polymath:0.1.0

packages/preview/polymath/0.1.0/LICENSE

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+MIT License
+
+Copyright (c) 2026 dhayer200
+
+Permission is hereby granted, free of charge, to any person obtaining a copy
+of this software and associated documentation files (the "Software"), to deal
+in the Software without restriction, including without limitation the rights
+to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
+copies of the Software, and to permit persons to whom the Software is
+furnished to do so, subject to the following conditions:
+
+The above copyright notice and this permission notice shall be included in all
+copies or substantial portions of the Software.
+
+THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
+IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
+FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
+AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
+LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
+OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
+SOFTWARE.

packages/preview/polymath/0.1.0/README.md

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+# polymath
+
+A Typst package for the Renaissance man. Write like Montaigne, typeset math like Axler — structured boxes for theorems, proofs, definitions, conjectures, essays, and notes.
+
+## Usage
+
+```typst
+#import "@preview/polymath:0.1.0": *
+
+#show: template
+
+#header(
+  name: "Kernel and Range",
+  author: "Your Name",
+  course: "Math 110 — Linear Algebra",
+  hw: "3",
+  date: "March 4, 2026",
+  professor: "Prof. Smith",
+)
+
+#qs(title: [Prove that the additive identity is unique.])[
+  #prf[
+    Suppose $0$ and $0'$ are both identities. Then $0 = 0 + 0' = 0'$.
+  ]
+]
+```
+
+See [`example.typ`](example.typ) for a full working document.
+
+## Functions
+
+### Setup
+
+| Function | Description |
+|----------|-------------|
+| `template` | Apply page, text, and math styling. Use with `#show: template`. |
+| `header(name?, note?, author?, course?, hw?, date?, professor?, ...)` | Page header with note title (`name` or `note`), author, and course metadata. |
+
+### Problem layer
+
+| Function | Description |
+|----------|-------------|
+| `qs(title?)[ ]` | Numbered question: **1.**, **2.**, … |
+| `ex(num?, loc?)[ ]` | Exercise with optional number and location. `#ex(num: 3, loc: [section 2.1])` → *Exercise 3, section 2.1* |
+| `eg(title?)[ ]` | Example box. |
+| `pt(title?)[ ]` | Lettered part: **a.**, **b.**, …; nests to **i.**, **ii.**, … |
+| `ans[ ]` | Answer / solution block. |
+
+### Knowledge layer
+
+| Function | Description |
+|----------|-------------|
+| `defn(title?)[ ]` | Definition — yellow. |
+| `notn(title?)[ ]` | Notation — yellow. |
+| `axiom(title?)[ ]` | Axiom — yellow. |
+| `postulate(title?)[ ]` | Postulate — yellow. |
+| `thm(title?)[ ]` | Theorem — blue. |
+| `lemma(title?)[ ]` | Lemma — blue. |
+| `prop(title?)[ ]` | Proposition — blue. |
+| `cor(title?)[ ]` | Corollary — blue. |
+| `conj(title?)[ ]` | Conjecture — amber (unproven). |
+| `prf[ ]` | Proof block with flush-right QED mark — green. |
+| `note[ ]` | Remark with blue left rule. |
+
+### Essay / Montaigne layer
+
+| Function | Description |
+|----------|-------------|
+| `aside[ ]` | Tangential digression with grey left rule. |
+| `epigraph(attribution?)[ ]` | Centered opening quote, italic, with optional attribution. |
+| `blockquote(attribution?)[ ]` | Left-ruled block quote with optional attribution. |
+
+## License
+
+MIT

packages/preview/polymath/0.1.0/example.typ

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+#import "lib.typ": *
+
+#show: template
+
+#header(
+  name: "Rank, Nullity, and Injectivity",
+  author: "Deep Hayer",
+  course: "Math 110 — Linear Algebra",
+  date: "March 25, 2026",
+  professor: "Prof. Sheldon Axler",
+)
+
+#epigraph(attribution: [Polymath example])[
+  A good set of notes should make room for proofs, calculations, and the side
+  remarks that explain why the proof was worth writing down.
+]
+
+#defn(title: [null space and range])[
+  Let $T : V -> W$ be linear. The _null space_ of $T$ is
+  $"null" T = {v in V : T(v) = 0}$ and the _range_ of $T$ is
+  $"range" T = {T(v) : v in V}$.
+]
+
+#notn(title: [coordinate maps])[
+  When $V = RR^n$ and $W = RR^m$, a linear map is often easiest to study by
+  writing explicit equations for its coordinates.
+]
+
+#lemma(title: [injectivity through the null space])[
+  A linear map $T : V -> W$ is injective if and only if $"null" T = {0}$.
+]
+
+#prf[
+  If $T$ is injective and $v in "null" T$, then $T(v) = 0 = T(0)$, so $v = 0$.
+  Conversely, if $"null" T = {0}$ and $T(u) = T(v)$, then
+  $T(u - v) = 0$, hence $u - v = 0$ and $u = v$.
+]
+
+#thm(title: [rank-nullity])[
+  If $V$ is finite-dimensional and $T : V -> W$ is linear, then
+  $dim V = dim "null" T + dim "range" T$.
+]
+
+#cor(title: [dimension blocks injectivity])[
+  If $dim V > dim W$, then no linear map $T : V -> W$ can be injective.
+]
+
+#prf[
+  If $T$ were injective, then $"null" T = {0}$ by the lemma, so
+  $dim "null" T = 0$. Rank-nullity would give
+  $dim V = dim "range" T <= dim W$, a contradiction.
+]
+
+#note[
+  This theorem is valuable because it turns a structural question such as
+  injective or surjective into a short dimension count.
+]
+
+#eg(title: [projection onto the first two coordinates])[
+  Let $P : RR^3 -> RR^2$ be given by $P(x, y, z) = (x, y)$. Then
+  $"null" P = {(0, 0, z) : z in RR}$, so $dim "null" P = 1$, and
+  $"range" P = RR^2$, so $dim "range" P = 2$.
+]
+
+#qs(title: [Let $T : RR^3 -> RR^2$ be defined by $T(x, y, z) = (x + y, y + z)$.])[
+  #pt(title: [Find the null space of $T$.])[
+    #ans[
+      We solve
+      $ x + y = 0 quad and quad y + z = 0. $
+      Hence $x = -y$ and $z = -y$, so every vector in the null space has the
+      form $(-t, t, -t)$ for some $t in RR$. Therefore
+      $ "null" T = {(-t, t, -t) : t in RR}. $
+    ]
+  ]
+
+  #pt(title: [Is $T$ injective? Is it surjective?])[
+    #ans[
+      The null space is not trivial, so $T$ is not injective.
+      To show surjectivity, let $(a, b) in RR^2$ and choose $(x, y, z) = (a, 0, b)$.
+      Then $T(a, 0, b) = (a, b)$, so every vector in $RR^2$ lies in the range.
+      Thus $T$ is surjective.
+    ]
+  ]
+]
+
+#ex(num: 7, loc: [section 3.2])[
+  #ans[
+    Suppose $S : RR^2 -> RR^2$ is injective. Then $"null" S = {0}$, so
+    rank-nullity gives $2 = 0 + dim "range" S$. Hence $dim "range" S = 2$,
+    which forces $"range" S = RR^2$. Therefore $S$ is surjective.
+  ]
+]
+
+#aside[
+  Coordinate computations are not a retreat from abstraction. They are often
+  the quickest way to see which dimensions are being forced.
+]
+
+#blockquote(attribution: [A margin note to future you])[
+  When the algebra starts to branch, count dimensions before expanding formulas.
+]