From 230946b2aaec52b371ff6b6bd920cc3ebb4732c7 Mon Sep 17 00:00:00 2001
From: Benjamin Barenblat <bbaren@mit.edu>
Date: Mon, 14 Mar 2011 23:46:17 -0400
Subject: Initial commit

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+%% 18.022 cheat sheet
+%%
+%% Copyright (C) 2009, 2010, 2011 Benjamin Barenblat
+%%                                http://benjamin.barenblat.name/
+%%
+%% This document is licensed under the Creative Commons
+%% Attribution-NonCommercial-ShareAlike 3.0 United States License.
+%% For more information, see
+%% http://creativecommons.org/licenses/by-nc-sa/3.0/us/.
+%%
+%% This document is designed to be typeset with pdfLaTeX.
+\documentclass[10pt,landscape]{article}
+\usepackage{eco}
+\usepackage{geometry}
+\usepackage{multicol}
+\usepackage{mathtools}
+\usepackage{color}
+\usepackage[colorlinks]{hyperref}
+\definecolor{darkred}{rgb}{0.5,0,0}
+\hypersetup{colorlinks,linkcolor=black,urlcolor=darkred}
+\usepackage{graphicx}
+\usepackage{wrapfig}
+\usepackage{18.022}
+\usepackage{esint}
+\usepackage{titlesec}
+
+% Only show equation numbers of referenced equations.
+\mathtoolsset{showonlyrefs=true}
+
+% Set margin.
+\geometry{margin=6mm}
+
+% Turn off header and footer.
+\pagestyle{empty}
+
+% Don't print subsection numbers.
+\setcounter{secnumdepth}{0}
+
+%% Squash paragraph headings.
+\titlespacing*{\paragraph}{0pt}{.94em}{*1}
+
+\setlength{\parindent}{0pt}
+\setlength{\parskip}{0pt}
+
+\makeatletter
+\newcommand{\@recitation}{}
+\newcommand{\recitation}[1]{\renewcommand{\@recitation}{#1}}
+\renewcommand{\maketitle}{%
+  \begin{center}
+    \LARGE{\textbf{\@title}}
+  \end{center}
+}
+\newcommand{\makeend}{%
+  \vfill
+  \rule{0.3\linewidth}{0.25pt}
+  \scriptsize
+  \begin{tabular}{@{}l}
+    Copyright \copyright\ \@date\ \@author.  \href{http://creativecommons.org/licenses/by-nc-sa/3.0/us/}{\textsc{cc by$\cdot$nc$\cdot$sa}}.  No warranty.\\
+    \url{http://benjamin.barenblat.name/}
+  \end{tabular}
+}
+\makeatother
+
+\title{18.022 Cheat Sheet}
+\author{Benjamin Barenblat}
+\date{2009--2011}
+
+\begin{document}
+\raggedright
+\begin{multicols*}{3}
+\maketitle
+
+\section{Vectors}
+\paragraph{Dot product}
+Let $\vec v, \vec w, \vec u \in \R^n$.
+Then, $\vec v \dot \vec w = \langle \vec v, \vec w \rangle = \langle\vec v \vert \vec w\rangle = \bra v_1w_1 & \cdots & v_nw_n \ket$.
+
+\paragraph{Cauchy-Schwarz inequality}
+Let $\vec v, \vec w \in \R^n$.
+Then, $(\vec v \dot \vec w)^2 \le (\vec v \dot \vec v)(\vec w \dot \vec w) \iff \abs{\vec v \dot \vec w} \le \norm{\vec v}\norm{\vec w}$.
+
+\paragraph{Triangle inequality}
+Let $\vec v, \vec w \in \R^n$.
+Then, $\norm{\vec v + \vec w} \le \norm{\vec x} + \norm{\vec y}$ and $\norm{\vec v - \vec w} \ge \big\lvert\norm{\vec v} - \norm{\vec w}\big\rvert$.
+
+\paragraph{Projection}
+For $\vec a, \vec b \in \R^n$, the projection of $\vec b$ onto $\vec a$
+\begin{equation}
+  \label{eq:projection}
+  \proj_{\vec a}\vec b = \frac{\vec a \dot \vec b}{\norm{\vec a}}\vec{\hat a} = \frac{\vec a \dot \vec b}{\norm{\vec a}^2}\vec a = \frac{\vec a \dot \vec b}{\vec a \dot \vec a}\vec a
+\end{equation}
+and $\norm{\proj_{\vec a}\vec b} = \norm{\vec b\cos\theta}$.
+
+\paragraph{Distance from a point to a line}
+Given a point $\vec p$ and a line $\vec l(t) = \vec vt + \vec q$, the shortest vector from $\vec p$ to $\vec l$ is
+\begin{equation}
+  \vec q - \vec p - \proj_{\vec v}(\vec q - \vec p) = \vec q - \vec p - \frac{(\vec q - \vec p) \dot \vec v}{\norm{\vec q - \vec p}^2}\vec v.
+\end{equation}
+
+\paragraph{Cross product}
+Let $\vec v, \vec w \in \R^3$.
+Then,
+\begin{equation}
+  \label{eq:crossproduct}
+  \vec v \cross \vec w = \bra v_1 \\ v_2 \\ v_3 \ket \cross \bra w_1 \\ w_2 \\ w_3 \ket = \bra v_2w_3 - v_3w_2 \\ v_3w_1 - v_1w_3 \\ v_1w_2 - v_2w_1 \ket
+\end{equation}
+and
+$\norm{\vec v \cross \vec w} = \norm{\vec v}\norm{\vec w}\abs{\sin\theta}$.
+
+\paragraph{Planes}
+Let $\vec x = \bra x_1 & \cdots & x_n \ket \in \R^n$.
+For a point $\vec p$ and normal vector $\vec n$, $(\vec x - \vec p) \dot \vec n = 0$.
+\begin{list}{\textbullet}{\setlength{\itemsep}{0pt}}
+\item Point and two vectors: $\vec n = \vec u \cross \vec v$.
+\item Three points: $\vec n = (\vec q - \vec p) \cross (\vec r - \vec p) = 0$.
+\item Function and point: $\vec n = \grad f(\vec p)$.
+\end{list}
+
+\paragraph{Triple scalar product / determinant}
+For $\vec u, \vec v, \vec w \in \R^3$,
+\begin{equation}
+  \vec u \dot \vec v \cross \vec w = \det \bra \vec u & \vec v & \vec w \ket
+\end{equation}
+is the volume of the parallelepiped spanned by $\vec u$, $\vec v$, and $\vec w$.
+$(\vec u, \vec v, \vec w)$ is right-handed iff $\det \bra \vec u & \vec v & \vec w \ket > 0$.
+
+\paragraph{Matrix Multiplication}
+Dimensionally, the number of columns in the first matrix must match the number of rows in the second.
+\begin{wrapfigure}[5]{r}{1.4in}
+  \includegraphics[width=1.4in]{matrix-multiplication}
+  \scriptsize Copyright \textcopyright\ 2009 \href{http://en.wikipedia.org/wiki/User:Fangfufu}{Fangfufu}.\\\textsc{\href{http://creativecommons.org/licenses/by-sa/3.0/}{cc by$\cdot$sa}}.
+\end{wrapfigure}
+\begin{multline}
+  \bra a & b & c \\ d & e & f \\ g & h & i \ket \bra \alpha & \delta \\ \beta & \epsilon \\ \gamma & \zeta \ket = \\ \bra a\alpha + b\beta + c\gamma & a\delta + b\epsilon + c\zeta \\ d\alpha + e\beta + f\gamma & d\delta + e\epsilon + f\zeta \\ g\alpha + h\beta + i\gamma & g\delta + h\epsilon + i\zeta\ket
+\end{multline}
+$a_{1,2}$ refers to the element in the $1$st row, $2$nd column.
+Generally, for two matrices $A$ and $B$,
+\begin{equation}
+  (AB)_{i,j} = A_{i,1} + B_{1,j} + A_{i,2}B{2,j} + \cdots + A_{i,n}B_{n,j}.
+\end{equation}
+
+\section{Differential calculus}
+\paragraph{Gradient}
+For a scalar field $f : \R^n \to \R$, the gradient of $f$
+\begin{equation}
+  \grad f = \bra D_1f \\ D_2f \\ D_3f \\ \vdots \\ D_nf \ket,
+\end{equation}
+a new vector field which consistently points in the direction of $f$'s greatest increase with magnitude equal to the rate of that increase.
+By corollary, $\grad f$ is always perpendicular to $f$'s level curves.
+
+\paragraph{Divergence}
+For a vector field $\vec F : \R^n \to \R^n$, the divergence of $\vec F$
+\begin{equation}
+  \div\vec F = D_1F_1 + D_2F_2 + D_3F_3 + \cdots + D_nF_n,
+\end{equation}
+a new scalar field.
+Positive values of $\div\vec F$ indicate field sources, while negative values indicate field sinks.
+
+\paragraph{Curl}
+For a vector field $\vec F : \R^3 \to \R^3$, the curl of $\vec F$
+\begin{equation}
+  \curl\vec F = \bra D_2F_3 - D_3F_2 \\ D_3F_1 - D_1F_3 \\ D_1F_2 - D_2F_1 \ket,
+\end{equation}
+a new vector field measuring the rate of rotation at each point.
+
+\paragraph{Laplacian}
+For a scalar field $f : \R^n \to \R$, the Laplacian of $f$
+\begin{equation}
+  \nabla^2 f = \div\grad f = D_1^2f + D_2^2f + D_3^2f + \cdots + D_n^2f.
+\end{equation}
+
+\paragraph{Conservative vector fields}
+Let $\vec F : U \subseteq \R^n \to \R^n$ ($U$ open) be a vector field
+of class $C^1$.  If there exists a class $C^2$ scalar field $f : U \to
+\R$ such that $\vec F = \grad f$ on $U$, then $\vec F$ is conservative
+on $U$.  If $\vec F$ is conservative on $U$, then $\vec F$ is also
+curl-free on $U$.  (The converses are true if $U$ is simply connected.)
+
+\paragraph{Chain rule}
+If $\vec f : \R^n \to \R^m$ is differentiable at $\vec x$ and $\vec g
+: \R^m \to \R^p$ is differentiable at $\vec f(\vec x)$, then $\vec g
+\circ \vec f$ is differentiable at $\vec x$ and
+\begin{equation}
+  \label{eq:chainrule}
+  D(\vec g \circ \vec f)_{\vec x} = \big(D\vec g_{\vec f(\vec x)}\big)(D\vec f_{\vec x}).
+\end{equation}
+
+\paragraph{Implicit function theorem}
+Let $\vec F : \R^{n+m} \to \R^m$ (i.e., $m$ functions in $n + m$ unknowns) be of class $C^1$ and let $\vec F(\vec x_0) = \0$ for some $\vec x_0 \in \R^{n+m}$.
+Write $\vec x = (\vec a, \vec b)$, where $\vec a \in \R^n$ and $\vec b \in \R^m$; write $\vec x_0 = (\vec a_0, \vec b_0)$, where $\vec a_0 \in \R^n$ and $\vec b_0 \in \R^m$.
+Note that $D\vec F = \bra D_{\vec a}\vec F & D_{\vec b}\vec F \ket$.
+If $D_{\vec b}\vec F(\vec b_0)$ is invertible (i.e., $\det D_{\vec b}\vec F(\vec b_0) \ne 0$), then there exists a neighborhood $U$ of $\vec a_0$ in $\R^n$ and a neighborhood $V$ of $\vec b_0$ in $\R^m$ and a function $\vec f : U \to V$ such that $F\big(\vec a_0, f(\vec a_0)\big) = 0$.
+$\vec f$ expresses $\vec b$ in terms of $\vec a$ in the neighborhood of $(\vec a_0, \vec b_0)$, and
+\begin{equation}
+  \label{eq:implicitdiff}
+  D\vec f_{\vec a_0} = -\big(D_{\vec b}\vec F(\vec b_0)\big)^{-1}\big(D_{\vec a}\vec F(\vec a_0)\big).
+\end{equation}
+
+\paragraph{Taylor's theorem}
+The $k$th-order Taylor polynomial of a class $C^k$ function $f : \R^2 \to \R$ at $\vec x \in \R^2$ near a point $\vec a \in \R^2$
+\begin{equation}
+  T^kf_{\vec a}(\vec x - \vec a) = \sum_{\substack{n,m\\n+m \le k}} \frac{D_1^nD_2^m f(a_1,a_2)}{n!m!}(x - a_1)^n(y - a_2)^m.
+\end{equation}
+
+\paragraph{Extrema}
+Consider a function $f : \R^n \to \R$ of class $C^2$.
+$f$ has critical points where $Df = \0$; if $f$ has local extrema, they will occur at critical points.
+At each critical point,
+\begin{itemize}
+\item If $D^2f$'s minors are all positive, $D^2f$ is positive definite and the point is a local minimum.
+\item If $-D^2f$'s minors are all positive, $D^2f$ is negative definite and the point is a local maximum.
+\item If neither of these are true, but $D^2f$ is invertible, $D^2f$ is indefinite and the point is a saddle point.
+\item If $D^2f$ is not invertible, then the point is degenerate.
+\end{itemize}
+If a function $f : K \to \R$ is of class $C^1$ on $K$ and continuous on the interior of $\partial K$ and $K$ is closed and bounded, then $f$ has at least one minimum and at least one maximum on $K$.
+The extrema will occur inside $K$ at critical points or somewhere on $\partial K$.
+
+\paragraph{Constrained optimization (one constraint)}
+Consider a differentiable function $f : \R^n \to \R$ on a set $\mathscr{C} \subseteq \R$ defined as the level set of some function -- i.e., for some $C^1$ function $g$, $\mathscr{C} = \{ \vec x \in \R^n : g(\vec x) = 0 \}$.
+Define $L(\vec x, \lambda) = f(\vec x) - \lambda g(\vec x)$.
+If, at a point $\vec x_0$, $\grad L(\vec x_0, \lambda_0) = 0$ for some constant $\lambda_0$ and $\grad g(\vec x_0) \ne 0$, then $f\vert_\mathscr{C}$ has a critical point at $\vec x_0$.
+
+\paragraph{Constrained optimization ($k$ constraints)}
+Consider a differentiable function $f : \R^n \to \R$ on a set $\mathscr{C} \subseteq \R$ defined as the level set of some family of functions -- i.e., for some $C^1$ function $\vec g = \bra g_1 & g_2 & \cdots & g_n \ket$, $\mathscr{C} = \{ \vec x \in \R^n : \vec g(\vec x) = \0 \}$.
+Define $L(\vec x, \boldsymbol\lambda) = f(\vec x) - \boldsymbol\lambda \dot \vec g(\vec x)$.
+If, at a point $\vec x_0$, $\grad L(\vec x_0, \boldsymbol\lambda_0) = 0$ for some constant $\boldsymbol\lambda_0$ and $\bra \grad g_1(\vec x_0) & \cdots & \grad g_k(\vec x_0) \ket$ has a $k\times k$ submatrix that is invertible, then $f\vert_\mathscr{C}$ has a critical point at $\vec x_0$.
+
+\section{Integral calculus}
+\paragraph{Double integrals}
+Let $f : \R^2 \to \R$ be Riemann integrable on some nice domain $D \subseteq \R^2$.
+Define $a, b, c, d \in \R$ such that $[a,b] \times [c,d]$ is $D$'s bounding box.
+Now define $f$'s extension
+\begin{equation}
+  f^{\text{ext}}(x,y,z) = \begin{cases}
+    f(x,y) & \text{if $(x,y) \in D$},\\
+    0 & \text{otherwise}.
+    \end{cases}
+\end{equation}
+Then,
+\begin{equation}
+  \iint_D f(x,y) dxdy = \int_a^b \int_c^d f^{\text{ext}}(x,y) dydx.
+\end{equation}
+In practice,
+\begin{equation}
+  \iint_D f(x,y) dxdy = \int_a^b \int_{g(x)}^{h(x)} f^{\text{ext}}(x,y) dydx
+\end{equation}j
+for some functions $g$ and $h$.
+
+\paragraph{Triple integrals}
+Let $f : \R^3 \to \R$ be Riemann integrable on some nice domain $D \subseteq \R^3$.
+Define $a, b, c, d, e, f \in \R$ such that $[a,b] \times [c,d] \times [e,f]$ is $D$'s bounding box.
+Now define $f$'s extension
+\begin{equation}
+  f^{\text{ext}}(x,y,z) = \begin{cases}
+    f(x,y,z) & \text{if $(x,y,z) \in D$},\\
+    0 & \text{otherwise}.
+    \end{cases}
+\end{equation}
+Then,
+\begin{equation}
+  \iiint_D f(x,y,z) dxdydz = \int_a^b \int_c^d \int_e^f f^{\text{ext}}(x,y,z) dzdydx.
+\end{equation}
+In practice,
+\begin{multline}
+  \iiint_D f(x,y,z) dxdydz\\ = \int_a^b \int_{g(x)}^{h(x)} \int_{p(x,y)}^{q(x,y)} f^{\text{ext}}(x,y,z) dzdydx
+\end{multline}
+for some functions $g$, $h$, $p$, and $q$.
+
+\paragraph{Change of variables}
+Let $f : D \subset \R^n \to \R$ be Riemann integrable on some nasty domain $D$ and let $\boldsymbol\Phi : D^* \subset \R^n \to D$ be such that $\boldsymbol\Phi$ is of class $C^1$, $\boldsymbol\Phi$ is one-to-one, $\boldsymbol\Phi$ is invertible on its domain (i.e., $\det D\boldsymbol\Phi_{\vec u} \ne 0$ for all $\vec u \in D^*$), and $\boldsymbol\Phi(D^*) = D$.  Then,
+\begin{equation}
+  \int_D f(\vec x) d\vec x = \int_{D^*} f\big(\boldsymbol\Phi(\vec u)\big)\abs{\det D\boldsymbol\Phi_{\vec u}} d\vec u.
+\end{equation}
+For well-known coordinate systems, $dxdy = rdrd\theta$; $dxdydz = \rho^2\sin\phi d\rho d\phi d\theta = rdrd\theta dz$.
+
+\paragraph{Scalar line integrals}
+Let $\vec x(t) : \R \to \R^n$ parametrize a curve $C$ in $\R^n$ with endpoints $\vec x(a)$ and $\vec x(b)$; let $f(\vec x) : \R^n \to \R$ be a function defined on $C$.
+Then,
+\begin{equation}
+  \int_C fds = \int_a^b f\big(\vec x(t)\big) \norm{\vec{\flux x}(t)} dt.
+\end{equation}
+
+\paragraph{Vector line integrals}
+Let $\vec x(t) : \R \to \R^n$ parametrize a curve $C$ in $\R^n$ with endpoints $\vec x(a)$ and $\vec x(b)$; let $\vec F(\vec x) : \R^n \to \R^m$ be a vector field defined on $C$.
+Then,
+\begin{equation}
+  \int_C \vec F \dot d\vec s = \int_a^b \vec F\big(\vec x(t)\big) \dot \vec{\flux x}(t) dt.
+\end{equation}
+
+\paragraph{Scalar surface integrals}
+Let $\vec X(u,v) : \R^2 \to \R^n$ be a piecewise smooth parametrization of a surface $\mathscr{S}$ in $\R^n$ such that $\vec X(D) = \mathscr{S}$; let $f(\vec X) : \R^n \to \R$ be a function defined on $\mathscr{S}$.
+Then,
+\begin{equation}
+  \iint_\mathscr{S} fdS = \iint_D f\big(\vec X(u,v)\big) \norm{\frac{\partial\vec X}{\partial u} \cross \frac{\partial\vec X}{\partial v}} dudv.
+\end{equation}
+
+\paragraph{Vector surface integrals}
+Let $\vec X(u,v) : \R^2 \to \R^n$ be a piecewise smooth parametrization of a surface $\mathscr{S}$ in $\R^n$ such that $\vec X(D) = \mathscr{S}$; let $\vec F(\vec X) : \R^n \to \R^m$ be a vector field defined on $\mathscr{S}$.
+Then,
+\begin{equation}
+  \iint_\mathscr{S} \vec F \dot d\vec S = \iint_D \vec F\big(\vec X(u,v)\big) \dot \frac{\partial\vec X}{\partial u} \cross \frac{\partial\vec X}{\partial v} dudv.
+\end{equation}
+The first member of this definition, $\iint_\mathscr{S} \vec F \dot d\vec S$, is also called the flux of $\vec F$ through $\mathscr{S}$.
+
+\section{Fundamental theorems}
+\paragraph{The first fundamental theorem of calculus}
+If $f : \R^n \to \R$ is of class $C^1$ and $C$ is a smooth curve in $\R^n$ with endpoints $\vec x_0$ and $\vec x_1$, then
+\begin{equation}
+  \int_C \grad f \dot d\vec s = f(\vec x_1) - f(\vec x_0).
+\end{equation}
+
+\paragraph{Green's theorem}
+Let $D$ be a closed set in $\R^2$ such that $\partial D$ is a collection of closed curves oriented such that $D$ is to the left.
+If $\vec F: D \to \R^2$ is of class $C^1$, then
+\begin{equation}
+  \oint_{\partial D} \vec F \dot d\vec s = \iint_D \curl \vec F \dot \vec k dxdy.
+\end{equation}
+This is a special case of Stokes' theorem.
+
+\paragraph{Gauss's theorem (the second fundamental theorem of calculus)}
+Let $\Omega \in \R^3$ be a closed domain whose boundary is a piecewise smooth surface $\partial\Omega$.
+Give $\partial\Omega$ outward-pointing orientation.
+If $\vec F$ is a $C^1$ vector field in $\Omega$, then
+\begin{equation}
+  \oiint_{\partial\Omega} \vec F \dot d\vec S = \iiint_\Omega \div\vec F dxdydz.
+\end{equation}
+
+\paragraph{Stokes' theorem (the third fundamental theorem of calculus)}
+Let $\mathscr{S}$ be a piecewise smooth surface in $\R^3$ with a given continuous normal vector field $\vec N$.
+Let $\partial\mathscr{S}$ be a collection of piecewise smooth curves.
+Orient the curves such that the outside of the surface is to the left.
+Now let $\vec F : \mathscr{S} \to \R$ be a $C^1$ vector field.
+Then,
+\begin{equation}
+  \oint_{\partial\mathscr{S}} \vec F \dot d\vec s = \iint_\mathscr{S} \curl \vec F \dot d\vec S.
+\end{equation}
+
+\makeend
+\end{multicols*}
+\end{document}
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