Generating Functions edits (#26)

Reviewed-on: #26

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/Mock a Mockingbird/main.tex

@@ -81,5 +81,6 @@
 	\input{parts/00 intro}
 	\input{parts/01 tmam}
 	\input{parts/02 kestrel}
+	\input{parts/03 bonus}
 
 \end{document}

src/Advanced/Mock a Mockingbird/parts/01 tmam.tex

@@ -23,7 +23,7 @@ Complete his proof.
 		\lineno{} let A           \cmnt{Let A be any any bird.}
 		\lineno{} let Cx = A(Mx)  \cmnt{Define C as the composition of A and M}
 		\lineno{} CC = A(MC)
-		\lineno{}    = A(CC)  \qed{}
+		\lineno{}    = A(CC)
 	\end{alltt}
 \end{solution}
 
@@ -46,7 +46,7 @@ Show that the laws of the forest guarantee that at least one bird is egocentric.
 		\lineno{}
 		\lineno{} ME = E     \cmnt{By definition of fondness}
 		\lineno{} ME = EE    \cmnt{By definition of M}
-		\lineno{} \thus{} EE = E  \qed{}
+		\lineno{} \thus{} EE = E
 	\end{alltt}
 \end{solution}
 
@@ -99,7 +99,7 @@ Show that if $C$ is agreeable, $A$ is agreeable.
 		\lineno{} let y so that Cy = Ey  \cmnt{Such a y must exist because C is agreeable}
 		\lineno{}
 		\lineno{} A(By) = Ey
-		\lineno{}       = D(By)  \qed{}
+		\lineno{}       = D(By)
 	\end{alltt}
 \end{solution}
 
@@ -129,6 +129,20 @@ Given three arbitrary birds $A$, $B$, and $C$, show that there exists a bird $D$
 We say two birds $A$ and $B$ are \textit{compatible} if there are birds $x$ and $y$ so that $Ax = y$ and $By = x$. \\
 Note that $x$ and $y$ may be the same bird. \\
 
+
+\problem{}
+Show that any bird that is fond of at least one bird is compatible with itself.
+
+\begin{solution}
+	\begin{alltt}
+		\lineno{} let A
+		\lineno{} let x so that Ax = x  \cmnt{A is fond of at least one other bird}
+		\lineno{} Ax = x
+	\end{alltt}
+\end{solution}
+
+\vfill
+
 \problem{}
 Show that any two birds in this forest are compatible. \\
 \begin{alltt}
@@ -144,7 +158,6 @@ Show that any two birds in this forest are compatible. \\
 \begin{solution}
 	\begin{alltt}
 		\lineno{} let A, B
-		\lineno{}
 		\lineno{} let Cx = A(Bx)  \cmnt{Composition}
 		\lineno{} let y = Cy      \cmnt{Let C be fond of y}
 		\lineno{}
@@ -152,24 +165,9 @@ Show that any two birds in this forest are compatible. \\
 		\lineno{}    = A(By)
 		\lineno{}
 		\lineno{} let x = By  \cmnt{Rename By to x}
-		\lineno{} Ax = y  \qed{}
+		\lineno{} Ax = y
 	\end{alltt}
 \end{solution}
 
-\vfill
-
-\problem{}
-Show that any bird that is fond of at least one bird is compatible with itself.
-
-\begin{solution}
-	\begin{alltt}
-		\lineno{} let A
-		\lineno{} let x so that Ax = x  \cmnt{A is fond of at least one other bird}
-		\lineno{} Ax = x  \qed{}
-	\end{alltt}
-
-	That's it.
-\end{solution}
-
 \vfill
 \pagebreak

src/Advanced/Mock a Mockingbird/parts/02 kestrel.tex

@@ -18,7 +18,7 @@ Say $A$ is fixated on $B$. Is $A$ fond of $B$?
 	\begin{alltt}
 		\lineno{} let A
 		\lineno{} let B so that Ax = B
-		\lineno{} \thus{} AB = B \qed{}
+		\lineno{} \thus{} AB = B
 	\end{alltt}
 \end{solution}
 \vfill
@@ -39,7 +39,7 @@ Show that an egocentric Kestrel is hopelessly egocentric.
 	\begin{alltt}
 		\lineno{} KK = K
 		\lineno{} \thus{} (KK)y = K  \cmnt{By definition of the Kestrel}
-		\lineno{} \thus{} Ky = K \qed{}  \cmnt{By 01}
+		\lineno{} \thus{} Ky = K     \cmnt{By 01}
 	\end{alltt}
 \end{solution}
 
@@ -64,7 +64,7 @@ Given the Law of Composition and the Law of the Mockingbird, show that at least
 	\begin{alltt}
 		\lineno{} let A so that KA = A   \cmnt{Any bird is fond of at least one bird}
 		\lineno{} (KA)y = y              \cmnt{By definition of the kestrel}
-		\lineno{} \thus{} Ay = A \qed{}  \cmnt{By 01}
+		\lineno{} \thus{} Ay = A              \cmnt{By 01}
 	\end{alltt}
 \end{solution}
 
@@ -90,7 +90,7 @@ Show that $Kx = Ky \implies x = y$.
 		\lineno{} (Kx)z = x
 		\lineno{} (Ky)z = y
 		\lineno{}
-		\lineno{} \thus{} x = (Kx)z = (Ky)z = y \qed{}
+		\lineno{} \thus{} x = (Kx)z = (Ky)z = y
 	\end{alltt}
 \end{solution}
 
@@ -128,7 +128,7 @@ An egocentric Kestrel must be extremely lonely. Why is this?
 		\lineno{} Ky = K
 		\lineno{} Kx = Ky
 		\lineno{} x = y for all x, y  \cmnt{By \ref{leftcancel}}
-		\lineno{} x = y = K \qed{}  \cmnt{By 10, and since K exists}
+		\lineno{} x = y = K  \cmnt{By 10, and since K exists}
 	\end{alltt}
 \end{solution}
 

src/Advanced/Mock a Mockingbird/parts/03 bonus.tex

@@ -0,0 +1,102 @@
+\section{Bonus Problems}
+
+\definition{}
+The identity bird has sometimes been maligned, owing to
+the fact that whatever bird x you call to $I$, all $I$ does is to echo
+$x$ back to you.
+
+\vspace{2mm}
+
+Superficially, the bird $I$ appears to have no intelligence or imagination; all it can do is repeat what it hears.
+For this reason, in the past, thoughtless students of ornithology
+referred to it as the idiot bird. However, a more profound or-
+nithologist once studied the situation in great depth and dis-
+covered that the identity bird is in fact highly intelligent! The
+real reason for its apparently unimaginative behavior is that it
+has an unusually large heart and hence is fond of every bird!
+When you call $x$ to $I$, the reason it responds by calling back $x$
+is not that it can't think of anything else; it's just that it wants
+you to know that it is fond of $x$!
+
+\vspace{2mm}
+
+Since an identity bird is fond of every bird, then it is also
+fond of itself, so every identity bird is egocentric. However,
+its egocentricity doesn't mean that it is any more fond of itself
+than of any other bird!.
+
+
+\problem{}
+The laws of the forest no longer apply.
+
+Suppose we are told that the forest contains an identity bird
+$I$ and that $I$ is agreeable. \
+Does it follow that every bird must be fond of at least one bird?
+
+\vfill
+
+
+\problem{}
+Suppose we are told that there is an identity bird $I$ and that
+every bird is fond of at least one bird. \
+Does it necessarily follow that $I$ is agreeable?
+
+\vfill
+\pagebreak
+
+
+\problem{}
+Suppose we are told that there is an identity bird $I$, but we are
+not told whether $I$ is agreeable or not.
+
+However, we are told that every pair of birds is compatible. \
+Which of the following conclusiens can be validly drawn?
+
+\begin{itemize}
+	\item Every bird is fond of at least one bird
+	\item $I$ is agreeable.
+\end{itemize}
+
+\vfill
+
+\problem{}
+The identity bird $I$, though egocentric, is in general not hope-
+lessly egocentric. Indeed, if there were a hopelessly egocentric
+identity bird, the situation would be quite sad. Why?
+
+\vfill
+
+\definition{}
+A bird $L$ is called a lark if the following
+holds for any birds $x$ and $y$:
+
+\[
+	(Lx)y = x(yy)
+\]
+
+\problem{}
+Prove that if the forest contains a lark $L$ and an identity bird
+$I$, then it must also contain a mockingbird $M$.
+
+\vfill
+\pagebreak
+
+
+\problem{}
+Why is a hopelessly egocentric lark unusually attractive?
+
+\vfill
+
+
+\problem{}
+Assuming that no bird can be both a lark and a kestrel---as
+any ornithologist knows!---prove that it is impossible for a
+lark to be fond of a kestrel.
+
+\vfill
+
+\problem{}
+It might happen, however, that a kestrel is fond of a lark. \par
+Show that in this case, \textit{every} bird is fond of the lark.
+
+\vfill

src/Warm-Ups/Bugs/main.typ

@@ -0,0 +1,11 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: Bugs on a Log],
+  by: "Mark",
+)
+
+#problem()
+2013 bugs are on a meter-long line. Each walks to the left or right at a constant speed. \
+If two bugs meet, both turn around and continue walking in opposite directions. \
+What is the longest time it could take for all the bugs to walk off the end of the log?

src/Warm-Ups/Bugs/meta.toml

@@ -0,0 +1,6 @@
+[metadata]
+title = "Bugs on a Log"
+
+[publish]
+handout = true
+solutions = true