Vincent Tam VincentTam

I was puzzled ats \expandafter, and I started with simple observations. Here're AI-generated explanations.

The command \expandafter is a low-level TeX primitive that controls the order of macro expansion. Its job is to force the expansion of a token that is not immediately next to it.

In the provided code, \expandafter is used to ensure that the recursive call \rrep receives a concrete number as its first argument, rather than an unexpanded macro sequence.

Let's break down the key line from the \rrep macro:

\expandafter\rrep\expandafter{\number\numexpr#1-1\relax}{\rep{#2}}

First attempt

Theoretical settings

$A = A_0 = \lbrace x^2 + y^2 \le 1 \rbrace$ ← unit circle
$B = B_0 = \lbrace \max(|x|,|y|) \le 1 \rbrace$ ← unit square
$f(x,y) = (x,y)/\sqrt2$ ← injection from $A$ to $B$ (shrinking by $1/\sqrt2$)
$g(x,y) = (x,y)/\varphi$ ← injection from $B$ to $A$ (shrinking by $1/\varphi$, where $\varphi$ is the golden ratio)

This gives a decreasing sequence of subsets of $A$ and $B$

$A_n = g(B_{n-1})$ is
a circle with radius $(\sqrt2)^{-k} \varphi^{-k}$ if $n = 2k$

从前，有位富翁，他的家产有 $1 （一整份家产）他的大儿子分走了当中的 1/2 （一半），富翁剩下 $1/2 他的二儿子分走了剩下的 1/2 ，富翁剩下 $1/4 他的三儿子分走了剩下的 1/2 ，富翁剩下 $1/8 他的四儿子分走了剩下的 1/2 ，富翁剩下 $1/16

如此类推

由於 (1 + 1/n)ⁿ 是二項式的 n 次方，要了解它的特性（如剛才展示的上下限），需要用上二項式定理。上面不等式最具技巧的一步在於藍色分母從 n! 變成 2ⁿ⁻¹。小問題：為何上標有 "-1"？重點在於 n! = 1 ⋅ 2 ⋅ 3 ⋯ (n - 1) n 中，當 n! 變成 (n + 1)! 時，額外乘上的數字 n + 1 是隨 n 而增長，但 2ⁿ 變成 2ⁿ⁺¹ 時，額外乘上的數字永遠是 2，所以當 n「足夠大」（n > 3）時，n! > 2ⁿ。取倒數後不等式反轉，於是出現 1/n! < 1/2ⁿ，然後就可以運用等比數列和公式，為 (1 + 1/n)ⁿ 給出上限。

上面實驗發現 n 越大，(1 + 1/n)ⁿ 也越大。即若 m < n，(1 + 1/m)ᵐ < (1 + 1/n)ⁿ。這種數列我們叫__嚴格遞增數列__。（「嚴格遞增」比「單調遞增」「嚴格」，只接受「>」，不接受「=」。）練習：試證 (1 + 1/n)ⁿ < (1 + 1/(n + 1))ⁿ⁺¹ 提示：

如上面般展開 (1 + 1/n)ⁿ，分離最簡單兩項。
其餘 n - 1 項中，調整分子分母配對，先把 1 / k! 放在左邊。

	#set page(height: auto, width: auto, margin: 1em)
	#import "@preview/cetz:0.4.1"

	#let n = 4
	#let vsep = 1
	#let width = 1
	#let lefts = range(2, n + 1).fold(
	((1/3,),),
	(a, e) => (..a, a.last().map(x => (x - 2calc.pow(1/3, e), x + 4calc.pow(1/3, e))).flatten())
	)

	\newcommand*\mytablecontents{}
	\gappto\mytablecontents{$n$ & $N=10^n$ & $(1+1/N)^N$}
	\gappto\mytablecontents{\\\hline}
	\foreach \i in {1,...,5}{
	\FPeval\ii{pow(\i,10)}
	\FPeval\ii{round(ii:0)}
	\FPeval\ieval{round((1+1/ii)^ii:8)}
	\xappto\mytablecontents{\i & $\ii$ & $\ieval$}
	\gappto\mytablecontents{\\\hline}
	}

	#import "@preview/touying:0.6.1": *
	#import themes.simple: *
	#show: simple-theme.with(aspect-ratio: "16-9")

	#import "@preview/theorion:0.3.3": *
	#show: show-theorion

	#import "@preview/cetz:0.3.4"
	#import "@preview/mannot:0.3.0"

	// format the paper
	#set page(
	paper: "us-letter",
	margin: (x: 1in, y: 1in),
	numbering: "1/1",
	header: [
	#grid(columns: (1fr, 1fr, 1fr),
	align(left)[BamDone Exmaple-002.typ],
	align(center)[Center],
	align(right)[Right]

	\documentclass[dvipsnames]{beamer}
	\usetheme{Madrid}
	\usepackage{annotate-equations}
	\renewcommand{\eqnhighlightshade}{30}
	\usepackage{gensymb}
	\usepackage{quiver} % for "swap pathways"
	\usepackage{cancel}
	\usetikzlibrary{graphs.standard, quotes} % for coins
	\usepackage{wrapfig} % for coins
	\usepackage{extpfeil} % for extensible arrows

	Hello world!

	I'm typing here so that inline equaitons like $x=3$ appears in real-time.

	$
	f(x) = sin(x)
	$

	$
	tan(x) (4x^2)/(3x+4)