Added "Tropical Polynomials" handout

2025-01-22 12:31:47 -08:00
parent 7597343fe3
commit ed3339829d
7 changed files with 1273 additions and 0 deletions
--- a/Polynomials/handout.typ
+++ b/Polynomials/handout.typ
@ -0,0 +1,280 @@
+
+/// 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: 8.5in,
+    height: 11in,
+    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/Polynomials/macros.typ
+++ b/Polynomials/macros.typ
@ -0,0 +1,70 @@
+#import "./handout.typ": *
+#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 "./handout.typ": *
+
+#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 +1,294 @@
+#import "../handout.typ": *
+#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 $x #tp y = #sym.infinity$? \
+  Remember that $#sym.infinity$ is the additive identity.
+])
+
+#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 $f(4)$ \
+#hint([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 +1,388 @@
+#import "../handout.typ": *
+#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_ states 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 you already know, 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 appropriate 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)
+    $,
+  ),
+)
+
+#note([Naturally, 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()
+Can you see the roots of this polynomial in the graph? \
+#hint([Yes, you can. What "features" do the roots correspond to?])
+
+#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 $accent(f, macron)$." \
+Also, $f$ 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 "../handout.typ": *
+#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)