Added "Tropical Polynomials" handout
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parent
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280
src/Advanced/Tropical Polynomials/handout.typ
Executable file
280
src/Advanced/Tropical Polynomials/handout.typ
Executable file
@ -0,0 +1,280 @@
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/// If false, hide instructor info.
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///
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/// Compile with the following command to hide solutions:
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/// `typst compile main.typ --input show_solutions=false`
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///
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/// Solutions are shown by default. This behavior
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/// is less surprising than hiding content by default.
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#let show_solutions = {
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if "show_solutions" in sys.inputs {
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// Show solutions unless they're explicitly disabled
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not (
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sys.inputs.show_solutions == "false" or sys.inputs.show_solutions == "no"
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)
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} else {
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// Show solutions by default
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true
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}
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}
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// Colors
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#let ored = rgb("D62121")
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#let ogrape = rgb("9C36B5")
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#let ocyan = rgb("2288BF")
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#let oteal = rgb("12B886")
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#let ogreen = rgb("37B26D")
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#let oblue = rgb("1C7ED6")
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//
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// MARK: header
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//
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#let make_title(
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group,
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quarter,
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title,
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subtitle,
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) = {
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align(
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center,
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block(
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width: 60%,
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height: auto,
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breakable: false,
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align(
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center,
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stack(
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spacing: 7pt,
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(
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text(size: 10pt, group) + h(1fr) + text(size: 10pt, quarter)
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),
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line(length: 100%, stroke: 0.2mm),
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(
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text(size: 20pt, title) + linebreak() + text(size: 10pt, subtitle)
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),
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line(length: 100%, stroke: 0.2mm),
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),
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),
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),
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)
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}
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#let warn = {
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set text(ored)
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align(
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center,
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block(
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width: 60%,
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height: auto,
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breakable: false,
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fill: rgb(255, 255, 255),
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stroke: ored + 2pt,
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inset: 3mm,
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(
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align(center, text(weight: "bold", size: 12pt, [Instructor's Handout]))
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+ parbreak()
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+ align(
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left,
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text(
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size: 10pt,
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[This handout contains solutions and notes.]
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+ linebreak()
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+ [Recompile without solutions before distributing.],
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),
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)
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),
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),
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)
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}
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#let preparedby(name) = (
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text(
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size: 10pt,
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[Prepared by ]
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+ name
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+ [ on ]
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+ datetime
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.today()
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.display("[month repr:long] [day padding:none], [year]"),
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)
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)
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//
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// MARK: Solutions
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//
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#let solution(content) = {
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if show_solutions {
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align(
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center,
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stack(
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block(
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width: 100%,
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breakable: false,
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fill: ored,
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stroke: ored + 2pt,
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inset: 1.5mm,
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(
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align(left, text(fill: white, weight: "bold", [Solution:]))
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),
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),
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block(
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width: 100%,
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height: auto,
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breakable: false,
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fill: ored.lighten(80%).desaturate(10%),
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stroke: ored + 2pt,
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inset: 3mm,
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align(left, content),
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),
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),
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)
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}
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}
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#let notsolution(content) = {
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if not show_solutions { content }
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}
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//
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// MARK: Sections
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//
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#let generic(t) = block(
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above: 8mm,
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below: 2mm,
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text(weight: "bold", t),
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)
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#let _generic_base(kind, ..args) = {
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counter("obj").step()
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if args.pos().len() == 0 {
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generic([
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#kind
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#context counter("obj").display():
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])
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} else {
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generic(
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[
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#kind
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#context counter("obj").display():
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]
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+ " "
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+ args.pos().at(0),
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)
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}
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}
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#let problem(..args) = _generic_base("Problem", ..args)
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#let definition(..args) = _generic_base("Definition", ..args)
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#let theorem(..args) = _generic_base("Theorem", ..args)
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//
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// MARK: Misc
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//
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#let hint(content) = {
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text(fill: rgb(100, 100, 100), style: "oblique", "Hint: ")
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text(fill: rgb(100, 100, 100), content)
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}
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#let note(content) = {
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text(fill: rgb(100, 100, 100), content)
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}
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#let examplesolution(content) = {
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let c = oblue
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align(
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center,
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stack(
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block(
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width: 100%,
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breakable: false,
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fill: c,
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stroke: c + 2pt,
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inset: 1.5mm,
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(
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align(left, text(fill: white, weight: "bold", [Example solution:]))
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),
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),
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block(
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width: 100%,
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height: auto,
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breakable: false,
|
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fill: c.lighten(80%).desaturate(10%),
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stroke: c + 2pt,
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inset: 3mm,
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align(left, content),
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),
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),
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)
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}
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//
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// MARK: wrapper
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//
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#let handout(
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doc,
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group: none,
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quarter: none,
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title: none,
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by: none,
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subtitle: none,
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) = {
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set par(leading: 0.55em, first-line-indent: 0mm, justify: true)
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set text(font: "New Computer Modern")
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set par(spacing: 0.5em)
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show list: set block(spacing: 0.5em, below: 1em)
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set heading(numbering: (..nums) => nums.pos().at(0))
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set page(
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margin: 20mm,
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width: 8.5in,
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height: 11in,
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footer: align(
|
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center,
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context counter(page).display(),
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),
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footer-descent: 5mm,
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)
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set list(
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tight: false,
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indent: 5mm,
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spacing: 3mm,
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)
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show heading.where(level: 1): it => {
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set align(center)
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set text(weight: "bold")
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block[
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Section #counter(heading).display(): #text(it.body)
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]
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}
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make_title(
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group,
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quarter,
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title,
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{
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if by == none { none } else { [#preparedby(by)\ ] }
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if subtitle == none { none } else { subtitle }
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},
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)
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if show_solutions {
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warn
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}
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doc
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}
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70
src/Advanced/Tropical Polynomials/macros.typ
Normal file
70
src/Advanced/Tropical Polynomials/macros.typ
Normal file
@ -0,0 +1,70 @@
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#import "./handout.typ": *
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#import "@preview/cetz:0.3.1"
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// Shorthand, we'll be using these a lot.
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#let tp = sym.plus.circle
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#let tm = sym.times.circle
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#let graphgrid(inner_content) = {
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align(
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center,
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box(
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inset: 3mm,
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cetz.canvas({
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import cetz.draw: *
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let x = 5.25
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|
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grid(
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(0, 0), (x, x), step: 0.75,
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stroke: luma(100) + 0.3mm
|
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)
|
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|
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if (inner_content != none) {
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inner_content
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}
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mark((0, x + 0.5), (0, x + 1), symbol: ">", fill: black, scale: 1)
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mark((x + 0.5, 0), (x + 1, 0), symbol: ">", fill: black, scale: 1)
|
||||
|
||||
line(
|
||||
(0, x + 0.25),
|
||||
(0, 0),
|
||||
(x + 0.25, 0),
|
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stroke: 0.75mm + black,
|
||||
)
|
||||
}),
|
||||
),
|
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)
|
||||
}
|
||||
|
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/// Adds extra padding to an equation.
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/// Used as follows:
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///
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/// #eqmbox($
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/// f(x) = -2(x #tp 2)(x #tp 8)
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/// $)
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///
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/// Note that there are newlines between the $ and content,
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/// this gives us display math (which is what we want when using this macro)
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#let eqnbox(eqn) = {
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align(
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center,
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box(
|
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inset: 3mm,
|
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eqn,
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),
|
||||
)
|
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}
|
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|
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#let dotline(a, b) = {
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cetz.draw.line(
|
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a,
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b,
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stroke: (
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dash: "dashed",
|
||||
thickness: 0.5mm,
|
||||
paint: ored,
|
||||
),
|
||||
)
|
||||
}
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22
src/Advanced/Tropical Polynomials/main.typ
Normal file
22
src/Advanced/Tropical Polynomials/main.typ
Normal file
@ -0,0 +1,22 @@
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#import "./handout.typ": *
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#show: doc => handout(
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doc,
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group: "Advanced 2",
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quarter: link(
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"https://betalupi.com/handouts",
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"betalupi.com/handouts",
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),
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title: [Tropical Polynomials],
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by: "Mark",
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subtitle: "Based on a handout by Bryant Mathews",
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)
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#include "parts/00 arithmetic.typ"
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#pagebreak()
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#include "parts/01 polynomials.typ"
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#pagebreak()
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#include "parts/02 cubic.typ"
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7
src/Advanced/Tropical Polynomials/meta.toml
Normal file
7
src/Advanced/Tropical Polynomials/meta.toml
Normal file
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[metadata]
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title = "Tropical Polynomials"
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[publish]
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handout = true
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solutions = true
|
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294
src/Advanced/Tropical Polynomials/parts/00 arithmetic.typ
Normal file
294
src/Advanced/Tropical Polynomials/parts/00 arithmetic.typ
Normal file
@ -0,0 +1,294 @@
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#import "../handout.typ": *
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#import "../macros.typ": *
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= Tropical Arithmetic
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#definition()
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The _tropical sum_ of two numbers is their minimum:
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$
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x #tp y = min(x, y)
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$
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|
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#definition()
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The _tropical product_ of two numbers is their sum:
|
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$
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x #tm y = x + y
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$
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#problem()
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- Is tropical addition commutative? \
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#note([i.e, does $x #tp y = y #tp x$?])
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- Is tropical addition associative? \
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#note([i.e, does $(x #tp y) #tp z = x #tp (y #tp z)$?])
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- Is there a tropical additive identity? \
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#note([i.e, is there an $i$ so that $x #tp i = x$ for all real $x$?])
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#solution([
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- Is tropical addition commutative?\
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Yes, $min(min(x,y),z) = min(x,y,z) = min(x,min(y,z))$
|
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- Is tropical addition associative? \
|
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Yes, $min(x,y) = min(y,x)$
|
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- Is there a tropical additive identity? \
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No. There is no $n$ where $x <= n$ for all real $x$
|
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])
|
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|
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#v(1fr)
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|
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#problem()
|
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Let's expand $#sym.RR$ to include a tropical additive identity.
|
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- What would be an appropriate name for this new number?
|
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- Give a reasonable definition for...
|
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- the tropical sum of this number and a real number $x$
|
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- the tropical sum of this number and itself
|
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- the tropical product of this number and a real number $x$
|
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- the tropical product of this number and itself
|
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|
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#solution([
|
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#sym.infinity makes sense, with
|
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$#sym.infinity #tp x = x$; #h(1em)
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$#sym.infinity #tp #sym.infinity = #sym.infinity$; #h(1em)
|
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$#sym.infinity #tm x = #sym.infinity$; #h(1em) and
|
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$#sym.infinity #tm #sym.infinity = #sym.infinity$
|
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])
|
||||
|
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#v(1fr)
|
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#pagebreak() // MARK: page
|
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|
||||
|
||||
|
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|
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|
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#problem()
|
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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)
|
||||
388
src/Advanced/Tropical Polynomials/parts/01 polynomials.typ
Normal file
388
src/Advanced/Tropical Polynomials/parts/01 polynomials.typ
Normal file
@ -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)
|
||||
212
src/Advanced/Tropical Polynomials/parts/02 cubic.typ
Normal file
212
src/Advanced/Tropical Polynomials/parts/02 cubic.typ
Normal file
@ -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)
|
||||
Loading…
x
Reference in New Issue
Block a user