From bd1a72b08909c9a51eaf6735d263b695df795cf9 Mon Sep 17 00:00:00 2001 From: Mark Date: Thu, 9 Jan 2025 11:09:27 -0800 Subject: [PATCH] Added generating functions --- Advanced/Generating Functions/main.tex | 22 +++ .../parts/00 introduction.tex | 106 +++++++++++ .../parts/01 fibonacci.tex | 168 ++++++++++++++++++ .../Generating Functions/parts/02 dice.tex | 102 +++++++++++ .../Generating Functions/parts/03 coins.tex | 102 +++++++++++ 5 files changed, 500 insertions(+) create mode 100755 Advanced/Generating Functions/main.tex create mode 100755 Advanced/Generating Functions/parts/00 introduction.tex create mode 100755 Advanced/Generating Functions/parts/01 fibonacci.tex create mode 100755 Advanced/Generating Functions/parts/02 dice.tex create mode 100755 Advanced/Generating Functions/parts/03 coins.tex diff --git a/Advanced/Generating Functions/main.tex b/Advanced/Generating Functions/main.tex new file mode 100755 index 0000000..dbe5e06 --- /dev/null +++ b/Advanced/Generating Functions/main.tex @@ -0,0 +1,22 @@ +% use [nosolutions] flag to hide solutions. +% use [solutions] flag to show solutions. +\documentclass[ + solutions, + singlenumbering +]{../../resources/ormc_handout} +\usepackage{../../resources/macros} + + +\uptitlel{Advanced 2} +\uptitler{\smallurl{}} +\title{Generating Functions} +\subtitle{Prepared by Mark on \today \\ Based on a handout by Aaron Anderson} + +\begin{document} + \maketitle + + \input{parts/00 introduction.tex} + \input{parts/01 fibonacci.tex} + \input{parts/02 dice.tex} + \input{parts/03 coins.tex} +\end{document} \ No newline at end of file diff --git a/Advanced/Generating Functions/parts/00 introduction.tex b/Advanced/Generating Functions/parts/00 introduction.tex new file mode 100755 index 0000000..96fca76 --- /dev/null +++ b/Advanced/Generating Functions/parts/00 introduction.tex @@ -0,0 +1,106 @@ +\section{Introduction} + +\definition{} +Say we have a sequence $a_0, a_1, a_2, ...$. \par +The \textit{generating function} of this sequence is defined as follows: +\begin{equation*} + A(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + a_3x+3 + ... +\end{equation*} + +Under some circumstances, this sum does not converge, and thus $A(x)$ is undefined. \par +However, we can still manipulate this infinite sum to get useful results even if $A(x)$ +diverges. + +\problem{} +Let $A(x)$ be the generating function of the sequence $a_n$, \par +and let $B(x)$ be the generating function of the sequence $b_n$. \par +Find the sequences that correspond to the following generating functions: +\begin{itemize}[itemsep=2mm] + \item $cA(x)$ + \item $xA(x)$ + \item $A(x) + B(x)$ + \item $A(x)B(x)$ +\end{itemize} + +\begin{solution} + \begin{itemize}[itemsep=2mm] + \item $cA(x)$ corresponds to $ca_n$ + \item $xA(x)$ corresponds to $0, a_0, a_1, ...$ + \item $A(x) + B(x)$ corresponds to $a_n+b_n$ + \item $A(x)B(x)$ is $a_0b_0 + (a_0b_1 + a_1b_0)x + (a_0b_2 + a_1b_1 + a_2b_0)x^2 + ...$ \par + Which corresponds to $c_n = \sum_{k=0}^n a_kb_{n-k}$ + \end{itemize} +\end{solution} + + +\vfill +\pagebreak + + + + +\problem{} +Assuming $|x| < 1$, show that +\begin{equation*} + \frac{1}{1-x} = 1 + x + x^2 + x^3 + ... +\end{equation*} + +\begin{solution} + Let $S = 1 + x + x^2 + ...$ \par + Then, $xS = x + x^2 + x^3 + ...$ \par + + \vspace{2mm} + + So, $xS = S - 1$ \par + and $1 = S - xS = S(1 - x)$ \par + and $S = \frac{1}{1-x}$. +\end{solution} + + +\vfill + + + + +\problem{} +Let $A(x)$ be the generating function of the sequence $a_n$. \par +Find the sequence that corresponds to the generating function $\frac{A(x)}{1-x}$ + +\begin{solution} + \begin{align*} + \frac{A(x)}{1-x} + &=~ A(x)(1 + x + x^2 + ...) \\ + &=~ (a_0 + a_1x + a_2x^2 + ...)(1 + x + x^2 + ...)\\ + &=~ a_0 + (a_0 + a_1)x + (a_0 + a_1 + a_2)x^2 + ... + \end{align*} + + Which corresponds to the sequence $c_n = \sum_{k=0}^n a_k$ +\end{solution} + +\vfill + +\problem{} +Find short expressions for the generating functions for the following sequences: +\begin{itemize} + \item $1, 0, 1, 0, ...$ + \item $1, 2, 4, 8, 16, ...$ + \item $1, 2, 3, 4, 5, ...$ +\end{itemize} + +\begin{solution} + \begin{itemize}[itemsep=2mm] + \item $1, 0, 1, 0, ...$ corresponds to $1 + x^2 + x^4 + ...$. \par + By \ref{xminusone}, this is $\frac{1}{1-x^2}$. + + \item $1, 2, 4, 8, 16, ...$ corresponds to $1 + 2x + (2x)^2 + ...$. \par + By \ref{xminusone}, this is $\frac{1}{1-2x}$. + + \item $1, 2, 3, 4, 5, ...$ corresponds to $1 + 2x + 3x^2 + 4x^3 + ...$.\par + This is equal to $(1 + x + x^2 + ...)^2$, and thus is $\left(\frac{1}{1-x}\right)^2$ + \end{itemize} +\end{solution} + + +\vfill + +\pagebreak diff --git a/Advanced/Generating Functions/parts/01 fibonacci.tex b/Advanced/Generating Functions/parts/01 fibonacci.tex new file mode 100755 index 0000000..af82307 --- /dev/null +++ b/Advanced/Generating Functions/parts/01 fibonacci.tex @@ -0,0 +1,168 @@ +\section{Fibonacci} + +\definition{} +The \textit{Fibonacci numbers} are defined by the following recurrence relation: +\begin{itemize} + \item $f_0 = 0$ + \item $f_1 = 1$ + \item $f_n = f_{n-1} + f_{n-2}$ +\end{itemize} + +\problem{} +Let $F(x)$ be the generating function that corresponds to the Fibonacci numbers. \par +Find the generating function of $0, f_0, f_1, ...$ in terms of $F(x)$. \par +Call this $G(x)$. + +\begin{solution} + \begin{equation*} + G(x) = xF(x) + \end{equation*} +\end{solution} + +\vfill + +\problem{} +Find the generating function of $0, 0, f_0, f_1, ...$ in terms of $F(x)$. +Call this $H(x)$. + +\begin{solution} + \begin{equation*} + H(x) = x^2F(x) + \end{equation*} +\end{solution} + +\vfill + +\problem{} +Calculate $F(x) - G(x) - H(x)$ using the recurrence relation that +we used to define the Fibonacci numbers. + +\begin{solution} + \begin{align*} + F(x) - G(x) - H(x) + &=~ f_0 + (f_1 - f_0)x + (f_2 - f_1 - f_0)x^2 + (f_3 - f_2 - f_1)x^3 + ... \\ + &=~ f_0 + (f_1 - f_0)x \\ + &=~ x + \end{align*} +\end{solution} + +\vfill +\pagebreak + + +\problem{} +Using the problems on the previous page, find $F(x)$ in terms of $x$. + +\begin{solution} + \begin{align*} + x + &=~ F(x) - G(x) - H(x) \\ + &=~ F(x) - xF(x) - x^2F(x) \\ + &=~ F(x)(1-x-x^2) + \end{align*} + + So, + \begin{equation*} + F(x) = \frac{x}{1-x-x^2} + \end{equation*} +\end{solution} + +\vfill + + + + +\definition{} +A \textit{rational function} $f$ is a function that can be written as a quotient of polynomials. \par +That is, $f(x) = \frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials. + +\problem{} +Solve this equation for $F(x)$, expressing it as a rational function. + +\begin{solution} + \begin{align*} + F(x) + &=~ \frac{-x}{x^2+x-1} = \frac{-x}{(x-a)(x-b)} \\ + &=~ \frac{1 - \sqrt{5}}{2\sqrt{5}}\frac{1}{x-a} + \frac{-1 - \sqrt{5}}{2\sqrt{5}}\frac{1}{x-b} + \end{align*} + + where + + \begin{equation*} + a = \frac{-1 + \sqrt{5}}{2} ;~~ + b = \frac{-1 - \sqrt{5}}{2} + \end{equation*} +\end{solution} + +\vfill +\pagebreak + + +\problem{} +Now that we have a rational function for $F(x)$, \par +find a closed-form expression for its coefficients. + +\vspace{2mm} + +Do this using \textit{partial fraction decomposition:} \par +We can break up a rational function $\frac{p(x)}{(x-a)(x-b)}$ as follows: +\begin{equation*} + F(x) = \frac{p(x)}{(x-a)(x-b)} = \frac{c}{x-a} + \frac{d}{x-b} +\end{equation*} +where $c$ and $d$ are constants. + + +\begin{solution} + \begin{align*} + F(x) + &=~ + \left(\frac{1 - \sqrt{5}}{2\sqrt{5}}\right)\left(\frac{-1}{a}\right)\left(\frac{1}{1-\frac{x}{a}}\right) + + \left(\frac{-1 - \sqrt{5}}{2\sqrt{5}}\right)\left(\frac{-1}{b}\right)\left(\frac{1}{1-\frac{x}{b}}\right) \\ + &=~ + \left(\frac{1}{\sqrt{5}}\right)\left(\frac{1}{1-\frac{x}{a}}\right) + + \left(\frac{-1}{\sqrt{5}}\right)\left(\frac{1}{1-\frac{x}{b}}\right) \\ + &=~ + \frac{1}{\sqrt{5}}\left(1 + \frac{x}{a} + \left(\frac{x}{a}\right)^2 + ...\right) + - \frac{1}{\sqrt{5}}\left(1 + \frac{x}{b} + \left(\frac{x}{b}\right)^2 + ...\right) + \end{align*} +\end{solution} + +\vfill + +\problem{} +Using problems from the introduction and \ref{pfd}, find an expression +for the coefficients of $F(x)$ (and this, for the Fibonacci numbers). + + +\begin{solution} + \begin{align*} + f_0 &= \frac{1}{\sqrt{5}} - \frac{1}{\sqrt{5}} = 0 \\ + f_1 &= \frac{1}{a\sqrt{5}} - \frac{1}{b\sqrt{5}} = 1 \\ + f_n &= \frac{1}{\sqrt{5}}\left(\frac{1}{a^n} - \frac{1}{b^n}\right) + = \frac{1}{\sqrt{5}}\left( + \left(\frac{1 + \sqrt{5}}{2}\right)^n + - \left(\frac{1-\sqrt{5}}{2}\right)^n + \right) + \end{align*} +\end{solution} + +\vfill +\pagebreak + +\problem{Bonus} +Repeat the method of recurrence, generating function, +partial fraction decomposition, and geometric series +to find a closed form for the following sequence: +\begin{equation*} + a_0 = 1 ;~~ a_{n+1} = 2a_n + n +\end{equation*} + +\hint{ + When doing partial fraction decomposition with a + denominator of the form $(x-a)^2(x-b)$, + you may need to express your expression as a sum of three fractions: + $\frac{c}{(x-a)^2} + \frac{d}{x-a} + \frac{e}{x-b}$.`' +} + +\vfill +\pagebreak \ No newline at end of file diff --git a/Advanced/Generating Functions/parts/02 dice.tex b/Advanced/Generating Functions/parts/02 dice.tex new file mode 100755 index 0000000..0196ef1 --- /dev/null +++ b/Advanced/Generating Functions/parts/02 dice.tex @@ -0,0 +1,102 @@ +\section{Dice} + +\definition{} +A \textit{die} is a device that randomly selects a positive integer from +a finite list of options. For example, the standard 6-sided die selects a value from +$[1,2,3,4,5,6]$. We may have many sides with the same value, as in $[1, 1, 2, 3]$. + +To describe a die with a generating function, let $a_k$ be the number of times +$k$ appears as a side of the die and consider $a_0 + a_1x + x_2x^2 + ... $. \par +A die has a finite number of sides, so this will be a regular polynomial. + +\problem{} +What is the generating function of the standard 6-sided die? + +\begin{solution} + $x + x^2 + x^3 + x^4 + x^5 + x^6$ +\end{solution} + +\vfill + +\problem{} +What is the generating function of the die with sides $[1, 2, 3, 5]$? + +\begin{solution} + $2x + x^2 + x^3 + x^5$ +\end{solution} + +\vfill + +\problem{} +Let $A(x)$ and $B(x)$ be the generating functions of two dice. \par +What is the significance of $A(1)$? +\begin{solution} + $A(1) = $ the number of sides on the die +\end{solution} + +\vfill + +\problem{} +Using formulas we found earlier, show that the $k^\text{th}$ coefficient +of $A(x)B(x)$ is the number of ways to roll $k$ as the sum of the two dice. +\begin{solution} + The $k^\text{th}$ coefficient of $A(x)B(x)$ is... + + \begin{align*} + a_0b_k + a_1b_{k+1} + ... + a_kb_0 \\ + &=~ \text{count}(A = 0; B = k) + ... + \text{count}(A = k; B = 0) \\ + &=~ \text{number of ways} A + B = k + \end{align*} +\end{solution} + + +\vfill +\pagebreak + +\problem{} +Find a generating function for the sequence $c_0, c_1, ...$, where $c_k$ is +the probability that the sum of the two dice is $k$. +\begin{solution} + + \begin{equation*} + c_k + = \frac{\text{number of ways sum } = k}{\text{number of total outcomes}} + = \frac{\text{number of ways sum } = k}{A(1)B(1)} + \end{equation*} + + So, + + \begin{equation*} + c_0 + c_1x + c_2x^2 = + \frac{A(x)B(x)}{A(1)B(1)} + \end{equation*} +\end{solution} + +\vfill + +\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 + +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 numer of ways the sum of +two standard dice rolls as $k$. +\hint{factor polynomials.} +\begin{solution} + We need a different factorization of + + \begin{equation*} + (x + x^2 + x^3 + x^4 + x^5 + x^6)^2 = A(x)B(x) + \end{equation*} + + We can use + + \begin{equation*} + (x + 2x^2 + 2x^3 + x^4) + (x + x^3 + x^4 + x^5 + x^6 + x^8) + \end{equation*} +\end{solution} + + +\vfill +\pagebreak \ No newline at end of file diff --git a/Advanced/Generating Functions/parts/03 coins.tex b/Advanced/Generating Functions/parts/03 coins.tex new file mode 100755 index 0000000..6ab2e22 --- /dev/null +++ b/Advanced/Generating Functions/parts/03 coins.tex @@ -0,0 +1,102 @@ +\section{Coins} + + +Consider the following problem: + +\say{How many different ways can you make change for \$0.50 \par +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: +how can we change our answer if we want to make change for +\$0.51, or \$1.05, or some other quantity? + +\vspace{2mm} + +We can use generating functions to solve this problem in a general way. + +\definition{} +Let $p_0, p_1, p_2, ...$ be such that $p_k$ is the number +of ways to make change for $k$ cents with only pennies. + +Similarly, let... +\begin{itemize} + \item $n_k$ be the number of ways to make change for $k$ cents with only nickels; + \item $d_k$ be the number of ways using only dimes; + \item $q_k$ be the number of ways using only quarters; + \item and $h_k$ be the number of ways using only half-dollars. +\end{itemize} + +\problem{} +Let $p(x)$ be the generating function that corresponds to $p_n$. \par +Express $p(x)$ as a rational function. + +\vfill + +\problem{} +Modify \ref{pcoins} to find expressions for $n(x)$, $d(x)$, $q(x)$, and $h(x)$. + +\vfill +\pagebreak + +\definition{} +Now, let $N(x)$ be the generating function for the sequence +$n_0, n_1, ...$, where $n_k$ is the number of ways to make +change for $k$ cents using pennies and nickels. + +Similarly, let... +\begin{itemize} + \item let $D(x)$ be the generating function for the sequence using pennies, nickels, and dimes; + \item let $Q(x)$ use pennies, nickels, dimes, and quarters; + \item and let $H(x)$ use all coins. +\end{itemize} + +\problem{} +Express $N(x)$ as a rational function. + +\vfill + +\problem{} +Using the previous problem, write $D(x)$, then $Q(x)$, then $H(x)$ +as rational functions. + + +\vfill + +\problem{} +Using these generating functions, find recurrence relations for +the sequences $N_k$, $D_k$, $Q_k$, and $H_k$. + +\hint{ + Your recurrence relation for $N_k$ should refer to the + previous values of itself and some values of $p_k$. + Your recurrence for $D_k$ should refer to itself and $N_k$; + the one for $Q_k$ should refer to itself $D_k$; + and the one for $H_k$ should refer to itself and $Q_k$. +} + +\vfill + + +\problem{} +Using these recurrence relations, fill following table +and solve the original problem. + +\begin{center} + \begin{tabular}{ c|cccc|cccc|ccc } + $n$ & 0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45 & 50 \\ + \hline + $p_k$ &&&&&&&&&& \\ + $N_k$ &&&&&&&&&& \\ + \hline + $D_k$ &&&&&&&&&& \\ + \hline + $Q_k$ &&&&&&&&&& \\ + $H_k$ &&&&&&&&&& + \end{tabular} +\end{center} + +\vspace{1cm} +\pagebreak \ No newline at end of file