
-If $p(x)$ is a polynomial and $a$ and $b$ are constants,
+If $p(x)$ is a polynomial of degree 1 and $a$ and $b$ are constants,
 we can rewrite the rational function $\frac{p(x)}{(x-a)(x-b)}$ as follows:
 \begin{equation*}
 	\frac{p(x)}{(x-a)(x-b)} = \frac{c}{x-a} + \frac{d}{x-b}
@@ -131,7 +131,7 @@ find a closed-form expression for its coefficients using partial fraction decomp
 
 \problem{}
 Using problems from the introduction and \ref{pfd}, find an expression
-for the coefficients of $F(x)$ (and this, for the Fibonacci numbers).
+for the coefficients of $F(x)$ (and thus, for the Fibonacci numbers).
 
 
 \begin{solution}

src/Advanced/Generating Functions/parts/02 dice.tex

@@ -76,7 +76,7 @@ the probability that the sum of the two dice is $k$.
 
 \problem{}
 Using generating functions, find two six-sided dice whose sum has the same
-distribution as the sum of two standard six-sided dice? \par
+distribution as the sum of two standard six-sided dice. \par
 
 That is, for any integer $k$, the number if ways that the sum of the two
 nonstandard dice rolls as $k$ is equal to the number of ways the sum of

src/Advanced/Generating Functions/parts/03 coins.tex

@@ -9,7 +9,7 @@ using pennies, nickels, dimes, quarters and half-dollars?}
 \vspace{2mm}
 
 Most ways of solving this involve awkward brute-force
-approache that don't reveal anything interesting about the problem:
+approaches that don't reveal anything interesting about the problem:
 how can we change our answer if we want to make change for
 \$0.51, or \$1.05, or some other quantity?
 

src/Advanced/Generating Functions/parts/04 bonus.tex

@@ -0,0 +1,57 @@
+\section{Extra Problems}
+
+
+\problem{USAMO 1996 Problem 6}
+Determine (with proof) whether there is a subset $X$ of
+the nonnegative integers with the following property: for any nonnegative integer $n$ there is exactly
+one solution of $a + 2b = n$ with $a, b \in X$.
+(The original USAMO question asked about all integers, not just nonnegative - this is harder,
+but still approachable with generating functions.)
+
+
+\vfill
+
+\problem{IMO Shortlist 1998}
+Let $a_0, a_1, ...$ be an increasing sequence of nonnegative integers
+such that every nonnegative integer can be
+expressed uniquely in the form $a_i + 2a_j + 4a_k$,
+where $i, j, k$ are not necessarily distinct.
+
+Determine $a_1998$.
+
+
+\vfill
+
+\problem{USAMO 1986 Problem 5}
+By a partition $\pi$ of an integer $n \geq 1$, we mean here a
+representation of $n$ as a sum of one or more positive integers where the summands must be put in
+nondecreasing order. (e.g., if $n = 4$, then the partitions $\pi$ are
+$1 + 1 + 1 + 1$, $1 + 1 + 2$, $1 + 3, 2 + 2$, and $4$).
+
+
+For any partition $\pi$, define $A(\pi)$ to be the number of ones which appear in $\pi$, and define $B(\pi)$
+to be the number of distinct integers which appear in $\pi$ (e.g, if $n = 13$ and $\pi$ is the partition
+$1 + 1 + 2 + 2 + 2 + 5$, then $A(\pi) = 2$ and $B(\pi) = 3$).
+
+Show that for any fixed $n$, the sum of $A(\pi)$ over all partitions of $\pi$ of $n$ is equal to the sum of
+$B(\pi)$ over all partitions of $\pi$ of $n$.
+
+\vfill
+
+\problem{USAMO 2017 Problem 2}
+Let $m_1, m_2, ..., m_n$ be a collection of $n$ distinct positive
+integers. For any sequence of integers $A = (a_1, ..., a_n)$ and any permutation $w = w_1, ..., w_n$ of
+$m_1, ..., m_n$, define an $A$-inversion of $w$ to be a pair of entries $w_i, w_j$ with $i < j$ for which one of the
+following conditions holds:
+\begin{itemize}
+	\item $ai \geq wi > wj$
+	\item $wj > ai \geq wi$
+	\item $wi > wj > ai$
+\end{itemize}
+
+Show that for any two sequences of integers $A = (a_1, ..., a_n)$ and $B = (b_1, ..., b_n)$ and for any
+positive integer $k$, the number of permutations of $m_1, ..., m_n$ having exactly $k$ $A$-inversions is equal
+to the number of permutations of $m_1, ..., m_n$ having exactly $k$ $B$-inversions.
+(The original USAMO problem allowed the numbers $m_1, ..., m_n$ to not necessarily be distinct.)
+
+\vfill

src/Advanced/Introduction to Quantum/src/parts/01 bits.tex

@@ -230,7 +230,7 @@ The \textit{tensor product} of two vectors is defined as follows:
 That is, we take our first vector, multiply the second
 vector by each of its components, and stack the result.
 You could think of this as a generalization of scalar
-mulitiplication, where scalar mulitiplication is a
+multiplication, where scalar multiplication is a
 tensor product with a vector in $\mathbb{R}^1$:
 \begin{equation*}
 	a

src/Advanced/Retrograde Analysis/parts/03 medium.tex

@@ -251,7 +251,7 @@ What is it, and what is its color? \par
 	\textbf{Part 4:}
 
 	The promoted black bishop on H2 must have been promoted on G1. The pawn which was promoted must have come from G7,
-	since neither of the pawns from F6 or H6 could make a capture to get to the G-file (all six missing white pieces have been accouted for).
+	since neither of the pawns from F6 or H6 could make a capture to get to the G-file (all six missing white pieces have been accounted for).
 	The Pawn from E7 has promoted to the bishop on A2.
 
 	What happened was this: the white pawn from G2 made its capture on H3 while the pawn on G3 was still on H2. This allowed the black pawn

src/Intermediate/Instant Insanity/main.tex

@@ -331,7 +331,7 @@
 	representing all four cubes. \\
 
 	\begin{center} \begin{small}
-	\begin{tikzpicture} \label{pic:II_comfiguration}
+	\begin{tikzpicture} \label{pic:II_configuration}
 		\filldraw [blue] (0,5) -- (1,5) -- (1,6) --
 		(0,6) -- (0,5);
 		\draw [line width = 1.5pt] (0,5) --