FIR draft

Add local typst packages
Added "Tropical Polynomials" handout
2025-01-23 10:34:06 -08:00 · 2025-01-23 10:29:13 -08:00 · 2025-01-23 10:29:07 -08:00
20 changed files with 26870 additions and 2 deletions
--- a/.vscode/settings.json
+++ b/.vscode/settings.json
@ -1,4 +1,5 @@
 {
 	"latex-workshop.latex.recipe.default": "latexmk (xelatex)",
-	"tinymist.formatterPrintWidth": 80
+	"tinymist.formatterPrintWidth": 80,
+	"tinymist.typstExtraArgs": ["--package-path=./lib/typst"]
 }
--- a/lib/typst/local/handout/0.1.0/handout.typ
+++ b/lib/typst/local/handout/0.1.0/handout.typ
@ -0,0 +1,282 @@
+/// Typst handout library, used for all documents in this repository.
+
+
+/// If false, hide instructor info.
+///
+/// Compile with the following command to hide solutions:
+/// `typst compile main.typ --input show_solutions=false`
+///
+/// Solutions are shown by default. This behavior
+/// is less surprising than hiding content by default.
+#let show_solutions = {
+  if "show_solutions" in sys.inputs {
+    // Show solutions unless they're explicitly disabled
+    not (
+      sys.inputs.show_solutions == "false" or sys.inputs.show_solutions == "no"
+    )
+  } else {
+    // Show solutions by default
+    true
+  }
+}
+
+// Colors
+#let ored = rgb("D62121")
+#let ogrape = rgb("9C36B5")
+#let ocyan = rgb("2288BF")
+#let oteal = rgb("12B886")
+#let ogreen = rgb("37B26D")
+#let oblue = rgb("1C7ED6")
+
+
+
+//
+// MARK: header
+//
+
+#let make_title(
+  group,
+  quarter,
+  title,
+  subtitle,
+) = {
+  align(
+    center,
+    block(
+      width: 60%,
+      height: auto,
+      breakable: false,
+      align(
+        center,
+        stack(
+          spacing: 7pt,
+          (
+            text(size: 10pt, group) + h(1fr) + text(size: 10pt, quarter)
+          ),
+          line(length: 100%, stroke: 0.2mm),
+          (
+            text(size: 20pt, title) + linebreak() + text(size: 10pt, subtitle)
+          ),
+          line(length: 100%, stroke: 0.2mm),
+        ),
+      ),
+    ),
+  )
+}
+
+#let warn = {
+  set text(ored)
+  align(
+    center,
+    block(
+      width: 60%,
+      height: auto,
+      breakable: false,
+      fill: rgb(255, 255, 255),
+      stroke: ored + 2pt,
+      inset: 3mm,
+      (
+        align(center, text(weight: "bold", size: 12pt, [Instructor's Handout]))
+          + parbreak()
+          + align(
+            left,
+            text(
+              size: 10pt,
+              [This handout contains solutions and notes.]
+                + linebreak()
+                + [Recompile without solutions before distributing.],
+            ),
+          )
+      ),
+    ),
+  )
+}
+
+#let preparedby(name) = (
+  text(
+    size: 10pt,
+    [Prepared by ]
+      + name
+      + [ on ]
+      + datetime
+        .today()
+        .display("[month repr:long] [day padding:none], [year]"),
+  )
+)
+
+//
+// MARK: Solutions
+//
+
+#let solution(content) = {
+  if show_solutions {
+    align(
+      center,
+      stack(
+        block(
+          width: 100%,
+          breakable: false,
+          fill: ored,
+          stroke: ored + 2pt,
+          inset: 1.5mm,
+          (
+            align(left, text(fill: white, weight: "bold", [Solution:]))
+          ),
+        ),
+        block(
+          width: 100%,
+          height: auto,
+          breakable: false,
+          fill: ored.lighten(80%).desaturate(10%),
+          stroke: ored + 2pt,
+          inset: 3mm,
+          align(left, content),
+        ),
+      ),
+    )
+  }
+}
+
+#let notsolution(content) = {
+  if not show_solutions { content }
+}
+
+
+//
+// MARK: Sections
+//
+
+#let generic(t) = block(
+  above: 8mm,
+  below: 2mm,
+  text(weight: "bold", t),
+)
+
+#let _generic_base(kind, ..args) = {
+  counter("obj").step()
+  if args.pos().len() == 0 {
+    generic([
+      #kind
+      #context counter("obj").display():
+    ])
+  } else {
+    generic(
+      [
+        #kind
+        #context counter("obj").display():
+      ]
+        + " "
+        + args.pos().at(0),
+    )
+  }
+}
+
+#let problem(..args) = _generic_base("Problem", ..args)
+#let definition(..args) = _generic_base("Definition", ..args)
+#let theorem(..args) = _generic_base("Theorem", ..args)
+
+
+//
+// MARK: Misc
+//
+
+
+#let hint(content) = {
+  text(fill: rgb(100, 100, 100), style: "oblique", "Hint: ")
+  text(fill: rgb(100, 100, 100), content)
+}
+
+#let note(content) = {
+  text(fill: rgb(100, 100, 100), content)
+}
+
+#let examplesolution(content) = {
+  let c = oblue
+
+  align(
+    center,
+    stack(
+      block(
+        width: 100%,
+        breakable: false,
+        fill: c,
+        stroke: c + 2pt,
+        inset: 1.5mm,
+        (
+          align(left, text(fill: white, weight: "bold", [Example solution:]))
+        ),
+      ),
+      block(
+        width: 100%,
+        height: auto,
+        breakable: false,
+        fill: c.lighten(80%).desaturate(10%),
+        stroke: c + 2pt,
+        inset: 3mm,
+        align(left, content),
+      ),
+    ),
+  )
+}
+
+
+//
+// MARK: wrapper
+//
+
+#let handout(
+  doc,
+  group: none,
+  quarter: none,
+  title: none,
+  by: none,
+  subtitle: none,
+) = {
+  set par(leading: 0.55em, first-line-indent: 0mm, justify: true)
+  set text(font: "New Computer Modern")
+  set par(spacing: 0.5em)
+  show list: set block(spacing: 0.5em, below: 1em)
+  set heading(numbering: (..nums) => nums.pos().at(0))
+
+  set page(
+    margin: 20mm,
+    width: 8in,
+    height: 11.5in,
+    footer: align(
+      center,
+      context counter(page).display(),
+    ),
+    footer-descent: 5mm,
+  )
+
+
+  set list(
+    tight: false,
+    indent: 5mm,
+    spacing: 3mm,
+  )
+
+  show heading.where(level: 1): it => {
+    set align(center)
+    set text(weight: "bold")
+    block[
+      Section #counter(heading).display(): #text(it.body)
+    ]
+  }
+
+  make_title(
+    group,
+    quarter,
+    title,
+    {
+      if by == none { none } else { [#preparedby(by)\ ] }
+      if subtitle == none { none } else { subtitle }
+    },
+  )
+
+  if show_solutions {
+    warn
+  }
+  doc
+}
+
--- a/lib/typst/local/handout/0.1.0/typst.toml
+++ b/lib/typst/local/handout/0.1.0/typst.toml
@ -0,0 +1,6 @@
+[package]
+name = "handout"
+version = "0.1.0"
+entrypoint = "handout.typ"
+authors = []
+license = "GPL"
--- a/src/Advanced/Fast
+++ b/src/Advanced/Fast
@ -0,0 +1,84 @@
+% use [nosolutions] flag to hide solutions.
+% use [solutions] flag to show solutions.
+\documentclass[
+	singlenumbering,
+	solutions
+]{../../resources/ormc_handout}
+\usepackage{../../resources/macros}
+
+
+\usepackage{pdfpages}
+\usepackage{sliderule}
+\usepackage{changepage}
+\usepackage{listings}
+
+% Args:
+% x, top scale y, label
+\newcommand{\slideruleind}[3]{
+	\draw[
+		line width=1mm,
+		draw=black,
+		opacity=0.3,
+		text opacity=1
+	]
+		({#1}, {#2 + 1})
+		--
+		({#1}, {#2 - 1.1})
+		node [below] {#3};
+}
+
+
+\uptitlel{Advanced 2}
+\uptitler{\smallurl{}}
+\title{Fast Inverse Square Root}
+\subtitle{Prepared by Mark on \today}
+
+\begin{document}
+
+	\maketitle
+
+
+	\begin{center}
+	\begin{minipage}{6cm}
+		Dad says that anyone who can't use
+		a slide rule is a cultural illiterate
+		and should not be allowed to vote.
+
+		\vspace{1ex}
+
+		\textit{Have Space Suit --- Will Travel, 1958}
+	\end{minipage}
+	\end{center}
+	\hfill
+
+	%\input{parts/0 logarithms.tex}
+	%\input{parts/1 intro.tex}
+	%\input{parts/2 multiplication.tex}
+	%\input{parts/3 division.tex}
+	\pagebreak
+	\input{parts/4 float.tex}
+	\input{parts/5 approximate.tex}
+	\input{parts/6 quake.tex}
+
+	% Make sure the slide rule is on an odd page,
+	% so that double-sided prints won't require
+	% students to tear off problems.
+	\checkoddpage
+	\ifoddpage\else
+		\vspace*{\fill}
+		\begin{center}
+		{
+			\Large
+			\textbf{This page unintentionally left blank.}
+		}
+		\end{center}
+		\vspace{\fill}
+		\pagebreak
+	\fi
+
+	%\includepdf[
+	%	pages=1,
+	%	fitpaper=true
+	%]{resources/rule.pdf}
+
+\end{document}
--- a/src/Advanced/Fast
+++ b/src/Advanced/Fast
@ -0,0 +1,72 @@
+\section{Logarithms}
+
+\definition{}<logdef>
+The \textit{logarithm} is the inverse of the exponent. That is, if $b^p = c$, then $\log_b{c} = p$. \par
+In other words, $\log_b{c}$ asks the question ``what power do I need to raise $b$ to to get $c$?''
+
+\vspace{2mm}
+
+In both $b^p$ and $\log_b{c}$, the number $b$ is called the \textit{base}.
+
+
+\problem{}
+Evaluate the following by hand:
+
+\begin{enumerate}
+	\item $\log_{10}{(1000)}$
+	\vfill
+	\item $\log_2{(64)}$
+	\vfill
+	\item $\log_2{(\frac{1}{4})}$
+	\vfill
+	\item $\log_x{(x)}$ for any $x$
+	\vfill
+	\item $log_x{(1)}$ for any $x$
+	\vfill
+\end{enumerate}
+
+\pagebreak
+
+
+
+\problem{}<logids>
+Prove the following identities:
+
+\begin{enumerate}[itemsep=2mm]
+	\item $\log_b{(b^x)} = x$
+	\item $b^{\log_b{x}} = x$
+	\item $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$
+	\item $\log_b{(\frac{x}{y})} = \log_b{(x)} - \log_b{(y)}$
+	\item $\log_b{(x^y)} = y \log_b{(x)}$
+\end{enumerate}
+
+\vfill
+
+\begin{instructornote}
+	A good intro to the following sections is the linear slide rule:
+
+	\begin{center}
+		\begin{tikzpicture}[scale=1]
+			\linearscale{2}{1}{}
+			\linearscale{0}{0}{}
+
+			\slideruleind
+				{5}
+				{1}
+				{2 + 3 = 5}
+	\end{tikzpicture}
+	\end{center}
+
+	Take two linear rulers, offset one, and you add. \par
+	If you do the same with a log scale, you multiply!
+
+	\vspace{2mm}
+
+	Note that the slide rules above start at 0.
+
+	\linehack{}
+
+	After assembling the paper slide rule, you can make a visor with some transparent tape. Wrap a strip around the slide rule, sticky side out, and stick it to itself to form a ring. Cover the sticky side with another layer of tape, and trim the edges to make them straight. Use the edge of the visor to read your slide rule!
+\end{instructornote}
+
+\pagebreak
--- a/src/Advanced/Fast
+++ b/src/Advanced/Fast
@ -0,0 +1,44 @@
+\section{Slide Rules}
+
+Mathematicians, physicists, and engineers needed to quickly solve complex equations even before computers were invented.
+
+\vspace{2mm}
+
+The \textit{slide rule} is an instrument that uses the logarithm to solve this problem. \par
+Before you continue, tear off the last page of this handout and assemble your slide rule.
+
+\vspace{2mm}
+
+There are four scales on your slide rule, each labeled with a letter on the left side:
+
+\def\sliderulewidth{13}
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\tscale{0}{9}{T}
+	\kscale{0}{8}{K}
+	\abscale{0}{7}{A}
+
+	\abscale{0}{5.5}{B}
+	\ciscale{0}{4.5}{CI}
+	\cdscale{0}{3.5}{C}
+
+	\cdscale{0}{2}{D}
+	\lscale{0}{1}{L}
+	\sscale{0}{0}{S}
+\end{tikzpicture}
+\end{center}
+
+Each scale's ``generating function'' is on the right:
+\begin{itemize}
+	\item T: $\tan$
+	\item K: $x^3$
+	\item A,B: $x^2$
+	\item CI: $\frac{1}{x}$
+	\item C, D: $x$
+	\item L: $\log_{10}(x)$
+	\item S: $\sin$
+\end{itemize}
+
+Once you understand the layout of your slide rule, move on to the next page.
+
+\pagebreak
--- a/multiplication.tex
+++ b/multiplication.tex
@ -0,0 +1,278 @@
+\section{Multiplication}
+
+We'll use the C and D scales of your slide rule to multiply. \par
+Say we want to multiply $2 \times 3$. First, move the \textit{left-hand index}
+of the C scale over the smaller number, $2$:
+
+\def\sliderulewidth{10}
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(2)}{1}{C}
+	\cdscale{0}{0}{D}
+\end{tikzpicture}
+\end{center}
+
+Then we'll find the second number, $3$ on the C scale, and read the D scale under it:
+
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(2)}{1}{C}
+	\cdscale{0}{0}{D}
+
+	\slideruleind
+		{\cdscalefn(6)}
+		{1}
+		{6}
+
+\end{tikzpicture}
+\end{center}
+
+Of course, our answer is 6.
+
+\problem{}
+What is $1.15 \times 2.1$? \par
+Use your slide rule.
+
+\begin{solution}
+	\begin{center}
+		\begin{tikzpicture}[scale=1]
+			\cdscale{\cdscalefn(1.15)}{1}{C}
+			\cdscale{0}{0}{D}
+
+			\slideruleind
+				{\cdscalefn(1.15)}
+				{1}
+				{1.15}
+
+			\slideruleind
+				{\cdscalefn(1.15) + \cdscalefn(2.1)}
+				{1}
+				{2.415}
+
+		\end{tikzpicture}
+		\end{center}
+\end{solution}
+
+\vfill
+
+Note that your answer isn't exact. $1.15 \times 2.1 = 2.415$, but an answer accurate within
+two decimal places is close enough for most practical applications.
+
+\pagebreak
+
+
+
+
+Look at your C and D scales again. They contain every number between 1 and 10, but no more than that.
+What should we do if we want to calculate $32 \times 210$? \par
+
+\problem{}
+Using your slide rule, calculate $32 \times 210$.
+
+
+\begin{solution}
+	\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\cdscale{\cdscalefn(2.1)}{1}{C}
+		\cdscale{0}{0}{D}
+
+		\slideruleind
+			{\cdscalefn(2.1)}
+			{1}
+			{2.1}
+
+		\slideruleind
+			{\cdscalefn(2.1) + \cdscalefn(3.2)}
+			{1}
+			{6.72}
+
+	\end{tikzpicture}
+	\end{center}
+
+		Placing the decimal point correctly is your job. \par
+		$10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$.
+\end{solution}
+
+\vfill
+
+\problem{}
+Compute the following:
+\begin{enumerate}
+	\item $1.44 \times 52$
+	\item $0.38 \times 1.24$
+	\item $\pi \times 2.35$
+\end{enumerate}
+
+\begin{solution}
+	\begin{enumerate}
+		\item $1.44 \times 52 = 74.88$
+		\item $0.38 \times 1.24 = 0.4712$
+		\item $\pi \times 2.35 = 7.382$
+	\end{enumerate}
+\end{solution}
+
+\vfill
+
+\problem{}<provemult>
+Note that the numbers on your C and D scales are logarithmically spaced.
+
+\def\sliderulewidth{13}
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{0}{1}{C}
+	\cdscale{0}{0}{D}
+\end{tikzpicture}
+\end{center}
+
+Why does our multiplication procedure work?
+
+\vfill
+\pagebreak
+
+Now we want to compute $7.2 \times 5.5$:
+
+\def\sliderulewidth{10}
+\begin{center}
+\begin{tikzpicture}[scale=0.8]
+	\cdscale{\cdscalefn(5.5)}{1}{C}
+	\cdscale{0}{0}{D}
+
+	\slideruleind
+		{\cdscalefn(5.5)}
+		{1}
+		{5.5}
+
+	\slideruleind
+		{\cdscalefn(5.5) + \cdscalefn(7.2)}
+		{1}
+		{???}
+
+\end{tikzpicture}
+\end{center}
+
+No matter what order we go in, the answer ends up off the scale. There must be another way.
+
+\vspace{2mm}
+
+Look at the far right of your C scale. There's an arrow pointing to the $10$ tick, labeled \textit{right-hand index}.
+Move it over the \textit{larger} number, $7.2$:
+
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
+	\cdscale{0}{0}{D}
+
+	\slideruleind
+		{\cdscalefn(7.2)}
+		{1}
+		{7.2}
+
+\end{tikzpicture}
+\end{center}
+
+Now find the smaller number, $5.5$, on the C scale, and read the D scale under it:
+
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
+	\cdscale{0}{0}{D}
+
+
+	\slideruleind
+		{\cdscalefn(7.2)}
+		{1}
+		{7.2}
+
+	\slideruleind
+		{\cdscalefn(3.96)}
+		{1}
+		{3.96}
+
+\end{tikzpicture}
+\end{center}
+
+Our answer should be about $7 \times 5 = 35$, so let's move the decimal point: $5.5 \times 7.2 = 39.6$.
+We can do this by hand to verify our answer.
+
+\vspace{2mm}
+
+\problem{}
+Why does this work?
+
+\begin{solution}
+	Consider the following picture, where we have two D scales next to each other:
+
+	\begin{center}
+	\begin{tikzpicture}[scale=0.7]
+		\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
+		\cdscale{0}{0}{}
+		\cdscale{-10}{0}{}
+
+		\draw[
+			draw=black,
+		]
+			(0, 0)
+			--
+			(0, -0.3)
+			node [below] {D};
+
+		\draw[
+			draw=black,
+		]
+			(-10, 0)
+			--
+			(-10, -0.3)
+			node [below] {D};
+
+		\slideruleind
+			{-10 + \cdscalefn(7.2)}
+			{1}
+			{7.2}
+
+		\slideruleind
+			{\cdscalefn(7.2)}
+			{1}
+			{7.2}
+
+		\slideruleind
+			{\cdscalefn(3.96)}
+			{1}
+			{3.96}
+
+	\end{tikzpicture}
+	\end{center}
+
+	\vspace{2mm}
+
+	The second D scale has been moved to the right by $(\log{10})$, so every value on it is $(\log{10})$
+	smaller than it should be.
+
+	\vspace{2mm}
+
+	\vspace{2mm}
+	In other words, the answer we get from reverse multiplication is $\log{a} + \log{b} - \log{10}$. \par
+	This reduces to $\log{(\frac{a \times b}{10})}$, and explains the misplaced decimal point in $7.2 \times 5.5$.
+\end{solution}
+
+
+\vfill
+\pagebreak
+
+\problem{}
+Compute the following using your slide rule:
+\begin{enumerate}
+	\item $9 \times 8$
+	\item $15 \times 35$
+	\item $42.1 \times 7.65$
+\end{enumerate}
+
+\begin{solution}
+	\begin{enumerate}
+		\item $9 \times 8 = 72$
+		\item $15 \times 35 = 525$
+		\item $42.1 \times 7.65 = 322.065$
+	\end{enumerate}
+\end{solution}
+
+\vfill
+\pagebreak
--- a/src/Advanced/Fast
+++ b/src/Advanced/Fast
@ -0,0 +1,91 @@
+\section{Division}
+
+Now that you can multiply, division should be easy. All you need to do is work backwards. \\
+Let's look at our first example again: $3 \times 2 = 6$.
+
+\medskip
+
+We can easily see that $6 \div 3 = 2$
+
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(2)}{1}{C}
+	\cdscale{0}{0}{D}
+
+	\slideruleind
+		{\cdscalefn(6)}
+		{1}
+		{Align here}
+
+	\slideruleind
+		{\cdscalefn(2)}
+		{1}
+		{2}
+\end{tikzpicture}
+\end{center}
+
+and that $6 \div 2 = 3$:
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(3)}{-3}{C}
+	\cdscale{0}{-4}{D}
+
+
+	\slideruleind
+		{\cdscalefn(6)}
+		{-3}
+		{Align here}
+
+	\slideruleind
+		{\cdscalefn(3)}
+		{-3}
+		{3}
+
+\end{tikzpicture}
+\end{center}
+
+If your left-hand index is off the scale, read the right-hand one. \\
+Consider $42.25 \div 6.5 = 6.5$:
+
+\begin{center}
+\begin{tikzpicture}[scale=1]
+	\cdscale{\cdscalefn(6.5) - \cdscalefn(10)}{1}{C}
+	\cdscale{0}{0}{D}
+
+
+	\slideruleind
+		{\cdscalefn(4.225)}
+		{1}
+		{Align here}
+
+	\slideruleind
+		{\cdscalefn(6.5)}
+		{1}
+		{6.5}
+
+\end{tikzpicture}
+\end{center}
+
+Place your decimal points carefully.
+
+\vfill
+
+\problem{}
+Compute the following using your slide rule. \par
+
+\begin{enumerate}
+	\item $135 \div 15$
+	\item $68.2 \div 0.575$
+	\item $(118 \times 0.51) \div 6.6$
+\end{enumerate}
+
+\begin{solution}
+	\begin{enumerate}
+		\item $135 \div 15 = 9$
+		\item $68.2 \div 0.575 = 118.609$
+		\item $(118 \times 0.51) \div 6.6 = 9.118$
+	\end{enumerate}
+\end{solution}
+
+\vfill
+\pagebreak
--- a/src/Advanced/Fast
+++ b/src/Advanced/Fast
@ -0,0 +1,219 @@
+
+\section{\texttt{int}s and \texttt{float}s}
+
+\definition{}
+A \textit{signed 32-bit integer} (equivalently, a \texttt{long int}) consists of thirty-two binary digits, \par
+and is used represent a subset of the integers.
+
+\vspace{2mm}
+
+The first bit of a \texttt{long} tells us its sign:
+\begin{itemize}
+	\item if the first bit of a \texttt{long} is \texttt{1}, it represents a negative number;
+	\item if the first bit is \texttt{0}, it represents a positive number.
+\end{itemize}
+We do not need negative numbers today, so we will assume that the first bit is always zero. \par
+\note{If you'd like to know how negative integers are written, look up \say{two's complement} after class.}
+
+\vspace{2mm}
+
+We'll denote binary strings with the prefix \texttt{0b}. \par
+Underscores are added between every eight digits for readability, and have no meaning.
+
+\vspace{2mm}
+
+
+The value of a positive signed \texttt{long} is simply the value of its binary digits. \par
+For example:
+\begin{itemize}
+	\item $\texttt{0b00000000\_00000000\_00000000\_00000000} = 0$
+	\item $\texttt{0b00000000\_00000000\_00000000\_00000011} = 3$
+	\item $\texttt{0b00000000\_00000000\_00000000\_00100000} = 32$
+	\item $\texttt{0b00000000\_00000000\_00000000\_10000010} = 130$
+\end{itemize}
+
+Remember---we only need positive integers today. Assume the \say{sign} bit is always \texttt{0}.
+
+\problem{}
+What is the largest number that can be represented with a \texttt{long}?
+
+\begin{solution}
+	$\texttt{0b01111111\_11111111\_11111111\_11111111} = 2^{31}$
+\end{solution}
+
+\vfill
+
+
+
+\problem{}
+What is the smallest possible number that can be represented with a \texttt{long}? \par
+\hint{
+	You do not need to know \textit{how} negative numbers are represented. \par
+	Assume that we do not skip any integers, and don't forget about zero.
+}
+
+\begin{solution}
+	There are $2^{64}$ possible 32-bit patterns,
+	of which 1 represents zero and $2^{31}$ represent positive numbers.
+
+	\vspace{2mm}
+
+	We therefore have access to $2^{64} - 1 - 2^{31}$ negative numbers,
+	giving us a minimum representable value of $-2^{31} + 1$.
+\end{solution}
+
+\problem{}
+What is the value of the following longs?
+\begin{itemize}
+	\item \texttt{0b00000000\_00000000\_00000101\_00111001}
+	\item \texttt{0b00000000\_00000000\_00000001\_00101100}
+	\item \texttt{0b00000000\_00000000\_00000100\_10110000}
+\end{itemize}
+\hint{The third conversion is easy---look carefully at the second.}
+
+\begin{solution}
+	\begin{itemize}
+		\item $\texttt{0b00000000\_00000000\_00000101\_00111001} = 1337$
+		\item $\texttt{0b00000000\_00000000\_00000001\_00101100} = 300$
+		\item $\texttt{0b00000000\_00000000\_00000010\_01011000} = 1200$
+	\end{itemize}
+
+	Notice that the third long is the second shifted left twice (i.e, multiplied by 4)
+\end{solution}
+
+\vfill
+\pagebreak
+
+
+\definition{}
+A \textit{signed 32-bit floating-point decimal} (equivalently, a \textit{float})
+consists of 32 binary digits, and is used to represent a subset of the real numbers.
+These 32 bits are interpreted as follows:
+
+\begin{center}
+	\begin{tikzpicture}
+
+
+		\node[anchor=south west] at (0, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (0.25, 0) {\texttt{\texttt{b}}};
+		\node[anchor=south west] at (0.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (0.75, 0) {\texttt{\texttt{\_}}};
+
+		\node[anchor=south west] at (1.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (1.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (1.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (1.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (2.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (2.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (2.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (2.75, 0) {\texttt{\texttt{0}}};
+
+		\node[anchor=south west] at (3.00, 0) {\texttt{\texttt{\_}}};
+		\node[anchor=south west] at (3.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (3.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (3.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (4.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (4.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (4.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (4.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (5.00, 0) {\texttt{\texttt{\_}}};
+
+		\node[anchor=south west] at (5.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (5.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (5.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (6.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (6.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (6.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (6.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (7.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (7.25, 0) {\texttt{\texttt{\_}}};
+
+		\node[anchor=south west] at (7.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (7.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (8.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (8.25, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (8.50, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (8.75, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (9.00, 0) {\texttt{\texttt{0}}};
+		\node[anchor=south west] at (9.25, 0) {\texttt{\texttt{0}}};
+
+
+		\draw (0.50, 0) -- (0.95, 0) node [midway, below=1mm] {sign};
+		\draw (1.05, 0) -- (3.15, 0) node [midway, below=1mm] {exponent};
+		\draw (3.30, 0) -- (9.70, 0) node [midway, below=1mm] {fraction};
+	\end{tikzpicture}
+\end{center}
+
+In other words:
+\begin{itemize}[itemsep = 1mm]
+	\item The first bit denotes the sign of the float's value. We'll label it $s$. \par
+	If $s = 1$, this float is negative; if $s = 0$, it is positive.
+
+	\item The next 8 bits represent the \textit{exponent} of this float. \par
+	We'll call the value of these eight bits $E$. \par
+	Naturally, $0 \leq E \leq 255$
+
+	\item The remaining 23 bits represent the \textit{fraction} of this float. \par
+	These 23 bits are interpreted as the fractional part of a binary decimal. \par
+	For example, the bits \texttt{0b1010000\_00000000\_00000000} represents $0.5 + 0.125 = 0.625$.
+\end{itemize}
+
+
+\vspace{2mm}
+
+The final value of a float with sign $s$, exponent $E$, and fraction $F$ is
+\begin{equation*}
+	(-1)^s ~\times~ 2^{E - 127} ~\times~ \left(1 + \frac{F}{2^{23}}\right)
+\end{equation*}
+
+
+Notice that this is very similar to decimal scientific notation, which is written as
+\begin{equation*}
+	(\pm 1) ~\times~ 10^{e} ~\times~ (f)
+\end{equation*}
+
+\vfill
+\pagebreak
+
+\problem{}
+What is the value of \texttt{0b01000001\_10101000\_00000000\_00000000} if it is interpreted as a float? \par
+\hint{$21 \div 16 = 1.3125$}
+
+\begin{solution}
+	This is 21:
+	\begin{align*}
+		&=~ 2^{131} \times \biggl(1 + \frac{2^{21} + 2^{19}}{2^{23}}\biggr) \\
+		&=~ 2^{4} \times (1 + 0.25 + 0.0625) \\
+		&=~ 16 \times (1.3125) \\
+		&=~ 21
+	\end{align*}
+\end{solution}
+
+\vfill
+
+\problem{}
+Encode $12.5$ as a float. \par
+\hint{$12.5 \div 8 = 1.5625$}
+
+\vspace{2mm}
+
+What is the value of the resulting 32 bits if they are interpreted as a long? \par
+\hint{A sum of powers of two is fine.}
+
+\begin{solution}
+	\begin{align*}
+		12.5
+		&=~ 8 \times 1.5625 \\
+		&=~ 2^{3} \times \biggl(1 + (0.5 + 0.0625)\biggr) \\
+		&=~ 2^{130} \times \biggl(1 + \frac{2^{22} + 2^{19}}{2^{23}}\biggr)
+	\end{align*}
+
+	\linehack{}
+
+	This is \texttt{0b01000001\_01001000\_00000000\_00000000}, \par
+	which is $2^{30} + 2^{24} + 2^{22} + 2^{19} = 11,095,237,632$
+\end{solution}
+
+
+\vfill
+\pagebreak
--- a/src/Advanced/Fast
+++ b/src/Advanced/Fast
@ -0,0 +1,47 @@
+\section{Approximation}
+
+\definition{}
+Say we have a bit string $x$. \par
+Let $x_f$ denote the value of $x$ interpreted as a float, \par
+and let $x_i$ denote the value of $x$ interpreted as an integer.
+
+\problem{}
+Let $x = \texttt{0b01000001\_01001000\_00000000\_00000000}$. \
+What are $x_f$ and $x_i$?
+
+\begin{solution}
+	$x_f = 12.5$ \par
+	\vspace{2mm}
+	$x_i = 2^{30} + 2^{24} + 2^{22} + 2^{19} = 11,095,237,632$
+\end{solution}
+
+
+\generic{Observation:}
+For small values of $a$, $\log_2(1 + a)$ is approximately equal to $a$. \par
+Note that this equality is exact for $a = 0$ and $a = 1$, since $\log_2(1) = 0$ and $\log_2(2) = 1$.
+
+\vspace{2mm}
+
+We'll add a \say{correction term} $\varepsilon$ to this approximation, so that $\log_2(1 + a) \approx a + \varepsilon$.
+
+TODO: why? Graphs.
+
+\problem{}
+Use the fact that $\log_2(1 + a) \approx a + \varepsilon$ to approximate $\log_2(x_f)$ in terms of $x_i$, \par
+
+
+\begin{solution}
+	Let $E$ and $F$ be the exponent and float bits of $x_f$. \par
+	We then have:
+	\begin{align*}
+		\log_2(x_f)
+		&=~ \log_2 \left( 2^{E-127} \times \left(1 + \frac{F}{2^{23}}\right) \right) \\
+		&=~ E-127 + \log_2\left(1 + \frac{F}{2^{23}}\right) \\
+		&\approx~ E-127 + \frac{F}{2^{23}} + \varepsilon \\
+		&=~ \frac{1}{2^{23}}(2^{23}E + F) - 127 + \varepsilon \\
+		&=~ \frac{1}{2^{23}}(x_i) - 127 + \varepsilon
+	\end{align*}
+\end{solution}
+
+\vfill
+\pagebreak
--- a/src/Advanced/Fast
+++ b/src/Advanced/Fast
@ -0,0 +1,71 @@
+\section{The Fast Inverse Square Root}
+
+The following code is present in \textit{Quake III Arena} (1999):
+
+\lstset{
+	breaklines=false,
+	numbersep=5pt,
+	xrightmargin=0in
+}
+
+\begin{lstlisting}[language=C]
+float Q_rsqrt( float number ) {
+    long i = * ( long * ) &number;
+    i = 0x5f3759df - ( i >> 1 );
+    return * ( float * ) &i;
+}
+\end{lstlisting}
+
+It defines a method \texttt{Q\_rsqrt} that consumes a float named \texttt{number} and quickly approximates its inverse
+square root (in other words, \texttt{Q\_rsqrt} computes $1/\sqrt{\texttt{number}}$).
+
+\vspace{8mm}
+
+If we rewrite this using the notation we're familiar with, we get the following:
+
+\begin{equation*}
+	\frac{1}{\sqrt{n_f}} \approx \kappa - (n_i \div 2)
+\end{equation*}
+Where $\kappa$ is the magic constant $6240089$ (which is \texttt{0x5f3759df} in hexadecimal)
+
+\problem{}
+Find the exact value of $\kappa$ in terms of $\varepsilon$. \par
+\hint{Remember that $\varepsilon$ is the correction term in the approximation $\log_2(1 + a) = a + \varepsilon$.}
+
+\problem{}
+Rewrite $\log_2(1 / \sqrt{x})$ in terms of $\log_2{x}$.
+
+\begin{solution}
+	Say $g_f = \frac{1}{\sqrt{n_f}}$---that is, $g_f$ is the value we want to compute. \par
+	Then:
+
+	\begin{align*}
+		\log_2(g_f)
+		&=\log_2(\frac{1}{\sqrt{n_f}}) \\
+		&=\frac{-1}{2}\log_2(n_f) \\
+		&=\frac{-1}{2}\left( \frac{n_i}{2^{23}} + \varepsilon - 127 \right)
+	\end{align*}
+
+	But we also know that
+
+	\begin{align*}
+		\log_2(g_f)
+		&=\frac{g_i}{2^{23}} + \varepsilon - 127
+	\end{align*}
+
+	So,
+
+	\begin{align*}
+		\frac{g_i}{2^{23}} + \varepsilon - 127
+		&=\frac{-1}{2}\left( \frac{n_i}{2^{23}} + \varepsilon - 127 \right) \\
+		\frac{g_i}{2^{23}}
+		&=\frac{-1}{2}\left( \frac{n_i}{2^{23}} \right) + \frac{3}{2}(\varepsilon - 127) \\
+		g_i
+		&=\frac{-1}{2}\left(n_i \right) + 2^{23}\frac{3}{2}(\varepsilon - 127) \\
+		&=2^{23}\frac{3}{2}(\varepsilon - 127) - \frac{n_i}{2}
+	\end{align*}
+
+	thus, $\kappa = 2^{23}\frac{3}{2}(\varepsilon - 127)$.
+\end{solution}
+
+\pagebreak
--- a/Root/resources/rule.svg
+++ b/Root/resources/rule.svg
--- a/Root/sliderule.sty
+++ b/Root/sliderule.sty
@ -0,0 +1,534 @@
+\NeedsTeXFormat{LaTeX2e}
+\ProvidesPackage{sliderule}[2022/08/22 Slide rule tools]
+
+\RequirePackage{tikz}
+\RequirePackage{ifthen}
+
+
+% Scale functions:
+% See https://sliderulemuseum.com/SR_Scales.htm
+%
+% l: length of the rule
+% n: the number on the rule
+%
+% A/B: (l/2) * log(n)
+% C/D: l / log(n)
+% CI: abs(l * log(10 / n) - l)
+% K: (l/3) * log(n)
+%
+% L: n * l
+% T: l * log(10 * tan(n))
+% S: l * log(10 * sin(n))
+
+\def\sliderulewidth{10}
+
+\def\abscalefn(#1){(\sliderulewidth/2) * log10(#1)}
+\def\cdscalefn(#1){(\sliderulewidth * log10(#1))}
+\def\ciscalefn(#1){(\sliderulewidth - \cdscalefn(#1))}
+\def\kscalefn(#1){(\sliderulewidth/3) * log10(#1)}
+\def\lscalefn(#1){(\sliderulewidth * #1)}
+\def\tscalefn(#1){(\sliderulewidth * log10(10 * tan(#1)))}
+\def\sscalefn(#1){(\sliderulewidth * log10(10 * sin(#1)))}
+
+
+% Arguments:
+% Label
+% x of start
+% y of start
+\newcommand{\linearscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks
+	\foreach \i in {0,..., 10}{
+		\draw[black]
+			({#1 + (\sliderulewidth / 10) * \i}, #2) --
+			({#1 + (\sliderulewidth / 10) * \i}, #2 + 0.3)
+			node[above] {\i};
+	}
+
+	% Submarks
+	\foreach \n in {0, ..., 9} {
+		\foreach \i in {1,..., 9} {
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2) --
+					({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2) --
+					({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2 + 0.1);
+			}
+		}
+	}
+}
+
+
+% Arguments:
+% Label
+% x of start
+% y of start
+\newcommand{\abscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks 1 - 9
+	\foreach \i in {1,..., 9}{
+		\draw[black]
+			({#1 + \abscalefn(\i)}, #2) --
+			({#1 + \abscalefn(\i)}, #2 + 0.3)
+			node[above] {\i};
+	}
+	% Numbers and marks 10 - 100
+	\foreach \i in {1,..., 10}{
+		\draw[black]
+			({#1 + \abscalefn(10 * \i)}, #2) --
+			({#1 + \abscalefn(10 * \i)}, #2 + 0.3)
+			node[above] {\ifthenelse{\i=10}{1}{\i}};
+	}
+
+	% Submarks 1 - 9
+	\foreach \n in {1, ..., 9} {
+		\ifthenelse{\n<5}{
+			\foreach \i in {1,..., 9}
+		} {
+			\foreach \i in {2,4,6,8}
+		}
+		{
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + \abscalefn(\n + \i / 10)}, #2) --
+					({#1 + \abscalefn(\n + \i / 10)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \abscalefn(\n + \i / 10)}, #2) --
+					({#1 + \abscalefn(\n + \i / 10)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 10 - 100
+	\foreach \n in {10,20,...,90} {
+		\ifthenelse{\n<50}{
+			\foreach \i in {1,..., 9}
+		} {
+			\foreach \i in {2,4,6,8}
+		}
+		{
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + \abscalefn(\n + \i)}, #2) --
+					({#1 + \abscalefn(\n + \i)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \abscalefn(\n + \i)}, #2) --
+					({#1 + \abscalefn(\n + \i)}, #2 + 0.1);
+			}
+		}
+	}
+}
+
+\newcommand{\cdscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks 1 - 10
+	\foreach \i in {1,..., 10}{
+		\draw[black]
+			({#1 + \cdscalefn(\i)}, #2) --
+			({#1 + \cdscalefn(\i)}, #2 + 0.3)
+			node[above] {\ifthenelse{\i=10}{1}{\i}};
+	}
+
+	% Submarks 1 - 9
+	\foreach \n in {1, ..., 9} {
+		\ifthenelse{\n<3}{
+			\foreach \i in {5,10,...,95}
+		} {
+			\foreach \i in {10,20,...,90}
+		}
+		{
+			\ifthenelse{\i=50}{
+				\draw[black]
+					({#1 + \cdscalefn(\n + \i / 100)}, #2) --
+					({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.2);
+				\ifthenelse{\n=1}{
+					\draw
+						({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.2)
+						node [above] {1.5};
+				}{}
+			} {
+				\ifthenelse{
+					\i=10 \OR \i=20 \OR \i=30 \OR \i=40 \OR
+					\i=60 \OR \i=70 \OR \i=80 \OR \i=90
+				}{
+					\draw[black]
+						({#1 + \cdscalefn(\n + \i / 100)}, #2) --
+						({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.15);
+				} {
+					\draw[black]
+						({#1 + \cdscalefn(\n + \i / 100)}, #2) --
+						({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.1);
+				}
+			}
+		}
+	}
+}
+
+\newcommand{\ciscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks
+	\foreach \i in {1,...,10}{
+		\draw[black]
+			({#1 + \ciscalefn(\i)}, #2) --
+			({#1 + \ciscalefn(\i)}, #2 + 0.3)
+			node[above] {\ifthenelse{\i=10}{1}{\ifthenelse{\i=0}{0}{.\i}}};
+	}
+
+	% Submarks 1 - 9
+	\foreach \n in {1, ..., 9} {
+		\ifthenelse{\n<3}{
+			\foreach \i in {5,10,...,95}
+		} {
+			\foreach \i in {10,20,...,90}
+		}
+		{
+			\ifthenelse{\i=50}{
+				\draw[black]
+					({#1 + \ciscalefn(\n + \i / 100)}, #2) --
+					({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.2);
+				\ifthenelse{\n=1}{
+					\draw
+						({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.2)
+						node [above] {1.5};
+				}{}
+			} {
+				\ifthenelse{
+					\i=10 \OR \i=20 \OR \i=30 \OR \i=40 \OR
+					\i=60 \OR \i=70 \OR \i=80 \OR \i=90
+				}{
+					\draw[black]
+						({#1 + \ciscalefn(\n + \i / 100)}, #2) --
+						({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.15);
+				} {
+					\draw[black]
+						({#1 + \ciscalefn(\n + \i / 100)}, #2) --
+						({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.1);
+				}
+			}
+		}
+	}
+}
+
+\newcommand{\kscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks 1 - 9
+	\foreach \i in {1,...,9}{
+		\draw[black]
+			({#1 + \kscalefn(\i)}, #2) --
+			({#1 + \kscalefn(\i)}, #2 + 0.3)
+			node[above] {\i};
+	}
+	% Numbers and marks 10 - 90
+	\foreach \i in {1,..., 9}{
+		\draw[black]
+			({#1 + \kscalefn(10 * \i)}, #2) --
+			({#1 + \kscalefn(10 * \i)}, #2 + 0.3)
+			node[above] {\ifthenelse{\i=10}{1}{\i}};
+	}
+	% Numbers and marks 100 - 1000
+	\foreach \i in {1,..., 10}{
+		\draw[black]
+			({#1 + \kscalefn(100 * \i)}, #2) --
+			({#1 + \kscalefn(100 * \i)}, #2 + 0.3)
+			node[above] {\ifthenelse{\i=10}{1}{\i}};
+	}
+
+	% Submarks 1 - 9
+	\foreach \n in {1, ..., 9} {
+		\ifthenelse{\n<4}{
+			\foreach \i in {1,..., 9}
+		} {
+			\foreach \i in {2,4,6,8}
+		}
+		{
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + \kscalefn(\n + \i / 10)}, #2) --
+					({#1 + \kscalefn(\n + \i / 10)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \kscalefn(\n + \i / 10)}, #2) --
+					({#1 + \kscalefn(\n + \i / 10)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 10 - 90
+	\foreach \n in {10,20,...,90} {
+		\ifthenelse{\n<40}{
+			\foreach \i in {1,..., 9}
+		} {
+			\foreach \i in {2,4,6,8}
+		}
+		{
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + \kscalefn(\n + \i)}, #2) --
+					({#1 + \kscalefn(\n + \i)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \kscalefn(\n + \i)}, #2) --
+					({#1 + \kscalefn(\n + \i)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 100 - 1000
+	\foreach \n in {100,200,...,900} {
+		\ifthenelse{\n<400}{
+			\foreach \i in {10,20,...,90}
+		} {
+			\foreach \i in {20,40,60,80}
+		}
+		{
+			\ifthenelse{\i=50}{
+				\draw[black]
+					({#1 + \kscalefn(\n + \i)}, #2) --
+					({#1 + \kscalefn(\n + \i)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \kscalefn(\n + \i)}, #2) --
+					({#1 + \kscalefn(\n + \i)}, #2 + 0.1);
+			}
+		}
+	}
+}
+
+\newcommand{\lscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks
+	\foreach \i in {0,..., 10}{
+		\draw[black]
+			({#1 + \lscalefn(\i / 10)}, #2) --
+			({#1 + \lscalefn(\i / 10)}, #2 + 0.3)
+			node[above] {\ifthenelse{\i=10}{1}{\ifthenelse{\i=0}{0}{.\i}}};
+	}
+
+	% Submarks
+	\foreach \n in {0, ..., 9} {
+		\foreach \i in {1,...,19} {
+			\ifthenelse{\i=10}{
+				\draw[black]
+					({#1 + \lscalefn((\n + (\i / 20))/10)}, #2) --
+					({#1 + \lscalefn((\n + (\i / 20))/10)}, #2 + 0.2);
+			} {
+				\ifthenelse{
+					\i=1 \OR \i=3 \OR \i=5 \OR \i=7 \OR
+					\i=9 \OR \i=11 \OR \i=13 \OR \i=15 \OR
+					\i=17 \OR \i=19
+				}{
+					\draw[black]
+						({#1 + \lscalefn((\n + (\i / 20))/10)}, #2) --
+						({#1 + \lscalefn((\n + (\i / 20))/10)}, #2 + 0.1);
+				} {
+					\draw[black]
+						({#1 + \lscalefn((\n + (\i / 20))/10)}, #2) --
+						({#1 + \lscalefn((\n + (\i / 20))/10)}, #2 + 0.15);
+				}
+			}
+		}
+	}
+}
+
+\newcommand{\tscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+	% First line
+	\draw[black] ({#1}, #2) -- ({#1}, #2 + 0.2);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks 6 - 10
+	\foreach \i in {6,...,9,10,15,...,45}{
+		\draw[black]
+			({#1 + \tscalefn(\i)}, #2) --
+			({#1 + \tscalefn(\i)}, #2 + 0.3)
+			node[above] {\i};
+	}
+
+	% Submarks 6 - 10
+	\foreach \n in {6, ..., 9} {
+		\foreach \i in {1,...,9}{
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + \tscalefn(\n + \i / 10)}, #2) --
+					({#1 + \tscalefn(\n + \i / 10)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \tscalefn(\n + \i / 10)}, #2) --
+					({#1 + \tscalefn(\n + \i / 10)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 15 - 45
+	\foreach \n in {10, 15, ..., 40} {
+		\foreach \i in {1,...,24}{
+			\ifthenelse{
+				\i=5 \OR \i=10 \OR \i=15 \OR \i=20
+			} {
+				\draw[black]
+					({#1 + \tscalefn(\n + \i / 5)}, #2) --
+					({#1 + \tscalefn(\n + \i / 5)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \tscalefn(\n + \i / 5)}, #2) --
+					({#1 + \tscalefn(\n + \i / 5)}, #2 + 0.1);
+			}
+		}
+	}
+}
+
+\newcommand{\sscale}[3]{
+	\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
+	\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
+	\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
+
+	% First line
+	\draw[black] ({#1}, #2) -- ({#1}, #2 + 0.2);
+
+
+	\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
+
+	% Numbers and marks
+	\foreach \i in {6,...,9,10,15,...,30,40,50,...,60,90}{
+		\draw[black]
+			({#1 + \sscalefn(\i)}, #2) --
+			({#1 + \sscalefn(\i)}, #2 + 0.3)
+			node[above] {\i};
+	}
+
+	% Submarks 6 - 10
+	\foreach \n in {6, ..., 9} {
+		\foreach \i in {1,...,9}{
+			\ifthenelse{\i=5}{
+				\draw[black]
+					({#1 + \sscalefn(\n + \i / 10)}, #2) --
+					({#1 + \sscalefn(\n + \i / 10)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i / 10)}, #2) --
+					({#1 + \sscalefn(\n + \i / 10)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 15 - 30
+	\foreach \n in {10, 15, ..., 25} {
+		\foreach \i in {1,...,24}{
+			\ifthenelse{
+				\i=5 \OR \i=10 \OR \i=15 \OR \i=20
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i / 5)}, #2) --
+					({#1 + \sscalefn(\n + \i / 5)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i / 5)}, #2) --
+					({#1 + \sscalefn(\n + \i / 5)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 30
+	\foreach \n in {30} {
+		\foreach \i in {1,...,19}{
+			\ifthenelse{
+				\i=2 \OR \i=4 \OR \i=6 \OR \i=8 \OR
+				\i=10 \OR \i=12 \OR \i=14 \OR \i=16 \OR
+				\i=18
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i / 2)}, #2) --
+					({#1 + \sscalefn(\n + \i / 2)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i / 2)}, #2) --
+					({#1 + \sscalefn(\n + \i / 2)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 40 - 50
+	\foreach \n in {40, 50} {
+		\foreach \i in {1,...,9}{
+			\ifthenelse{
+				\i=5 \OR \i=10 \OR \i=15 \OR \i=20
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i)}, #2) --
+					({#1 + \sscalefn(\n + \i)}, #2 + 0.2);
+			} {
+				\draw[black]
+					({#1 + \sscalefn(\n + \i)}, #2) --
+					({#1 + \sscalefn(\n + \i)}, #2 + 0.1);
+			}
+		}
+	}
+
+	% Submarks 60
+	\foreach \i in {1,...,10}{
+		\ifthenelse{
+			\i=5 \OR \i=10
+		} {
+			\draw[black]
+				({#1 + \sscalefn(60 + \i * 2)}, #2) --
+				({#1 + \sscalefn(60 + \i * 2)}, #2 + 0.2);
+		} {
+			\draw[black]
+				({#1 + \sscalefn(60 + \i * 2)}, #2) --
+				({#1 + \sscalefn(60 + \i * 2)}, #2 + 0.1);
+		}
+	}
+}
--- a/Polynomials/macros.typ
+++ b/Polynomials/macros.typ
@ -0,0 +1,70 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.3.1"
+
+
+// Shorthand, we'll be using these a lot.
+#let tp = sym.plus.circle
+#let tm = sym.times.circle
+
+#let graphgrid(inner_content) = {
+  align(
+    center,
+    box(
+      inset: 3mm,
+      cetz.canvas({
+        import cetz.draw: *
+        let x = 5.25
+
+        grid(
+          (0, 0), (x, x), step: 0.75,
+          stroke: luma(100) + 0.3mm
+        )
+
+        if (inner_content != none) {
+          inner_content
+        }
+
+        mark((0, x + 0.5), (0, x + 1), symbol: ">", fill: black, scale: 1)
+        mark((x + 0.5, 0), (x + 1, 0), symbol: ">", fill: black, scale: 1)
+
+        line(
+          (0, x + 0.25),
+          (0, 0),
+          (x + 0.25, 0),
+          stroke: 0.75mm + black,
+        )
+      }),
+    ),
+  )
+}
+
+/// Adds extra padding to an equation.
+/// Used as follows:
+///
+/// #eqmbox($
+///   f(x) = -2(x #tp 2)(x #tp 8)
+/// $)
+///
+/// Note that there are newlines between the $ and content,
+/// this gives us display math (which is what we want when using this macro)
+#let eqnbox(eqn) = {
+  align(
+    center,
+    box(
+      inset: 3mm,
+      eqn,
+    ),
+  )
+}
+
+#let dotline(a, b) = {
+  cetz.draw.line(
+    a,
+    b,
+    stroke: (
+      dash: "dashed",
+      thickness: 0.5mm,
+      paint: ored,
+    ),
+  )
+}
--- a/src/Advanced/Tropical
+++ b/src/Advanced/Tropical
@ -0,0 +1,22 @@
+#import "@local/handout:0.1.0": *
+
+#show: doc => handout(
+  doc,
+  group: "Advanced 2",
+  quarter: link(
+    "https://betalupi.com/handouts",
+    "betalupi.com/handouts",
+  ),
+
+  title: [Tropical Polynomials],
+  by: "Mark",
+  subtitle: "Based on a handout by Bryant Mathews",
+)
+
+#include "parts/00 arithmetic.typ"
+#pagebreak()
+
+#include "parts/01 polynomials.typ"
+#pagebreak()
+
+#include "parts/02 cubic.typ"
--- a/Polynomials/meta.toml
+++ b/Polynomials/meta.toml
@ -0,0 +1,7 @@
+[metadata]
+title = "Tropical Polynomials"
+
+
+[publish]
+handout = true
+solutions = true
--- a/src/Advanced/Tropical
+++ b/src/Advanced/Tropical
@ -0,0 +1,293 @@
+#import "@local/handout:0.1.0": *
+#import "../macros.typ": *
+
+= Tropical Arithmetic
+
+#definition()
+The _tropical sum_ of two numbers is their minimum:
+$
+  x #tp y = min(x, y)
+$
+
+#definition()
+The _tropical product_ of two numbers is their sum:
+$
+  x #tm y = x + y
+$
+
+
+#problem()
+- Is tropical addition commutative? \
+  #note([i.e, does $x #tp y = y #tp x$?])
+- Is tropical addition associative? \
+  #note([i.e, does $(x #tp y) #tp z = x #tp (y #tp z)$?])
+- Is there a tropical additive identity? \
+  #note([i.e, is there an $i$ so that $x #tp i = x$ for all real $x$?])
+
+#solution([
+  - Is tropical addition commutative?\
+    Yes, $min(min(x,y),z) = min(x,y,z) = min(x,min(y,z))$
+  - Is tropical addition associative? \
+    Yes, $min(x,y) = min(y,x)$
+  - Is there a tropical additive identity? \
+    No. There is no $n$ where $x <= n$ for all real $x$
+])
+
+#v(1fr)
+
+#problem()
+Let's expand $#sym.RR$ to include a tropical additive identity.
+- What would be an appropriate name for this new number?
+- Give a reasonable definition for...
+  - the tropical sum of this number and a real number $x$
+  - the tropical sum of this number and itself
+  - the tropical product of this number and a real number $x$
+  - the tropical product of this number and itself
+
+#solution([
+  #sym.infinity makes sense, with
+  $#sym.infinity #tp x = x$; #h(1em)
+  $#sym.infinity #tp #sym.infinity = #sym.infinity$; #h(1em)
+  $#sym.infinity #tm x = #sym.infinity$; #h(1em) and
+  $#sym.infinity #tm #sym.infinity = #sym.infinity$
+])
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+
+
+
+
+#problem()
+Do tropical additive inverses exist? \
+#note([
+  Is there an inverse $y$ for every $x$ so that $x #tp y = #sym.infinity$?
+])
+
+#solution([
+  No. Unless $x = #sym.infinity$, there is no x where $min(x, y) = #sym.infinity$
+])
+
+#v(1fr)
+
+
+
+#problem()
+Is tropical multiplication associative? \
+#note([Does $(x #tm y) #tm z = x #tm (y #tm z)$ for all $x,y,z$?])
+
+#solution([Yes, since (normal) addition is associative])
+
+#v(1fr)
+
+
+#problem()
+Is tropical multiplication commutative? \
+#note([Does $x #tm y = y #tm x$ for all $x, y$?])
+
+#solution([Yes, since (normal) addition is commutative])
+
+#v(1fr)
+
+
+#problem()
+Is there a tropical multiplicative identity? \
+#note([Is there an $i$ so that $x #tm i = x$ for all $x$?])
+
+#solution([Yes, it is 0.])
+
+#v(1fr)
+
+
+#problem()
+Do tropical multiplicative inverses always exist? \
+#note([
+  For every $x != #sym.infinity$, does there exist an inverse $y$ so that $x #tm y = i$, \
+  where $i$ is the additive identity?
+])
+
+#solution([Yes, it is $-x$. For $x != 0$, $x #tm (-x) = 0$])
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+
+#problem()
+Is tropical multiplication distributive over addition? \
+#note([Does $x #tm (y #tp z) = x #tm y #tp x #tm z$?])
+
+#solution([Yes, $x + min(y,z) = min(x+y, x+z)$])
+
+#v(1fr)
+
+#problem()
+Fill the following tropical addition and multiplication tables
+
+#let col = 10mm
+
+#notsolution(
+  table(
+    columns: (1fr, 1fr),
+    align: center,
+    stroke: none,
+    table(
+      columns: (col, col, col, col, col, col),
+      align: center,
+      table.header(
+        [$#tp$],
+        [$1$],
+        [$2$],
+        [$3$],
+        [$4$],
+        [$#sym.infinity$],
+      ),
+
+      box(inset: 3pt, $1$), [], [], [], [], [],
+      box(inset: 3pt, $2$), [], [], [], [], [],
+      box(inset: 3pt, $3$), [], [], [], [], [],
+      box(inset: 3pt, $4$), [], [], [], [], [],
+      box(inset: 3pt, $#sym.infinity$), [], [], [], [], [],
+    ),
+    table(
+      columns: (col, col, col, col, col, col),
+      align: center,
+      table.header(
+        [$#tm$],
+        [$0$],
+        [$1$],
+        [$2$],
+        [$3$],
+        [$4$],
+      ),
+
+      box(inset: 3pt, $0$), [], [], [], [], [],
+      box(inset: 3pt, $1$), [], [], [], [], [],
+      box(inset: 3pt, $2$), [], [], [], [], [],
+      box(inset: 3pt, $3$), [], [], [], [], [],
+      box(inset: 3pt, $4$), [], [], [], [], [],
+    ),
+  ),
+)
+
+#solution(
+  table(
+    columns: (1fr, 1fr),
+    align: center,
+    stroke: none,
+    table(
+      columns: (col, col, col, col, col, col),
+      align: center,
+      table.header(
+        [$#tp$],
+        [$1$],
+        [$2$],
+        [$3$],
+        [$4$],
+        [$#sym.infinity$],
+      ),
+
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $1$),
+
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $2$),
+
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $3$),
+
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $4$),
+
+      box(inset: 3pt, $#sym.infinity$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $#sym.infinity$),
+    ),
+    table(
+      columns: (col, col, col, col, col, col),
+      align: center,
+      table.header(
+        [$#tm$],
+        [$0$],
+        [$1$],
+        [$2$],
+        [$3$],
+        [$4$],
+      ),
+
+      box(inset: 3pt, $0$),
+      box(inset: 3pt, $0$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $4$),
+
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $5$),
+
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $5$),
+      box(inset: 3pt, $6$),
+
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $5$),
+      box(inset: 3pt, $6$),
+      box(inset: 3pt, $7$),
+
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $5$),
+      box(inset: 3pt, $6$),
+      box(inset: 3pt, $7$),
+      box(inset: 3pt, $8$),
+    ),
+  ),
+)
+
+
+#v(2mm)
+
+#problem()
+Expand and simplify $f(x) = (x #tp 2)(x #tp 3)$, then evaluate $f(1)$ and $f(4)$ \
+Adjacent parenthesis imply tropical multiplication
+
+#solution([
+  $
+    (x #tp 2)(x #tp 3)
+    &= x^2 #tp 2x #tp 3x #tp (2 #tm 3) \
+    &= x^2 #tp (2 #tp 3)x #tp (2 #tm 3) \
+    &= x^2 #tp 2x #tp 5
+  $
+
+  Also, $f(1) = 2$ and $f(4) = 5$.
+])
+
+#v(1fr)
--- a/src/Advanced/Tropical
+++ b/src/Advanced/Tropical
@ -0,0 +1,387 @@
+#import "@local/handout:0.1.0": *
+#import "../macros.typ": *
+#import "@preview/cetz:0.3.1"
+
+= Tropical Polynomials
+
+#definition()
+A _polynomial_ is an expression formed by adding and multiplying numbers and a variable $x$. \
+Every polynomial can be written as
+#align(
+  center,
+  box(
+    inset: 3mm,
+    $
+      c_0 + c_1 x + c_2 x^2 + ... + c_n x^n
+    $,
+  ),
+)
+for some nonnegative integer $n$ and coefficients $c_0, c_1, ..., c_n$. \
+The _degree_ of a polynomial is the largest $n$ for which $c_n$ is nonzero.
+
+#theorem()
+The _fundamental theorem of algebra_ implies that any non-constant polynomial with real coefficients
+can be written as a product of polynomials of degree 1 or 2 with real coefficients.
+
+#v(2mm)
+
+For example, the polynomial $-160 - 64x - 2x^2 + 17x^3 + 8x^4 + x^5$ \
+can be written as $(x^2 + 2x+5)(x-2)(x+4)(x+4)$
+
+#v(2mm)
+A similar theorem exists for polynomials with complex coefficients. \
+These coefficients may be found using the roots of this polynomial. \
+As it turns out, there are formulas that determine the roots of quadratic, cubic, and quartic #note([(degree 2, 3, and 4)]) polynomials. There are no formulas for the roots of polynomials with larger degrees---in this case, we usually rely on approximate roots found by computers.
+
+#v(2mm)
+In this section, we will analyze tropical polynomials:
+- Is there a fundamental theorem of tropical algebra?
+- Is there a tropical quadratic formula? How about a cubic formula?
+- Is it difficult to find the roots of tropical polynomials with large degrees?
+
+
+#definition()
+A _tropical_ polynomial is a polynomial that uses tropical addition and multiplication. \
+In other words, it is an expression of the form
+#align(
+  center,
+  box(
+    inset: 3mm,
+    $
+      c_0 #tp (c_1 #tm x) #tp (c_2 #tm x^2) #tp ... #tp (c_n #tm x^n)
+    $,
+  ),
+)
+where all exponents represent repeated tropical multiplication.
+
+#pagebreak() // MARK: page
+
+
+
+
+#problem()
+Draw a graph of the tropical polynomial $f(x) = x^2 #tp 1x #tp 4$. \
+#hint([$1x$ is not equal to $x$.])
+
+#notsolution(graphgrid(none))
+
+#solution([
+  $f(x) = min(2x , 1+x, 4)$, which looks like:
+
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 0), (4 * step, 8 * step))
+    dotline((0, 1 * step), (7 * step, 8 * step))
+    dotline((0, 4 * step), (8 * step, 4 * step))
+
+    line(
+      (0, 0),
+      (1 * step, 2 * step),
+      (3 * step, 4 * step),
+      (7.5 * step, 4 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
+
+
+#problem()
+Now, factor $f(x) = x^2 #tp 1x #tp 4$ into two polynomials with degree 1. \
+In other words, find $r$ and $s$ so that
+#align(
+  center,
+  box(
+    inset: 3mm,
+    $
+      x^2 #tp 1x #tp 4 = (x #tp r)(x #tp s)
+    $,
+  ),
+)
+
+we will call $r$ and $s$ the _roots_ of $f$.
+
+#solution([
+  Because $(x #tp r)(x #tp s) = x^2 #tp (r #tp s)x #tp s r$, we must have $r #tp s = 1$ and $r #tm s = 4$. \
+  In standard notation, we need $min(r, s) = 1$ and $r + s = 4$, so we take $r = 1$ and $s = 3$:
+
+  #v(2mm)
+
+  $
+    f(x) = x^2 #tp 1x #tp 4 = (x #tp 1)(x #tp 3)
+  $
+])
+
+#v(1fr)
+
+#problem()
+How can we use the graph to determine these roots?
+
+#solution([The roots are the corners of the graph.])
+
+#v(0.5fr)
+#pagebreak() // MARK: page
+
+
+
+
+
+
+#problem()
+Graph $f(x) = -2x^2 #tp x #tp 8$. \
+#hint([Use half scale. 1 box = 2 units.])
+
+#notsolution(graphgrid(none))
+
+#solution([
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 0), (8 * step, 8 * step))
+    dotline((0.5 * step, 0), (4 * step, 8 * step))
+    dotline((0, 4 * step), (8 * step, 4 * step))
+
+    line(
+      (0.5 * step, 0),
+      (1 * step, 1 * step),
+      (4 * step, 4 * step),
+      (7.5 * step, 4 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
+
+#problem()
+Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$.
+
+#solution([
+  We (tropically) factor out $-2$ to get
+
+  #eqnbox($
+    f(x) = -2(x^2 #tp 2x #tp 10)
+  $)
+
+
+  by the same process as the previous problem, we get
+  #eqnbox($
+    f(x) = -2(x #tp 2)(x #tp 8)
+  $)
+])
+
+#v(1fr)
+
+#problem()
+Can you see the roots $r$ and $s$ in the graph? \
+How are the roots related to the coefficients of $f$? \
+#hint([look at consecutive coefficients: $0 - (-2) = 2$])
+
+#solution([
+  The roots are the differences between consecutive coefficients of $f$:
+  - $0-(-2) = 2$
+  - $8 - 0 = 8$
+])
+
+#v(0.5fr)
+
+
+#problem()
+Find a tropical polynomial that has roots $4$ and $5$ \
+and always produces $7$ for sufficiently large inputs.
+
+#solution([
+  We are looking for $f(x) = a x^2 #tp b x #tp c$. \
+  Since $f(#sym.infinity) = 7$, we know that $c = 7$. \
+  Using the pattern from the previous problem, we'll subtract $5$ from $c$ to get $b = 2$, \
+  and $4$ from $b$ to get $a = -2$.
+
+  And so, $f(x) = -2x^2 #tp 2x #tp 7$
+
+  #v(2mm)
+
+  Subtracting roots in the opposite order does not work.
+])
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+
+#problem()
+Graph $f(x) = 1x^2 #tp 3x #tp 5$.
+
+#notsolution(graphgrid(none))
+
+#solution([
+  The graphs of all three terms intersect at the same point:
+
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 1 * step), (3.5 * step, 8 * step))
+    dotline((0, 5 * step), (8 * step, 5 * step))
+    dotline((0, 3 * step), (5 * step, 8 * step))
+
+    line(
+      (0, 1 * step),
+      (2 * step, 5 * step),
+      (7.5 * step, 5 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
+
+
+#problem()
+Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$.
+
+#solution(
+  eqnbox($
+    f(x) = 1x^2 #tp 3 x #tp 5 = 1(x #tp 2)^2
+  $),
+)
+
+#v(1fr)
+
+#problem()
+How is this graph different from the previous two? \
+How is this polynomial's factorization different from the previous two? \
+How are the roots of $f$ related to its coefficients?
+
+#solution([
+  The factorization contains the same term twice. \
+  Also note that the differences between consecutive coefficients of $f$ are both two.
+])
+
+#v(0.5fr)
+#pagebreak() // MARK: page
+
+
+#problem()
+Graph $f(x) = 2x^2 #tp 4x #tp 4$.
+
+#notsolution(graphgrid(none))
+
+#solution(
+  graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 2 * step), (3 * step, 8 * step))
+    dotline((0, 4 * step), (5 * step, 8 * step))
+    dotline((0, 4 * step), (8 * step, 4 * step))
+
+    line(
+      (0, 2 * step),
+      (1 * step, 4 * step),
+      (7.5 * step, 4 * step),
+      stroke: 1mm + oblue,
+    )
+  }),
+)
+
+
+#problem()
+Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$, or show that one does not exist.
+
+#solution([
+  We can factor out a 2 to get $f(x) = 2(x^2 #tp 2x #tp 2)$,
+  but $x^2 #tp 2x #tp 2$ does not factor. \
+  There are no $a$ and $b$ with minimum 2 and sum 2.
+])
+
+#v(1fr)
+
+#problem()
+Find a polynomial that has the same graph as $f$, but can be factored.
+
+#solution([
+  $
+    2x^2 #tp 3x #tp 4 = 2(x #tp 1)^2
+  $
+])
+
+
+#v(1fr)
+
+#pagebreak() // MARK: page
+
+
+#theorem()
+The _fundamental thorem of tropical algebra_ states that for every tropical polynomial $f$, \
+there exists a _unique_ tropical polynomial $accent(f, macron)$ with the same graph that can be factored \
+into linear factors.
+
+#v(2mm)
+
+Whenever we say "the roots of $f$", we really mean "the roots of $accent(f, macron)$." \
+$f$ and $accent(f, macron)$ might be the same polynomial.
+
+#problem()
+If $f(x) = a x^2 #tp b x #tp c$, then $accent(f, macron)(x) = a x^2 #tp B x #tp c$ for some $B$. \
+Find a formula for $B$ in terms of $a$, $b$, and $c$. \
+#hint([there are two cases to consider.])
+
+#solution([
+  If we want to factor $a(x^2 #tp (b-a)x #tp (c-a))$, we need to find $r$ and $s$ so that
+  - $min(r,s) = b-a$, and
+  - $r + s = c - a$
+
+  #v(2mm)
+
+  This is possible if and only if $2(b-a) <= c-a$, \
+  or equivalently if $b <= (a+c) #sym.div 2$
+
+  #v(8mm)
+
+  *Case 1:* If $b <= (a + c #sym.div) 2$, then $accent(f, macron) = f$ and $b = B$.
+
+  #v(2mm)
+
+  *Case 2:* If $b > (a + c #sym.div) 2$, then
+  $
+    accent(f, macron)(x)
+    &= a x^2 #tp ((a+c)/2)x #tp c \
+    &= a(x #tp (c-a)/2)^2
+  $
+  has the same graph as $f$, and thus $B = (a+c) #sym.div 2$
+
+  #v(8mm)
+
+  We can combine these results as follows:
+  $
+    B = min(b, (a+c)/2)
+  $
+])
+
+
+#v(1fr)
+
+#problem()
+Find a tropical quadratic formula in terms of $a$, $b$, and $c$ \
+for the roots $x$ of a tropical polynomial $f(x) = a x^2 #tp b x #tp c$. \
+#hint([
+  again, there are two cases. \
+  Remember that "roots of $f$" means "roots of $accent(f, macron)$".
+])
+
+#solution([
+  *Case 1:* If $b <= (a+c) #sym.div 2$, then $accent(f, macron) = f$ has roots $b-a$ and $c-b$, so
+  $
+    accent(f, macron)(x) = a(x #tp (b-a))(x #tp (c-b))
+  $
+
+  #v(8mm)
+
+  *Case 2:* If $b > (a+c) #sym.div 2$, then $accent(f, macron)$ has root $(c-a) #sym.div$ with multiplicity 2, so
+  $
+    accent(f, macron)(x) = a(x #tp (c-a)/2)^2
+  $
+
+  #v(8mm)
+
+  It is interesting to note that the condition $2b < a+ c$ for there to be two distinct roots becomes $b^2 > a c$ in tropical notation. This is reminiscent of the discriminant condition for standard polynomials!
+])
+
+#v(1fr)
--- a/src/Advanced/Tropical
+++ b/src/Advanced/Tropical
@ -0,0 +1,212 @@
+#import "@local/handout:0.1.0": *
+#import "../macros.typ": *
+#import "@preview/cetz:0.3.1"
+
+= Tropical Cubic Polynomials
+
+#problem()
+Consider the polynomial $f(x) = x^3 #tp x^2 #tp 3x #tp 6$. \
+- sketch a graph of this polynomial
+- use this graph to find the roots of $f$
+- write (and expand) a product of linear factors with the same graph as $f$.
+
+#notsolution(graphgrid(none))
+
+#solution([
+  - Roots are 1, 2, and 3.
+  - $accent(f, macron)(x) = x^3 #tp 1x^2 #tp 3x #tp 6 = (x #tp 1)(x #tp 2)(x #tp 3)$
+
+
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 0), (2.66 * step, 8 * step))
+    dotline((0, 1 * step), (3.5 * step, 8 * step))
+    dotline((0, 3 * step), (5 * step, 8 * step))
+    dotline((0, 6 * step), (8 * step, 6 * step))
+
+    line(
+      (0, 0),
+      (1 * step, 3 * step),
+      (2 * step, 5 * step),
+      (3 * step, 6 * step),
+      (7.5 * step, 6 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
+
+
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+#problem()
+Consider the polynomial $f(x) = x^3 #tp x^2 #tp 6x #tp 6$. \
+- sketch a graph of this polynomial
+- use this graph to find the roots of $f$
+- write (and expand) a product of linear factors with the same graph as $f$.
+
+#notsolution(graphgrid(none))
+
+#solution([
+  - Roots are 1, 2.5, and 2.5.
+  - $accent(f, macron)(x) = x^3 #tp 1x^2 #tp 3.5x #tp 6 = (x #tp 1)(x #tp 2.5)^2$
+
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 0), (2.66 * step, 8 * step))
+    dotline((0, 1 * step), (3.5 * step, 8 * step))
+    dotline((0, 6 * step), (2 * step, 8 * step))
+    dotline((0, 6 * step), (8 * step, 6 * step))
+
+    line(
+      (0, 0),
+      (1 * step, 3 * step),
+      (2.5 * step, 6 * step),
+      (7.5 * step, 6 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
+
+
+#v(1fr)
+
+#problem()
+Consider the polynomial $f(x) = x^3 #tp 6x^2 #tp 6x #tp 6$. \
+- sketch a graph of this polynomial
+- use this graph to find the roots of $f$
+- write (and expand) a product of linear factors with the same graph as $f$.
+
+#notsolution(graphgrid(none))
+
+#solution([
+  - Roots are 2, 2, and 2.
+  - $accent(f, macron)(x) = x^3 #tp 2x^2 #tp 4x #tp 6 = (x #tp 2)^3$
+
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 0), (2.66 * step, 8 * step))
+    dotline((0, 6 * step), (1 * step, 8 * step))
+    dotline((0, 6 * step), (2 * step, 8 * step))
+    dotline((0, 6 * step), (8 * step, 6 * step))
+
+    line(
+      (0, 0),
+      (2 * step, 6 * step),
+      (7.5 * step, 6 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
+
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+
+#problem()
+If $f(x) = a x^3 #tp b x^2 #tp c x #tp d$, then $accent(f, macron)(x) = a x^3 #tp B x^2 #tp C x #tp d$ for some $B$ and $C$. \
+Using the last three problems, find formulas for $B$ and $C$ in terms of $a$, $b$, $c$, and $d$.
+
+#solution([
+
+  $
+    B = min(b, (a+c)/2, (2a+d)/2)
+  $
+  $
+    C = min(c, (b+d)/2, (a+2d)/2)
+  $
+])
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+#problem()
+What are the roots of the following polynomial?
+
+#align(
+  center,
+  box(
+    inset: 3mm,
+    $
+      3 x^6 #tp 4 x^5 #tp 2 x^4 #tp x^3 #tp x^2 #tp 4 x #tp 5
+    $,
+  ),
+)
+
+#solution([
+  We have
+  $
+    accent(f, macron)(x) = 3x^6 #tp 2x^5 #tp 1x^4 #tp x^3 #tp 1x^2 #tp 3x #tp 5
+  $
+  which has roots $-1$, $-1$, $-1$, $1$, $2$, $2$
+])
+
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+#problem()
+If
+$
+  f(x) = c_0 #tp c_1 x #tp c_2 x^2 #tp ... #tp c_n x^n
+$
+then
+$
+  accent(f, macron)(x) = c_0 #tp C_1 x #tp C_2 x^2 #tp ... #tp C_(n-1) x^(n-1) #tp c_n x^n
+$
+
+#v(2mm)
+
+Find a formula for each $C_i$ in terms of $c_0, c_1, ..., c_n$.
+
+#solution([
+  $
+    A_j
+    &= min_(l<=j<k)( (a_l - a_k) / (k-l) (k-j) + a_k ) \
+    &= min_(l<=j<k)( a_l (k-j) / (k-l) + a_k (j-l) / (k-l) )
+  $
+
+  #v(2mm)
+
+  which is a weighted average of some $a_l$ and $a_k$, with $l<=j<k$
+])
+
+#v(1fr)
+
+
+#problem()
+With the same setup as the previous problem, \
+find formulas for the roots $r_1, r_2, ..., r_n$.
+
+#solution([
+  The roots are the differences between consecutive coefficients of $accent(f, macron)$:
+  $
+    r_i = A_i - A_(i-1)
+  $
+  where we set $A_n = a_n$ and $A_0 = a_0$.
+])
+
+#v(1fr)
+
+#problem()
+Can you find a geometric interpretation of these formulas \
+in terms of the points $(-i, c_i)$ for $0 <= i <= n$?
+
+#solution([
+  The inequality #note([(for $l <= k < k$)])
+  $
+    A_j <= (a_l - a_k) / (k-l) (k-j)+a_k
+  $
+  states that the point $(-j,A_j)$ must lie on or below the line segment between the points $(-k, a_k)$ and $(-l, a_l)$.
+  This makes it easy to find the $A_j$ using a graph of the points $(-i, a_i)$ for $0 <= i <=n$.
+])
+
+#v(0.5fr)
--- a/tools/scripts/build.py
+++ b/tools/scripts/build.py
@ -143,10 +143,12 @@ def build_typst(source_dir: Path, out_subdir: Path) -> IndexEntry | None:
 		[
 			TYPST_PATH,
 			"compile",
+			"--package-path",
+			f"{ROOT}/lib/typst",
 			"--ignore-system-fonts",
-			"main.typ",
 			"--input",
 			"show_solutions=false",
+			"main.typ",
 			f"{out}/{handout_file}",
 		],
 		cwd=source_dir,
@ -164,6 +166,8 @@ def build_typst(source_dir: Path, out_subdir: Path) -> IndexEntry | None:
 			[
 				TYPST_PATH,
 				"compile",
+				"--package-path",
+				f"{ROOT}/lib/typst",
 				"--ignore-system-fonts",
 				"main.typ",
 				f"{out}/{solutions_file}",
Author	SHA1	Message	Date
Mark	c5d14cb917	FIR draft All checks were successful CI / Typst formatting (pull_request) Successful in 29s Details CI / Typos (pull_request) Successful in 34s Details CI / Build (pull_request) Successful in 13m33s Details	2025-01-23 10:34:06 -08:00
Mark	ede934369b	Add local typst packages All checks were successful CI / Typst formatting (push) Successful in 15s Details CI / Typos (push) Successful in 18s Details CI / Build (push) Successful in 13m16s Details	2025-01-23 10:29:13 -08:00
Mark	3b41ea714a	Added "Tropical Polynomials" handout	2025-01-23 10:29:07 -08:00