Make it compile

Makefile

@@ -21,6 +21,3 @@ build:
 	cd $(BUILD_DIR) && ln -sf ../img
 	cd $(BUILD_DIR) && ln -sf ../mathematics.sty
 	cd $(BUILD_DIR) && for f in ../*.{tex,pdf}; do ln -sf $$f; done
-
-
-test:

discrete-math.tex

@@ -174,7 +174,7 @@ \section{Combinatorics}
 \end{proof}
 
 \begin{claim}
-  Consider the assignment of $n$ items $x_1, x_2, \ldots, x_n$ to $k$ buckets. Define $S_i$ to be the sum of items assigned to bucket $i$. The assignments for which $\max_i S_i$ is minimized is the assignment for which $\var S_i$ is minimized.
+  Consider the assignment of $n$ items $x_1, x_2, \ldots, x_n$ to $k$ buckets. Define $S_i$ to be the sum of items assigned to bucket $i$. The assignments for which $\max_i S_i$ is minimized is the assignment for which $\Var S_i$ is minimized.
 \end{claim}
 
 Not true?

foundations.tex

@@ -336,13 +336,11 @@ \subsection*{\url{https://www.reddit.com/r/explainlikeimfive/comments/61siyw/eli
   then $\gcd(a, b) = 2^2 = 4$ and $\lcm(a, b) = 2^3 \cdot 3 \cdot 5 = 120$.
 
   This can be written as a general theorem involving mins and maxes in the exponents of a product of primes.
-
-
 \end{remark*}
 
 \begin{theorem*}
   $\gcd(a,b) \times \lcm(a,b) = ab$
-\end{theorem}
+\end{theorem*}
 
 
 \section{$\sqrt{2}$ is irrational}

linear-algebra.tex

@@ -937,7 +937,7 @@ \section{Computing the n-th Fibonacci number: generating function}
   \varphi^n &= f_{n-1} + f_n\varphi \\
   \psi^n &= f_{n-1} + f_n\psi.
 \end{align*}
-\end{claim}
+\end{lemma}
 
 \begin{proof}
   If we define $f_0 = 0$ then the $\varphi$ claim is true for $n=1$ ($\varphi^1 = 0 + 1\varphi$). It's also true for $n=2$ ($\varphi^2 = 1  + 1\varphi$).
@@ -1100,19 +1100,31 @@ \section{Computing the n-th Fibonacci number: generating function}
 
 \begin{mdframed}
 \includegraphics[width=400pt]{img/linear-algebra--vector-spaces-and-fields--computing-the-n-th-fibonacci-number-generating-function-26b7.png}
+\end{mdframed}
+
 \begin{mdframed}
 \includegraphics[width=400pt]{img/linear-algebra--vector-spaces-and-fields--computing-the-n-th-fibonacci-number-generating-function-03cb.png}
+\end{mdframed}
+
 \begin{mdframed}
 \includegraphics[width=400pt]{img/linear-algebra--vector-spaces-and-fields--computing-the-n-th-fibonacci-number-generating-function-f048.png}
+\end{mdframed}
+
 \begin{mdframed}
 \includegraphics[width=400pt]{img/linear-algebra--vector-spaces-and-fields--computing-the-n-th-fibonacci-number-generating-function-535e.png}
+\end{mdframed}
+
 \begin{mdframed}
 \includegraphics[width=400pt]{img/linear-algebra--vector-spaces-and-fields--computing-the-n-th-fibonacci-number-generating-function-63d4.png}
+\end{mdframed}
+
 \begin{mdframed}
 \includegraphics[width=400pt]{img/linear-algebra--vector-spaces-and-fields--computing-the-n-th-fibonacci-number-generating-function-eaaa.png}
+\end{mdframed}
+
 \begin{mdframed}
 \includegraphics[width=400pt]{img/linear-algebra--vector-spaces-and-fields--computing-the-n-th-fibonacci-number-generating-function-e844.png}
-\end{mdframed}\end{mdframed}\end{mdframed}\end{mdframed}\end{mdframed}\end{mdframed}\end{mdframed}
+\end{mdframed}
 
 
 {\bf Check:}
@@ -1573,7 +1585,7 @@ \subsubsection{2. Find inverse of $V$}
 \end{align*}
 
 
-\subsubsection{3. Find the matrix product $V^{-1}AV$
+\subsubsection{3. Find the matrix product $V^{-1}AV$}
 
 Before getting lost in calculations, let's remember what this is. It's a
 matrix that does the $A$ transformation, but \textit{in the coordinate system defined

mathematics.sty

@@ -14,7 +14,8 @@
 \usepackage{amssymb}
 \usepackage{amsthm}
 \usepackage{array}
-\usepackage{bbm}
+% \usepackage{bbm}
+\usepackage{braket}
 \usepackage{caption}
 \usepackage{centernot}
 \usepackage{xcolor}
@@ -32,11 +33,12 @@
 \usepackage{mathrsfs}
 \usepackage{mathtools}
 \usepackage{mdframed}
-%\usepackage[outputdir=.build]{minted} \setminted{fontsize=\footnotesize}
-%\usepackage{MnSymbol} % \upmodels
+\usepackage[outputdir=.build]{minted} \setminted{fontsize=\footnotesize}
+\usepackage{MnSymbol} % \upmodels
 \usepackage{parskip}
 \usepackage{relsize} % \mathlarger
 \usepackage{ulem} % sout
+% \usepackage{upgreek}
 \usepackage{verbatim}
 \usepackage{tikz}
 \usetikzlibrary{graphs,graphdrawing}
@@ -407,6 +409,9 @@
 \newcommand{\correct}{\red{\checkmark}}
 \newcommand{\grad}{\nabla}
 
+\newcommand{\va}[1]{\overrightarrow{#1}}
+
+
 % https://tex.stackexchange.com/questions/2233/whats-the-best-way-make-an-augmented-coefficient-matrix$
 \newenvironment{amatrix}[1]{%
   \left(\begin{array}{@{}*{#1}{c}|c@{}}

mathematics.tex

@@ -42,7 +42,7 @@ \chapter{Differential Equations}
 \chapter{Complex Analysis}
 \include{analysis--complex-analysis--berg}
 \include{analysis--complex-analysis--wittens}
-% \include{analysis--complex-analysis-math185}
+\include{analysis--complex-analysis-math185}
 
 \chapter{Calculus of variations}
 \include{calculus-of-variations}
@@ -58,8 +58,8 @@ \chapter{Classical Mechanics}
 \include{physics--classical-mechanics--taylor}
 \include{physics--classical-mechanics--morin--questions}
 
-% \chapter{Quantum Mechanics}
-% \include{physics--quantum-mechanics}
+\chapter{Quantum Mechanics}
+\include{physics--quantum-mechanics}
 
 \chapter{Probability, Statistical Inference and Machine Learning}
 \include{statistics-and-machine-learning--pearl--causality}

statistics-and-machine-learning--cs189-hw.tex

@@ -4,8 +4,8 @@
 \renewcommand{\N}{\mathcal{N}}
 \newcommand{\Pbf}{\textbf{P}}
 \renewcommand{\R}{\mathbb{R}}
-\newcommand{\Var}{\text{Var}}
-\newcommand{\Cov}{\text{Cov}}
+\renewcommand{\Var}{\text{Var}}
+\renewcommand{\Cov}{\text{Cov}}
 \renewcommand{\mat}{\mathbf}
 
 \section{Homework 2}
@@ -609,8 +609,8 @@ \subsection{Gaussian Classification}
 
 \newpage
 \subsection{Maximum Likelihood Estimation}
-Let $X$ be a discrete random variable which takes values in $\{1, 2, 3\}$ with probabilities $P(X = 1) = p_1, P(X=2) = p_2,$ and $P(X = 3) = p_3$, where $p_1 + p_2 + p_3 = 1$.  Show how to use the method of maximum likelihood to estimate $p_1, p_2,$ and $p_3$ from $n$ observations of $X: x_1, \ldots, x_n$. Express your answer in terms of the counts $$k_1 = \sum_{i=1}^n \mathbbm{1}(x_i = 1), k_2 = \sum_{i=1}^n \mathbbm{1}(x_i = 2), \text{ and }k_3 = \sum_{i=1}^n \mathbbm{1}(x_i = 3),$$ where
-$$\mathbbm{1}(x = a) =
+Let $X$ be a discrete random variable which takes values in $\{1, 2, 3\}$ with probabilities $P(X = 1) = p_1, P(X=2) = p_2,$ and $P(X = 3) = p_3$, where $p_1 + p_2 + p_3 = 1$.  Show how to use the method of maximum likelihood to estimate $p_1, p_2,$ and $p_3$ from $n$ observations of $X: x_1, \ldots, x_n$. Express your answer in terms of the counts $$k_1 = \sum_{i=1}^n \mathbb{1}(x_i = 1), k_2 = \sum_{i=1}^n \mathbb{1}(x_i = 2), \text{ and }k_3 = \sum_{i=1}^n \mathbb{1}(x_i = 3),$$ where
+$$\mathbb{1}(x = a) =
 \begin{cases}
 1 & \text{if } x = a \\
 0 & \text{if } x \neq a.
@@ -2217,4 +2217,3 @@ \subsection{Gradient with respect to weight matrix $\V$}
 \begin{mdframed}
   kaggle: \texttt{dandavison7} 0.88577
 \end{mdframed}
-

statistics-and-machine-learning--pearl--causality.tex

@@ -215,7 +215,7 @@ \subsection{Conditional independence and graphoids}
 
 Conditional independence between subsets obeys some fairly unsurprising laws:
 
-\begin{itemize}
+
   \begin{table}[!h]
     \centering
     \begin{tabular}{|l|l l l l}
@@ -227,7 +227,7 @@ \subsection{Conditional independence and graphoids}
   \end{tabular}
     \caption{Properties of conditional independence relation between subsets}
   \end{table}
-\end{itemize}
+
 
 \begin{itemize}
 \item Graph >