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0
.gitea/workflows/ci.yml → .github/workflows/ci.yml vendored
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4
.gitignore vendored
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@ -3,16 +3,14 @@ venv
__pycache__
*.ignore
.mypy_cache
.DS_Store
# Output files
/output
/output.zip
*.pdf
/manual
# TeX build files
*.synctex*
*.synctex.gz*
*.latexmk
*.aux
*.out

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.vscode/settings.json vendored
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@ -1,6 +1,5 @@
{
"latex-workshop.latex.recipe.default": "latexmk (xelatex)",
"tinymist.formatterPrintWidth": 80,
"tinymist.typstExtraArgs": ["--package-path=./lib/typst"],
"tinymist.formatterMode": "typstyle"
"tinymist.typstExtraArgs": ["--package-path=./lib/typst"]
}

128
README.md
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@ -1,128 +0,0 @@
[tinymist]: https://marketplace.visualstudio.com/items?itemName=myriad-dreamin.tinymist
[latex-workshop]: https://marketplace.visualstudio.com/items?itemName=James-Yu.latex-workshop
[CC BY-NC-SA 4.0]: https://creativecommons.org/licenses/by-nc-sa/4.0
[betalupi.com/handouts]: https://betalupi.com/handouts
[ORMC]: https://circles.math.ucla.edu/circles/
[Overleaf]: https://overleaf.com
[Typst.app]: https://typst.app
[vscode]: https://code.visualstudio.com
[vscodium]: https://vscodium.com
[homebrew]: https://brew.sh
# Mark's Handout Library
This is a collection of math circle handouts that I (and many others) have written. \
They are used regularly at the [ORMC].
For more information, visit [betalupi.com/handouts]. \
The latest version of each handout is available at that page.
## License
Unless otherwise stated, all documents in this repository are licensed under [CC BY-NC-SA 4.0]. \
Each document has its own authors. See `meta.toml` in each project directory for details.
By submitting or editing a handout in this repository, you agree to release it under this license.
## 🛠️ Contributing
If you want to use one of these handouts for a class, see [betalupi.com/handouts]. \
You only need to read this section if you want to edit these handouts.
### Setup
Use git to clone this repository, then open the root folder in [vscode] or [vscodium].
We use the [latex-workshop] and [tinymist] extensions. Install them before continuing.
[`./vscode/settings.json`](./vscode/settings.json) will automatically configure them to work with this repository. \
You may need to install texlive and typst:
- If you use Linux, you'll figure it out.
- On macos, use [homebrew]: \
`brew install texlive typst typstyle`
- On Windows, I don't know. I may write instructions later.
### Editing
This repository is organized as follows:
- All handouts are in [`./src`](./src). \
Every handout is stored in its own directory, even if it only consists of one file. \
Handouts are organized by group (see [betalupi.com/handouts] for details).
- Packages are stored in [`./lib`](./lib) \
You shouldn't need to modify any library files, but you may want to read them to see how they work.
- [`./tools`](./tools) contains build scripts, [`./.github`](./.github) configures automation. \
You can ignore everything in these directories.
All handouts in this repository are based on `handout.cls` or `handout@0.1.0`.
- If you're using Typst (preferred), read [`docs-typst.md`](./docs-typst.md)
- If you're still using LaTeX, read [`docs-latex.md`](./docs-latex.md).
### Metadata
Every handout directory should contain a file called `meta.toml` with the following contents:
```toml
# This is a sample `meta.toml`.
# A copy of this file should exist in every handout directory.
# All keys are required.
[metadata]
title = "title of this handout"
[publish]
# Should we publish this handout?
# If `false`, no part of this handout is published.
handout = true
# Should we publish an "instructor's" version of this handout?
# This key has no effect if `publish.handout` is false.
#
# If `true`, publish a second version of this handout with solutions.
# Set this to `false` if solutions haven't been written.
solutions = true
```
## 💾 Out-of-band compilation
If you want to compile these handouts _without_ this repository (e.g, on [Overleaf] or [Typst.app]), do the following: \
_(I do not recommend this. The default toolchain makes it easier to share improvements to these handouts.)_
### For LaTeX:
1. Get the handout's directory (i.e, download the whole repo as a zip and extract the folder you want.)
2. Download [`./resources/handout.cls`](./resources/handout.cls)
3. Put this `handout.cls` in the same directory as the handout.
4. Fix the include path at the top of `main.tex`:
You'll need to replace
```latex
\documentclass[
...
]{../../../lib/tex/handout}
```
with
```latex
\documentclass[
...
]{handout}
```
5. Make a new overleaf project with the resulting directory.
6. **Do not use pdflatex**, it misbehaves with `handout`. Tell Overleaf to use XeLaTeX.
### For Typst:
Out-of-band typst compilation isn't supported. Clone the repository and use vscode. \
This is because typst can't import packages from a relative path.
If you _really_ want it, standalone typst compilation _is_ possible. \
Follow the LaTeX instructions, but fix `handout@0.1.0` instead of `handout`. \
You'll figure it out.

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authors.toml Normal file
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[authors."mark"]
email = "mark@betalupi.com"
webpage = "betalupi.com"

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docs-latex.md
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@ -1,90 +0,0 @@
# LaTeX documentation
All LaTeX handouts are based on [`handout.cls`](./lib/tex/handout.cls). \
This class is based on `article.cls`, and should work with most LaTeX packages.
The best way to start a new document is to make a copy of an existing one.
- [Advanced/Cryptography](./src/Advanced/Cryptography) is a good example of a simple handout.
- [Advanced/DFAs](./src/Advanced/DFAs) is a good example of a handout with graphs.
- [Advanced/Geometric Optimization](./src/Advanced/Geometric%20Optimization) is a good example of a handout with geometry.
## Notes
- Compile your handouts with XeLaTeX. \
`pdflatex` is known to misbehave with `handout.cls`. \
This will happen by default if you use vscode. \
If you use Overleaf, you'll have to configure it manually (see document settings).
## Document Options
Document options are passed to `\documentclass`, as follows:
```latex
\documentclass[
% Show solutions is `solutions` is provided,
% hide them if `nosolutions` is provided.
%
% You should set only ONE of these flags at a time.
% Solutions are shown by default.
% All handouts are stored with `solutions` enabled.
solutions,
nosolutions,
% Enable this option if you need more space on the handout's first page.
% We use a long warning by default.
shortwarning,
% If present, hide page numbers.
% This should only be used for single-page handouts
% (e.g, warm-ups)
nopagenumber
]{handout}
```
Use `geometry` to change margins and page dimensions. US letter is the default.
## Utilities
- `\say{text}`: Puts text in quotes, handling details like period spacing. Courtesy of `dirtytalk`.
- `\note[Type]{text}`: Makes a note.
- `\hint{text}`: Shorthand for `\note[Hint]{text}`
## Sections
The usual LaTeX title-customization techniques _WILL NOT WORK_ with this class. \
Don't even try to load `titlesec`.
`handout.cls` supports two levels of sections:
- `\section`, for large parts of the handout
- `\definition`, `\theorem`, `\proposition`, `\example`, `\remark`, `\problem`, and `\problempart`
All these macros have the following syntax: `\problem{title}<label>`
- `title` is the problem's title, and may be empty.
- `label` is the problem's label. This is optional. \
If a label is provided, this section may be referenced with `\ref{label}`.
Examples:
- `\problem{}`
- `\problem{Bonus}`
- `\problem{}<gcd>`, which may be referenced with `\ref{gcd}`
Do **not** use `\begin{problem} ... \end{problem}`. \
Sections are macros, not environments.
## Environments:
- `\begin{solution}`: A fancy red for solutions to problems. \
This is hidden if the `nosolutions` is provided.
- `\begin{instrutornote}`: A fancy blue box for instructor notes. \
This is hidden if the `nosolutions` is provided.
- `\begin{examplesolution}`: A fancy gray for sample solutions. \
This is never hidden.
All the above environments break across pages and may safely be nested.
Each of these environments also provides the `\linehack` macro, which draws a line across the box. \
This is useful for, say, solutions to multipart problems.

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docs-typst.md
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@ -1,103 +0,0 @@
# Typst documentation
See [typst.app/docs](https://typst.app/docs) for typst's documentation. \
All typst handouts are based on [`handout@0.1.0`](./lib/typst/local/handout/0.1.0).
The best way to start a new document is to make a copy of an existing one.
- [Advanced/Tropical Polynomials](./src/Advanced/Tropical%20Polynomials) is a good place to start.
- [Warm-Ups/Painting](./src/Warm-Ups/Painting) is a good example of tikz-like pictures.
## Notes
- Typst's equivalent of tikz is cetz ([homepage](https://cetz-package.github.io), [docs](https://cetz-package.github.io/docs/api))
- Typst handouts are always compiled with solutions. \
Handouts without solutions are automatically compiled and published at [betalupi.com/handouts](https://betalupi.com/handouts). \
If you'd like to compile a student handout manually, run the following command in a handout directory:
```bash
typst compile main.typ --package-path ../../../lib/typst --input show_solutions=false
```
Where `package_path` is a relative path to [./lib/typst](./lib/typst).
## Document Options
All typst handouts start with the following:
```typst
#show: handout.with(
// Should match `meta.toml`
title: [handout title],
// Authors
by: "Mark",
// Subtitle (optional)
subtitle: "Based on a handout by Bryant Mathews",
// Group (optional)
group: "Advanced 2",
)
```
## Notable commands
- `#v(1fr)`: Like LaTeX's `\vfill`. Creates whitespace that grows automatically. \
`fr` means "fraction". `#v(2fr)` will fill twice as much space as `#v(1fr)` on the same page.
## Utilities
- `#note([content], type: "Note type")`: Makes a note. `type` is optional.
- `#hint([content])`: Shorthand for `#note([content], type: "Hint")`
- `#solution([content])`: A pretty box for solutions. Hidden in student handouts.
- `#examplesolution([content])`: Like `#solution()`, but is never hidden.
- `#if_solutions([content])`: Shows content only if we are showing solutions.
- `#if_no_solutions([content])`: Shows content only if we **aren't** showing solutions.
## Sections
High-level sections are denoted with `=`. \
Subsections start with `==`, subsubsections with `===`, and so on. \
**`handout@0.1.0` is only designed to use `=`, subsections might be ugly.**
`handout@0.1.0` also provides the following commands:
- `problem`
- `definition`
- `theorem`
- `example`
- `remark`
These all have the same syntax: `#problem("title", label: "label")`
- `title` is the problem's title, and may be omitted.
- `label` is the problem's label. This is optional. \
If a label is provided, this problem can be referenced with `@label`
**Examples:**
- `#problem()`
- `#problem("Bonus")`
- `#problem(label: "gcd")`, which may be referenced with `@gcd`
### Complete example:
```typst
// Import definition(), problem(), etc.
// Must be at the top of each file.
#import "@local/handout:0.1.0": *
// Make a section called "Tropical Cubic Polynomials"
= Tropical Cubic Polynomials
// Make a problem with a label
#problem(label: "imaproblem")
Consider the polynomial $f(x) = x^3 + 1x^2 + 3x + 6$.
- sketch a graph of this polynomial
// Make an untitled problem that references `problem`.
#problem()
Recall @imaproblem.
- use this graph to find the roots of $f$
```

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lib/tex/macros.sty
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@ -1,5 +1,5 @@
\NeedsTeXFormat{LaTeX2e}
\ProvidesPackage{../../../lib/tex/macros}[2025/03/02 Handout macros]
\ProvidesPackage{../../../lib/tex/macros}[2023/10/16 ORMC Macros]
\RequirePackage{hyperref}
\RequirePackage{pgf}

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lib/tex/handout.cls → lib/tex/ormc_handout.cls
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@ -1,5 +1,20 @@
% Copyright (C) 2023 Mark (mark@betalupi.com)
%
% This program is free software: you can redistribute it and/or modify
% it under the terms of the GNU General Public License as published by
% the Free Software Foundation, either version 3 of the License, or
% (at your option) any later version.
%
% This program is distributed in the hope that it will be useful,
% but WITHOUT ANY WARRANTY; without even the implied warranty of
% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
% GNU General Public License for more details.
%
% You should have received a copy of the GNU General Public License
% along with this program. If not, see <https://www.gnu.org/licenses/>.
\NeedsTeXFormat{LaTeX2e}
\ProvidesClass{../../../lib/tex/handout}[2025/03/02 2.0.2 Mark's handout class]
\ProvidesClass{../../../lib/tex/ormc_handout}[2023/05/29 2.0.2 ORMC Handout]
@ -13,7 +28,7 @@
\@twocolumnfalse
\@twosidefalse
\@mparswitchfalse
% Handout-specific
% ORMC-specific
\newif{\if@solutions} % If false, solutions and instructor notes are hidden.
\newif{\if@singlenumbering} % If true, the same counter is used for all objects.
\newif{\if@nopagenumber} % If true, don't number pages.
@ -25,7 +40,7 @@
\DeclareOption{10pt}{\renewcommand\@ptsize{0}}
\DeclareOption{11pt}{\renewcommand\@ptsize{1}}
\DeclareOption{12pt}{\renewcommand\@ptsize{2}}
% Handout-specific options
% ORMC-specific options
\DeclareOption{solutions}{\@solutionstrue}
\DeclareOption{nosolutions}{\@solutionsfalse}
\DeclareOption{multinumbering}{\@singlenumberingfalse}
@ -37,13 +52,13 @@
\DeclareOption{showwarning}{\@nowarningfalse}
\DeclareOption{hidewarning}{\@nowarningtrue}
\DeclareOption{unfinished}{\@unfinishedtrue}
\DeclareOption*{\ClassWarning{handout}{\CurrentOption ignored}}
\DeclareOption*{\ClassWarning{ormc_handout}{\CurrentOption ignored}}
\@unfinishedfalse
\ExecuteOptions{
10pt,
solutions,
singlenumbering,
multinumbering,
pagenumber,
longwarning,
yeswarning
@ -629,12 +644,12 @@
% Keep track of the current background color.
% Useful for transparent tikz drawings.
\def\bgcolor{white}
\def\ORMCbgcolor{white}
% Make a box environment.
% These can safely be nested.
% Args: title, back color, frame color.
\newenvironment{hobox}[3]{
\newenvironment{ORMCbox}[3]{
% \linehack draws a line across a tcolorbox.
% tcolorbox only supports two sections, but
% this hack allows us to have more.
@ -653,7 +668,7 @@
% Keep track of the current background color.
% Useful for transparent tikz drawings.
\def\bgcolor{#2}
\def\ORMCbgcolor{#2}
\begin{tcolorbox}[
enhanced,
@ -675,21 +690,21 @@
}
\newenvironment{examplesolution}{
\begin{hobox}{Example Solution}{black!10!white}{black!65!white}
\begin{ORMCbox}{Example Solution}{black!10!white}{black!65!white}
} {
\end{hobox}
\end{ORMCbox}
}
\if@solutions
\newenvironment{solution}{
\begin{hobox}{Solution}{ored!10!white}{ored}
\begin{ORMCbox}{Solution}{ored!10!white}{ored}
} {
\end{hobox}
\end{ORMCbox}
}
\newenvironment{instructornote}{
\begin{hobox}{Note for Instructors}{ocyan!10!white}{ocyan}
\begin{ORMCbox}{Note for Instructors}{ocyan!10!white}{ocyan}
} {
\end{hobox}
\end{ORMCbox}
}
\else
\excludecomment{solution}

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lib/typst/local/handout/0.1.0/handout.typ Executable file
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@ -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
}

98
lib/typst/local/handout/0.1.0/header.typ
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@ -1,98 +0,0 @@
#import "misc.typ": ored
#let solution_warning() = {
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. \
Recompile without solutions before distributing.
],
),
)
),
),
)
}
#let short_solution_warning() = {
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()
),
),
)
}
#let make_header(
title,
subtitle: none,
by: none,
top_left: "",
top_right: "",
) = {
let date = datetime
.today()
.display("[month repr:long] [day padding:none], [year]")
if (by != none) {
by = text(size: 10pt, [Prepared by #by on #date])
}
let sub = ()
if (by != none) {
sub.push(by)
}
if (subtitle != none) {
sub.push(subtitle)
}
// Main title
align(
center,
block(
width: 60%,
height: auto,
breakable: false,
align(
center,
stack(
spacing: 7pt,
// Top
text(size: 10pt, top_left) + h(1fr) + text(size: 10pt, top_right),
line(length: 100%, stroke: 0.2mm),
// Title
text(size: 20pt, title),
// Subtitle
..sub,
line(length: 100%, stroke: 0.2mm),
),
),
),
)
}

137
lib/typst/local/handout/0.1.0/lib.typ
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@ -1,137 +0,0 @@
/// Typst handout library, used for all documents in this repository.
// Re-exports
// All functions that maybe used by client code are listed here
#import "misc.typ": *
#import "object.typ": problem, definition, theorem, example, remark, generic
#import "solution.typ": (
if_solutions,
if_no_solutions,
if_solutions_else,
solution,
instructornote,
)
/// Main handout wrapper.
/// Use this as follows:
///
/// ```
/// #show: handout.with(
/// group: "Advanced 2",
/// title: [Handout Title],
/// by: "author",
/// subtitle: "optional",
/// )
///
/// <rest of document>
/// ```
#let handout(
doc,
group: none,
title: none,
by: none,
subtitle: none,
short_warning: false,
) = {
set page(
margin: 20mm,
width: 8.5in,
height: 11in,
footer: align(
center,
context counter(page).display(),
),
footer-descent: 5mm,
)
//
// Text style
set text(font: "New Computer Modern")
set par(
leading: 0.5em,
spacing: 0.5em,
first-line-indent: 0mm,
hanging-indent: 0mm,
justify: true,
)
//
// List style
set list(
indent: 4mm,
body-indent: 1.5mm,
// Manually set spacing,
// `tight` has no effect.
spacing: 2mm,
)
set enum(
indent: 4mm,
body-indent: 1.5mm,
spacing: 2mm,
)
//
// Heading style
set heading(numbering: (..nums) => nums.pos().at(0))
show heading.where(level: 1): it => {
set align(center)
set text(weight: "bold")
block[
Section #counter(heading).display(): #text(it.body)
]
}
//
// Hack for custom references
show ref: it => {
import "object.typ": ref_obj
let x = ref_obj(it) // Custom impl for object references
if (x != none) { return x }
return it // Use default `ref` implementation otherwise
}
//
// Begin content
//
// Make handout title
{
import "header.typ": make_header, solution_warning, short_solution_warning
import "solution.typ": solutions_state, reset_solutions
reset_solutions()
let url = link(
"https://betalupi.com/handouts",
`betalupi.com/handouts`,
)
make_header(
title,
subtitle: subtitle,
by: by,
top_left: group,
top_right: url,
)
context {
if solutions_state.get() {
if short_warning {
short_solution_warning()
} else {
solution_warning()
}
}
}
}
// Include rest of document
doc
}

49
lib/typst/local/handout/0.1.0/misc.typ
View File

@ -1,49 +0,0 @@
/// Miscellaneous utilities
#let ored = rgb("D62121")
#let oorange = rgb("#ffaa3b")
#let ogrape = rgb("9C36B5")
#let ocyan = rgb("2288BF")
#let oteal = rgb("12B886")
#let ogreen = rgb("37B26D")
#let oblue = rgb("1C7ED6")
#let note(content, type: none) = {
set text(fill: rgb(100, 100, 100))
if type != none {
text(style: "oblique", [#type: ])
}
text(content)
}
#let hint = note.with(type: "Hint")
#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),
),
),
)
}

108
lib/typst/local/handout/0.1.0/object.typ
View File

@ -1,108 +0,0 @@
/// This module defines all handout "objects"
/// (problems, theorems, definitions, etc)
/// Core render code for all objects (problems, theorems, etc)
/// This should never be used directly by client code.
///
/// Args:
/// - kind: the kind of object to make ("Problem", "Definition", etc)
/// - label_name: a string. If provided, generate metadata for this object
/// under the given label. Labels must be unique.
/// This label can then be used to reference this object.
///
/// For example:
/// ```
/// #problem(label: "problem1")
/// This is @problem1
/// ```
#let _obj_base(kind, ..args, label_name: none) = {
counter("obj").step()
let n = context counter("obj").get().first()
// The complete title text of this object,
// like "Problem 5:" or "Theorem: "
let obj_content = if args.pos().len() == 0 {
[#kind #n:]
} else {
[#kind #n: #args.pos().at(0)]
}
// Render the object
block(
above: 8mm,
below: 2mm,
text(weight: "bold", obj_content),
)
// Generate labeled metadata for this object.
//
// This can be viewed directly with `#context query(<label>).first().value`,
// Or referenced with `@label` (we define a custom renderer for this metadata later)
if label_name != none {
let meta = (
"obj_meta_ref_kind": kind,
// "obj_content": obj_content,
"label": label(label_name),
"counter": counter("obj"),
)
[ #metadata(meta) #label(label_name) ]
}
}
// `ref` implementation for object meta-references.
// Returns `none` if `it` is not object metadata.
#let ref_obj(it) = {
let magic_key = "obj_meta_ref_kind"
if not (
it.element != none
and it.element.has("value")
and type(it.element.value) == "dictionary"
and it.element.value.keys().contains(magic_key)
) {
// This label is not attached to object metadata
return none
}
let v = it.element.value
let obj_type = v.at(magic_key)
// The value of this object's counter at its label
let obj_count = v.counter.at(v.label).first()
// Produces text like "Problem 2",
// which takes you to the referenced object when clicked.
return link(v.label, [#obj_type #obj_count])
}
/// Factory function for objects.
/// Provided for convenience, lets us define objects in one line.
#let _mkobj(kind) = {
let out(..args, label: none) = _obj_base(
kind,
..args,
label_name: label,
)
return out
}
//
// MARK: export
//
// Functions for client code are defined below
#let problem = _mkobj("Problem")
#let definition = _mkobj("Definition")
#let theorem = _mkobj("Theorem")
#let example = _mkobj("Example")
#let remark = _mkobj("Remark")
#let generic(obj_content) = {
block(
above: 8mm,
below: 2mm,
text(weight: "bold", obj_content),
)
}

117
lib/typst/local/handout/0.1.0/solution.typ
View File

@ -1,117 +0,0 @@
#import "misc.typ": ored, oblue
/// If false, hide instructor info.
/// If true, show it.
///
/// Solutions are shown by default. This behavior
/// is less surprising than hiding content by default.
#let solutions_state = state("solutions_state", true)
/// Force solutions to be hidden after this point.
///
/// This function produces content that must be
/// included in the document. If it is not included,
/// this function will have no effect.
#let hide_solutions() = {
return solutions_state.update(x => false)
}
/// Force solutions to be shown after this point.
///
/// This function produces content that must be
/// included in the document. If it is not included,
/// this function will have no effect.
#let show_solutions() = {
return solutions_state.update(x => true)
}
/// Reset the solution flag to its default value.
/// This value is determined by compile flags:
/// Compile with the following command to hide solutions:
/// `typst compile main.typ --input show_solutions=false`
///
/// Solutions are shown by default.
///
/// This function produces content that must be
/// included in the document. If it is not included,
/// this function will have no effect.
#let reset_solutions() = {
if "show_solutions" in sys.inputs {
if (
sys.inputs.show_solutions == "false" or sys.inputs.show_solutions == "no"
) {
return solutions_state.update(x => false)
}
}
}
#let if_solutions(content) = context {
if solutions_state.get() { content }
}
#let if_no_solutions(content) = context {
if not solutions_state.get() { content }
}
#let if_solutions_else(if_yes, if_no) = context {
if solutions_state.get() { if_yes } else { if_no }
}
#let solution(content) = {
if_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 instructornote(content) = {
if_solutions(
align(
center,
stack(
block(
width: 100%,
breakable: false,
fill: oblue,
stroke: oblue + 2pt,
inset: 1.5mm,
align(left, text(fill: white, weight: "bold", [Instructor note:])),
),
block(
width: 100%,
height: auto,
breakable: false,
fill: oblue.lighten(80%).desaturate(10%),
stroke: oblue + 2pt,
inset: 3mm,
align(left, content),
),
),
),
)
}

12
lib/typst/local/handout/0.1.0/typst.toml
View File

@ -1,12 +1,6 @@
[package]
name = "handout"
description = "A library for math circle handouts"
version = "0.1.0"
entrypoint = "lib.typ"
homepage = "https://betalupi.com/handouts"
repository = "https://git.betalupi.com/Mark/handouts"
authors = ["Mark <mark@betalupi.com>"]
license = "GPL-3.0-only "
disciplines = ["education", "mathematics"]
categories = ["layout", "components"]
entrypoint = "handout.typ"
authors = []
license = "GPL"

2
src/Advanced/Compression/main.tex
View File

@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\input{tikzset.tex}

4
src/Advanced/Compression/tikzset.tex
View File

@ -24,7 +24,7 @@
label/.style = {
rectangle,
% For automatic red background in solutions
fill = \bgcolor,
fill = \ORMCbgcolor,
draw = none,
rounded corners = 0mm
},
@ -32,7 +32,7 @@
% Nodes
edg/.style = {
midway,
fill = \bgcolor,
fill = \ORMCbgcolor,
text = gray
},
int/.style = {},

2
src/Advanced/Continued Fractions/main.tex
View File

@ -5,7 +5,7 @@
shortwarning,
singlenumbering,
unfinished
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{multicol}

2
src/Advanced/Cryptography/main.tex
View File

@ -4,7 +4,7 @@
solutions,
singlenumbering,
shortwarning
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{multicol}

2
src/Advanced/DFAs/main.tex
View File

@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\input{tikzset.tex}

2
src/Advanced/DFAs/tikzset.tex
View File

@ -24,7 +24,7 @@
label/.style = {
rectangle,
% For automatic red background in solutions
fill = \bgcolor,
fill = \ORMCbgcolor,
draw = none,
rounded corners = 0mm
},

2
src/Advanced/De Bruijn/main.tex
View File

@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\input{tikzset.tex}

2
src/Advanced/De Bruijn/tikzset.tex
View File

@ -24,7 +24,7 @@
label/.style = {
rectangle,
% For automatic red background in solutions
fill = \bgcolor,
fill = \ORMCbgcolor,
draw = none,
rounded corners = 0mm
},

2
src/Advanced/Definable Sets/main.tex
View File

@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

7
src/Advanced/Error-Correcting Codes/main.tex
View File

@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
@ -16,11 +16,6 @@
\title{Error-Correcting Codes}
\subtitle{Prepared by Mark on \today}
% TODO:
% ISBN section is tedious, could use some work.
% Check problem 5
\begin{document}
\maketitle

35
src/Advanced/Error-Correcting Codes/parts/00 detection.tex
View File

@ -1,8 +1,6 @@
\section{Error Detection}
An ISBN\footnote{International Standard Book Number} is a unique identifier publishers assign to their books. \par
It comes in two forms: ISBN-10 and ISBN-13. Naturally, ISBN-10s have ten digits, and ISBN-13s have thirteen.
The final digit in both versions is a \textit{check digit}.
An ISBN\footnote{International Standard Book Number} is a unique numeric book identifier. It comes in two forms: ISBN-10 and ISBN-13. Naturally, ISBN-10s have ten digits, and ISBN-13s have thirteen. The final digit in both versions is a \textit{check digit}.
\vspace{3mm}
@ -17,25 +15,23 @@ If $n_{10}$ is equal to 10, it is written as \texttt{X}.
\problem{}
Only one of the following ISBNs is valid. Which one is it? \par
\note[Note]{Dashes are meaningless.}
Only one of the following ISBNs is valid. Which one is it?
\begin{itemize}
\item \texttt{0-134-54896-2}
\item \texttt{0-895-77258-2} % oliver twist
\item \texttt{0-895-77258-2}
\end{itemize}
\begin{solution}
The second is valid.
The first has an inconsistent check digit.
\end{solution}
\vfill
\pagebreak
\problem{}
Take a valid ISBN-10 and change one digit. \par
Is it possible that you get another valid ISBN-10? \par
Provide an example or a proof.
Take a valid ISBN-10 and change one digit. Is it possible that you get another valid ISBN-10? \par
Provide a proof.
\begin{solution}
Let $S$ be the sum $10n_1 + 9n_2 + ... + 2n_9 + n_{10}$, before any digits are changed.
@ -54,8 +50,9 @@ Provide an example or a proof.
\vfill
\problem{}
Take a valid ISBN-10 and swap two adjacent digits. This is called a \textit{transposition error}. \par
When will the result be a valid ISBN-10?
Take a valid ISBN-10 and swap two adjacent digits. When will the result be a valid ISBN-10? \par
This is called a \textit{transposition error}.
\begin{solution}
Let $n_1n_2...n_{10}$ be a valid ISBN-10. \par
@ -70,20 +67,18 @@ When will the result be a valid ISBN-10?
\vfill
\pagebreak
\definition{}
ISBN-13 error checking is slightly different. \par
Given a partial ISBN-13 with digits $n_1 n_2 n_3 ... n_{12}$, the final digit is given by
\problem{}
ISBN-13 error checking is slightly different. Given a partial ISBN-13 $n_1 n_2 n_3 ... n_{12}$, the final digit is given by
$$
n_{13} = \Biggr[ \sum_{i=1}^{12} n_i \times (2 + (-1)^i) \Biggl] \text{ mod } 10
$$
\problem{}
What is the last digit of the following ISBN-13? \par
\texttt{978-030-7292-06*} % foundation
\texttt{978-0-380-97726-?}
\begin{solution}
The final digit is 3.
The final digit is 0.
\end{solution}
\vfill
@ -131,8 +126,8 @@ Take a valid ISBN-13 and swap two adjacent digits. When will the result be a val
\vfill
\problem{}<isbnnocorrect>
\texttt{978-008-2066-466} was a valid ISBN until I changed a single digit. \par
\problem{}<isbn-nocorrect>
\texttt{978-0-08-2066-46-6} was a valid ISBN until I changed a single digit. \par
Can you find the digit I changed? Can you recover the original ISBN?
\begin{solution}

8
src/Advanced/Error-Correcting Codes/parts/01 correction.tex
View File

@ -1,6 +1,6 @@
\section{Error Correction}
As we saw in \ref{isbnnocorrect}, the ISBN check-digit scheme does not allow us to correct errors. \par
As we saw in \ref{isbn-nocorrect}, the ISBN check-digit scheme does not allow us to correct errors. \par
QR codes feature a system that does. \par
\vspace{1mm}
@ -8,8 +8,7 @@ QR codes feature a system that does. \par
Odds are, you've seen a QR code with an image in the center. Such codes aren't \say{special}---they're simply missing their central pixels. The error-correcting algorithm in the QR specification allows us to read the code despite this damage.
\begin{figure}[h]
\centering
%\href{https://youtube.com/watch?v=dQw4w9WgXcQ}{\includegraphics[width = 3cm]{qr-rick}}
\href{https://betalupi.com}{\includegraphics[width = 3cm]{qr-betalupi}}
\href{https://youtube.com/watch?v=dQw4w9WgXcQ}{\includegraphics[width = 3cm]{qr}}
\end{figure}
\definition{Repeating codes}
@ -48,8 +47,7 @@ What is the efficiency of a $k$-repeat code?
\vfill
Repeat codes are not efficient. We need to inflate our message by
\textit{three times} if we want to correct even a single error. We need a better system.
As you just saw, repeat codes are not a good solution. You need many extra bits for even a small amount of redundancy. We need a better system.
\pagebreak

59
src/Advanced/Error-Correcting Codes/parts/02 hamming.tex
View File

@ -1,6 +1,6 @@
\section{Hamming Codes}
Say we have an $n$-bit message, for example \texttt{1011 0101 1101 1001}. \par
Say we have a 16-bit message, for example \texttt{1011 0101 1101 1001}. \par
We will number its bits in binary, from left to right:
@ -62,14 +62,13 @@ We will number its bits in binary, from left to right:
\problem{}
If we number the bits of a 16-bit message as described above, \par
how many message bits have an index with a one as the last digit? \par
In this 16-bit message, how many message bits have an index with a one as the last digit? \par
(i.e, an index that looks like \texttt{***1})
\vspace{2cm}
\problem{}
Now consider a 32-bit message. \par
Say we number the bits in a 32-bit message as above. \par
How many message bits have an index with a one as the $n^\text{th}$ digit? \par
\vspace{2cm}
@ -77,10 +76,7 @@ How many message bits have an index with a one as the $n^\text{th}$ digit? \par
Now, let's come up with a way to detect errors in our 16-bit message.
To do this, we'll replace a few data bits with parity bits.
This will reduce the amount of information we can send,
but will allow the receiver to detect errors in the received message.
We now want a way to detect errors in our 16-bit message. To do this, we'll replace a few data bits with parity bits. This will reduce the amount of information we can send, but will also improve our error-detection capabilities.
\vspace{1mm}
@ -178,8 +174,7 @@ Can this coding scheme correct a single-bit error?
\vfill
\pagebreak
We'll now add four more parity bits, in positions \texttt{0001}, \texttt{0010}, \texttt{0100}, and \texttt{1000}: \par
This error-detection scheme is called the \textit{Hamming code}.
We'll now add four more parity bits, in positions \texttt{0001}, \texttt{0010}, \texttt{0100}, and \texttt{1000}:
\begin{center}
\begin{tikzpicture}[scale = 1.25]
@ -251,7 +246,7 @@ Bits \texttt{0100} and \texttt{1000} work in the same way. \par
When counting bits in binary numbers, go from right to left.}
\problem{}
Which message bits does each parity bit \say{cover}? \par
Which message bits does each parity bit cover? \par
In other words, which message bits affect the value of each parity bit? \par
\vspace{1mm}
@ -363,7 +358,7 @@ Analyze this coding scheme.
\vfill
\problem{}
Each of the following messages has either one, two, or no errors. \par
Each of the following messages has either 0, 1, or two errors. \par
Find the errors and correct them if possible. \par
\hint{Bit \texttt{0000} should tell you how many errors you have.}
@ -575,46 +570,8 @@ If we know which parity bits are inconsistent, how can we find where the error i
\vfill
\problem{}<generalize-hamming>
Generalize this system for messages of 4, 64, or 256 bits. \par
\begin{itemize}
\item How does the resilience of this scheme change if we use a larger message size?
\item How does the efficiency of this scheme change if we send larger messages?
\end{itemize}
\vfill
\pagebreak
\definition{}
A \textit{deletion} error occurs when one bit of the message is deleted. \par
Likewise, an \textit{insertion} error consists of a random inserted bit. \par
\definition{}
A \textit{message stream} is an infinite string of binary digits.
\problem{}
Show that Hamming codes do not reliably detect bit deletions: \par
\hint{
Create a 17-bit message whose first 16 bits are a valid Hamming-coded message, \par
and which is still valid when a bit (chosen by you; not the $17^\text{th}$) is deleted.
}
Can you generalize this system for messages of 4, 64, or 256 bits?
\vfill
\problem{}
Convince yourself that Hamming codes cannot correct insertions. \par
Then, create a 16-bit message that:
\begin{itemize}
\item is itself a valid Hamming code, and
\item incorrectly "corrects" a single bit error when it encounters an insertion error. \par
You may choose where the insertion occurs.
\end{itemize}
\vfill
As we have seen, Hamming codes effectively handle substitutions, but cannot reliably
detect (or correct) insertions and deletions. Correcting those errors is a bit more difficult:
if the number of bits we receive is variable, how can we split a stream into a series of messages? \par
\note{This is a rhetorical question, which we'll discuss another day.}
\pagebreak

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2
src/Advanced/Esoteric Languages/main.tex
View File

@ -3,7 +3,7 @@
\documentclass[
nosolutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

2
src/Advanced/Estimathon/main.tex
View File

@ -4,7 +4,7 @@
solutions,
singlenumbering,
nopagenumber
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\uptitlel{Advanced 2}

31
src/Advanced/Fast Inverse Root/main.typ
View File

@ -1,31 +0,0 @@
#import "@local/handout:0.1.0": *
// Bonus:
// - Floats vs fixed point
// - Float density
// - Find non-floatable rational numbers
// - What if we use `n`-bit floats?
#show: doc => handout(
doc,
group: "Advanced 2",
title: [Fast Inverse Square Root],
by: "Mark",
)
#include "parts/00 intro.typ"
#pagebreak()
#include "parts/01 int.typ"
#pagebreak()
#include "parts/02 float.typ"
#pagebreak()
#include "parts/03 approx.typ"
#pagebreak()
#include "parts/04 quake.typ"
#pagebreak()
#include "parts/05 bonus.typ"

7
src/Advanced/Fast Inverse Root/meta.toml
View File

@ -1,7 +0,0 @@
[metadata]
title = "Fast Inverse Square Root"
[publish]
handout = true
solutions = true

45
src/Advanced/Fast Inverse Root/parts/00 intro.typ
View File

@ -1,45 +0,0 @@
#import "@local/handout:0.1.0": *
= Introduction
In 2005, ID Software published the source code of _Quake III Arena_, a popular game released in 1999. \
This caused quite a stir: ID Software was responsible for many games popular among old-school engineers (most notably _Doom_, which has a place in programmer humor even today).
#v(2mm)
Naturally, this community immediately began dissecting _Quake_'s source. \
One particularly interesting function is reproduced below, with original comments: \
#v(3mm)
```c
float Q_rsqrt( float number ) {
long i;
float x2, y;
const float threehalfs = 1.5F;
x2 = number * 0.5F;
y = number;
i = * ( long * ) &y; // evil floating point bit level hacking
i = 0x5f3759df - ( i >> 1 ); // [redacted]
y = * ( float * ) &i;
y = y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration
// y = y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed
return y;
}
```
#v(3mm)
This code defines a function `Q_sqrt`, which was used as a fast approximation of the inverse square root in graphics routines. (in other words, `Q_sqrt` efficiently approximates $1 div sqrt(x)$)
#v(3mm)
The key word here is "fast": _Quake_ ran on very limited hardware, and traditional approximation techniques (like Taylor series)#footnote[Taylor series aren't used today, and for the same reason. There are better ways.] were too computationally expensive to be viable.
#v(3mm)
Our goal today is to understand how `Q_sqrt` works. \
To do that, we'll first need to understand how computers represent numbers. \
We'll start with simple binary integers---turn the page.

102
src/Advanced/Fast Inverse Root/parts/01 int.typ
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@ -1,102 +0,0 @@
#import "@local/handout:0.1.0": *
= Integers
#definition()
A _bit string_ is a string of binary digits. \
In this handout, we'll denote bit strings with the prefix `0b`. \
#note[This prefix is only notation---it is _not_ part of the string itself.] \
For example, $1001$ is the number "one thousand and one," while $#text([`0b1001`])$ is the string of bits "1 0 0 1".
#v(2mm)
We will separate long bit strings with underscores for readability. \
Underscores have no meaning: $#text([`0b1111_0000`]) = #text([`0b11110000`])$.
#problem()
What is the value of the following bit strings, if we interpret them as integers in base 2?
- `0b0001_1010`
- `0b0110_0001`
#solution([
- $#text([`0b0001_1010`]) = 2 + 8 + 16 = 26$
- $#text([`0b0110_0001`]) = 1 + 32 + 64 = 95$
])
#v(1fr)
#definition()
We can interpret a bit string in any number of ways. \
One such interpretation is the _unsigned integer_, or `uint` for short. \
`uint`s allow us to represent positive (hence "unsigned") integers using 32-bit strings.
#v(2mm)
The value of a `uint` is simply its value as a binary number:
- $#text([`0b00000000_00000000_00000000_00000000`]) = 0$
- $#text([`0b00000000_00000000_00000000_00000011`]) = 3$
- $#text([`0b00000000_00000000_00000000_00100000`]) = 32$
- $#text([`0b00000000_00000000_00000000_10000010`]) = 130$
#problem()
What is the largest number we can represent with a 32-bit `uint`?
#solution([
$#text([`0b11111111_11111111_11111111_11111111`]) = 2^(32)-1$
])
#v(1fr)
#pagebreak()
#problem()
Find the value of each of the following 32-bit unsigned integers:
- `0b00000000_00000000_00000101_00111001`
- `0b00000000_00000000_00000001_00101100`
- `0b00000000_00000000_00000100_10110000`
#hint([The third conversion is easy---look carefully at the second.])
#instructornote[
Consider making a list of the powers of two $>= 1024$ on the board.
]
#solution([
- $#text([`0b00000000_00000000_00000101_00111001`]) = 1337$
- $#text([`0b00000000_00000000_00000001_00101100`]) = 300$
- $#text([`0b00000000_00000000_00000010_01011000`]) = 1200$
Notice that the third int is the second shifted left twice (i.e, multiplied by 4)
])
#v(1fr)
#definition()
In general, fast division of `uints` is difficult#footnote([One may use repeated subtraction, but this isn't efficient.]). \
Division by powers of two, however, is incredibly easy: \
To divide by two, all we need to do is shift the bits of our integer right.
#v(2mm)
For example, consider $#text[`0b0000_0110`] = 6$. \
If we insert a zero at the left end of this string and delete the zero at the right \
(thus "shifting" each bit right), we get `0b0000_0011`, which is 3. \
#v(2mm)
Of course, we lose the remainder when we right-shift an odd number: \
$9$ shifted right is $4$, since `0b0000_1001` shifted right is `0b0000_0100`.
#problem()
Right shifts are denoted by the `>>` symbol: \
$#text[`00110`] #text[`>>`] n$ means "shift `0b0110` right $n$ times." \
Find the value of the following:
- $12 #text[`>>`] 1$
- $27 #text[`>>`] 3$
- $16 #text[`>>`] 8$
#note[Naturally, you'll have to convert these integers to binary first.]
#solution[
- $12 #text[`>>`] 1 = 6$
- $27 #text[`>>`] 3 = 3$
- $16 #text[`>>`] 8 = 0$
]
#v(1fr)

211
src/Advanced/Fast Inverse Root/parts/02 float.typ
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@ -1,211 +0,0 @@
#import "@local/handout:0.1.0": *
#import "@preview/cetz:0.3.1"
= Floats
#definition()
_Binary decimals_#footnote([Note that "binary decimal" is a misnomer---"deci" means "ten"!]) are very similar to base-10 decimals.\
In base 10, we interpret place value as follows:
- $0.1 = 10^(-1)$
- $0.03 = 3 times 10^(-2)$
- $0.0208 = 2 times 10^(-2) + 8 times 10^(-4)$
#v(5mm)
We can do the same in base 2:
- $#text([`0.1`]) = 2^(-1) = 0.5$
- $#text([`0.011`]) = 2^(-2) + 2^(-3) = 0.375$
- $#text([`101.01`]) = 5.125$
#v(5mm)
#problem()
Rewrite the following binary decimals in base 10: \
#note([You may leave your answer as a fraction.])
- `1011.101`
- `110.1101`
#v(1fr)
#pagebreak()
#definition(label: "floatbits")
Another way we can interpret a bit string is as a _signed floating-point decimal_, or a `float` for short. \
Floats represent a subset of the real numbers, and are interpreted as follows: \
#note([The following only applies to floats that consist of 32 bits. We won't encounter any others today.])
#align(
center,
box(
inset: 2mm,
cetz.canvas({
import cetz.draw: *
let chars = (
`0`,
`b`,
`0`,
`_`,
`0`,
`0`,
`0`,
`0`,
`0`,
`0`,
`0`,
`0`,
`_`,
`0`,
`0`,
`0`,
`0`,
`0`,
`0`,
`0`,
`_`,
`0`,
`0`,
`0`,
`0`,
`0`,
`0`,
`0`,
`0`,
`_`,
`0`,
`0`,
`0`,
`0`,
`0`,
`0`,
`0`,
`0`,
)
let x = 0
for c in chars {
content((x, 0), c)
x += 0.25
}
let y = -0.4
line((0.3, y), (0.65, y))
content((0.45, y - 0.2), [s])
line((0.85, y), (2.9, y))
content((1.9, y - 0.2), [exponent])
line((3.10, y), (9.4, y))
content((6.3, y - 0.2), [fraction])
}),
),
)
- The first bit denotes the sign of the float's value
We'll label it $s$. \
If $s = #text([`1`])$, this float is negative; if $s = #text([`0`])$, it is positive.
- The next eight bits represent the _exponent_ of this float.
#note([(we'll see what that means soon)]) \
We'll call the value of this eight-bit binary integer $E$. \
Naturally, $0 <= E <= 255$ #note([(since $E$ consist of eight bits)])
- The remaining 23 bits represent the _fraction_ of this float. \
They are interpreted as the fractional part of a binary decimal. \
For example, the bits `0b10100000_00000000_00000000` represent $0.5 + 0.125 = 0.625$. \
We'll call the value of these bits as a binary integer $F$. \
Their value as a binary decimal is then $F div 2^23$. #note([(convince yourself of this)])
#problem(label: "floata")
Consider `0b01000001_10101000_00000000_00000000`. \
#hint([The underscores here do _not_ match those in @floatbits])
#v(2mm)
Find the $s$, $E$, and $F$ we get if we interpret this bit string as a `float`. \
#note([Leave $F$ as a sum of powers of two.])
#solution([
$s = 0$ \
$E = 131$ \
$F = 2^21+2^19$
])
#v(1fr)
#definition(label: "floatdef")
The final value of a float with sign $s$, exponent $E$, and fraction $F$ is
$
(-1)^s times 2^(E - 127) times (1 + F / (2^(23)))
$
Notice that this is very similar to base-10 scientific notation, which is written as
$
(-1)^s times 10^(e) times (f)
$
#note[
We subtract 127 from $E$ so we can represent positive and negative numbers. \
$E$ is an eight bit binary integer, so $0 <= E <= 255$ and thus $-127 <= (E - 127) <= 127$.
]
#problem()
Consider `0b01000001_10101000_00000000_00000000`. \
This is the same bit string we used in @floata. \
#v(2mm)
What value do we get if we interpret this bit string as a float? \
#hint([$21 div 16 = 1.3125$])
#solution([
This is 21:
$
2^4 times (1 + (2^(21) + 2^(19)) / (2^(23)))
= 2^(4) times (1 + 2^(-2) + 2^(-4))
= 16 + 4 + 1
= 21
$
])
#v(1fr)
#pagebreak()
#problem()
Encode $12.5$ as a float. \
#hint([$12.5 div 8 = 1.5625$])
#solution([
$
12.5
= 8 times 1.5625
= 2^(3) times (1 + (0.5 + 0.0625))
= 2^(130) times (1 + (2^(22) + 2^(19)) / (2^(23)))
$
which is `0b01000001_01001000_00000000_00000000`. \
])
#v(1fr)
#definition()
Say we have a bit string $x$. \
We'll let $x_f$ denote the value we get if we interpret $x$ as a float, \
and we'll let $x_i$ denote the value we get if we interpret $x$ an integer.
#problem()
Let $x = #text[`0b01000001_01001000_00000000_00000000`]$. \
What are $x_f$ and $x_i$? #note([As always, you may leave big numbers as powers of two.])
#solution([
$x_f = 12.5$
#v(2mm)
$x_i = 2^30 + 2^24 + 2^22 + 2^19 = 11,095,237,632$
])
#v(1fr)

173
src/Advanced/Fast Inverse Root/parts/03 approx.typ
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@ -1,173 +0,0 @@
#import "@local/handout:0.1.0": *
#import "@preview/cetz:0.3.1"
#import "@preview/cetz-plot:0.1.0": plot, chart
= Integers and Floats
#generic("Observation:")
If $x$ is smaller than 1, $log_2(1 + x)$ is approximately equal to $x$. \
Note that this equality is exact for $x = 0$ and $x = 1$, since $log_2(1) = 0$ and $log_2(2) = 1$.
#v(5mm)
We'll add the _correction term_ $epsilon$ to our approximation: $log_2(1 + a) approx a + epsilon$. \
This allows us to improve the average error of our linear approximation:
#table(
stroke: none,
align: center,
columns: (1fr, 1fr),
inset: 5mm,
[$log_2(1+x)$ and $x + 0$]
+ cetz.canvas({
import cetz.draw: *
let f1(x) = calc.log(calc.abs(x + 1), base: 2)
let f2(x) = x
// Set-up a thin axis style
set-style(axes: (stroke: .5pt, tick: (stroke: .5pt)))
plot.plot(
size: (7, 7),
x-tick-step: 0.2,
y-tick-step: 0.2,
y-min: 0,
y-max: 1,
x-min: 0,
x-max: 1,
legend: none,
axis-style: "scientific-auto",
x-label: none,
y-label: none,
{
let domain = (0, 1)
plot.add(
f1,
domain: domain,
label: $log(1+x)$,
style: (stroke: ogrape),
)
plot.add(
f2,
domain: domain,
label: $x$,
style: (stroke: oblue),
)
},
)
})
+ [
Max error: 0.086 \
Average error: 0.0573
],
[$log_2(1+x)$ and $x + 0.045$]
+ cetz.canvas({
import cetz.draw: *
let f1(x) = calc.log(calc.abs(x + 1), base: 2)
let f2(x) = x + 0.0450466
// Set-up a thin axis style
set-style(axes: (stroke: .5pt, tick: (stroke: .5pt)))
plot.plot(
size: (7, 7),
x-tick-step: 0.2,
y-tick-step: 0.2,
y-min: 0,
y-max: 1,
x-min: 0,
x-max: 1,
legend: none,
axis-style: "scientific-auto",
x-label: none,
y-label: none,
{
let domain = (0, 1)
plot.add(
f1,
domain: domain,
label: $log(1+x)$,
style: (stroke: ogrape),
)
plot.add(
f2,
domain: domain,
label: $x$,
style: (stroke: oblue),
)
},
)
})
+ [
Max error: 0.041 \
Average error: 0.0254
],
)
A suitiable value of $epsilon$ can be found using calculus or with computational trial-and-error. \
We won't bother with this---we'll simply leave the correction term as an opaque constant $epsilon$.
#v(1fr)
#note(
type: "Note",
[
"Average error" above is simply the area of the region between the two graphs:
$
integral_0^1 abs( #v(1mm) log(1+x)_2 - (x+epsilon) #v(1mm))
$
Feel free to ignore this note, it isn't a critical part of this handout.
],
)
#pagebreak()
#problem(label: "convert")
Use the fact that $log_2(1 + a) approx a + epsilon$ to approximate $log_2(x_f)$ in terms of $x_i$. \
Namely, show that
$
log_2(x_f) = (x_i) / (2^23) - 127 + epsilon
$
#note([
In other words, we're finding an expression for $x$ as a float
in terms of $x$ as an int.
])
#solution([
Let $E$ and $F$ be the exponent and float bits of $x_f$. \
We then have:
$
log_2(x_f)
&= log_2 ( 2^(E-127) times (1 + (F) / (2^23)) ) \
&= E - 127 + log_2(1 + F / (2^23)) \
& approx E-127 + F / (2^23) + epsilon \
&= 1 / (2^23)(2^23 E + F) - 127 + epsilon \
&= 1 / (2^23)(x_i) - 127 + epsilon
$
])
#v(1fr)
#problem()
Using basic log rules, rewrite $log_2(1 / sqrt(x))$ in terms of $log_2(x)$.
#solution([
$
log_2(1 / sqrt(x)) = (-1) / (2)log_2(x)
$
])
#v(1fr)

210
src/Advanced/Fast Inverse Root/parts/04 quake.typ
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@ -1,210 +0,0 @@
#import "@local/handout:0.1.0": *
= The Fast Inverse Square Root
A simplified version of the _Quake_ routine we are studying is reproduced below.
#v(2mm)
```c
float Q_rsqrt( float number ) {
long i = * ( long * ) &number;
i = 0x5f3759df - ( i >> 1 );
return * ( float * ) &i;
}
```
#v(2mm)
This code defines a function `Q_rsqrt` that consumes a float `number` and approximates its inverse square root.
If we rewrite this using notation we're familiar with, we get the following:
$
#text[`Q_sqrt`] (n_f) =
6240089 - (n_i div 2)
#h(10mm)
approx 1 / sqrt(n_f)
$
#note[
`0x5f3759df` is $6240089$ in hexadecimal. \
Ask an instructor to explain if you don't know what this means. \
It is a magic number hard-coded into `Q_sqrt`.
]
#v(2mm)
Our goal in this section is to understand why this works:
- How does Quake approximate $1 / sqrt(x)$ by simply subtracting and dividing by two?
- What's special about $6240089$?
#v(1fr)
#remark()
For those that are interested, here are the details of the "code-to-math" translation:
- "`long i = * (long *) &number`" is C magic that tells the compiler \
to set `i` to the `uint` value of the bits of `number`. \
#note[
"long" refers to a "long integer", which has 32 bits. \
Normal `int`s have 16 bits, `short int`s have 8.
] \
In other words, `number` is $n_f$ and `i` is $n_i$.
#v(2mm)
- Notice the right-shift in the second line of the function. \
We translated `(i >> 1)` into $(n_i div 2)$.
#v(2mm)
- "`return * (float *) &i`" is again C magic. \
Much like before, it tells us to return the value of the bits of `i` as a float.
#pagebreak()
#generic("Setup:")
We are now ready to show that $#text[`Q_sqrt`] (x)$ effectively approximates $1/sqrt(x)$. \
For convenience, let's call the bit string of the inverse square root $r$. \
In other words,
$
r_f := 1 / (sqrt(n_f))
$
This is the value we want to approximate. \
#problem(label: "finala")
Find an approximation for $log_2(r_f)$ in terms of $n_i$ and $epsilon$ \
#note[Remember, $epsilon$ is the correction constant in our approximation of $log_2(1 + x)$.]
#solution[
$
log_2(r_f)
= log_2(1 / sqrt(n_f))
= (-1) / 2 log_2(n_f)
approx (-1) / 2 ( (n_i) / (2^23) + epsilon - 127 )
$
]
#v(1fr)
#problem(label: "finalb")
Let's call the "magic number" in the code above $kappa$, so that
$
#text[`Q_sqrt`] (n_f) = kappa - (n_i div 2)
$
Use @convert and @finala to show that $#text[`Q_sqrt`] (n_f) approx r_i$ \
#note(type: "Note")[
If we know $r_i$, we know $r_f$. \
We don't even need to convert between the two---the underlying bits are the same!
]
#solution[
From @convert, we know that
$
log_2(r_f) approx (r_i) / (2^23) + epsilon - 127
$
Combining this with the result from @finala, we get:
$
(r_i) / (2^23) + epsilon - 127
&approx (-1) / (2) ( (n_i) / (2^23) + epsilon - 127) \
(r_i) / (2^23)
&approx (-1) / (2) ( (n_i) / (2^23)) + 3 / 2 (127 - epsilon) \
r_i
&approx (-1) / 2 (n_i) + 2^23 3 / 2(127 - epsilon)
= 2^23 3 / 2 (127 - epsilon) - (n_i) / 2
$
#v(2mm)
This is exactly what we need! If we set $kappa$ to $(3 times 2^22) (127-epsilon)$, then
$
r_i approx kappa - (n_i div 2) = #text[`Q_sqrt`] (n_f)
$
]
#v(1fr)
#problem(label: "finalc")
What is the exact value of $kappa$ in terms of $epsilon$? \
#hint[Look at @finalb. We already found it!]
#solution[
This problem makes sure our students see that
$kappa = (3 times 2^22) (127 - epsilon)$. \
See the solution to @finalb.
]
#v(2cm)
#pagebreak()
#remark()
In @finalc we saw that $kappa = (3 times 2^22) (127 - epsilon)$. \
Looking at the code again, we see that $kappa = #text[`0x5f3759df`]$ in _Quake_:
#v(2mm)
```c
float Q_rsqrt( float number ) {
long i = * ( long * ) &number;
i = 0x5f3759df - ( i >> 1 );
return * ( float * ) &i;
}
```
#v(2mm)
Using a calculator and some basic algebra, we can find the $epsilon$ this code uses: \
#note[Remember, #text[`0x5f3759df`] is $6240089$ in hexadecimal.]
$
(3 times 2^22) (127 - epsilon) &= 6240089 \
(127 - epsilon) &= 126.955 \
epsilon &= 0.0450466
$
So, $0.045$ is the $epsilon$ used by Quake. \
Online sources state that this constant was generated by trial-and-error, \
though it is fairly close to the ideal $epsilon$.
#remark()
And now, we're done! \
We've shown that `Q_sqrt(x)` approximates $1/sqrt(x)$ fairly well. \
#v(2mm)
Notably, `Q_sqrt` uses _zero_ divisions or multiplications (`>>` doesn't count). \
This makes it _very_ fast when compared to more traditional approximation techniques (i.e, Taylor series).
#v(2mm)
In the case of _Quake_, this is very important. 3D graphics require thousands of inverse-square-root calculations to render a single frame#footnote[e.g, to generate normal vectors], which is not an easy task for a Playstation running at 300MHz.
#instructornote[
Let $x$ be a bit string. If we assume $x_f$ is positive and $E$ is even, then
$
(x #text[`>>`] 1)_f = 2^((E div 2) - 127) times (1 + (F div 2) / (2^(23)))
$
Notably: a right-shift divides the exponent of $x_f$ by two, \
which is, of course, a square root!
#v(2mm)
This intuition is hand-wavy, though: \
If $E$ is odd, its lowest-order bit becomes the highest-order bit of $F$ when we shift $x$ right. \
Also, a right shift doesn't divide the _entire_ exponent, skipping the $-127$ offset. \
#v(2mm)
Remarkably, this intuition is still somewhat correct. \
The bits align _just so_, and our approximation still works.
#v(8mm)
One can think of the fast inverse root as a "digital slide rule": \
The integer representation of $x_f$ already contains $log_2(x_f)$, offset and scaled. \
By subtracting and dividing in "log space", we effectively invert and root $x_f$!
After all,
$
- 1 / 2 log_2(n_f) = 1 / sqrt(n_f)
$
]

36
src/Advanced/Fast Inverse Root/parts/05 bonus.typ
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@ -1,36 +0,0 @@
#import "@local/handout:0.1.0": *
= Bonus -- More about Floats
#problem()
Convince yourself that all numbers that can be represented as a float are rational.
#problem()
Find a rational number that cannot be represented as a float.
#v(1fr)
#problem()
What is the smallest positive 32-bit float?
#v(1fr)
#problem()
What is the largest positive 32-bit float?
#v(1fr)
#problem()
How many floats are between $-1$ and $1$?
#v(1fr)
#problem()
How many floats are between $1$ and $2$?
#v(1fr)
#problem()
How many floats are between $1$ and $128$?
#v(1fr)

2
src/Advanced/Generating Functions/main.tex
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@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

2
src/Advanced/Geometric Optimization/main.tex
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@ -2,7 +2,7 @@
% use [solutions] flag to show solutions.
\documentclass[
solutions
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\uptitlel{Advanced 2}

2
src/Advanced/Geometry of Masses/main.tex
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@ -4,7 +4,7 @@
solutions,
singlenumbering,
shortwarning
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{tikz}

13
src/Advanced/Gods, Demons, Mortals/main.typ
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@ -1,13 +0,0 @@
#import "@local/handout:0.1.0": *
#show: handout.with(
title: [Gods, Demons, and Mortals],
by: "Mark",
subtitle: [Based on Raymond Smullyan's _To Mock a Mockingbird_.],
short_warning: true,
)
#include "parts/00 warmup.typ"
#pagebreak()
#include "parts/01 gods.typ"

6
src/Advanced/Gods, Demons, Mortals/meta.toml
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@ -1,6 +0,0 @@
[metadata]
title = "Gods, Demons, and Mortals"
[publish]
handout = true
solutions = true

62
src/Advanced/Gods, Demons, Mortals/parts/00 warmup.typ
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@ -1,62 +0,0 @@
#import "@local/handout:0.1.0": *
= Warm-Up
#problem("The Flower Garden")
In a certain garden, each flower was either red, yellow, or blue, and all three colors were represented. A statistician once visited the garden and made the observation that whatever three flowers you picked, at least one of them was bound to be red.
#v(2mm)
A second statistician visited the garden and made the observation that whatever three flowers you picked, at least one was bound to be yellow.
#v(2mm)
Two logic students heard about this and got into an argument. \
The first student said: "It therefore follows that whatever three flowers you pick, at least one is bound to be blue,
doesn't it?" The second student said: "Of course not!"
Which student was right, and why?
#solution[
The first student was right.
#v(2mm)
From the first statistician's report it follows that there cannot be more than one yellow flower, because if there were two yellows, you could pick two yellows and one blue, thus having a group of three flowers that contained no red. This is contrary to the report that every group of three is bound to contain at least one red flower. Therefore there cannot be more than one yellow flower.
#v(2mm)
Similarly, there cannot be more than one blue flower, because if there were two blues, you could pick two blue flowers and one yellow and again have a group of three that contained no red. And so from the first statistician's report it follows that there is at most one yellow flower and one blue.
#v(2mm)
And it follows from the report of the second statistician that there is at most one red flower, for if there were two reds,
you could pick two reds and one blue, thus obtaining a group of three that contained no yellow. It also follows from the
second report that there cannot be more than one blue, although we have already deduced this from the first report.
#v(2mm)
The upshot of all this is that there are only three flowers in the entire garden---one red, one yellow, and one blue! And so it is of course true that whatever three flowers you pick, one of them must be blue.
]
#v(1fr)
#problem("What Question")
There is a question I could ask you that has a definite correct
answer---either yes or no---but it is logically impossible for
you to give the correct answer. You might know what the
correct answer is, but you cannot give it. Anybody other than
you might possibly be able to give the correct answer, but
you cannot!
Can you figure out what question I could have in mind?
#solution[
Suppose I ask you: "Is no your answer to this question?"
If you answer yes, then you are affirming that no is your answer to the question,
which is of course wrong. If you answer
no, then you are denying that no is your answer, although no
was your answer.
]
#v(1fr)

316
src/Advanced/Gods, Demons, Mortals/parts/01 gods.typ
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@ -1,316 +0,0 @@
#import "@local/handout:0.1.0": *
= Gods, Demons, and Mortals
#generic("Setup:")
One night, Inspector Craig had a curious dream.
He had been browsing that day in a library specializing in rare
books on mythology, another of his many interests. His head
was filled with gods and demons, and so his dream was perhaps not so surprising.
#v(2mm)
Time sometimes passes in unusual ways in the course of a
dream. Craig dreamed that he spent nine days in a region in
which dwelled gods, demons, and mortals.
#v(2mm)
The gods, of course, always told the truth, and the demons always lied. As to the mortals, half were knights and half were knaves. As usual, the knights told the truth and the knaves lied.
#problem("The First Day")
Craig dreamed that on the first day he met a dweller of the
region who looked as if he might be a god, though Craig could
not be sure. The dweller evidently guessed Craig's thoughts,
smiled, and made a statement to reassure him. From this statement, Craig knew that he was in the presence of a god.
#v(2mm)
Can you supply such a statement?
#solution[
One statement that works is: "I am not a knight."
#v(2mm)
If the speaker were a knave or a demon, then it would be true that he was not a knight, but knaves and demons don't make true statements.
#v(2mm)
Therefore the speaker was neither a knave nor a
demon, hence he was a knight or a god and his statement was
true.
#v(2mm)
Since it was true, then he really is not a knight; hence he must be a god.
]
#v(1fr)
#problem("The Second Day")
In this episode of the dream, Craig met a terrifying being who had every appearance of being a demon.
"What sort of being are you?" asked Craig, in some alarm.
The being answered, and Craig then realized that he was confronting not a demon, but a knave. \
What could the being have answered?
#solution[
A statement that works is: "I am a demon." Obviously
no demon can claim he is a demon, so the speaker is not a
demon. Therefore his statement was false and since he is not
a demon, he must be a knave.
]
#v(1fr)
#pagebreak()
#problem("The Third Day")
In this episode, Craig met a totally nondescript-looking being
who from appearances could have been anything at all. The
being then made a statement from which Craig could deduce
that he was either a god or a demon, but Craig could not tell
which.
Can you supply such a statement?
#solution[
This is a bit more tricky: A statement that works is: "I
am either a god or a knave." That could be said by a god,
since a god is either a god or a knave; it could also be falsely said by a demon.
#v(2mm)
It couldn't be said by a knight, because a knight would never lie and claim that he is either a god or a knave, and it couldn't be said by a knave, because a knave would never admit to the true fact that he is either a god or
a knave.
#v(2mm)
And so the speaker must be either a god or a demon,
but there is no way to tell which.
]
#v(1fr)
#problem("The Fourth Day")
Craig next met a being who made the following two statements:
- A god once claimed that I am a demon.
- No knight has ever claimed that I am a knave.
What sort of being was he?
#solution[
The speaker's first statement was obviously false, for if
it were true, a god would have once claimed that the speaker
was a demon, which would mean that the speaker really was
a demon, but no one who tells the truth can be a demon. Since
the first statement was false, so was the second statement, since it was made by the same speaker. Therefore a knight did once claim that the speaker was a knave, hence the speaker really
is a knave.
]
#v(1fr)
#problem("The Fifth Day")
A being made the following two statements to Craig:
- I never claim to be a knave.
- I sometimes claim that I am a demon.
What sort of being are we now dealing with?
#solution[
The speaker's second statement was obviously a lie, because no truth-teller would ever say that he sometimes claims
to be a demon.
#v(2mm)
Therefore the first statement was also a lie,
hence the speaker does sometimes claim to be a knave, hence
he must be a demon.
]
#v(1fr)
#pagebreak()
#problem("The Sixth Day")
In this episode, Craig came across two beings, each of whom
made a statement. \
Craig could then infer that at least one of
them must be a god, but he could not tell which one. \
From neither statement alone could Craig have deduced this. \
What statements could the beings have made?
#solution[
Many solutions are possible; here is one.
#v(2mm)
Let us call the two beings A and B. \
Now, suppose A and B make the following two statements:
- A: B is a knight.
- B: A is not a knight.
A is either telling the truth or lying.
#v(2mm)
*Case 1: A is telling the truth.* \
Then B really is a knight,
hence his statement is true, hence A is not a knight, therefore A must be a god, since he is telling the truth.
#v(2mm)
*Case 2: A is lying.* \
Then B is not a knight, since A says he is. Also, since A is lying, then A is certainly not a knight, hence B's statement is true. Therefore B is telling the truth, but is not a knight, hence B is a god.
#v(2mm)
So if Case 1 is true, A is a god; if Case 2 is true, then B is a god. \
There is no way to tell whether A is telling the truth or lying.
]
#v(1fr)
#problem("The Seventh Day")
On the next day, Craig again met two beings each of whom
made a statement. Craig could then infer that one of them was
a knave and the other a demon, though he could not tell which
was which. Again, from neither statement alone could Craig
have inferred this. Can you supply two such statements?
#solution[
Again, let us call the two beings A and B. \
The following statements would work:
- A: Both of us are knaves.
- B: Both of us are demons.
It is obvious that both are lying. Since A is lying, they are
not both knaves. Since B is lying, they are not both demons.
Therefore one is a knave and one is a demon, but there is no
way to tell which one is which.
]
#v(1fr)
#problem("Introducing Thor")
On the eighth day, Craig met a being who had every ap-
pearance of being the god Thor. The being made a statement,
and Craig then knew he must be Thor.
What statement could Thor have made?
#solution[
A statement that works is: "I am either a knave or a demon
or the god Thor."
#v(2mm)
If the speaker were either a knave or a demon, then it would
be true that he is either a knave or a demon or the god Thor.
#v(2mm)
This would mean that a knave or a demon made a true statement, which is not possible. Therefore the speaker is neither a knave nor a demon, hence his statement is true.
#v(2mm)
Hence he must be the god Thor.
]
#v(1fr)
#pagebreak()
#problem("A Perplexity Resolved")
Craig and Thor became fast friends. In fact, on the evening of
the ninth day, Thor gave a magnificent banquet in Craig's
honor. "I propose a toast to our illustrious guest!" said Thor,
as he raised his glass of nectar.
#v(2mm)
After a round of cheers, Craig was asked to speak.
"I am very perplexed!" said Craig as he rose. "I wonder
if this may not all be a dream!"
"Why do you think you may be dreaming?" asked Thor.
"Because," said Craig, "two incidents have occurred today
that seem totally inexplicable. This morning I met someone
who made a statement which no knight, knave, god, or demon
could possibly make. Then this afternoon I met someone else
who also made a statement which no dweller of this region
could possibly make. That is why I suspect that I may be
dreaming. "
#v(2mm)
"Oh!" said Thor. "Be reassured; you are not dreaming.
The two incidents have a perfectly rational explanation. You
see, we have had two visitors here from another realm. Both
of them are mortal. One is Cyrus, who always tells the truth,
although he is not called a knight since he is not from this
region. The other is Alexander, who sometimes tells the truth
and sometimes lies. It must have been those two whom you
met today. What statements did they make?"
#v(2mm)
Craig then told the company what each had said.
"That explains it perfectly!" said Thor. "Moreover, it follows from their having said what they did that Cyrus was the
one you met in the morning. And interestingly enough, if you
hadn't met Alexander in the afternoon, you could never have
known whether the one you met in the morning was Cyrus
or Alexander."
#v(2mm)
Craig thought the matter over and realized that Thor was right. \
What statements could these two outsiders have made which fulfill all of the above conditions?
#solution[
Here is one possible solution:
- morning speaker: "I am neither a knight nor a god."
- afternoon speaker: "I am either a knave or a demon."
No inhabitant of the region could make either of those
statements. No knight or god could claim that he is neither a
knight nor a god; no knave or demon could make the true
statement that he is neither a knight nor a god.
#v(2mm)
As for the second statement, obviously no knight or god would claim to be either a knave or a demon and no knave or demon would admit to being a knave or a demon. Therefore both were outsiders; namely, Cyrus and Alexander.
#v(2mm)
The statement of the morning speaker was true and the statement of the afternoon speaker was false. Since Cyrus never makes false statements, he couldn't have been the afternoon speaker. Thus he was the morning speaker.
]
#v(1fr)
#pagebreak()
#problem("A Philosophical Puzzle")
The next morning, Craig was wide awake and recalling
his dream. He wondered whether he had been logically inconsistent in his sleep. "The trouble is this," thought Craig: "In my dream I believed that Thor was a god and that gods always tell the truth. Yet Thor told me that I wasn't dreaming. Now how could Thor, who tells the truth, say that I wasn't dreaming when in fact I was? Wasn't this an inconsistency on my part?"
#v(2mm)
Would you say that Craig's dream was logically inconsistent?
#solution[
As I see it, Craig's dream was not necessarily inconsistent. If Craig had actually believed in the dream that he was dreaming, then the set of his beliefs during his dream would have been inconsistent, since the following propositions are indeed logically contradictory:
- Thor is a god
- Gods make only true statements
- Thor stated that Craig was not dreaming;
- Craig was dreaming.
#v(2mm)
The contradiction is obvious. However, there is no evidence that Craig at any time of his dream believed that he was dreaming, although at one point he wondered whether he might be dreaming. Craig presumably believed that he was awake, and this belief, though false, was perfectly consistent with the other beliefs of his dream.
#v(2mm)
Curiously enough, if Craig had formulated the belief that he was dreaming, then this belief, though correct, would have created a logical inconsistency!
]

231
src/Advanced/Gods, Demons, Mortals/parts/01 logician.typ
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@ -1,231 +0,0 @@
#import "@local/handout:0.1.0": *
= The Absentminded Logician
#problem("Only Three Words?")
We are given three brothers named John, James, and William.
John and James always lie, but William always
tells the truth. The three are indistinguishable in appearance.
You meet one of the three brothers on the street one day and
wish to find out whether he is John (because John owes you
money). You are allowed to ask him one question answerable
by yes or no, but the question may not contain more than
three words! What question would you ask?
#solution[
The only three-word question I can think of that works
is: "Are you James?" If you are addressing John, he will answer
yes, since John lies, whereas both James and William would
answer no-James because he lies, and William because he tells
the truth. So a yes answer means that he is John and a no
answer means that he is not John.
]
#v(1fr)
#problem("A Variant")
Suppose we change the above conditions by making John and
James both truthful and William a liar. Again you meet one
of the three and wish to find out if he is John. Is there now a
three-word yes/no question that can accomplish this?
#solution[
The very same question---"Are you James?"---works,
only a yes answer now indicates that he isn't John and a no
answer indicates that he is John.
]
#v(1fr)
#pagebreak()
#problem("A More Subtle Puzzle")
We now have only two brothers (identical twins). One of them
is named Arthur and the other has a different name. One of
the two always lies and the other always tells the truth, but
we are not told whether Arthur is the liar or the truth-teller.
One day you meet the two brothers together, and you wish
to find out which one is Arthur. Note that you are not inter-
ested in finding out which one lies and which one tells the
truth, but only in finding out which one is Arthur. You are
allowed to ask just one of them a question answerable by yes
or no, and again the question may not contain more than three
words. What question would you ask?
#solution[
A common wrong guess is: "Are you Arthur?" This question is quite useless here; the answer you get could be the truth
or a lie, and you would still have no idea which one is really
Arthur.
A question that works is: "Is Arthur truthful?" Arthur will
surely answer yes to this question, because if Arthur is truthful,
he will truthfully claim that Arthur is truthful, and if Arthur
is not truthful, then he will falsely claim that Arthur is truthful.
So regardless of whether Arthur is truthful or whether he lies,
he will certainly claim that Arthur is truthful.
On the other hand, Arthur's brother---call him Henry---will claim that Arthur is not truthful, because if Henry is truthful, then Arthur is really not truthful and Henry will truthfully claim that Arthur is not. And if Henry lies, then Arthur really is truthful,inwhich case Henry will falsely claim that Arthur is not truthful. So whether Henry is truthful or not, he will surely claim that Arthur is not truthful.
In summary, Arthur will claim that
Arthur is truthful and Arthur's brother will claim that Arthur
is not truthful. So if you ask one of the brothers whether Arthur is truthful, and if you get yes for an answer, you will
know that you are speaking to Arthur; if you get no for an
answer, you will know that you are speaking to Arthur's
brother.
Incidentally, there is another three-word question that
works: "Does Arthur lie?" A yes answer to that question
would indicate that you are not speaking to Arthur, and a no
answer would indicate that you are speaking to Arthur. I leave
the verification of this to the reader.
]
#v(1fr)
#pagebreak()
#problem()
Suppose that instead of wanting to find out which one is Arthur, you want to find out whether Arthur is the liar or the
truth-teller. Again there is a three-word question that will do
this. What three-word question will work? There is a pretty
symmetry between the solutions of this and the last problem!
#solution[
To find out whether Arthur is truthful, all you need to
ask is: "Are you Arthur?" Suppose you get the answer yes. If
it is a truthful answer, then the one addressed really is Arthur,
in which case Arthur is the truthful brother.
If the answer is
a lie, then the answerer is not really Arthur, in which case
Arthur must be the other one, again the truthful brother. So
regardless of whether the answer is truthful or a lie, a yes
answer indicates that Arthur-whichever one he is-must be
truthful. What if you get no for an answer?
Well, if it is a
truthful answer, then the speaker is not Arthur, but since he
is truthful, Arthur must be the brother who lies. On the other
hand, if the no answer was a lie, then the speaker really is
Arthur, in which case Arthur just told a lie. So a no answer,
whether it is the truth or a lie, indicates that Arthur is the liar.
]
#v(1fr)
#problem()
This time, all you are interested in finding out is which of the
two brothers you meet is the liar and which is the truth-teller.
You don't care which one is Arthur, or whether Arthur is the
liar or the truth-teller. What three-word question will accomplish this?
#solution[
Just ask him: "Do you exist?
]
#v(1fr)
#problem()
Next you are told to ask one of the brothers just one three-
word question. If he answers yes, you will get a prize; if he
answers no, then you get no prize. What question would you
ask?
#solution[
Just ask: "Are you truthful?" Both constant truth-tellers
and constant liars will answer yes to that question.
]
#v(1fr)
#pagebreak()
#problem("The Absentminded Logician")
A certain logician, though absolutely brilliant in theoretical
matters, was extremely unobservant and highly absent-
minded. He met two beautiful identical-twin sisters named
Teresa and Lenore. The two were indistinguishable ill ap-
pearance, but Teresa always told the truth and Lenore always
lied. The logician fell in love with one of them and married
her, but unfortunately he forgot to find out her first name!
The other sister didn't get married till a couple of years later.
Quite shortly after the wedding, the logician had to go
away for a logic conference. He returned a few days later. He
then met one of the two sisters at a cocktail party and, of
course, had no idea whether or not it was his wife. "I can find
out in only one question," he thought proudly. "I'll simply
use the Nelson Goodman principle and ask her if she is the
type who could claim that she is my wife!" Then he had an
even better idea: "I don't really have to be that elaborate and
ask such a convoluted question. Why, I can find out if she is
my wife by asking a much simpler question-in fact, one having only three words!"
The logician was right! What three-word question answerable
by yes or no should he ask to find out whether the lady he was addressing was his wife?
#solution[
We recall that his wife's sister was not married at the time.
A three-word question that works is: "Is Teresa married?"
Suppose the lady answers yes. She is either Teresa or Lenore.
Suppose she is Teresa. Then the answer is truthful, hence Teresa is really married,
and the lady addressed is married and his
wife. If she is Lenore, the answer is a lie; Teresa is not really
married, so Lenore-who is the lady addressed-is married,
hence again the lady addressed is his wife. So a yes answer
indicates that he is speaking to his wife, regardless of whether
the answer is the truth or a lie. I leave it to the reader to verify
that a no answer indicates that he is speaking to his wife's sister.
]
#v(1fr)
#problem()
A few days later the logician again met one of the two sisters
at another cocktail party. He again didn't know whether it was
his wife or his sister-in-law. "It's high time I find out once
and for all my wife's first name," he thought. "I can ask this
lady just one three-word yes/no question, and then I'll know!"
What three-word question could he ask?
#solution[
The question to ask now is: "Are you married?" Suppose
she answers yes. Again, she is either Teresa or Lenore. Suppose
she is Teresa. Then the answer is truthful, hence the lady ad-
dressed is married, and since she is Teresa, he is married to
Teresa. But ,what if the lady addressed is Lenore? Then the
answer is a lie, hence the lady addressed is not really married,
and he is married to the other lady, again Teresa. So in either
case, a yes answer indicates that his wife's name is Teresa.
I again leave it to the reader to verify that a no answer
indicates that his wife's name is Lenore.
]
#v(1fr)
#pagebreak()
#problem()
Suppose that in the last problem, the logician had wanted to
know both the identity of the lady he met and the first name
of his wife. He is again restricted to asking only one question
answerable by yes or no, but this time there is no restriction
on the number of words in the question.
Can you find a question that will work?
#solution[
No, because no such question exists!
You see, in each of the preceding problems, we were trying
to find out which of two possibilities holds, but in this problem,
we are trying to find out which ofJour possibilities holds.
(The four possibilities are that the lady addressed is Teresa,
his wife; that she is Lenore, his wife; that she is Teresa, his
sister-in-law; and that she is Lenore, his sister-in-law.)
However,
a yes/no question can elicit only two possible responses,
and with only two possible responses it is impossible to
determine which of four possibilities holds.
]
#v(1fr)

2
src/Advanced/Graph Algorithms/main.tex
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@ -2,7 +2,7 @@
% use [solutions] flag to show solutions.
\documentclass[
solutions
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\input{tikxset}

2
src/Advanced/Graph Algorithms/tikxset.tex
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@ -22,7 +22,7 @@
label/.style = {
circle,
% For automatic red background in solutions
fill = \bgcolor,
fill = \ORMCbgcolor,
draw = none
},
%

2
src/Advanced/Intro to Proofs/main.tex
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@ -3,7 +3,7 @@
\documentclass[
nosolutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

2
src/Advanced/Introduction to Quantum/main.tex
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@ -1,7 +1,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{units}
\input{src/tikzset}

17
src/Advanced/Introduction to Quantum/src/main.tex
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@ -1,9 +1,24 @@
% Copyright (C) 2023 <Mark (mark@betalupi.com)>
%
% This program is free software: you can redistribute it and/or modify
% it under the terms of the GNU General Public License as published by
% the Free Software Foundation, either version 3 of the License, or
% (at your option) any later version.
%
% You may have received a copy of the GNU General Public License
% along with this program. If not, see <https://www.gnu.org/licenses/>.
%
%
%
% If you edit this, please give credit!
% Quality handouts take time to make.
% use the [nosolutions] flag to hide solutions,
% use the [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering
]{../../../../lib/tex/handout}
]{../../../../lib/tex/ormc_handout}
\usepackage{../../../../lib/tex/macros}
\usepackage{units}
\input{tikzset}

4
src/Advanced/Introduction to Quantum/src/parts/01 bits.tex
View File

@ -462,7 +462,7 @@ Thus,
\end{equation*}
\begin{hobox}{Review: Multiplying Vectors by Matrices}{black!10!white}{black!65!white}
\begin{ORMCbox}{Review: Multiplying Vectors by Matrices}{black!10!white}{black!65!white}
\begin{equation*}
Av =
\begin{bmatrix}
@ -480,7 +480,7 @@ Thus,
\end{equation*}
Note that each element of $Av$ is the dot product of a row in $A$ and a column in $v$.
\end{hobox}
\end{ORMCbox}
\problem{}
Compute the following product:

4
src/Advanced/Introduction to Quantum/src/parts/05 quantum gates.tex
View File

@ -154,7 +154,7 @@ The \textit{Hadamard Gate} is given by the following matrix: \par
\end{equation*}
\note{Note that we divide by $\sqrt{2}$, since $H$ must be orthonormal.}
\begin{hobox}{Review: Matrix Multiplication}{black!10!white}{black!65!white}
\begin{ORMCbox}{Review: Matrix Multiplication}{black!10!white}{black!65!white}
Matrix multiplication works as follows:
\begin{equation*}
@ -196,7 +196,7 @@ The \textit{Hadamard Gate} is given by the following matrix: \par
This is exactly the first column of the matrix product. \par
Also, note that each element of $Ac_0$ is the dot product of a row in $A$ and a column in $c_0$.
\end{hobox}
\end{ORMCbox}
\problem{}

2
src/Advanced/Introduction to Quantum/src/tikzset.tex
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@ -13,7 +13,7 @@
line width = 0.35mm
},
qubit/.style = {
fill = \bgcolor,
fill = \ORMCbgcolor,
line width = 0.35mm
},
wire/.style = {

2
src/Advanced/Lambda Calculus/main.tex
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@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

2
src/Advanced/Lattices/main.tex
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@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{ifthen}

2
src/Advanced/Mock a Mockingbird/main.tex
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@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

18
src/Advanced/Nonstandard Analysis/main.tex
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@ -1,9 +1,25 @@
% Copyright (C) 2023 <Mark (mark@betalupi.com)>
%
% This program is free software: you can redistribute it and/or modify
% it under the terms of the GNU General Public License as published by
% the Free Software Foundation, either version 3 of the License, or
% (at your option) any later version.
%
% You may have received a copy of the GNU General Public License
% along with this program. If not, see <https://www.gnu.org/licenses/>.
%
%
%
% If you edit this, please give credit!
% Quality handouts take time to make.
% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
nosolutions,
singlenumbering,
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{units}

16
src/Advanced/Nonstandard Analysis/parts/dual.tex
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@ -1,3 +1,19 @@
% Copyright (C) 2023 <Mark (mark@betalupi.com)>
%
% This program is free software: you can redistribute it and/or modify
% it under the terms of the GNU General Public License as published by
% the Free Software Foundation, either version 3 of the License, or
% (at your option) any later version.
%
% You may have received a copy of the GNU General Public License
% along with this program. If not, see <https://www.gnu.org/licenses/>.
%
%
%
% If you edit this, please give credit!
% Quality handouts take time to make.
\section{Dual Numbers}
\definition{}

15
src/Advanced/Nonstandard Analysis/parts/extensions.tex
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@ -1,3 +1,18 @@
% Copyright (C) 2023 <Mark (mark@betalupi.com)>
%
% This program is free software: you can redistribute it and/or modify
% it under the terms of the GNU General Public License as published by
% the Free Software Foundation, either version 3 of the License, or
% (at your option) any later version.
%
% You may have received a copy of the GNU General Public License
% along with this program. If not, see <https://www.gnu.org/licenses/>.
%
%
%
% If you edit this, please give credit!
% Quality handouts take time to make.
\section{Extensions of $\mathbb{R}$}
\definition{}

16
src/Advanced/Nonstandard Analysis/parts/supremum.tex
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@ -1,3 +1,19 @@
% Copyright (C) 2023 <Mark (mark@betalupi.com)>
%
% This program is free software: you can redistribute it and/or modify
% it under the terms of the GNU General Public License as published by
% the Free Software Foundation, either version 3 of the License, or
% (at your option) any later version.
%
% You may have received a copy of the GNU General Public License
% along with this program. If not, see <https://www.gnu.org/licenses/>.
%
%
%
% If you edit this, please give credit!
% Quality handouts take time to make.
\section*{The supremum \& the infimum}
\definition{}

2
src/Advanced/Origami/main.tex
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@ -3,7 +3,7 @@
\documentclass[
solutions,
%shortwarning
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\graphicspath{ {./images/} }

2
src/Advanced/Pidgeonhole Problems/main.tex
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@ -2,7 +2,7 @@
% use [solutions] flag to show solutions.
\documentclass[
solutions
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{amsmath}

2
src/Advanced/Random Walks/main.tex
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@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\input{tikxset.tex}

2
src/Advanced/Random Walks/tikxset.tex
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@ -24,7 +24,7 @@
label/.style = {
rectangle,
% For automatic red background in solutions
fill = \bgcolor,
fill = \ORMCbgcolor,
draw = none,
rounded corners = 0mm
},

249
src/Advanced/Relativity/diagram.tex
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@ -1,249 +0,0 @@
% Diagrams adapted from https://tikz.net/relativity_minkowski_diagram/
%
% MARK: Matt's tweaks
%
\newcommand{\emptydiagram}[1]{
\begin{center}
\begin{tikzpicture}[scale=2]
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{6} % number of world lines (at constant x/t)
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {#1: $t$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
\end{tikzpicture}
\end{center}
}
\newcommand{\emptydiagramc}[1]{
\begin{center}
\begin{tikzpicture}[scale=2]
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{6} % number of world lines (at constant x/t)
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {#1: $ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
\end{tikzpicture}
\end{center}
}
\newcommand{\halfdiagramc}[1]{
\begin{center}
\begin{tikzpicture}[scale=2]
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{6} % number of world lines (at constant x/t)
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
% WORLD LINE GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] (-\x,-0.2) -- (-\x,\xmax);
\draw[world line] ( \x,-0.2) -- ( \x,\xmax);
%\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% AXES
\draw[->,thick] (0,-0.2) -- (T) node[left=-1] {#1: $ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
\end{tikzpicture}
\end{center}
}
\newcommand{\halfdiagramcwide}[1]{
\begin{center}
\begin{tikzpicture}[scale=2]
\def\xmax{3.5}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{10} % number of world lines (at constant x/t)
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,1.95);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line] (-\x,-0.2) -- (-\x,1.95);
\draw[world line] ( \x,-0.2) -- ( \x,1.95);
%\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
%\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,5}{
% \draw[world line] (-\x,-0.2) -- (-\x,\xmax);
% \draw[world line] ( \x,-0.2) -- ( \x,\xmax);
%\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% AXES
\draw[->,thick] (0,-0.2) -- (T) node[left=-1] {#1: $ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
\end{tikzpicture}
\end{center}
}
%
% MARK: tikz setup
%
\usepackage{tikz}
\usetikzlibrary{patterns}
\usepackage{hyperref}
\usepackage{graphicx}
\usepackage{caption}
\usepackage{etoolbox} % ifthen
\usepackage[outline]{contour} % glow around text
\usetikzlibrary{calc} % for adding up coordinates
\usetikzlibrary{decorations.markings,decorations.pathmorphing}
\usetikzlibrary{angles,quotes} % for pic (angle labels)
\usetikzlibrary{arrows.meta} % for arrow size
\usepackage{xfp} % higher precision (16 digits?)
\contourlength{1.1pt}
\tikzset{>=latex} % for LaTeX arrow head
\colorlet{myred}{red!85!black}
\colorlet{mydarkred}{red!55!black}
\colorlet{mylightred}{red!85!black!12}
\colorlet{myfieldred}{mydarkred!5} % for S' background
\colorlet{myredhighlight}{myred!20} % highlights simultaneity in ladder paradox
\colorlet{myblue}{blue!80!black}
\colorlet{mydarkblue}{blue!50!black}
\colorlet{mylightblue}{blue!50!black!30}
\colorlet{mylightblue2}{myblue!10}
\colorlet{mygreen}{green!80!black}
\colorlet{mypurple}{blue!40!red!80!black}
\colorlet{mydarkgreen}{green!50!black}
\colorlet{mydarkpurple}{blue!40!red!50!black}
\colorlet{myorange}{orange!40!yellow!95!black}
\colorlet{mydarkorange}{orange!40!yellow!85!black}
\colorlet{mybrown}{brown!20!orange!90!black}
\colorlet{mydarkbrown}{brown!20!orange!55!black}
\colorlet{mypurplehighlight}{mydarkpurple!20} % highlights simultaneity in ladder paradox
\tikzstyle{world line}=[myblue!40,line width=0.3]
\tikzstyle{world line t}=[mypurple!50!myblue!40,line width=0.3]
\tikzstyle{world line'}=[mydarkred!40,line width=0.3]
\tikzstyle{mysmallarr}=[-{Latex[length=3,width=2]},thin]
\tikzstyle{mydashed}=[dash pattern=on 3 off 3]
\tikzstyle{rod}=[mydarkbrown,draw=mydarkbrown,double=mybrown,double distance=2pt,
line width=0.2,line cap=round,shorten >=1pt,shorten <=1pt]
%\tikzstyle{rod'}=[rod,draw=mydarkbrown!80!red!85,double=mybrown!80!red!85]
\tikzstyle{vector}=[->,line width=1,line cap=round]
\tikzstyle{vector'}=[vector,shorten >=1.2]
\tikzstyle{particle}=[mygreen,line width=0.9]
\tikzstyle{photon}=[-{Latex[length=5,width=4]},myorange,line width=0.8,decorate,
decoration={snake,amplitude=1.0,segment length=5,post length=5}]
\def\tick#1#2{\draw[thick] (#1) ++ (#2:0.06) --++ (#2-180:0.12)}
\def\tickp#1#2{\draw[thick,mydarkred] (#1) ++ (#2:0.06) --++ (#2-180:0.12)}
\def\Nsamples{100} % number samples in plot
% COMMON AXES
\pgfdeclarelayer{back} % to draw on background
\pgfsetlayers{back,main} % set order
\def\xmin{0.23}
\def\xmax{2}
\def\Nlines{6} % number of world lines (at constant x/t)
\def\DNxp{0} % difference in number of world lines of x' axis
\def\DNyp{0} % difference in number of world lines of ct' axis
\def\DNy{0} % difference in number of world lines of ct axis
\def\ang{20} % angle between x and x' axes
\def\xplabelang{180} % anchor angle of x' axis label
%\pgfmathsetmacro\ang{atan(0.44)} % angle between x and x' axes
\def\axes{
\pgfmathsetmacro\d{\xmax/(\Nlines+0.4)} % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\pgfmathsetmacro\ymax{\xmax+\DNy*\d} % maximum of y = ct axis
\pgfmathsetmacro\xmaxp{(\xmax/\d+\DNxp)*\D} % maximum of x' axis
\pgfmathsetmacro\ymaxp{(\xmax/\d+\DNyp)*\D} % maximum of y' = ct' axis
\pgfmathsetmacro\Nylines{\Nlines+\DNy} % number of world lines at constant ct'
\pgfmathsetmacro\Nxplines{\Nlines+\DNxp} % number of world lines at constant x'
\pgfmathsetmacro\Nyplines{\Nlines+\DNyp} % number of world lines at constant ct'
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.15,0);
\coordinate (T) at (0,\ymax+0.15);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\ymaxp+0.2);
% FILL
\begin{pgfonlayer}{back} % draw on back
\fill[myfieldred]
(\ang:-\xmin) -- (\ang:\xmaxp) --++ (90-\ang:\ymaxp) --++ (\ang:-\xmaxp)
-- (90-\ang:-\xmin) -- cycle;
\end{pgfonlayer}
% WORLD LINE GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line] (\x,0) -- (\x,\ymax);
}
\foreach \i [evaluate={\t=\i*\d;}] in {1,...,\Nylines}{
\draw[world line t] (0,\t) -- (\xmax,\t);
}
% BOOSTED WORLD LINE GRID
\foreach \i [evaluate={\x=\i*\D;}] in {1,...,\Nxplines}{
\draw[world line'] (\ang:\x) --++ (90-\ang:\ymaxp);
}
\foreach \i [evaluate={\t=\i*\D;}] in {1,...,\Nyplines}{
\draw[world line'] (90-\ang:\t) --++ (\ang:\xmaxp);
}
% AXES
\draw[->,thick] (0,-\xmin) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmin,0) -- (X) node[below=0] {$x$};
\draw[->,thick,mydarkred] (90-\ang:-\xmin) -- (T')
node[right=5,above=-1] {$ct'$};
\draw[->,thick,mydarkred] (\ang:-\xmin) -- (X')
node[anchor=\xplabelang,inner sep=2] {$x'$};
}

32
src/Advanced/Relativity/main.tex
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@ -1,32 +0,0 @@
% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering,
shortwarning
]{../../../lib/tex/handout}
\usepackage{../../../lib/tex/macros}
\input{diagram}
\uptitlel{Advanced 2}
\uptitler{\smallurl{}}
\title{Special Relativity}
\subtitle{
Prepared by Matthew Kowalski on \today{} \\
}
\begin{document}
\maketitle
%\input{parts/relative velocity}
\input{parts/01 spacetime diagrams}
\input{parts/02 galilean}
\input{parts/03 special}
%\input{parts/relative velocity}
\input{parts/04 simultaneity}
\input{parts/05 proper}
\input{parts/06 contraction}
\end{document}

8
src/Advanced/Relativity/meta.toml
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@ -1,8 +0,0 @@
[metadata]
title = "Special Relativity"
# Matthew Kowalski, 2025-02-02
[publish]
handout = false
solutions = true

196
src/Advanced/Relativity/parts/01 spacetime diagrams.tex
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@ -1,196 +0,0 @@
\section{Spacetime Diagrams}
We are going to derive the principles and consequences of special relativity using basic geometry. \\
o help with our visualization, we will be using spacetime diagrams (called {\em Minkowski diagrams}).
To make our models simpler, we will only be considering {\em one spatial dimension}.
We plot space, which we denote by $x$, as the horizontal axis, and time, which we denote by $t$ as the vertical axis. For a given object, we can then plot its position at any given time. \\
This will give a (potentially curvy) line that we call the object's {\em world line}.
\example{}
Suppose that at time $t = 0$, you are standing still with your cat at your feet.
Your cat walks away from you at speed $1$. We can represent this with a spacetime diagram:
\begin{center}
% SPACETIME DIAGRAM with WORLD LINES
\begin{tikzpicture}[scale=2.0]
%\message{Worldlines^^J}
\def\ymin{0.2}
\def\xmin{1.6}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
%\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% VECTORS
\draw[vector,myred, very thick] (O) -- (4*\d,4*\d)
node[mydarkred,right=10,above] {\contour{white}{cat: $x(t)=t$}};
\draw[vector,myblue, very thick] (O) -- (0,4*\d)
node[mydarkblue,below left=0] {\contour{white}{you: $x(t)=0$}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
\end{tikzpicture}
\end{center}
\note[Note]{
The horizontal axis is space and the vertical axis is time. \par
We are only working with one spatial dimension.
}
\vfill
\pagebreak
\problem{}<pets scatter>
Suppose that you are standing still at time $t = 0$ and your many pets lie at your feet.
\begin{itemize}
\item Your cat, unhappy that she is not fed, begins walking away to your right at speed $2$.
\item Your dog, distracted by a squirrel, walks away to your left at speed $1$.
\item Your hamster, just wanting to feel included, waits a second and then follows the dog at speed $2$.
After reaching your dog, your hamster turns around and sprints after the cat at speed $3$.
\end{itemize}
Draw this situation in the provided spacetime diagram.
\makeatletter
\if@solutions\else
\emptydiagram{Alice}
\fi
\makeatother
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=2.0]
\def\ymin{0.2}
\def\xmin{2}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
\draw[world line] (-4*\d,-\ymin) -- (-4*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% VECTORS
\draw[vector,myred] (O) -- (0,4*\d)
node[mydarkred,below left=0] {\contour{white}{you: $x(t)=0$}};
\draw[vector,myblue] (O) -- (4*\d,2*\d)
node[mydarkblue,above left=0] {\contour{white}{cat: $x(t)=2t$}};
\draw[vector,mygreen] (O) -- (-4*\d,4*\d)
node[mydarkgreen,below left=0] {\contour{white}{dog: $x(t)=-t$}};
\draw[vector,black] (O) -- (0,\d) -- (-2*\d, 2*\d) -- (4*\d, 4*\d)
node[black,below right=0] {\contour{white}{hamster}};
% \draw[vector,myblue]
% (O) to[out=35,in=-100] (O)
% to[out=80,in=-80,looseness=1.5] (0.3*\xmax,4*\d)
% node[mydarkblue,above=-3] {\contour{white}{cat: $x(t)$}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
\end{tikzpicture}
\end{center}
\end{solution}
\vfill
\problem{Event}
Any single point $(x,t)$ on a spacetime diagram is considered an {\em event}
because it describes a time and place. For instance, what is the event that
corresponds to your hamster catching up to your dog?
\begin{solution}
$(-2,2)$
\end{solution}
\vfill
\pagebreak
\problem{}<pets scatter train>
Suppose that the situation of \ref{pets scatter} occurred while you were riding on
a train moving to the right at speed $1$.
Everything occurs relative to you in the same way.
Draw the same diagram in this new situation.
Are any of your pets staying still in this new situation?
\makeatletter
\if@solutions\else
\emptydiagram{Train}
\fi
\makeatother
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=2.0]
\def\ymin{0.2}
\def\xmin{2}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
\draw[world line] (-4*\d,-\ymin) -- (-4*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% VECTORS
\draw[vector,myred] (O) -- (4*\d,4*\d)
node[mydarkred,below left=0] {\contour{white}{you: $x(t)=t$}};
\draw[vector,myblue] (O) -- (6*\d,2*\d)
node[mydarkblue,above left=0] {\contour{white}{cat: $x(t)=3t$}};
\draw[vector,mygreen] (O) -- (0,4*\d)
node[mydarkgreen,below left=0] {\contour{white}{dog: $x(t)=0$}};
\draw[vector,black] (O) -- (\d,\d) -- (0, 2*\d) -- (8*\d, 4*\d)
node[black,below right=0] {\contour{white}{hamster}};
% \draw[vector,myblue]
% (O) to[out=35,in=-100] (O)
% to[out=80,in=-80,looseness=1.5] (0.3*\xmax,4*\d)
% node[mydarkblue,above=-3] {\contour{white}{cat: $x(t)$}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
\end{tikzpicture}
\end{center}
The dog remains stationary in this reference frame.
\end{solution}
\vfill
\pagebreak

195
src/Advanced/Relativity/parts/02 galilean.tex
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@ -1,195 +0,0 @@
\section{Galilean Relativity}
Much like you can watch your pets scatter from the perspective of a train, we can watch the world from anyone's perspective. When we shift perspective like this, just using our normal intuition, we call this {\em Galilean relativity}.
\example{}
Consider the situation of Example $1$ again, but now from the perspective of your cat.
From your cat's perspective, she's the one staying still and you're the one walking away,
only now you're walking away at speed $1$ to the left. We'll denote our new spatial variable with $x'$.
\begin{center}
% SPACETIME DIAGRAM with WORLD LINES
\begin{tikzpicture}[scale=2.0]
\def\ymin{0.2}
\def\xmin{2}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
\draw[world line] (-4*\d,-\ymin) -- (-4*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x'$};
% VECTORS
\draw[vector,myred, very thick] (O) -- (0,4*\d)
node[mydarkred,above right] {\contour{white}{cat: $x'(t)=0$}};
\draw[vector,myblue, very thick] (O) -- (-4*\d,4*\d)
node[mydarkblue,below left=0] {\contour{white}{you: $x'(t)=-t$}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
\end{tikzpicture}
\end{center}
\problem{}
Draw the situation from \ref{pets scatter} in your cat's perspective. \par
What if we drew the situation from \ref{pets scatter train} in your cat's perspective? \par
Would there be any change when the cat is on the train? Why or why not?
\makeatletter
\if@solutions\else
\emptydiagram{Cat}
\fi
\makeatother
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=1.8]
\def\ymin{0.2}
\def\xmin{4}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
\draw[world line] (-4*\d,-\ymin) -- (-4*\d,\xmax);
\draw[world line] (-5*\d,-\ymin) -- (-5*\d,\xmax);
\draw[world line] (-6*\d,-\ymin) -- (-6*\d,\xmax);
\draw[world line] (-7*\d,-\ymin) -- (-7*\d,\xmax);
\draw[world line] (-8*\d,-\ymin) -- (-8*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$t$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% VECTORS
\draw[vector,myred] (O) -- (-8*\d,4*\d)
node[mydarkred,below left=0] {\contour{white}{you: $x(t)=-2t$}};
\draw[vector,myblue] (O) -- (0,4*\d)
node[mydarkblue,above left=0] {\contour{white}{cat: $x(t)=0$}};
\draw[vector,mygreen] (O) -- (-9*\d,3*\d)
node[mydarkgreen,below left=0] {\contour{white}{dog: $x(t)=-3t$}};
\draw[vector,black] (O) -- (-2*\d,1*\d) -- (-6*\d, 2*\d) -- (-4*\d, 4*\d)
node[black,below right=0] {\contour{white}{hamster}};
% \draw[vector,myblue]
% (O) to[out=35,in=-100] (O)
% to[out=80,in=-80,looseness=1.5] (0.3*\xmax,4*\d)
% node[mydarkblue,above=-3] {\contour{white}{cat: $x(t)$}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
\end{tikzpicture}
\end{center}
\end{solution}
\vfill
\pagebreak
\definition{Reference Frame}
When we view the world from the perspective of different objects,
we say that we are working in different {\em reference frames}.
The original Example 1, where you are stationary, is your reference frame.
The new plot in Example 5, where your cat is stationary, is your cat's reference frame.
If we want to compare what is happening in multiple reference frames at once, we can graph
multiple spacetime grids on one plot. If we overlay the cat's reference frame onto your reference frame,
we can visualize everything in Example 1 as:
% SPACETIME DIAGRAM - GALILEAN TRANSFORMATION
\begin{center}
\begin{tikzpicture}[scale=1.8]
\def\xmax{2}
\def\xmaxp{2.1} % maximum of rotated axis
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\pgfmathsetmacro\ang{atan(1)} % angle
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
% WORLD LINES GRID
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] ({-\xmax-tan(\ang)*\x},-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax,\x) -- ({\xmax+tan(\ang)*\x},\x);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=0] {$t$};
\draw[->,thick] (-\xmax,0) -- (X) node[right=6,below=-1] {$x={\color{mydarkred}x'}$};
\draw[->,thick,mydarkred, very thick] (90-\ang:-\xmaxp) -- (T')
node[left=-1] {$t'$}
node[right=2,below right=-2] {cat: $x(t) = t$};
% VECTORS
\draw[vector,myblue, very thick] (O) -- (0,4*\d)
node[mydarkblue,below left=0] {\contour{white}{you: $x(t)=0$}};
% WORLD LINES GRID - BOOSTED
\message{ Making world lines, boosted...^^J}
\fill[mydarkred,opacity=0.05]
(O) --++ (90-\ang:\xmax) --++ (\xmax,0) --++ (90-\ang:-\xmax) -- cycle;
\fill[mydarkred,opacity=0.05]
(O) --++ (90-\ang:-\xmax) --++ (-\xmax,0) --++ (90-\ang:\xmax) -- cycle;
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line'] (\x,0) --++ (90-\ang:\xmax);
\draw[world line'] (-\x,0) --++ (90-\ang:-\xmax);
}
\node[right] at (5.6*\d, \d) {$x=\color{mydarkred}x' + 1$};
%\draw pic[<-,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = T'--O--T};
\end{tikzpicture}
\end{center}
Here $x,t$ are the spacetimes coordinates in your perspective and $x',t'$ are
the spacetime coordinates in your cat's perspective. Note that $t = t'$ for any point while $x = x' + t$.
\problem{}
What does it mean for two events to lie on the same vertical (blue) line from your perspective? \par
What does it mean for two events to lie on the same slanted (red) line from your cat's perspective?
\vfill
\begin{solution}
Two events lie on the same vertical line if they occur at the same location in your reference frame. \par
Two events lie on the same slanted line if they occur at the same location in your cat's reference frame.
\end{solution}
% \remark{}
% Here we are forcing time in the cat's reference frame to behave the same as in our reference frame. As in, one second for the cat is one second for us and vice versa. However, looking at the plot and measuring distances, it almost looks like one second for us is longer than one second for the cat... Suspicious...
\problem{}
In the situation from Problem 2, when will your hamster catch up to your cat? \\
Choose the most convenient reference frame to work in, you shouldn't have to do much math.
\begin{solution}
Using the cat's reference frame drawn in Problem 6, the hamster will catch up to the cat at $t = 8$.
\end{solution}
\vfill
\pagebreak

301
src/Advanced/Relativity/parts/03 special.tex
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@ -1,301 +0,0 @@
\section{Special Relativity}
Galilean relativity is nice until we start going really, {\em really} fast.
Since most of us are terribly slow, we can use it without any issues.
However, in reality, things are much weirder. In particular, there is a maximum speed: the speed of light,
$$c = 299,792,458 \tfrac{m}{s}.$$
Nothing can move faster than the speed of light and {\bf in every reference frame, light will move at this speed}.
Let's see if this is consistent with Galilean relativity.
We are going to making things easier for ourselves now and change units. Instead of measuring time $t$, we are now going to measure $ct$.
\problem{}<photon diagram>
Suppose you are sitting still and you send one photon to your right. Draw this photon on a spacetime diagram, with horizontal axis $x$ and vertical axis $ct$.
\makeatletter
\if@solutions\else
\halfdiagramc{You}
\fi
\makeatother
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=2.0]
\def\ymin{0.2}
\def\xmin{1.6}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% VECTORS
\draw[vector,myred] (O) -- (0,4*\d)
node[mydarkred,below left=0] {\contour{white}{you: $x(t)=0$}};
\draw[photon,shorten >=2] (O) -- (4*\d, 4*\d)
node[black, above right] {\contour{white}{photon}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
\end{tikzpicture}
\end{center}
\end{solution}
\vfill
\problem{}
Suppose you are now sitting on a train that is moving to the right at $c/2$
and again send one photon to your right.
Draw this diagram in the reference frame of the ground.
Draw your (Galilean) reference frame on top of this diagram
What is the speed of the photon in your reference frame? Is that a problem?
\makeatletter
\if@solutions\else
\halfdiagramc{Train}
\fi
\makeatother
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=1.5]
\pgfmathsetmacro\ang{atan(1/2)} % angle
\def\ymin{0.2}
\def\xmin{1.6}
\def\xmax{2}
\def\Nlines{4} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\coordinate (O) at (0,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINES GRID
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line] ( \x,-\ymin) -- ( \x,\xmax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line] (-\d,-\ymin) -- (-\d,\xmax);
\draw[world line] (-2*\d,-\ymin) -- (-2*\d,\xmax);
\draw[world line] (-3*\d,-\ymin) -- (-3*\d,\xmax);
% AXES
\draw[->,thick] (0,-\ymin) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x=\color{mydarkred}x'$};
% VECTORS
\draw[vector,myred] (O) -- (2*\d,4*\d)
node[mydarkred,below left=0] {\contour{white}{you: $x(t)=ct/2$}};
\draw[photon,shorten >=2] (O) -- (4*\d, 4*\d)
node[black, above right] {\contour{white}{photon}};
%\node[right=8,above,mydarkpurple] at (T) {$x(t)=0$};
% WORLD LINES GRID - BOOSTED
\fill[mydarkred,opacity=0.05]
(O) --++ (90-\ang:\xmax) --++ (\xmax,0) --++ (90-\ang:-\xmax) -- cycle;
% \fill[mydarkred,opacity=0.05]
% (O) --++ (90-\ang:-\xmax) --++ (-\xmax,0) --++ (90-\ang:\xmax) -- cycle;
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line'] (\x,0) --++ (90-\ang:\xmax);
% \draw[world line'] (-\x,0) --++ (90-\ang:-\xmax);
}
\end{tikzpicture}
\end{center}
In your reference frame, the light is only moving at speed $c/2$. Uh oh.
\end{solution}
\vfill
\pagebreak
\problem{}
Clearly, Galilean relativity and the absolute speed of light do not mix well together. Having noticed this, you are now in the same boat as early 20th century physicists.
Can you brainstorm any ways to fix Galilean relativity to account for this absolute speed of light?
{\em Hint 1}: Try different methods of drawing the axes of your reference frame that would maintain the speed of light in both the rest frame and in your reference frame.
{\em Hint 2}: The photon worldline always bisects the angle between the space and time axes. \\ Is there a way that you can make that happen in your reference frame?
{\em \color{gray} Don't worry if you don't have any ideas! It took physicists a while to figure this out. Whenever you want to move on, we have the solution on the new page.}
\makeatletter
\if@solutions\else
\emptydiagramc{Train}
\fi
\makeatother
\vfill
\pagebreak
\definition{Lorentz Boost}
Looking at our spacetime diagram from \ref{photon diagram}, we see that the photon worldline bisects the angle between the $ct$ axis and $x$ axis. So if we want to maintain this speed in all reference frames, we just need to make sure that photons bisect our new time axis $ct'$ and our new space axis $x'$.
In order to do this, we're going to to rotate our space axis $x'$ by the same angle that our $ct'$ axis is rotated. Rotating both axes like this is called a {\em Lorentz boost} and is best visualized in the following diagram:
% SPACETIME DIAGRAM - LORENTZ BOOST
\begin{center}
\begin{tikzpicture}[scale=1.8]
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{5} % number of world lines (at constant x/t)
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
% WORLD LINE GRID
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% BOOSTED WORLD LINE GRID
\message{ Making world lines for boosted frame...^^J}
\fill[mydarkred,opacity=0.05]
(O) --++ (\ang:\xmaxp) --++ (90-\ang:\xmaxp) --++ (\ang:-\xmaxp) -- cycle;
\fill[mydarkred,opacity=0.05]
(O) --++ (\ang:-\xmaxp) --++ (90-\ang:-\xmaxp) --++ (\ang:\xmaxp) -- cycle;
\foreach \i [evaluate={\x=\i*\D;}] in {1,...,4}{
\draw[world line'] (\ang:-\x) --++ (90-\ang:-\xmaxp);
\draw[world line'] (90-\ang:-\x) --++ (\ang:-\xmaxp);
\draw[world line'] (\ang:\x) --++ (90-\ang:\xmaxp);
\draw[world line'] (90-\ang:\x) --++ (\ang:\xmaxp);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
\draw[->,thick,mydarkred] (90-\ang:-\xmaxp) -- (T')
node[right=5,above=-1] {you: $ct'$};
\draw[->,thick,mydarkred] (\ang:-\xmaxp) -- (X') node[right=-1] {$x'$};
% ANGLES
\draw pic[->,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = X--O--X'};
\draw pic[<-,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = T'--O--T};
% PHOTON
\draw[photon] (O) --++ (5*\d, 5*\d)
node[black, above right] {\contour{white}{photon}};
\end{tikzpicture}
\end{center}
As before, the slanted (red) axes are your reference frame as you're moving. \\
Note that you (and anything moving the same speed as you) are stationary in this reference frame.
\remark{}
In this diagram, we've not only rotated your space axis ($x'$), we've also adjusted the scale of $ct'$ and $x'$ relative to the rest frame. This scaling comes from physical experiments which we will conduct later.
\problem{}
Please verify that in the diagram above, if you shoot a photon behind you, it still moves at speed $c$ in both your reference frame and the ground's reference frame. You can do this by drawing directly on the diagram.
\begin{solution}
This follows by just extending the boosted axes to the second quadrant and drawing the photon's worldline.
\end{solution}
\vfill
\pagebreak
% \begin{solution}\begin{center}\begin{tikzpicture}[scale=1.8]
% \message{Lorentz boost^^J}
% \def\xmax{2}
% \def\xmaxp{2.2} % maximum of rotated axis
% \def\Nlines{5} % number of world lines (at constant x/t)
% \pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
% \pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
% \pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
% \coordinate (O) at (0,0);
% \coordinate (X) at (\xmax+0.2,0);
% \coordinate (T) at (0,\xmax+0.2);
% \coordinate (X') at (\ang:\xmaxp+0.2);
% \coordinate (T') at (90-\ang:\xmaxp+0.2);
% % WORLD LINE GRID
% \message{ Making world lines...^^J}
% \foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
% \message{ Running i/N=\i/\Nlines, x=\x...^^J}
% \draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
% \draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
% \draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
% \draw[world line t] (-\xmax, \x) -- (\xmax, \x);
% }
% % BOOSTED WORLD LINE GRID
% \message{ Making world lines for boosted frame...^^J}
% \fill[mydarkred,opacity=0.05]
% (O) --++ (\ang:\xmaxp) --++ (90-\ang:\xmaxp) --++ (\ang:-\xmaxp) -- cycle;
% \fill[mydarkred,opacity=0.05]
% (O) --++ (\ang:-\xmaxp) --++ (90-\ang:-\xmaxp) --++ (\ang:\xmaxp) -- cycle;
% \foreach \i [evaluate={\x=\i*\D;}] in {1,...,4}{
% \message{ Running i/N=\i/\Nlines, x=\x...^^J}
% \draw[world line'] (\ang:-\x) --++ (90-\ang:-\xmaxp);
% \draw[world line'] (90-\ang:-\x) --++ (\ang:-\xmaxp);
% \draw[world line'] (\ang:\x) --++ (90-\ang:\xmaxp);
% \draw[world line'] (90-\ang:\x) --++ (\ang:\xmaxp);
% }
% % AXES
% \draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {$ct$};
% \draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
% \draw[->,thick,mydarkred] (90-\ang:-\xmaxp) -- (T')
% node[right=5,above=-1] {you: $ct'$};
% \draw[->,thick,mydarkred] (\ang:-\xmaxp) -- (X') node[right=-1] {$x'$};
% % ANGLES
% \draw pic[->,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = X--O--X'};
% \draw pic[<-,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = T'--O--T};
% % PHOTON
% \draw[photon] (O) --++ (4*\d,4*\d);
% \end{tikzpicture}\end{center}\end{solution}
\problem{}
A caveat to Lorentz boosts is that we cannot boost to reference frames which are at the speed of light or
faster. Based on the diagram given, why can't we do that? \\
{\em "If my calculations are correct, when this baby hits $c$, you're gonna see some serious stuff."}
\begin{solution}
If boosted to $c$, our axes would overlap, compressing time and space into one.
If we boosted past $c$, our axes would flip, making the past the future and the future the past.
\end{solution}
\vfill
\problem{}
This diagram implies some strange things. We'll spend the next sections discussing
some consequences of this, but take a minute to note anything weird that you notice.
In particular, look at a unit of time (or a unit of length) in your frame vs the rest frame.
Which is longer, one unit of time in your reference frame or in the rest frame?
Which is longer, one unit of distance in your reference frame or in the rest frame?
Consider the implications of a slanted space line.
What does it mean if two events both lie on this line?
\vfill
\pagebreak

257
src/Advanced/Relativity/parts/04 simultaneity.tex
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@ -1,257 +0,0 @@
\section{Simultaneity}
Let's look now at the consequences of a rotated spatial coordinate.
To help ourselves switch between different perspectives,
we'll bring in some professionals: Alice and Bob.
\ref{simultaneity setup} to \ref{Bob overreacted} can all be completed on one spacetime diagram. Feel free to use the one provided below, or draw your own if it gets too crowded.
\makeatletter
\if@solutions\else
\emptydiagramc{Alice}
\fi
\makeatother
\problem{}<simultaneity setup>
Draw a spacetime diagram from Alice's reference frame or use the one provided.
What do the horizontal gridlines represent?
If two events (remember, points in spacetime) lie on the same horizontal line,
what does that imply about the events?
\begin{solution}
Horizontal lines have constant $t$ value.
Therefore, if two events lie on the same horizontal
line then they occur at the same time.
\end{solution}
\vfill
\pagebreak
\problem{}
Suppose that Bob is walking to the right at speed $c/3$, relative to Alice.
Add Bob's worldline to your spacetime diagram. Superimpose Bob's reference frame onto the diagram
What do the "horizontal" lines in Bob's reference frame represent to Bob?
\begin{solution}
Same solution, horizontal lines in Bob's reference frame represent events at a particular time.
\end{solution}
\vfill
\problem{}
Suppose that Alice is passing time, snapping to some music with her arms out.\\
Both of her arms are length $1$ and she snaps both hands every unit time.
So there is a snap at location $x = \pm 1$ at $ct = 1$, $ct = 2$, and so on.
Draw this situation on your spacetime diagram.
Consider this from Bob's perspective. What does Bob hear?
(We'll assume that sound propagates instantly, so if a sound occurs at $ct' = 3.5$ then Bob hears it instantly at $ct' = 3.5$.)
\begin{solution}
The snaps will be out of time in Bob's reference frame.
He will hear Alice's right hand first and then her left hand after.
\end{solution}
\vfill
\problem{}<Bob overreacted>
Bob is deeply annoyed at Alice because she {keeps}.~{\em snapping}.~{\em out}.~{\bf\em of}.~{\bf\em TIME}.
In a fit of frustration he starts running faster and faster to the right.
Eventually, he notices that the snaps are changing.
What happens to the timing of the snaps? Do the snaps ever come back into time?
At what speed is Bob running if they do?
\begin{solution}
We can line up new sets of snaps with lines of slope $1/2$.
So if Bob is running at speed $c/2$then he will hear the snaps in time again.
\end{solution}
\vfill
\pagebreak
\problem{}
Done with music, Alice and Bob decide to have a race and they agree to race from $x = 0$ to $x = 2$.
Suppose that Alice can run at an impressive speed of $c/2$ while Bob can only run at a measly speed of $c/4$.
\begin{enumerate}
\item Who wins the race?
\begin{solution}
Alice
\end{solution}
\vspace{20pt}
\item Is there any reference frame in which Bob wins?
\makeatletter
\if@solutions\else
\halfdiagramcwide{Ground}
\fi
\makeatother
\begin{solution}
No
\end{solution}
\vfill
\item Suppose instead that Alice starts at $x = 0$ and is racing to $x = 2$, while Bob starts at $x = 8$ and is racing to $x = 10$. Is there now a reference frame where Bob wins? Why?\\
{\em Hint: Plot these events on a spacetime diagram.}
\begin{solution}
Bob finishes at $(2, 8)$ and Alice finishes at $(10,4)$.
If you consider a reference frame moving to the left at speed $c/2$, then Alice and Bob will tie. Anything faster, Bob will win. Anything slower, Alice will win.
\end{solution}
\end{enumerate}
\makeatletter
\if@solutions\else
\halfdiagramcwide{Ground}
\fi
\makeatother
\vfill
\pagebreak
% \problem{}
% We've seen that by changing your reference frame, events that were simultaneous may now occur at different times.
% Which of the following pairs of events (points in spacetime $(x,ct)$) are simultaneous in {\em some} inertial reference frame?
% % Remember, we can only consider reference frames which are moving less than the speed of light.
% \begin{enumerate}
% \item $(1,1)$ and $(0,1)$
% \item $(2,1)$ and $(0,0)$
% \item $(2,3)$ and $(0,1)$
% \item $(1,2)$ and $(0,0)$
% \item $(1,1)$ and $(1,3)$
% \end{enumerate}
% \begin{solution}
% \begin{enumerate}
% \item $(1,1)$ and $(0,1)$ - yes
% \item $(2,1)$ and $(0,0)$ - yes
% \item $(2,3)$ and $(0,1)$ - no, would require a reference frame at the speed of light
% \item $(1,2)$ and $(0,0)$ - no, would require a reference frame moving faster than the speed of light
% \item $(1,1)$ and $(1,3)$ - definitely not
% \end{enumerate}
% \end{solution}
\problem{}
This weirdness with simultaneity might make us question what the past, present, and future are.
To make sense of this, suppose that we want to define the {\em present} to be every event which is simultaneous to you, right here, right now --- a.k.a.~the spacetime point $(0,0)$ --- in at least one reference frame.
On the given spacetime diagram, draw the region which represents the present.
Draw a region which represents the future: events that are later in time in every reference frame.
Draw a region which represents the past: events that are earlier in time in every reference frame.
\begin{center}
\begin{tikzpicture}[scale=2]
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{5} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\pgfmathsetmacro\ang{atan(1/3)} % angle between x and x' axes
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINE GRID
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
% LABELS
%\draw pic[->,"$45^\circ$",draw=black,angle radius=23,angle eccentricity=1.38] {angle = X--O--C};
% \node[mydarkorange,above right] at (0.1*\xmax,\xmax) {future light cone};
% \node[mydarkorange,below] at (0,-\xmax) {past light cone};
% FILLS
% \fill[myblue,opacity=0.05] % SPACELIKE
% (\xmax,\xmax) -- (-\xmax,-\xmax) -- (-\xmax,\xmax) -- (\xmax,-\xmax) -- cycle;
% \fill[myorange,opacity=0.05] % TIMELIKE
% (\xmax,\xmax) -- (-\xmax,\xmax) -- (\xmax,-\xmax) -- (-\xmax,-\xmax) -- cycle;
% \node[mydarkblue,right,align=center] at (-\xmax,0.18*\xmax)
% {\contour{myblue!5}{present}\\[-2]\contour{myblue!5}{}};
% \node[mydarkblue,left,align=center] at (\xmax,0.18*\xmax)
% {\contour{myblue!5}{present}\\[-2]\contour{myblue!5}{}};
% \node[mydarkorange,align=center] at (-0.22*\xmax,0.67*\xmax)
% {\contour{myorange!5}{future}\\[-2]\contour{myorange!5}{}};
% \node[mydarkorange,align=center] at (0.22*\xmax,-0.67*\xmax)
% {\contour{myorange!5}{past}\\[-2]\contour{myorange!5}{}};
% PHOTON
\draw[photon] ( \xmax,-\xmax) -- ( 0.02*\xmax,-0.02*\xmax);
\draw[photon] (-\xmax,-\xmax) -- (-0.02*\xmax,-0.02*\xmax);
\draw[photon] ( 0.02*\xmax,0.02*\xmax) -- ( \xmax,\xmax)
node[mydarkorange,above right] {$x=ct$};
\draw[photon] (-0.02*\xmax,0.02*\xmax) -- (-\xmax,\xmax);
% % PARTICLE WORLDLINE
% \draw[particle,decoration={markings,mark=at position 0.27 with {\arrow{latex}},
% mark=at position 0.76 with {\arrow{latex}}},postaction={decorate}]
% (-0.5*\xmax,-\xmax) to[out=80,in=-110] (O) to[out=70,in=-100] (0.45*\xmax,\xmax);
% \fill[mydarkgreen] (O) circle(0.04); % event
\end{tikzpicture}
\end{center}
Another way to think of this is thinking of causality. In a reference frame, the {\em future} is every event that can be affected by an event at $(0,0)$. The {\em past} is every event that could have affected an event at $(0,0)$. The {\em present} is every event that is causally independent of $(0,0)$.
\begin{solution}
% SPACETIME DIAGRAM - LIGHT CONE
The photons split the diagram into four regions.
The top region is the future, the bottom region is the past, and the left/right regions are the present.
\end{solution}
\vfill
\pagebreak
\problem{Bell's Spaceship Paradox}
Suppose that we have two spaceships, distance $L$ apart, tied together with floss of length $L$. The floss is so weak that any stretching at all will cause it to disintegrate.
The spaceships are at rest and then simultaneously accelerate to speed $c/2$.
Draw the spacetime diagram for this situation. Include the reference frame of the spaceships {\em after} they start moving (i.e. the reference frame moving at speed $c/2$).\\
What happens to the floss in the boosted frame? Does it break? Why?
\makeatletter
\if@solutions\else
\emptydiagramc{Rest}
\fi
\makeatother
\vfill
\problem{Bell's Spaceship Paradox (continued)}
The same outcome has to occur in all reference frames, so we know that the floss breaks in the stationary reference frame. However, in the stationary reference frame, the two spaceships accelerate at the same time, so our explanation no longer seems accurate.
Can you come up with a hypothesis for why the floss breaks in the stationary reference frame?
{\em Hint: The floss breaks if the ships are farther apart than the length of the floss.}
\begin{solution}
The floss will break as soon as the spaceships accelerate. \par
In the boosted frame, the spaceship on the right accelerates first,
stretching the floss, causing it to break. \par
In the rest frame, this paradox motivates length contraction.
Either the ships get farther apart (doesn't happen) or the floss gets shorter.
This implies that the length of moving objects must get smaller.
\end{solution}
\vfill
\pagebreak

391
src/Advanced/Relativity/parts/05 proper.tex
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@ -1,391 +0,0 @@
\section{Proper Time and Proper Length}
We mentioned that the space and time axes also get rescaled during a Lorentz boost. Let's figure out by how much! To do this, we're going to use the only consistent tool at our disposal, the speed of light.
\problem{Time dilation}
Suppose that Alice remains stationary.
In her reference frame, Bob moves to the right at speed $v$.
Every quantity in Alice's reference frame will be denoted normally,
i.e.~$x,t$. Every quantity in Bob's reference frame will be given a prime, i.e.~$x',t'$.
Bob holds onto a very special clock.
The clock consists of a light bulb, a mirror distance $L$ away from the light
(perpendicular to the direction Bob is moving) and a photosensor.
\begin{enumerate}
\item Draw the experimental setup.
Note again that the light bulb and mirror are separated in the $y$ direction, not the $x$.
\vfill
\item In Bob's reference frame (where the clock isn't moving), how long does it take the light to be emitted, travel to the mirror, bounce back, and be reabsorbed? Write this as $t'$.
\begin{solution}
$ct' = 2L \implies t' = 2L/c$
\end{solution}
\vfill\pagebreak
\item In Alice's reference frame, she still sees the light move, reflect off the mirror, then come back. How long does this take?
{\em Note: The mirror and photosensor are moving at speed $v$ during this process. Also, the light is now moving at an angle, but it still moves at $c$.}
\begin{solution}
Pythagorean theorem:
$$(ct)^2 = (vt)^2 + 4L^2 \implies t^2 = \frac{4L^2}{c^2 - v^2} \implies t = \frac{2L}{\sqrt{c^2 - v^2}}$$
\end{solution}
\vfill
\item Combine these two formulas to find a relationship between $t$ and $t'$.
If one unit of time elapses in Bob's reference frame, does more or less time elapse in Alice's reference frame? Who ages faster?
\begin{solution}
$$ct' = \sqrt{c^2 - v^2}t \implies t' = \sqrt{1 - v^2/c^2} t$$
Since $v < c$, $0 \leq v^2/c^2 <1$. Then $t' < t$.
Alice ages faster. The time that Bob experiences is less than Alice.
\end{solution}
\end{enumerate}
\vfill
\pagebreak
\problem{Space dilation}
Since the speed of light is always constant, we can use time dilation to see what happens to our notion of distance. Suppose that a photon is emitted from one location and absorbed in another.
\begin{enumerate}
\item Suppose that in Alice's reference frame, the photon takes time $t$ to be emitted and absorbed. In this time, what distance $x$ does the photon travel?
\begin{solution}
$x = ct$
\end{solution}
\vfill
\item Suppose that Bob is traveling at speed $v$ relative to Alice. Using time dilation, how long does the photon take to be emitted and absorbed in Bob's reference frame? What distance $x'$ does the photon travel in that time?
{\em This will give you a formula relating distance $x'$ in Bob's reference frame to distance $x$ in Alice's reference frame.}
\begin{solution}
$x' = ct' = ct \sqrt{1 - v^2/c^2} = x \sqrt{1 - v^2/c^2}$
\end{solution}
\vfill
\end{enumerate}
\vfill \pagebreak
\remark{}
Switching back and forth between reference frames gets confusing quickly,
especially when you are consideration time and space dilation. To avoid this,
we will define a new notion of time and space that doesn't care about which reference frame we're in.
Since this will be universal, we'll call these {\em proper time}, denoted by $\tau$, and {\em proper distance},
denoted by $\chi$.
Recall our notion of past and future from earlier:
% SPACETIME DIAGRAM - LIGHT CONE
\begin{center}
\begin{tikzpicture}[scale=1.8]
\message{Light cone^^J}
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{5} % number of world lines (at constant x/t)
\pgfmathsetmacro\d{0.9*\xmax/\Nlines} % grid size
\pgfmathsetmacro\ang{atan(1/3)} % angle between x and x' axes
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
% WORLD LINE GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {$ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
% LABELS
%\draw pic[->,"$45^\circ$",draw=black,angle radius=23,angle eccentricity=1.38] {angle = X--O--C};
\node[mydarkorange,above right] at (0.1*\xmax,\xmax) {future light cone};
\node[mydarkorange,below] at (0,-\xmax) {past light cone};
% FILLS
\fill[myblue,opacity=0.05] % SPACELIKE
(\xmax,\xmax) -- (-\xmax,-\xmax) -- (-\xmax,\xmax) -- (\xmax,-\xmax) -- cycle;
\fill[myorange,opacity=0.05] % TIMELIKE
(\xmax,\xmax) -- (-\xmax,\xmax) -- (\xmax,-\xmax) -- (-\xmax,-\xmax) -- cycle;
\node[mydarkblue,right,align=center] at (-\xmax,0.18*\xmax)
{\contour{myblue!5}{present}\\[-2]\contour{myblue!5}{}};
\node[mydarkblue,left,align=center] at (\xmax,0.18*\xmax)
{\contour{myblue!5}{present}\\[-2]\contour{myblue!5}{}};
\node[mydarkorange,align=center] at (-0.22*\xmax,0.67*\xmax)
{\contour{myorange!5}{future}\\[-2]\contour{myorange!5}{}};
\node[mydarkorange,align=center] at (0.22*\xmax,-0.67*\xmax)
{\contour{myorange!5}{past}\\[-2]\contour{myorange!5}{}};
\node at (2*\d, 4*\d){\textbullet};
\node[black, above] at (2*\d, 4*\d){future event};
\node at (4*\d, 2*\d){\textbullet};
\node[black, above right] at (4*\d, 2*\d){present event};
% PHOTON
\draw[photon] ( \xmax,-\xmax) -- ( 0.02*\xmax,-0.02*\xmax);
\draw[photon] (-\xmax,-\xmax) -- (-0.02*\xmax,-0.02*\xmax);
\draw[photon] ( 0.02*\xmax,0.02*\xmax) -- ( \xmax,\xmax)
node[mydarkorange,above right] {$x=ct$};
\draw[photon] (-0.02*\xmax,0.02*\xmax) -- (-\xmax,\xmax);
% % PARTICLE WORLDLINE
% \draw[particle,decoration={markings,mark=at position 0.27 with {\arrow{latex}},
% mark=at position 0.76 with {\arrow{latex}}},postaction={decorate}]
% (-0.5*\xmax,-\xmax) to[out=80,in=-110] (O) to[out=70,in=-100] (0.45*\xmax,\xmax);
% \fill[mydarkgreen] (O) circle(0.04); % event
\end{tikzpicture}
\end{center}
\problem{Proper Time}
For any event $(x,ct)$ in the past or future, we can choose a reference frame to make it occur at $x' = 0$.
\begin{enumerate}
\item What velocity do we need to boost to so that the event $(x,ct)$ now occurs at $x' = 0$?
\begin{solution}
We need to boost to $v = x/t$.
\end{solution}
\vfill
\item After boosting to this reference frame, what time $ct'$ does the event occur at? \\
This will be our proper time $c\tau$.
\begin{solution}
Using time dilation, the event now occurs at $$c\tau = ct' = ct\sqrt{1 - x^2/c^2 t^2} = \sqrt{c^2 t^2 - x^2}$$
\end{solution}
\vfill
\end{enumerate}
\pagebreak
\problem{Proper Distance}
Similarly, for any event $(x,ct)$ in the present, we can choose a reference frame to make it occur at $t' = 0$.
\begin{enumerate}
\item What velocity do we need to boost to so that the event $(x,ct)$ now occurs at $t' = 0$?
\begin{solution}
We need to boost to $v = c^2t/x$.
\end{solution}
\vfill
\item After boosting to that velocity, what spatial position $x'$ does the event occur at? This will be our proper distance $\chi$.
\begin{solution}
Using space dilation, the event now occurs at $$\chi = x' = x\sqrt{1 - c^2t^2/x^2} = \sqrt{x^2 - c^2t^2}$$.
\end{solution}
\end{enumerate}
\vfill
Defined in this way, proper time and proper distance give a quick formula to
figure out how old an object is, or how far an object traveled, in its own reference frame!
We'll now use this proper time and proper space to quickly solve one of
the most famous paradoxes in special relativity.
\pagebreak
% \remark{}
% To visualize the consequences of this change, let's see what lines of constant proper time and constant proper space look like.
% Lines in green are constant proper time and lines in blue are constant proper space.
% % SPACETIME DIAGRAM - MULTIPLE INVARIANT HYPERBOLOIDS
% % Inspiration: https://commons.wikimedia.org/wiki/File:Spacelike_and_Timelike_Invariant_Hyperbolas.png
% \begin{center}\begin{tikzpicture}[scale=1.8]
% \message{Multiple invariant hyperboloids^^J}
% \def\xmax{2}
% \def\Nlines{4} % number of world lines (at constant x/t)
% \pgfmathsetmacro\w{\xmax/(\Nlines+1)}
% % AXES
% \draw[->,thick] (0,-\xmax) -- (0,\xmax+0.2) node[left=-1] {$ct$};
% \draw[->,thick] (-\xmax,0) -- (\xmax+0.2,0) node[right=-1] {$x$};
% % LIGHTCONE
% \draw[myorange,thick] (-\xmax,-\xmax) -- (\xmax, \xmax);
% \draw[myorange,thick] (-\xmax, \xmax) -- (\xmax,-\xmax);
% \foreach \i [evaluate={\s=\xmax*\i/(\Nlines+1); \xm=sqrt(\xmax^2-\s^2);}] in {1,...,\Nlines}{
% % SPACELIKE HYPERBOLOIDS
% \draw[mygreen,thick,samples=\Nsamples,smooth,variable=\x,domain=-\xm:\xm]
% plot(\x,-{sqrt(\s^2+(\x)^2)})
% plot(\x,{sqrt(\s^2+(\x)^2)});
% % TIMELIKE HYPERBOLOIDS
% \draw[myblue,thick,samples=\Nsamples,smooth,variable=\y,domain=-\xm:\xm]
% plot(-{sqrt(\s^2+(\y)^2)},\y)
% plot({sqrt(\s^2+(\y)^2)},\y);
% }
% % LABELS
% \node[mydarkgreen,above left=2,align=center] at (-0.2*\xmax,\xmax)
% {timelike separation\\[-1]$s^2 = c^2t^2 - x^2 > 0$};
% \node[mydarkorange,left=2,above right=-2,align=center] at (\xmax,\xmax)
% {lightlike separation\\[-1]$s^2 = c^2t^2 - x^2 = 0$};
% \node[mydarkblue,right=0,align=center] at (0.88*\xmax,-0.24*\xmax)
% {spacelike separation\\[-1]$s^2 = c^2t^2 - x^2 < 0$};
% % % VECTORS
% % \def\xa{0.5}
% % \def\xb{2.7}
% % \def\ta{-0.7}
% % \def\tb{1.7}
% % \draw[mydarkgreen,very thick,decoration={markings,mark=at position 0.55 with {\arrow{latex}}},
% % postaction={decorate},samples=20,variable=\x,domain=\xa:\xb]
% % plot({\w*\x},{\w*sqrt((\x)^2+3^2)});
% % \draw[mydarkblue,very thick,decoration={markings,mark=at position 0.6 with {\arrow{latex}}},
% % postaction={decorate},samples=20,variable=\x,domain=\ta:\tb]
% % plot({\w*sqrt((\x)^2+3^2)},{\w*\x});
% % \fill[mydarkgreen] ({\w*\xa},{\w*sqrt(\xa^2+3^2)}) coordinate (A) circle(0.03);
% % \fill[mydarkgreen] ({\w*\xb},{\w*sqrt(\xb^2+3^2)}) coordinate (A') circle(0.03);
% % \fill[mydarkblue] ({(\w*sqrt((\ta)^2+3^2)},{\w*\ta}) coordinate (B) circle(0.03);
% % \fill[mydarkblue] ({(\w*sqrt((\tb)^2+3^2)},{\w*\tb}) coordinate (B') circle(0.03);
% % \draw[vector',mydarkgreen] (0,0) -- (A)
% % node[pos=0.53,right=-2] {$s$};
% % \draw[vector',mydarkgreen] (0,0) -- (A')
% % node[pos=0.57,right=-2] {$s$};
% % \draw[vector',mydarkblue] (0,0) -- (B)
% % node[pos=0.53,below=-1] {$s$};
% % \draw[vector',mydarkblue] (0,0) -- (B')
% % node[pos=0.53,above=-1] {$s$};
% \end{tikzpicture}\end{center}
\problem{Twin paradox}
Suppose that you have two twins, Mark and Lucas, born at the exact same instant and location. \\
One day, Mark sends Lucas off in a spaceship traveling at $c/2$ to a planet that is one lightyear away ($c*(\text{1 year})$). Upon reaching the planet, Lucas realizes that he's all alone! Saddened, he turns around immediately, traveling back at speed $c/2$.
The question is: Which twin is older? We'll break this down step by step.
\begin{enumerate}
\item From Mark's perspective, what time does Lucas reach the planet? What time does he return?
\begin{solution}
Lucas reaches the planet at $t = 2$ years.
Lucas reaches home at $t = 4$ years.
\end{solution}
\vfill
\item From Lucas' perspective, how long does it take him to reach the planet? Hint: use proper time.
\begin{solution}
The proper time of Lucas arriving at the planet is $$c\tau = \sqrt{4 - 1}\text{ lightyears} \implies \tau = \sqrt{3}\text{ years}.$$
\end{solution}
\vfill
\item How long does it take him to return?
\begin{solution}
$\tau = \sqrt{3}\text{ years}.$
\end{solution}
\vfill
\item Which twin is older?
\begin{solution}
Mark
\end{solution}
\end{enumerate}
\vfill
\pagebreak
\problem{Twin paradox (continued)}
The weird part is, from Lucas' perspective, it was Mark who flew off and then came back! So why is Mark not younger than Lucas?
Let's break this down step by step.
\begin{enumerate}
\item[\bf E:]
To help answer this, draw a spacetime diagram of this situation. In Lucas' reference frames (one for going out to the planet and one for coming back), draw lines of constant time. What happens to Mark from Lucas' perspective?\\
{\em Hint: changing speed is weird. Focus on where Lucas changes speed.}
\makeatletter
\if@solutions\else
\halfdiagramc{Mark}
\fi
\makeatother
\begin{solution}
% SPACETIME DIAGRAM of TWIN PARADOX
\begin{tikzpicture}[scale=2.0]
\def\xmin{0.2}
\def\xmax{2}
\def\ymax{2.3}
\def\Nlines{5} % number of world lines (at constant x/t)
\def\ang{60} % angle between ct and ct' axes
\pgfmathsetmacro\d{0.94*\xmax/\Nlines} % grid size
\pgfmathsetmacro\dt{3*\d} % time of half trip
\pgfmathsetmacro\D{\dt/tan(\ang)} % distance between observers
\pgfmathsetmacro\h{\dt-\D/tan(\ang)} % half time gap of return
\coordinate (A) at (0,0); % observer A at t=0
\coordinate (B) at (\D,0); % observer B at t=0
\coordinate (C) at (\D,\dt); % point of return
\coordinate (T1) at (0,\dt); % time of return
\coordinate (T2) at (0,2*\dt); % time of arrival
% WORLD LINES GRID
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] ( \x,-\xmin) -- ( \x,\ymax);
\draw[world line t] (-\xmin, \x) -- (\xmax, \x);
}
\draw[world line t] (-\xmin,{(\Nlines+1)*\d}) -- (\xmax,{(\Nlines+1)*\d});
% AXES
\draw[->,thick] (0,-\xmin) -- (0,\ymax+0.2) node[above left=-2] {$ct$};
\draw[->,thick] (-\xmin,0) -- (\xmax+0.2,0) node[below=0] {$x$};
% VECTORS
\draw[vector,myred,shorten >=1] (A) -- (T2);
\draw[vector,mygreen,shorten >=2] (A) -- (C);
\draw[vector,mygreen,shorten >=1] (C) -- (T2);
% PLANES OF SIMULTANEITY
\fill[mydarkred,opacity=0.06]
(0,\h) -- (C) -- (0,2*\dt-\h) -- cycle;
\pgfmathsetmacro\ystep{\h/3}
\foreach \i [evaluate={\dy=(\i-1)*\ystep; \ya=\i*\ystep; \yb=2*\dt-\i*\ystep;}] in {1,...,3}{
\draw[mydarkred,dashed,line width=0.6]
(0,\ya)++(90-\ang:-0.8*\xmin) --++ (90-\ang:{1.2*\xmin+\D/sin(\ang)});
\draw[mydarkblue,dashed,line width=0.6]
(0,\yb)++(\ang-90:-0.8*\xmin) --++ (\ang-90:{1.2*\xmin+\D/sin(\ang)});
\fill[mydarkred] (0,\ya) circle(0.02);
\fill[mydarkblue] (0,\yb) circle(0.02);
%\fill[mydarkblue] ({\D-\dy*cot(\ang)},\dt+\dy) circle(0.02);
%\fill[mydarkred] ({\D-\dy*cot(\ang)},\dt-\dy) circle(0.02);
\fill[mydarkblue] (C)++(-\ang:{\dy*sin(\ang)/cos(2*\ang)}) circle(0.02);
\fill[mydarkred] (C)++( \ang:{\dy*sin(\ang)/cos(2*\ang)}) circle(0.02);
}
\fill[mydarkred] (A) circle(0.04) node[below left=-1] {Mark}; % observer A
\fill[mydarkgreen] (C) circle(0.04)
node[right=4] {\contour{white}{Lucas turns around}}; % observer B returns
\node[mydarkblue,above right=0,align=left] at (2*\d,1.15*\dt)
{\contour{white}{planes of}\\[-2]\contour{white}{simultaneity}};
\node[mydarkred,below right=0,align=left] at (2*\d,0.85*\dt)
{\contour{white}{planes of}\\[-2]\contour{white}{simultaneity}};
% TICKS
\node[fill=white,inner sep=1,above=1,left=3] at (T1) {$\dfrac{ct_2}{2}=ct_1$};
\node[fill=white,inner sep=1,above=2,left=3] at (T2) {$ct_2$};
\tick{T1}{0};
\tick{T2}{0};
\end{tikzpicture}
\end{solution}
\vfill
\item[\bf F:] We said that when Lucas landed on the planet, he immediately took off back towards Earth. From Mark's perspective, how long was Lucas on that planet?
\begin{solution}
$4 - 2 \sqrt{3}$ years.
\end{solution}
\vfill
\item[\bf G:] What is physically different between Lucas' perspective and Mark's perspective?
\begin{solution}
Lucas is the one who accelerates. His reference frame is non-inertial (not defined in this packet) and would require additional care.
\end{solution}
\vfill
\end{enumerate}
\pagebreak

366
src/Advanced/Relativity/parts/06 contraction.tex
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@ -1,366 +0,0 @@
\section{Length Contraction}
With proper time and distance done, we can now tackle length contraction easily! Length contraction is weird because different parts of our object will now be experiencing different times.
\problem{}
Suppose that you (at rest) see a rod of length $L$ moving at speed $v$.
\makeatletter
\if@solutions\else
\halfdiagramc{you}
\fi
\makeatother
\begin{enumerate}
\item Using the provided grid, draw a spacetime diagram where the left side of the rod is at $x = 0$ at $t = 0$.
\item When we switch to the rod's reference frame, space gets rotated. Draw a line in the rod's reference frame which represents the rod at time $t' = 0$, when the left side is at $x' = 0$. Call the right side of the rod at this time $P$.
\item Switching back to your reference frame, what are the spacetime coordinates $(ct,x)$ of $P$?
\vfill
\item Use the formula for proper distance to compute the length $L'$ of the rod in its own reference frame.
\vfill
\item Which is larger, $L'$ or $L$? Do moving objects shrink?
\vspace{30pt}
\end{enumerate}
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=1.5]
%\def\R{2*\xmax} % radius of clip
%\clip (-\xmin,\R) |- (\R,-\xmin) arc(0:90:\xmin+\R);
% AXES
\def\xmin{0.2}
\def\xmax{2.6}
\def\ang{33.5} % angle between x and x' axes
\def\Nlines{12} % number of world lines (at constant x/t)
\def\DNy{1} % difference in number of world lines of y axis (lengthen)
\def\DNxp{-6} % difference in number of world lines of x' axis (shorten)
\def\DNyp{-2} % difference in number of world lines of y' axis (shorten)
\def\xplabelang{170} % anchor angle of x' axis label
\axes
% SETTINGS
\pgfmathsetmacro\Lz{4*\D} % proper/rest length L0 of ladder in S'
\pgfmathsetmacro\L{cos(2*\ang)/cos(\ang)*\Lz} % contracted length L in S
\pgfmathsetmacro\yminb{-0.7*\xmin} % ymin of barn in S
\pgfmathsetmacro\xb{4.96*\d} % x coordinate of barn in S
\pgfmathsetmacro\wb{3.08*\d} % width of barn in S
\pgfmathsetmacro\yA{(\xb+0.04*\d)/tan(\ang)} % y = ct coordinate when ladder is fully in barn in S
\coordinate (L) at (\L,0); % ladder end in S
\coordinate (L') at (\ang:\Lz); % ladder end in S'
\coordinate (A) at (90-\ang:{(\xb+0.04*\d)/sin(\ang)}); % left end of ladder when fully in barn
\coordinate (B) at ($(A)+(\L,0)$); % right end of ladder when fully in barn
\coordinate (C) at (90-\ang:{(\xb+\wb+0.08*\d)/sin(\ang)}); % left end of ladder when fully passed through barn
% FILL
\begin{pgfonlayer}{back} % draw on back (behind axes)
% \fill[mydarkblue!22] % barn frame
% (\xb,\yminb) rectangle (\xb+\wb,\ymax);
\fill[mylightred] % ladder frame
(90-\ang:-\xmin) -- (90-\ang:\ymaxp) --++ (\ang:\Lz) -- (L) --++ (90-\ang:-\xmin) -- cycle;
% \begin{scope}
% \clip (0,0) rectangle(1.3*\xmax,\ymax+0.2);
% \draw[myredhighlight,line width=3.1] % highlight simultaneity in S'
% (A)++(\ang:{-\xb/cos(\ang)-0.05}) --++ (\ang:\xmax+3.6*\D)
% (B)++(\ang:{-\xb/cos(\ang)-0.05-\Lz}) --++ (\ang:\xmax+3.8*\D);
% \draw[mypurplehighlight,line width=3.1] % highlight simultaneity in S
% (0,\yA) --++ (\xmax+0.8*\d,0);
% \end{scope}
\end{pgfonlayer}
% \draw[->,thick,mydarkblue] % barn left door
% (\xb,\yminb) -- (\xb,\ymax+0.15);
% \draw[->,thick,mydarkblue] % barn right door
% (\xb+\wb,\yminb) -- (\xb+\wb,\ymax+0.15);
\draw[->,thick,mydarkbrown] % rod left end
(L)++(90-\ang:-\xmin) -- (L) -- (L') --++ (90-\ang:\ymaxp+0.2);
% LADDER
\draw[rod] (O) -- (L')
node[pos=0.55,above=2,scale=0.8] {\contour{mylightred}{$L'$}};
\draw[rod] (O) -- (L)
node[pos=0.46,below=1,scale=0.8] {$L$};
% % LADDER IN BARN
% \draw[rod] (A) --++ (L');
% \draw[rod] (B) --++ (\ang:-\Lz);
% \draw[rod] (A) --++ (L);
% % LADDER RIGHT OF BARN
% \draw[rod] (C) --++ (L');
% \draw[rod] (C) --++ (L);
% LABELS
% \node[mydarkblue,below=0,align=center,scale=0.8,yshift=1] at (\xb+\wb/2,0)
% {barn\\$L$};
% \node[mydarkpurple,right,align=left,scale=0.65,yshift=1.2] at (\xb+3.6*\d,\yA)
% {both doors closed in S};
% %{both doors\\[-3]close in S};
% \node[mydarkred,right,scale=0.65,yshift=1.8,rotate=\ang] at ($(A)+(\ang:\Lz+0.8*\D)$)
% {left door closed in S$'$};
% \node[mydarkred,right,scale=0.65,yshift=0.7,rotate=\ang] at ($(B)+(\ang:0.8*\D)$)
% {right door closed in S$'$};
\end{tikzpicture}
\end{center}
We want to calculate the spacetime position of the right side side of the rod when the left side is at the origin.
In the rest frame, the equation for the right edge of the rod is $x = vt + L$.
The equation for the spatial axis of the rod's rest frame is $c^2t = vx$.
This implies that they intersect at $ct = \frac{cvL}{c^2 - v^2}$ and $x = \frac{L}{1-v^2/c^2}$.
Calculating the proper length between the left and right side of the rod, we find that
\begin{align*}
L' = \chi = \sqrt{\frac{c^4L^2}{(c^2 - v^2)^2} - \frac{c^2 v^2 L^2}{(c^2 - v^2)}}
= L\sqrt{\frac{c^2(c^2 - v^2)}{(c^2 - v^2)^2}}
= \frac{L}{\sqrt{1 - v^2/c^2}}
\end{align*}
We note that this implies $L' > L$. i.e., moving objects get shorter
\end{solution}
\pagebreak
\problem{Ladder paradox}
Aiden and Matt have a ladder of length $2L$ that they are trying to squeeze into a barn of length $L$. Suppose that the barn has a front door and back door which can be open/closed simultaneously.
Now, Aiden is particular smart, so he gives Matt the ladder and has Matt run at the barn at speed $v = \sqrt{3}c/2$. Does the ladder fit in the barn? We'll analyze this in the new few questions.
\begin{enumerate}
\item
What is the length of the ladder from Aiden's perspective when Matt is running? Does the ladder fit in the barn?
\begin{solution}
$L$
\end{solution}
\vfill
\item As soon as Matt and the ladder are inside the barn, Aiden quickly closes and opens the doors of the barn. Success!
However, consider this from Matt's perspective. From Matt's perspective, he's holding a ladder of length $2L$ and a barn is flying at him at speed $v = \sqrt{3}c/2$. By length contraction, what is the length of the barn?
\begin{solution}
$L/2$
\end{solution}
\vfill\pagebreak
\item Despite the barn being too short, we know that the ladder has to fit! Using the provided grid, draw a spacetime diagram of the situation. Include Matt's reference frame on your diagram.
\makeatletter
\if@solutions\else
\emptydiagramc{Aiden}
\fi
\makeatother
\vfill
\item From Matt's perspective, why don't the doors of the barn crush the ladder?
\end{enumerate}
\begin{solution}
In short, from Matt's perspective, the barn doors do not close at the same time. The back door closes right when the front of the ladder reaches it, then opens again. Later, the front door closes right when the back of the ladder passes it, then opens again.
% SPACETIME DIAGRAM - LADDER PARADOX
\begin{tikzpicture}[scale=2.5]
\message{Ladder paradox^^J}
%\def\R{2*\xmax} % radius of clip
%\clip (-\xmin,\R) |- (\R,-\xmin) arc(0:90:\xmin+\R);
% AXES
\def\xmin{0.2}
\def\xmax{2.6}
\def\ang{33.5} % angle between x and x' axes
\def\Nlines{12} % number of world lines (at constant x/t)
\def\DNy{1} % difference in number of world lines of y axis (lengthen)
\def\DNxp{-6} % difference in number of world lines of x' axis (shorten)
\def\DNyp{-2} % difference in number of world lines of y' axis (shorten)
\def\xplabelang{170} % anchor angle of x' axis label
\axes
% SETTINGS
\pgfmathsetmacro\Lz{4*\D} % proper/rest length L0 of ladder in S'
\pgfmathsetmacro\L{cos(2*\ang)/cos(\ang)*\Lz} % contracted length L in S
\pgfmathsetmacro\yminb{-0.7*\xmin} % ymin of barn in S
\pgfmathsetmacro\xb{4.96*\d} % x coordinate of barn in S
\pgfmathsetmacro\wb{3.08*\d} % width of barn in S
\pgfmathsetmacro\yA{(\xb+0.04*\d)/tan(\ang)} % y = ct coordinate when ladder is fully in barn in S
\coordinate (L) at (\L,0); % ladder end in S
\coordinate (L') at (\ang:\Lz); % ladder end in S'
\coordinate (A) at (90-\ang:{(\xb+0.04*\d)/sin(\ang)}); % left end of ladder when fully in barn
\coordinate (B) at ($(A)+(\L,0)$); % right end of ladder when fully in barn
\coordinate (C) at (90-\ang:{(\xb+\wb+0.08*\d)/sin(\ang)}); % left end of ladder when fully passed through barn
% FILL
\begin{pgfonlayer}{back} % draw on back (behind axes)
\fill[mydarkblue!22] % barn frame
(\xb,\yminb) rectangle (\xb+\wb,\ymax);
\fill[mylightred] % ladder frame
(90-\ang:-\xmin) -- (90-\ang:\ymaxp) --++ (\ang:\Lz) -- (L) --++ (90-\ang:-\xmin) -- cycle;
\begin{scope}
\clip (0,0) rectangle(1.3*\xmax,\ymax+0.2);
\draw[myredhighlight,line width=3.1] % highlight simultaneity in S'
(A)++(\ang:{-\xb/cos(\ang)-0.05}) --++ (\ang:\xmax+3.6*\D)
(B)++(\ang:{-\xb/cos(\ang)-0.05-\Lz}) --++ (\ang:\xmax+3.8*\D);
\draw[mypurplehighlight,line width=3.1] % highlight simultaneity in S
(0,\yA) --++ (\xmax+0.8*\d,0);
\end{scope}
\end{pgfonlayer}
\draw[->,thick,mydarkblue] % barn left door
(\xb,\yminb) -- (\xb,\ymax+0.15);
\draw[->,thick,mydarkblue] % barn right door
(\xb+\wb,\yminb) -- (\xb+\wb,\ymax+0.15);
\draw[->,thick,mydarkbrown] % rod left end
(L)++(90-\ang:-\xmin) -- (L) -- (L') --++ (90-\ang:\ymaxp+0.2);
% LADDER
\draw[rod] (O) -- (L')
node[pos=0.55,above=2,scale=0.8] {\contour{mylightred}{$2L$}};
\draw[rod] (O) -- (L)
node[pos=0.46,below=1,scale=0.8] {$L$};
% LADDER IN BARN
\draw[rod] (A) --++ (L');
\draw[rod] (B) --++ (\ang:-\Lz);
\draw[rod] (A) --++ (L);
% LADDER RIGHT OF BARN
\draw[rod] (C) --++ (L');
\draw[rod] (C) --++ (L);
% LABELS
\node[mydarkblue,below=0,align=center,scale=0.8,yshift=1] at (\xb+\wb/2,0)
{barn\\$L$};
\node[mydarkpurple,right,align=left,scale=0.65,yshift=1.2] at (\xb+3.6*\d,\yA)
{both doors closed in S};
%{both doors\\[-3]close in S};
\node[mydarkred,right,scale=0.65,yshift=1.8,rotate=\ang] at ($(A)+(\ang:\Lz+0.8*\D)$)
{left door closed in S$'$};
\node[mydarkred,right,scale=0.65,yshift=0.7,rotate=\ang] at ($(B)+(\ang:0.8*\D)$)
{right door closed in S$'$};
\end{tikzpicture}
% SPACETIME DIAGRAM - LADDER PARADOX from perspective of S' (i.e. in the S' frame)
% \begin{tikzpicture}[scale=2.5]
% \message{Ladder paradox from the perspective of S'^^J}
% % SETTINGS
% \def\ang{-33.5} % angle between x and x' axes
% \def\Nxlines{9} % number of world lines (at constant x)
% \def\Nylines{13} % number of world lines (at constant t)
% \def\Nxplines{6} % number of world lines (at constant x')
% \def\Nyplines{10} % number of world lines (at constant t')
% \def\xmin{0.2}
% \pgfmathsetmacro\D{2.6/13} % grid size
% \pgfmathsetmacro\d{\D/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
% \pgfmathsetmacro\xmax{(\Nxlines+0.4)*\d} % maximum of x axis in S
% \pgfmathsetmacro\ymax{(\Nylines+0.4)*\d} % maximum of y = ct axis in S
% \pgfmathsetmacro\xmaxp{(\Nxplines+0.4)*\D} % maximum of x' axis in S'
% \pgfmathsetmacro\ymaxp{(\Nyplines+0.4)*\D} % maximum of y' = ct' axis in S'
% \pgfmathsetmacro\Lz{4*\D} % proper/rest length L0 of ladder in S'
% \pgfmathsetmacro\L{\Lz/cos(\ang)} % contracted length L in S
% \pgfmathsetmacro\xb{4.96*\d} % x coordinate of barn in S
% \pgfmathsetmacro\wb{3.08*\d} % width of barn in S
% \pgfmathsetmacro\yAp{(\xb+0.04*\d)*sin(\ang)*(1-cot(\ang)^2)} % y' = ct' coordinate when ladder is fully in barn in S
% \pgfmathsetmacro\yBp{\yAp+\L*sin(\ang)} % y' = ct' coordinate when ladder is fully in barn in S
% \coordinate (O) at (0,0);
% \coordinate (X) at (\ang:\xmax+0.05);
% \coordinate (T) at (90-\ang:\ymax+0.05);
% \coordinate (X') at (\xmaxp+0.15,0);
% \coordinate (T') at (0,\ymaxp+0.15);
% \coordinate (L) at (\ang:\L); % ladder end in S
% \coordinate (L') at (\Lz,0); % ladder end in S'
% \coordinate (A) at (0,\yAp); % left end of ladder when fully in barn in S
% \coordinate (B) at ($(A)+(L)$); % right end of ladder when fully in barn in S
% \coordinate (C) at (0,{(\xb+\wb+0.08*\d)*sin(\ang)*(1-cot(\ang)^2)}); % left end of ladder when fully passed through barn
% % FILL
% \fill[myfieldred]
% (-\xmin,0) -| (\xmaxp,\ymaxp) -| (0,-\xmin) -| cycle;
% \fill[mylightred] % ladder frame
% (0,-\xmin) |- (\Lz,\ymaxp) -- ($(L)+(0,-\xmin)$) -- cycle;
% \fill[mydarkblue!22] % barn frame
% (\ang:\xb)++(90-\ang:-\xmin) --++ (90-\ang:\xmin+\ymax)
% --++ (\ang:\wb) --++ (90-\ang:-\xmin-\ymax) -- cycle;
% % HIGHLIGHT DOORS OPEN/CLOSED
% \begin{scope}
% \clip (0,0) --++ (90-\ang:\ymax) -- (\xmaxp+1.8*\d,\ymaxp) --++ (0,-1.1*\ymaxp) -- cycle;
% \draw[myredhighlight,line width=3.1] % highlight simultaneity in S'
% ({\yAp*tan(\ang)-0.1},\yAp) -- (\xmaxp+1.6*\d,\yAp)
% ({\yBp*tan(\ang)-0.1},\yBp) -- (\xmaxp+1.8*\d,\yBp);
% \draw[mypurplehighlight,line width=3.1] % highlight simultaneity in S
% (A)++(\ang:-\xb-0.1) --++ (\ang:{\xb+\L+3.75*\d});
% \end{scope}
% % BOOSTED WORLD LINE GRID
% \message{ Making world lines for boosted frame...^^J}
% \foreach \i [evaluate={\x=\i*\D;}] in {1,...,\Nxplines}{
% \draw[world line] (\x,0) -- (\x,\ymaxp);
% }
% \foreach \i [evaluate={\t=\i*\D;}] in {1,...,\Nyplines}{
% \draw[world line t] (0,\t) -- (\xmaxp,\t);
% }
% % WORLD LINE GRID
% \message{ Making world lines...^^J}
% \foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nxlines}{
% \draw[world line'] (\ang:\x) --++ (90-\ang:\ymax);
% }
% \foreach \i [evaluate={\t=\i*\d;}] in {1,...,\Nylines}{
% \draw[world line'] (90-\ang:\t) --++ (\ang:\xmax);
% }
% % WORLD LINES BARN & ROD
% \draw[->,thick,mydarkblue] % barn left door
% (\ang:\xb+\wb)++(90-\ang:-\xmin) --++ (90-\ang:\xmin+\ymax+0.2);
% \draw[->,thick,mydarkblue] % barn right door
% (\ang:\xb)++(90-\ang:-\xmin) --++ (90-\ang:\xmin+\ymax+0.2);
% \draw[->,thick,mydarkbrown] % rod right end
% (L)++(0,-\xmin) -- (\Lz,\ymaxp+0.15);
% % AXES
% \draw[->,thick] (90-\ang:-\xmin) -- (T) node[below left=-1] {$ct$};
% \draw[->,thick] (\ang:-\xmin) -- (X) node[below left=0] {$x$};
% \draw[->,thick,mydarkred] (0,-\xmin) -- (T')
% node[right=3,above=-1] {$ct'$};
% \draw[->,thick,mydarkred] (-\xmin,0) -- (X')
% node[anchor=140,inner sep=0.5] {$x'$};
% % LADDER
% \draw[rod] (O) -- (L)
% node[pos=0.45,below=2,scale=0.8] {$L$};
% \draw[rod] (O) -- (L')
% node[pos=0.45,above=0.6,scale=0.8] {\contour{mylightred}{$2L$}};
% \draw[rod] (O) -- (L');
% % LADDER IN BARN
% \draw[rod] (A) --++ (L);
% \draw[rod] (A) --++ (L');
% \draw[rod] (B) --++ (-\Lz,0);
% % LADDER RIGHT OF BARN
% \draw[rod] (C) --++ (L');
% \draw[rod] (C) --++ (L);r
% % LABELS
% \node[mydarkbrown,above=1,scale=0.8] at (\Lz/2,\ymaxp)
% {rod};
% \node[mydarkblue,below=0,scale=0.8]
% at ({(\xb+\wb/2)*cos(\ang)*(1-tan(\ang)^2)+0.07},0)
% {\contour{mydarkblue!22}{barn}};
% %\node[mydarkblue,anchor=90-\ang,inner sep=2,scale=0.8,rotate=\ang] at (\ang:\xb+\L/2)
% % {barn}; %{barn\\$L$};
% \node[mydarkpurple,right,align=left,scale=0.65,yshift=1.6,rotate=\ang]
% at ($(B)+(\ang:0.4*\d)$) {both doors closed in S};
% \node[mydarkred,right,scale=0.65,yshift=1.8] at ($(A)+(\Lz+0.3*\d,0)$)
% {left door closed in S$'$};
% \node[mydarkred,right,scale=0.65,yshift=0.7] at ($(B)+(0.3*\d,0)$)
% {right door closed in S$'$};
% \end{tikzpicture}
\end{solution}
\vfill
\pagebreak

79
src/Advanced/Relativity/parts/relative velocity.tex
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@ -1,79 +0,0 @@
\section{Relative Velocity}
The first weirdness occurs if we consider multiple people running in multiple directions and speeds. Let's explore this.
You shouldn't have to do any math here besides thinking about angles.
\problem{}
Using the given diagram, verify that if, in Bob's reference frame, Alice is running away from Bob at speed $v$, then in Alice's reference frame, Bob is running away at speed $v$.
{\em Hint: slopes}
% SPACETIME DIAGRAM - LORENTZ BOOST
\begin{center}\begin{tikzpicture}[scale=1.8]
\message{Lorentz boost^^J}
\def\xmax{2}
\def\xmaxp{2.2} % maximum of rotated axis
\def\Nlines{5} % number of world lines (at constant x/t)
\pgfmathsetmacro\ang{atan(1/2)} % angle between x and x' axes
\pgfmathsetmacro\d{0.9*\xmax/\Nlines}refer % grid size
\pgfmathsetmacro\D{\d/cos(\ang)/sqrt(1-tan(\ang)^2)} % grid size, boosted
\coordinate (O) at (0,0);
\coordinate (X) at (\xmax+0.2,0);
\coordinate (T) at (0,\xmax+0.2);
\coordinate (X') at (\ang:\xmaxp+0.2);
\coordinate (T') at (90-\ang:\xmaxp+0.2);
% WORLD LINE GRID
\message{ Making world lines...^^J}
\foreach \i [evaluate={\x=\i*\d;}] in {1,...,\Nlines}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line] (-\x,-\xmax) -- (-\x,\xmax);
\draw[world line] ( \x,-\xmax) -- ( \x,\xmax);
\draw[world line t] (-\xmax,-\x) -- (\xmax,-\x);
\draw[world line t] (-\xmax, \x) -- (\xmax, \x);
}
% BOOSTED WORLD LINE GRID
\message{ Making world lines for boosted frame...^^J}
\fill[mydarkred,opacity=0.05]
(O) --++ (\ang:\xmaxp) --++ (90-\ang:\xmaxp) --++ (\ang:-\xmaxp) -- cycle;
\fill[mydarkred,opacity=0.05]
(O) --++ (\ang:-\xmaxp) --++ (90-\ang:-\xmaxp) --++ (\ang:\xmaxp) -- cycle;
\foreach \i [evaluate={\x=\i*\D;}] in {1,...,4}{
\message{ Running i/N=\i/\Nlines, x=\x...^^J}
\draw[world line'] (\ang:-\x) --++ (90-\ang:-\xmaxp);
\draw[world line'] (90-\ang:-\x) --++ (\ang:-\xmaxp);
\draw[world line'] (\ang:\x) --++ (90-\ang:\xmaxp);
\draw[world line'] (90-\ang:\x) --++ (\ang:\xmaxp);
}
% AXES
\draw[->,thick] (0,-\xmax) -- (T) node[left=-1] {Alice: $ct$};
\draw[->,thick] (-\xmax,0) -- (X) node[below=0] {$x$};
\draw[->,thick,mydarkred] (90-\ang:-\xmaxp) -- (T')
node[right=5,above=-1] {Bob: $ct'$};
\draw[->,thick,mydarkred] (\ang:-\xmaxp) -- (X') node[right=-1] {$x'$};
% ANGLES
\draw pic[->,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = X--O--X'};
\draw pic[<-,"$\theta$",draw=black,angle radius=34,angle eccentricity=1.2] {angle = T'--O--T};
% % PHOTON
% \draw[photon] (O) --++ (4*\d, 4*\d);
\end{tikzpicture}\end{center}
\begin{solution}
Extend the graph to the left and look at the slope of Alice's world line in Bob's reference frame. You'll find that Alice's slope in Bob's reference frame is the opposite of Bob's slope in Alice's reference frame.
\end{solution}
\problem{}
What if Alice runs to the left at speed $v$, Bob stays still, and Charlie runs to the right at speed $v$?
From Charlie's reference frame, is Alice running away at speed $2v$? Why or why not?
{\em Think about the consequences.}
\begin{solution}
If $v > c/2$, then this poses an immediate problem. Alice has to move at a speed below $2v$. So velocities do not just add.
\end{solution}
\vfill
\pagebreak

2
src/Advanced/Retrograde Analysis/main.tex
View File

@ -3,7 +3,7 @@
\documentclass[
solutions,
shortwarning
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{chessfss}

2
src/Advanced/Size of Sets/main.tex
View File

@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

2
src/Advanced/Stopping Problems/main.tex
View File

@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{units}

2
src/Advanced/Symmetric Groups/main.tex
View File

@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usetikzlibrary{calc}

9
src/Advanced/Tropical Polynomials/main.typ
View File

@ -1,6 +1,13 @@
#import "@local/handout:0.1.0": *
#show: handout.with(
#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",

2
src/Advanced/Tropical Polynomials/meta.toml
View File

@ -3,5 +3,5 @@ title = "Tropical Polynomials"
[publish]
handout = false
handout = true
solutions = true

4
src/Advanced/Tropical Polynomials/parts/00 arithmetic.typ
View File

@ -104,7 +104,7 @@ Is there a tropical multiplicative identity? \
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 multiplicative identity?
where $i$ is the additive identity?
])
#solution([Yes, it is $-x$. For $x != 0$, $x #tm (-x) = 0$])
@ -126,7 +126,7 @@ Fill the following tropical addition and multiplication tables
#let col = 10mm
#if_no_solutions(
#notsolution(
table(
columns: (1fr, 1fr),
align: center,

10
src/Advanced/Tropical Polynomials/parts/01 polynomials.typ
View File

@ -63,7 +63,7 @@ where all exponents represent repeated tropical multiplication.
Draw a graph of the tropical polynomial $f(x) = x^2 #tp 1x #tp 4$. \
#hint([$1x$ is not equal to $x$.])
#if_no_solutions(graphgrid(none))
#notsolution(graphgrid(none))
#solution([
$f(x) = min(2x , 1+x, 4)$, which looks like:
@ -132,7 +132,7 @@ How can we use the graph to determine these roots?
Graph $f(x) = -2x^2 #tp x #tp 8$. \
#hint([Use half scale. 1 box = 2 units.])
#if_no_solutions(graphgrid(none))
#notsolution(graphgrid(none))
#solution([
#graphgrid({
@ -140,7 +140,7 @@ Graph $f(x) = -2x^2 #tp x #tp 8$. \
let step = 0.75
dotline((0, 0), (8 * step, 8 * step))
dotline((0.5 * step, 0), (4.5 * step, 8 * step))
dotline((0.5 * step, 0), (4 * step, 8 * step))
dotline((0, 4 * step), (8 * step, 4 * step))
line(
@ -210,7 +210,7 @@ and always produces $7$ for sufficiently large inputs.
#problem()
Graph $f(x) = 1x^2 #tp 3x #tp 5$.
#if_no_solutions(graphgrid(none))
#notsolution(graphgrid(none))
#solution([
The graphs of all three terms intersect at the same point:
@ -261,7 +261,7 @@ How are the roots of $f$ related to its coefficients?
#problem()
Graph $f(x) = 2x^2 #tp 4x #tp 4$.
#if_no_solutions(graphgrid(none))
#notsolution(graphgrid(none))
#solution(
graphgrid({

18
src/Advanced/Tropical Polynomials/parts/02 cubic.typ
View File

@ -5,12 +5,12 @@
= Tropical Cubic Polynomials
#problem()
Consider the polynomial $f(x) = x^3 #tp 1x^2 #tp 3x #tp 6$. \
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 $f$.
#if_no_solutions(graphgrid(none))
#notsolution(graphgrid(none))
#solution([
- Roots are 1, 2, and 3.
@ -43,12 +43,12 @@ Consider the polynomial $f(x) = x^3 #tp 1x^2 #tp 3x #tp 6$. \
#pagebreak() // MARK: page
#problem()
Consider the polynomial $f(x) = x^3 #tp 1x^2 #tp 6x #tp 6$. \
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 $f$.
#if_no_solutions(graphgrid(none))
#notsolution(graphgrid(none))
#solution([
- Roots are 1, 2.5, and 2.5.
@ -82,7 +82,7 @@ Consider the polynomial $f(x) = x^3 #tp 6x^2 #tp 6x #tp 6$. \
- use this graph to find the roots of $f$
- write (and expand) a product of linear factors with the same graph as $f$.
#if_no_solutions(graphgrid(none))
#notsolution(graphgrid(none))
#solution([
- Roots are 2, 2, and 2.
@ -118,10 +118,10 @@ Using the last three problems, find formulas for $B$ and $C$ in terms of $a$, $b
#solution([
$
B = min(b, (a+c)/2, (2a+d)/3)
B = min(b, (a+c)/2, (2a+d)/2)
$
$
C = min(c, (b+d)/2, (a+2d)/3)
C = min(c, (b+d)/2, (a+2d)/2)
$
])
@ -153,7 +153,7 @@ What are the roots of the following polynomial?
#v(1fr)
#pagebreak() // MARK: page
#problem(label: "findci")
#problem()
If
$
f(x) = c_0 #tp c_1 x #tp c_2 x^2 #tp ... #tp c_n x^n
@ -183,7 +183,7 @@ Find a formula for each $C_i$ in terms of $c_0, c_1, ..., c_n$.
#problem()
With the same setup as @findci, \
With the same setup as the previous problem, \
find formulas for the roots $r_1, r_2, ..., r_n$.
#solution([

2
src/Intermediate/An Introduction to Graph Theory/main.tex
View File

@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

2
src/Intermediate/An Introduction to Graph Theory/tikxset.tex
View File

@ -24,7 +24,7 @@
label/.style = {
rectangle,
% For automatic red background in solutions
fill = \bgcolor,
fill = \ORMCbgcolor,
draw = none,
rounded corners = 0mm
},

2
src/Intermediate/Combinatorics/main.tex
View File

@ -2,7 +2,7 @@
% use [solutions] flag to show solutions.
\documentclass[
solutions
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

2
src/Intermediate/Instant Insanity/main.tex
View File

@ -2,7 +2,7 @@
% use [solutions] flag to show solutions.
\documentclass[
solutions
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

2
src/Intermediate/Newton's Laws/main.tex
View File

@ -2,7 +2,7 @@
% use [solutions] flag to show solutions.
\documentclass[
solutions
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

2
src/Intermediate/Probability/main.tex
View File

@ -1,6 +1,6 @@
% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[solutions]{../../../lib/tex/handout}
\documentclass[solutions]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

2
src/Intermediate/Proof Techniques/main.tex
View File

@ -3,7 +3,7 @@
\documentclass[
solutions,
singlenumbering
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\usepackage{units}

4
src/Intermediate/Proof Techniques/parts/2 induction.tex
View File

@ -190,7 +190,7 @@ For example, see the proof of the statement in \ref{binomsum} on the next page.
\pagebreak
\begin{hobox}{Alternative Proof}{ogrape!10!white}{ogrape}
\begin{ORMCbox}{Alternative Proof}{ogrape!10!white}{ogrape}
Consider the following problem: \par
How many ways are there to write a number $x$ as an ordered sum of positive integers? \par
\note{
@ -234,7 +234,7 @@ For example, see the proof of the statement in \ref{binomsum} on the next page.
We've found that the number of ways to split $x$ can be written as either
$\sum_{n = 1}^{x-1} \binom{x-1}{n}$ or $2^{x-1}$,
and therefore $\sum_{n = 1}^{x-1} \binom{x-1}{n} = 2^{x-1}$.
\end{hobox}
\end{ORMCbox}
\pagebreak

2
src/Intermediate/Slide Rules/main.tex
View File

@ -2,7 +2,7 @@
% use [solutions] flag to show solutions.
\documentclass[
solutions
]{../../../lib/tex/handout}
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

2
src/Intermediate/Vectors 1/main.tex
View File

@ -1,6 +1,6 @@
% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[solutions]{../../../lib/tex/handout}
\documentclass[solutions]{../../../lib/tex/ormc_handout}
\usepackage{adjustbox}
\usepackage{../../../lib/tex/macros}

2
src/Intermediate/Vectors 2/main.tex
View File

@ -1,6 +1,6 @@
% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[solutions]{../../../lib/tex/handout}
\documentclass[solutions]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}

35
src/Warm-Ups/A Familiar Concept/main.tex Executable file
View File

@ -0,0 +1,35 @@
\documentclass[
solutions,
hidewarning,
singlenumbering,
nopagenumber
]{../../../lib/tex/ormc_handout}
\usepackage{../../../lib/tex/macros}
\title{Warm-Up: A Familiar Concept}
\uptitler{\smallurl{}}
\subtitle{Prepared by Mark on \today}
\begin{document}
\maketitle
\problem{}<one>
Let $v = [-5, -2, 0, 1, 4, 1000]$. Find all $x$ that minimize the following metric. \par
$$
\sum_{\forall i} |v_i - x| = |v_1 - x| + |v_2 - x| + ... + |v_6 - x|
$$
\vfill
\problem{}
Let $v = [-5, -2, 0, 1, 4, 1000, 1001]$. Find all $x$ that minimize the metric in \ref{one}.
\vfill
\problem{}
What is this metric usually called?
\end{document}

33
src/Warm-Ups/A Familiar Concept/main.typ
View File

@ -1,33 +0,0 @@
#import "@local/handout:0.1.0": *
#show: handout.with(
title: [Warm-Up: A Familiar Concept],
by: "Mark",
)
#problem()
Let $v = [-5, -2, 0, 1, 4, 1000]$. Find all $x$ that minimize the following metric:
#align(
center,
box(
inset: 3mm,
$
sum_(#sym.forall i) |v_i - x| = |v_1 - x| + |v_2 - x| + ... + |v_6 - x|
$,
),
)
#v(1fr)
#problem()
Let $v = [-5, -2, 0, 1, 4, 1000, 1001]$. Find all $x$ that minimize the metric in the previous problem.
#v(1fr)
#problem()
What is this metric usually called?
#v(0.25fr)

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