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+%% cheat-sheet.tex
+%% Copyright 2021 Rebecca B. Turner.
+%
+% This work may be distributed and/or modified under the
+% conditions of the LaTeX Project Public License, either version 1.3
+% of this license or (at your option) any later version.
+% The latest version of this license is in
+% http://www.latex-project.org/lppl.txt
+% and version 1.3 or later is part of all distributions of LaTeX
+% version 2005/12/01 or later.
+%
+% This work has the LPPL maintenance status `maintained'.
+%
+% The Current Maintainer of this work is Rebecca B. Turner.
+%
+% This work consists of the files:
+% README.md
+% rbt-mathnotes.tex
+% rbt-mathnotes.sty
+% rbt-mathnotes.cls
+% rbt-mathnotes-util.sty
+% rbt-mathnotes-messages.sty
+% rbt-mathnotes-hw.cls
+% rbt-mathnotes-formula-sheet.cls
+% examples/cheat-sheet.tex
+% examples/multivar.tex
+% examples/topology-hw-1.tex
+% and the derived files:
+% rbt-mathnotes.pdf
+% examples/cheat-sheet.pdf
+% examples/multivar.pdf
+% examples/topology-hw-1.pdf
+
+\documentclass{rbt-mathnotes-formula-sheet}
+\usepackage{nicefrac}
+\ExplSyntaxOn
+\NewDocumentCommand \normalized { m }
+ { \frac { #1 } { \| #1 \| } }
+\let \gr \grad
+\def \ddx { \frac{d}{dx} }
+% VL = vector literal
+\NewDocumentCommand \vl { m } { \left\langle #1 \right\rangle }
+\ExplSyntaxOff
+
+\title{Formula Sheet}
+\author{Rebecca Turner}
+\date{2019-11-12}
+
+% "The most common size for index cards in North America and UK is 3 by 5
+% inches (76.2 by 127.0 mm), hence the common name 3-by-5 card. Other sizes
+% widely available include 4 by 6 inches (101.6 by 152.4 mm), 5 by 8 inches
+% (127.0 by 203.2 mm) and ISO-size A7 (74 by 105 mm or 2.9 by 4.1 in)."
+\mathnotes{
+ height = 4in ,
+ width = 6in ,
+}
+\begin{document}
+\maketitle
+\begin{gather*}
+% 12.2: Vectors
+% 12.3: Dot product
+ \textstyle\vec a \cdot \vec b = \sum_i a_i b_i = |\vec a| |\vec b| \cos \theta. \\
+% 12.4: Cross product
+ \vec a \times \vec b
+ % = \left| \begin{array}{rrr}
+ % \hat{i} & \hat{j} & \hat{k} \\
+ % a_1 & a_2 & a_3 \\
+ % b_1 & b_2 & b_3 \\
+ % \end{array} \right| \\
+ = \langle a_2 b_3 - a_3 b_2,
+ \quad a_3b_1 - a_1b_3, \\
+ a_1b_2 - a_2b_1 \rangle.\quad
+ |\vec a \times \vec b| = |\vec a| |\vec b| \sin \theta.
+% 12.5: Equations of lines and planes.
+\shortintertext{Param.\ eqns.\ of line through $\langle x_0,y_0,z_0 \rangle$
+par.\ to $\langle a,b,c \rangle$:}
+ x = x_0 + at,
+ \quad y = y_0 + bt,
+ \quad z = z_0 + ct. \\
+\text{Symm.\ eqns.: }
+ \frac{x-x_0}{a}
+ = \frac{y-y_0}{b}
+ = \frac{z-z_0}{c}. \\
+\shortintertext{Vec.\ eqn.\ of plane through $\vec r$ with $\vec n$ normal:}
+ \vec n \cdot (\vec r - \vec r_0) = 0,
+ \quad \vec n \cdot \vec r = \vec n \cdot \vec r_0. \\
+% 13.1: Vector functions
+% 13.2: Derivatives/integrals of vector functions
+% 13.3: Arc length and curvature
+\shortintertext{Length along a vec.\ fn.\ $\vec r(t)$:}
+ \textstyle\int_a^b \left|\vec r'(t)\right|\,dt = \int_a^b \sqrt{\sum_i
+ r_i'(t)^2}\,dt, \\
+\shortintertext{Unit tang.\ $\vec T(t) = \vec r'(t)/\left|\vec
+r'(t)\right|$, so curvature of $\vec r(t)$ w/r/t the arc len.\ fn. $s$:}
+ \kappa = \left|\frac{d\vec T}{ds}\right|
+ = \frac{\left| \vec T'(t) \right|}{\left| \vec r'(t) \right|}
+ = \frac{\left| \vec r'(t) \times \vec r''(t) \right|}{\left| \vec r'(t)
+ \right|^3}. \\
+\text{Unit normal:}\quad
+ \vec N(t) = \vec T'(t)\,/\,\left| \vec T'(t) \right| \\
+% 14.1: Functions of several variables
+% 14.2: Limits and continuity
+% 14.3: Partial derivatives
+\text{Clairaut's thm.:}\quad
+ f_{xy}(a,b) = f_{yx}(a,b) \\
+% 14.4: Tangent planes & linear approximations
+\shortintertext{Tan.\ plane to $z = f(x,y)$ at $\langle x_0, y_0,
+z_0\rangle$:}
+ z - z_0 = f_x(x_0, y_0) (x-x_0) \\
+ + f_y(x_0, y_0) (y-y_0). \\
+% Partial derivatives of f for each variable exist near a point and are
+% continuous => f is differentiable at the po\int.
+% 14.6: Directional derivatives and the gradient vector
+\text{Grad.:}\quad
+ \grad f(x,y) = \pd[f]x \hat{i} + \pd[f]y \hat{j}. \\
+\shortintertext{Dir.\ deriv.\ towards $\vec u$ at $\langle x_0, y_0 \rangle$:}
+ D_{\langle a,b\rangle} f(x_0, y_0) = f_x(x,y) a + f_y(x,y) b \\
+ = \grad f(x,y) \cdot \vec u. \\
+\shortintertext{Max of $D_{\vec u} f(\vec x) = \left|\grad f(\vec
+x)\right|$. Tan.\ plane of $f$ at $\vec p$:}
+ 0 =
+ f_x(\vec p)(x-\vec p_x)
+ + f_y(\vec p)(y-\vec p_y) \\
+ + f_z(\vec p)(z-\vec p_z).
+% 14.7: Maximum and minimum values
+\shortintertext{If $f$ has loc.\ extrem.\ at $\vec p$, then $f_x(\vec p) =
+0$ (\& $f_y$, etc). If so, let}
+ D = \left| \begin{array}{ll}
+ f_{xx} & f_{xy} \\
+ f_{yx} & f_{yy}
+ \end{array}\right|
+ = f_{xx} f_{yy} - (f_{xy})^2.
+\shortintertext{%
+ $D = 0$: no information.
+ $D < 0$: saddle pt.
+ $D > 0$: $f_{xx}(\vec p) > 0 \implies$ loc.\ min;
+ $f_{xx}(\vec p) < 0 \implies$ loc.\ max.
+ ($D$ is the \textbf{Hessian mat.})
+\endgraf
+ Set of possible abs. min and max vals of $f$ in reg.\ $D$: $f$ at critical
+ pts.\ and extreme vals.\ on the boundary of $D$.
+% 14.8: Lagrange multipliers
+\endgraf
+ Lagrange mults.: extreme vals of $f(\vec p)$ when $g(\vec p) = k$.
+ Find all $\vec x, \lambda$ s.t.
+}
+ \grad f(\vec x) = \lambda \grad g(\vec x),\quad g(\vec x) = k.
+\shortintertext{i.e.\ $f_x = \lambda g_x$, etc.}
+% 15.1: Double integrals over rectangles
+% 15.2: Iterated integrals
+% 15.3: Double integrals over general regions
+ \iint f(r\cos\theta, r\sin\theta)r\,dr\,d\theta. \\
+ A = \iint_D \left(\sqrt{f_x(x,y)^2 + f_y(x,y)^2 + 1}\right) \,dA. \\
+\shortintertext{Line int.s}
+ \int_C f(x,y)\,ds = \\
+ \int_a^b f(x(t), y(t))\sqrt{\left(\pd[x]t\right)^2 + \left(\pd[y]t\right)^2}\,dt \\
+\shortintertext{If $C$ is a smooth curve given by $\vec r(t)$ from $a \le t
+\le b$,}
+ \int_C \grad f \cdot d\vec r = f(\vec r(b)) - f(\vec r(a)) \\
+\text{Spherical coords:}\quad
+ x = \rho \sin \phi \cos \theta \\
+ y = \rho \sin \phi \sin \theta, z = \rho \cos \phi \\
+ \curl \vec F = \\ \left< \pd[R]y - \pd[Q]z, \pd[P]z - \pd[R]x, \pd[Q]x -
+ \pd[P]y\right>. \\
+ \vec F = \langle P,Q,R \rangle,\quad
+ \curl \vec F = \grad \times \vec F \\
+ \vec F \text{ ``conservative''} \implies \exists f, \vec F = \grad f. \\
+ \dive \vec F = \grad \cdot \vec F = \pd[P]x + \pd[Q]y + \pd[R]z. \\
+ \curl(\grad f) = \vec 0,\quad \dive \curl \vec F = 0 \\
+\shortintertext{If $C$ is a positively-oriented (ccw) closed curve, $D$
+is bounded by $C$, and $\vec n$ represents the normal,}
+ % \int_C P\,dx + Q\,dy = \iint_D\left( \pd[Q]{x} - \pd[P]{y} \right). \\
+ \oint_C \vec F \cdot \vec n\,ds = \iint_D \dive \vec F(x,y)\,dA.
+\end{gather*}
+
+\pagebreak
+\raggedright Common derivs:
+$f(g(x)) \to g'(x) f'(g(x))$,
+$b^x \to b^x \ln b$,
+$f^{-1}(x) \to 1/f'(f^{-1}(x))$,
+$\ln x \to 1/x$,
+$\sin x \to \cos x$, $\cos x \to -\sin x$,
+$\tan x \to \sec^2 x$,
+$\sin^{-1} x \to 1/\sqrt{1-x^2}$,
+$\cos^{-1} x \to -(\sin^{-1}x)'$ (etc.),
+$\tan^{-1} x \to 1/(1+x^2)$,
+$\sec^{-1} x \to 1/(|x|\sqrt{x^2-1})$.
+
+Common ints (don't forget $+C$):
+\begin{gather*}
+ x^n \to \frac{x^{n + 1}}{n + 1} + C \quad \text{when } n \ne -1 \\
+ 1/x \to \ln |x| \\
+ \tan x \to -\ln(\cos x) \\
+ \int uv'\,dx = uv - \int u'v\,dx \quad\text{(Int.\ by parts)} \\
+ \int u\,dv = uv-\int v\,du \\
+ \int_{g(a)}^{g(b)} f(u)\,du = \int_a^b f(g(x))g'(x)\,dx
+ \quad\text{$u$-substitution.}
+\intertext{E.x.\ in $\int 2x \cos x^2\,dx$, let $u=x^2$, find $du/dx=2x
+\implies du = 2x\,dx$, subs.\ $\int \cos u\,du = \sin u + C = \sin x^2 +
+C$.}
+ \iint_R f(x,y)\,dA = \int_\alpha^\beta \int_a^b f(r\cos\theta,
+ r\sin\theta)r\,dr\,d\theta
+\end{gather*}
+\begin{itemize}
+ \item Integrand contains $a^2-x^2$, let $x = a\sin\theta$ and use $1 -
+ \sin^2 \theta = \cos^2 \theta$.
+ \item $a^2 + x^2$, let $x = a\tan\theta$, use $1 + \tan^2 \theta = \sec^2
+ \theta$.
+ \item $x^2 - a^2$, let $x = a\sec\theta$, use $\sec^2\theta - 1 = \tan^2
+ \theta$.
+\end{itemize}
+
+\begin{gather*}
+ \lim_{x \to 0} \sin x/x = 1 \\
+ \lim_{x \to 0} (1-\cos x)/x = 0 \\
+ \lim_{x \to \infty} x \sin(1/x) = 1 \\
+ \lim_{x \to 0} (1+x)^{1/x} = e \\
+ \lim_{x \to 0} (e^{ax}-1)/(bx) = a/b \\
+ \lim_{x \to 0^+} x^x = 1 \\
+ \lim_{x \to 0^+} x^{-n} = \infty \\
+ \text{For $0/0$ or $\pm\infty/\infty$,}\quad
+ \lim_{x \to c} f(x)/g(x) = \lim_{x \to c} f'(x)/g'(x) \\
+ \text{For $g(x)$ cont.\ at $L$,}
+ \lim_{x \to c} f(x) = L \implies \lim_{x \to c} g(L)
+\end{gather*}
+
+\end{document}
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+%% multivar.tex
+%% Copyright 2021 Rebecca B. Turner.
+%
+% This work may be distributed and/or modified under the
+% conditions of the LaTeX Project Public License, either version 1.3
+% of this license or (at your option) any later version.
+% The latest version of this license is in
+% http://www.latex-project.org/lppl.txt
+% and version 1.3 or later is part of all distributions of LaTeX
+% version 2005/12/01 or later.
+%
+% This work has the LPPL maintenance status `maintained'.
+%
+% The Current Maintainer of this work is Rebecca B. Turner.
+%
+% This work consists of the files:
+% README.md
+% rbt-mathnotes.tex
+% rbt-mathnotes.sty
+% rbt-mathnotes.cls
+% rbt-mathnotes-util.sty
+% rbt-mathnotes-messages.sty
+% rbt-mathnotes-hw.cls
+% rbt-mathnotes-formula-sheet.cls
+% examples/cheat-sheet.tex
+% examples/multivar.tex
+% examples/topology-hw-1.tex
+% and the derived files:
+% rbt-mathnotes.pdf
+% examples/cheat-sheet.pdf
+% examples/multivar.pdf
+% examples/topology-hw-1.pdf
+
+\documentclass[knowledge]{rbt-mathnotes}
+\title{Multivariable Calculus}
+\mathnotes{
+ instructor = Prof.~Corey Bregman ,
+ name = Rebecca Turner ,
+ email = rebeccaturner@brandeis.edu ,
+ course = \textsc{math} 20a (multivariable calculus) ,
+ institution = Brandeis University ,
+ semester = Fall 2019 ,
+}
+
+\ExplSyntaxOn
+\NewDocumentCommand \normalized { m }
+ { \frac { #1 } { \| #1 \| } }
+\let \gr \grad
+\def \ddx { \frac{d}{dx} }
+% VL = vector literal
+\NewDocumentCommand \vl { m } { \left\langle #1 \right\rangle }
+\ExplSyntaxOff
+
+% \makeatletter
+% \@ifpackageloaded{knowledge}
+% {
+\knowledge{notion, index={Derivative!Partial}}
+ | partial derivative
+\AtBeginDocument{\index{Partial derivative|see{Derivative, partial}}}
+
+\knowledge{notion, index={Derivative!Directional}}
+ | directional derivative
+\AtBeginDocument{\index{Directional derivative|see{Directional, partial}}}
+
+\knowledge{notion, index=Unit vector}
+ | unit vector
+
+\knowledge{notion, index=Limit}
+ | limit
+
+\knowledge{notion, index={Differentiable functions}}
+ | differentiable
+
+\knowledge{notion, index=Gradient}
+ | gradient
+
+\knowledge{notion, index={Neighborhood (topology)}}
+ | neighborhood
+
+\knowledge{notion, index=Local maximum}
+ | local maximum
+ | local maxima
+
+\knowledge{notion, index=Local minimum}
+ | local minimum
+ | local minima
+
+\knowledge{notion, index=Absolute maximum}
+ | absolute maximum
+ | absolute maxima
+
+\knowledge{notion, index=Absolute minimum}
+ | absolute minimum
+ | absolute minima
+
+\knowledge{notion, index=Local extremum}
+ | local extremum
+ | local extrema
+
+\knowledge{notion, index=Critical point}
+ | critical point
+
+\knowledge{notion, index=Saddle point}
+ | saddle point
+
+\knowledge{notion, index={Lagrange multiplier}}
+ | Lagrange multiplier
+ | Lagrange multipliers
+ | the method of Lagrange multipliers
+
+\knowledge{url={https://en.wikipedia.org/wiki/Joseph-Louis_Lagrange}}
+ | Joseph-Louis Lagrange
+ | Lagrange
+
+\knowledge{url={https://en.wikipedia.org/wiki/Marquis_de_Condorcet}}
+ | Marquis de Condorcet
+
+ \knowledge{url={https://en.wikipedia.org/wiki/Adrien-Marie_Legendre}}
+ | Adrien-Marie Legendre
+% }
+% {}
+% \makeatother
+\date{2019-10-23}
+\begin{document}
+\maketitle
+\tableofcontents
+
+\chapter{Vectors}
+I already know about vectors --- I've been taught them in about five
+different courses so far. I'm skipping this.
+
+\chapter{Partial derivatives}
+
+If we have a function of multiple variables, say
+\[f(a_1, a_2, a_3, \dots),\]
+we might care about the change of $f$ with respect to only one variable. By
+picking a fixed value for all but one of the variables, we can determine
+this.
+
+Say that we want to find the "partial derivative" of $f$ with respect to
+$a_2$; then, by constructing $g(a_2) = f(c_1, a_2, c_3, \dots)$, we've
+created a function of \emph{one} variable, which we can differentiate as
+usual.
+\begin{notation}
+ We write the "partial derivative" of a function $f$ at a point $\vec p$
+ with respect to a basis element $a$ of $\vec p$ as \fbox{$f_a(\vec p)$.}
+
+ We may also use much more common notation
+ \[\pd[f]{a},\]
+ using the "partial derivative" symbol $\partial$, a stylized cursive
+ ``d''.\footnote{Introduced by "Marquis de Condorcet" in 1770, who used it
+ to represent a partial \emph{differential}, i.e.~the $dy$ or $dx$ in
+ $dy/dx$, and then adapted in 1786 by "Adrien-Marie Legendre" for use as the
+ partial derivative.}
+
+ In the interest of completeness, I'll exhaustedly note that the book also
+ uses, on occasion, the notation $D_a f$.
+\end{notation}
+
+We can also calculate ""higher partial derivatives"" --- similarly to the
+higher ordinary derivatives. The notation is a fairly clear extension:
+\[(f_x)_x = f_{xx} = \pd x \left( \pd[f]{x} \right) = \pd[^2 f]{x^2}.\]
+
+\begin{thm}[Clairaut's Theorem]
+ Suppose $f$ is defined on a neighborhood $N$ about a point $\vec p$.
+ If $f_{xy}$ and $f_{yx}$ are continuous in $N$, then $f_{xy}(\vec p) =
+ f_{yx}(\vec p)$.
+\end{thm}
+
+\section{Gradients}
+\begin{notation}
+ This ridiculous textbook denotes the "partial derivative" of a function
+ $f(x, y) = z$ with respect to $x$ as $f_x(x, y)$.
+\end{notation}
+
+\begin{defn}
+ The ""directional derivative"" of a function $f$ at $(x_0, y_0)$ in the
+ direction of a "unit vector" $\vec u = \langle a, b \rangle$ is
+ \[D_{\vec u} f(x_0, y_0) = \lim_{h \to 0} \frac{f(x_0 + ha, y_0 + hb) -
+ f(x_0, y_0)}{h},\]
+ if the "limit" exists.
+\end{defn}
+
+If $f : \Re^2 \mapsto \Re$ is a "differentiable" function, then $f$ has a
+"directional derivative" in the direction of any "unit vector" $\vec u =
+\langle a, b \rangle$ of
+\begin{alignat*}{1}
+ D_{\vec u} f(x, y) &= f_x(x, y) a + f_y(x, y) b.
+\intertext{Or, if $\vec u = \langle \cos \theta, \sin \theta \rangle$, then}
+ D_{\vec u} f(x, y) &= f_x(x, y) \cos \theta + f_y(x, y) \sin \theta.
+\end{alignat*}
+
+Noticing that the "directional derivative" of a function can be written as
+the dot product of two vectors,
+\begin{alignat*}{1}
+ D_{\vec u} f(x, y) &= f_x(x, y) a + f_y(x, y) b \\
+ &= \langle f_x(x, y), f_y(x, y) \rangle \cdot \langle a, b \rangle \\
+ &= \langle f_x(x, y), f_y(x, y) \rangle \cdot \vec u,
+\end{alignat*}
+we call the first vector $\langle f_x(x, y), f_y(x, y) \rangle$ the
+""gradient"" of $f$ and denote it as $\grad f$.
+
+\begin{defn}
+ The "gradient" of a function $f$ of two variables is defined as
+ \[\grad f(x, y) = \langle f_x(x, y), f_y(x, y) \rangle
+ = \frac{\partial f}{\partial x} \vec i + \frac{\partial f}{\partial y}
+ \vec j.\]
+\end{defn}
+
+Therefore, we can rewrite the "directional derivative" of a function $f$
+as
+\[D_{\vec u} f(x, y) = \grad f(x, y) \cdot \vec u.\]
+It's intuitive, then, that the maximum value of the "directional derivative"
+is $|\grad f(x, y)|$, when $\vec u$ is parallel to $\grad f(x, y)$.
+
+\section{Maximum and minimum values}
+\begin{defn}
+ $f : A^k \mapsto B$ has a ""local maximum"" at $\vec a$ if for some
+ "neighborhood" $N \subset A$ about $\vec a$, for all $\vec x \in N$,
+ $f(\vec x) \le f(\vec a)$.
+
+ Conversely, if $f(\vec x) \ge f(\vec a)$, then $f(\vec a)$ is a ""local
+ \emph{minimum}@local minimum"".
+
+ If the statement also holds true for $N = A$, then $\vec a$ is an
+ ""absolute maximum"" (or ""absolute minimum"").
+\end{defn}
+
+If $f$ has a "local maximum" or "minimum@local minimum" at $\vec a$ and the
+partials of $f$ exist at $\vec a$, then $\partial f/\partial x (\vec a) = 0$
+and $f_y(a, b) = 0$; geometrically, the tangent plane to a maximum or
+minimum must be horizontal.
+
+\begin{defn}
+ A point $\vec a$ is called a ""critical point"" of $f$ if $f_x(\vec a) =
+ 0$ or $f_x(\vec a)$ doesn't exist for all variables of $f$.
+\end{defn}
+
+\begin{defn}
+ A ""saddle point"" of a function is a "critical point" which is not a
+ "local extremum" of the function.
+\end{defn}
+
+If $(a, b)$ is a critical point of $f$, then let
+\[D = D(a, b) = f_{xx} (a,b) \, f_{yy} (a,b) - (f_{xy}(a,b))^2.\]
+If $D < 0$, then $(a, b)$ is a "saddle point" of $f$.
+
+\section{Lagrange multipliers}
+Often we want to find the "local extrema" of a function subject to
+constraints, i.e.~maximizing the volume of an object while keeping its
+surface area constant. The method of "Lagrange multipliers"\footnote{After
+"Joseph-Louis Lagrange" (1736--1813), ``an Italian Enlightenment Era
+mathematician and astronomer [who] made significant contributions to the
+fields of analysis, number theory, and both classical and celestial
+mechanics.''} is a strategy for doing this.
+
+To find extrema of $f(\vec p)$ constrained with $g(\vec p) = k$,
+we look for extrema of $f$ that are restricted to lie on the level curve
+$g(\vec p) = k$; it happens that the largest $c$ such that $f(\vec p) = c$
+intersects with $g(\vec p) = k$ when the two level curves are tangent with
+each other, i.e.~they have identical normals. In other words, for some
+scalar $\lambda$, $\grad f(\vec p) = \lambda \grad g(\vec p)$.
+
+More formally, suppose $f$ has an extrema at $\vec p_0$. Then, let the level
+surface generated by the constraint $g(\vec p) = k$ be called $S$, where
+$\vec p_0 \in S$. Then, let $C$ be the set of points given by $\vec r(t)$ such
+that $C \subset S$ and $\vec p_0 \in C$. Further, let $t_0$ be a point such
+that $\vec r(t_0) = \vec p_0$.
+
+Then, $f \after \vec r$ gives the values of $f$ on the curve $C$. $f$ has an
+extrema at $\vec p$, so $f \after \vec r$ must also, and $(f \after \vec
+r)'(t_0) = 0$. If $f$ is "differentiable", we can use the chain rule to
+write
+\begin{alignat*}{1}
+ 0 &= (f \after \vec r)'(t_0) \\
+ &= \grad f(\vec p_0) \cdot \vec r'(t_0).
+\end{alignat*}
+Therefore, the gradient of $f$ is orthogonal to the tangent of every such
+curve $C$. We also know that $\grad g(\vec p_0)$ is orthogonal to $\vec
+r'(t_0)$, so the gradients of $f$ and $g$ at $\vec p_0$ must be parallel.
+Therefore, if $\grad g(\vec p_0) \ne 0$, there exists some $\lambda$ such
+that
+\begin{equation}
+ \grad f(\vec p_0) = \lambda \grad g(\vec p_0),
+\end{equation}
+where the constant $\lambda$ is called a "Lagrange multiplier".
+
+Then, the ""method of Lagrange multipliers"" gives us a process to
+find the maximum and minimum values of a function $f(\vec p)$ subject to the
+constraint $g(\vec p) = k$, where $\vec p \in \Re^n$. To use the method of
+Lagrange multipliers, we assume that the extreme values exist and that
+$\grad g \ne 0$ on the level surface $g(\vec p) = k$.
+\begin{enumerate}
+ \item Find all values of $\vec p$ and $\lambda$ such that
+ \begin{alignat*}{1}
+ \grad f(\vec p) &= \lambda \grad g(\vec p) \\
+ \text{and}\qquad g(\vec p) &= k. \\
+ \end{alignat*}
+
+ \item Next, evaluate $f$ at all of the points found in the first step. The
+ largest of these values is the maximum value of $f$, and the smallest of
+ them is the minimum value.
+\end{enumerate}
+
+\chapter{Multiple integrals}
+Single integrals are good for functions of one variable. To integrate
+functions of multiple variables, we use multiple integrals. Straightforward
+enough.
+
+Multiple integrals allow us to calculate things like surface areas and
+volumes of geometric objects.
+
+In general, for some double integral
+\[\underbrace{\int_a^b \overbrace{\int_c^d f(x,y)\,dx}^{\mathclap{\text{We
+treat $y$ as constant while evaluating this.}}}\,dy,}_{\mathclap{\text{We've
+eliminated $x$ from the equation before evaluating this.}}}\]
+we do the opposite of partial differentiation and treat all variables other
+than the one we're integrating for as constant, repeatedly, until we've
+integrated with respect to all variables; each step in this process is
+called, predictably, ""partial integration"".
+
+\section{Double integrals}
+For an axis-aligned rectangle $R$ on the $xy$-plane from $(x_0, y_0)$ to $(x_1,
+y_1)$, the area of a function $f(x, y)$ under $R$ is given by the double
+integral
+\begin{alignat*}{1}
+ \iint_{Y} f(x,y)\,dA &= \underbrace{\int_{x_0}^{x_1} \int_{y_0}^{y_1}
+ f(x,y)\,dy\,dx}_{\mathclap{\text{This is the iterated form of the
+ integral.}}} \\
+ &= \int_{y_0}^{y_1} \int_{x_0}^{x_1} f(x,y)\,dx\,dy, \\
+\end{alignat*}
+where we use $\iint_R$ to mean ``integrating over the area of $R$'' and
+``$dA$'' to mean ``with respect to area.''
+
+The right-hand side of the equation above is called the ""iterated form"",
+or an \reintro*"iterated integral".
+
+We can also iterate over funkier regions if we're willing to play with the
+limits of integration a bit. The easiest regions to integrate over are the
+ones that are easily expressible as the region bounded above and below by
+functions of one variable, e.g. ``the region under the line $y = 2x$ and
+above the line $y = x^2$'' (note that this is bounded on the left at $x = 0$
+and on the right at $x = 2$).
+
+% Area[ImplicitRegion[x^2 < y \[And] y < 2 x, {x, y}
+The area of that region is expressed by the integral
+\begin{alignat*}{1}
+ A &= \int_0^2 \int_{x^2}^{2x}\,dy\,dx \\
+ &= \int_0^2 {\bigg[ x \bigg]}_{x^2}^{2x}\,dx \\
+ &= \int_0^2 {\bigg( 2x-x^2 \bigg)}\,dx \\
+ &= {\left[ x^2 - \frac{x^3}{3} \right]}_0^2 \\
+ &= 4 - \frac{8}{3} = \frac{4}{3}.
+\end{alignat*}
+
+A more complicated region might be ``the region under the paraboloid $z =
+x^2 + y^2$ and above the region in the $xy$-plane bounded by $y = \sqrt{x}$
+and $y=1-\cos x$.''
+
+We can build larger regions out of pieces, by summing smaller integrals.
+
+\section{Polar coordinates}
+Use the conversions
+\begin{alignat*}{1}
+ r &= \sqrt{x^2+y^2} \\
+ x &= r\cos\theta \\
+ y &= r\sin\theta \\
+\intertext{for the coordinates and then we have that if $R$ is a ``polar
+rectangle'' (arc-shaped region bounded by angles and radii) from $r=a$ to
+$r=b$ and $\theta = \alpha$ to $\theta = \beta$, we have}
+ \iint_R f(x,y)\,dA &= \int_\alpha^\beta \int_a^b f(r\cos\theta,
+ r\sin\theta)r\,dr\,d\theta, \\
+\intertext{which makes our lives easier for circly areas and volumes. Don't
+forget to multiply by $r$.
+\endgraf
+For squiggly and varying radii, we can use functions $h_1(\theta)$ and
+$h_2(\theta)$ instead of constants $a$ and $b$:}
+ \iint_D f(x,y)\,dA &= \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta,
+ r\sin\theta)r\,dr\,d\theta. \\
+\end{alignat*}
+
+\section{Cylindrical coordinates}
+Just add $z$.
+
+\section{Spherical coordinates}
+I can never remember how these work. If we have a point $P$, and we drop it
+down to the $xy$-plane, the angle between the positive $x$-axis and the
+segment from the origin to $P$ is $\theta$.
+
+Next, the angle between the positive $z$-axis and the segment from the origin
+to $P$ is $\phi$.
+
+Finally, the length of the segment from the origin to $P$ is $\rho$.
+
+The conversions
+\begin{alignat*}{1}
+ x &= \rho \sin \phi \cos \theta \\
+ y &= \rho \sin \phi \sin \theta \\
+ z &= \rho \cos \phi \\
+\intertext{give us the integral-conversion for the spherical wedge bounded
+by $a \le \rho \le b, \alpha \le \theta \le \beta, c \le \phi \le d$ as}
+ \iiint_E f(x,y,z)\,dV &= \int_c^d \int_\alpha^\beta \int_a^b
+ f(\rho \sin\phi \cos\theta,\,
+ \rho \sin\phi \sin\theta,\,
+ \rho \cos \theta)
+ [\rho^2 \sin\phi]
+ \,d\rho\,d\theta\,d\phi. \\
+\end{alignat*}
+Very gross!
+
+\section{Surface area}
+
+For $f(x,y)$ with $f_x$, $f_y$ continuous, the surface area of $f$ within
+a region $D$ is
+\[A = \iint_D \left(\sqrt{f_x(x,y)^2 + f_y(x,y)^2 + 1}\right) \,dA.\]
+
+% Need notes on:
+% - Change of variables (in multiple integrals)
+
+\chapter{Vector calculus}
+A vector field is a mapping $\R^k \mapsto \R^n$; for each point in
+$k$-dimensional Euclidean space, we associate an $n$-dimensional vector.
+These vectors can represent velocity, distance, or anything else, and come
+up in all sorts of applied fields.
+
+We'll be mostly concerned with vector fields $\R^2\mapsto\R^2$ and
+$\R^3\mapsto\R^3$.
+
+If we have a plane curve given by the vector equation
+\begin{alignat*}{1}
+ \vec r(t) &= \left< x(t), y(t) \right> \qquad a \le t \le b, \\
+\intertext{then the line integral of $f$ along $\vec r(t)$ from $a$ to $b$
+is}
+ & \int_a^b f(x(t), y(t)) \sqrt{{\left(\dd[x]{t}\right)}^2 +
+ {\left(\dd[y]{t}\right)}^2}\,dt,
+\end{alignat*}
+i.e.\ the length of the curve multiplied, at each point, by the value of the
+vector field $f$ at that point.
+
+% Need notes on:
+% 16.1, 16.2, 16.3, 16.4, 16.5,
+% Need to update cheat sheet.
+
+\appendix
+\chapter{Common formulas for derivatives and integrals}
+\backmatter
+\section{Derivatives}
+
+\begin{alignat*}{2}
+ \ddx&\;& (f + g) &= f' + g' \\
+ \ddx&& x^n &= nx^{n - 1} \\
+ \ddx&& (fg) &= fg' + f'g \\
+ \ddx&& \frac{h}{l} &= \frac{l h' - h l'}{l^2} \\
+ \ddx&& f(g(x)) &= f'(g(x)) g'(x) \quad\text{(Chain rule.)} \\
+ \ddx&& b^x &= b^x \ln b \\
+ \ddx&& f^{-1}(x) &= \frac{1}{f'(f^{-1}(x)} \\
+ \ddx&& c &= 0 \\
+ \ddx&& c\,f &= c\,f' \\
+ \ddx&& e^x &= e^x \\
+ \ddx&& e^{f(x)} &= f'(x) e^{f(x)} \quad\text{(By the chain rule.)} \\
+ \ddx&& \ln x &= \frac{1}{x} \\
+ \ddx&& \log_b x &= \frac{1}{x \ln b} \\
+ \ddx&& [\vec u \cdot \vec v] &= \vec u' \cdot \vec v + \vec u \cdot \vec v' \\
+ \ddx&& [\vec u \times \vec v] &= \vec u' \times \vec v + \vec u \times \vec v' \\
+\end{alignat*}
+
+\subsection{Trigenometric}
+\begin{alignat*}{2}
+ \ddx&\;& \sin x &= \cos x \\
+ \ddx&& \cos x &= -\sin x \\
+ \ddx&& \tan x &= \sec^2 x \\
+ \ddx&& \cot x &= -\csc^2 x \\
+ \ddx&& \sec x &= \sec x \tan x \\
+ \ddx&& \csc x &= -\csc x \cot x \\
+ % inverse
+ \ddx&& \sin^{-1} x &= \frac{ 1}{\sqrt{1 - x^2}} \\
+ \ddx&& \cos^{-1} x &= \frac{-1}{\sqrt{1 - x^2}} \\
+ \ddx&& \tan^{-1} x &= \frac{ 1}{1 + x^2} \\
+ \ddx&& \cot^{-1} x &= \frac{-1}{1 + x^2} \\
+ \ddx&& \sec^{-1} x &= \frac{ 1}{|x| \sqrt{x^2 - 1}} \\
+ \ddx&& \csc^{-1} x &= \frac{-1}{|x| \sqrt{x^2 - 1}} \\
+\end{alignat*}
+
+\section{Integrals}
+See also:
+\emph{\href{https://www.whitman.edu/mathematics/calculus/calculus_08_Techniques_of_Integration.pdf}{Techniques
+of Integration}}.
+\begin{alignat*}{1}
+ \int x^n\,dx &= \frac{x^{n + 1}}{n + 1} + C \quad \text{when } n \ne -1 \\
+ \int x^{-1}\,dx &= \ln|x| + C \\
+ \int e^x\,dx &= e^x + C \\
+ \dd{t} \int_{a(t)}^{b(t)} g(s)\,ds &= b'(t) g(b(t)) - a'(t) g(a(t))
+ \quad\text{(Leibniz' rule.)} \\
+ \int uv'\,dx &= uv - \int u'v\,dx \\
+\end{alignat*}
+
+\subsection{Trigenometric}
+\begin{alignat*}{1}
+ \int \sin x\,dx &= -\cos x + C \\
+ \int \cos x\,dx &= \sin x + C \\
+ \int \sec^2 x\,dx &= \tan x + C \\
+ \int \sec x \tan x\,dx &= \sec x + C \\
+ \int \frac{1}{1 + x^2}\,dx &= \tan^{-1} x + C \\
+ \int \frac{1}{\sqrt{1 + x^2}}\,dx &= \sin^{-1} x + C \\
+\end{alignat*}
+
+\printindex
+\end{document}
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+%% topology-hw-1.tex
+%% Copyright 2021 Rebecca B. Turner.
+%
+% This work may be distributed and/or modified under the
+% conditions of the LaTeX Project Public License, either version 1.3
+% of this license or (at your option) any later version.
+% The latest version of this license is in
+% http://www.latex-project.org/lppl.txt
+% and version 1.3 or later is part of all distributions of LaTeX
+% version 2005/12/01 or later.
+%
+% This work has the LPPL maintenance status `maintained'.
+%
+% The Current Maintainer of this work is Rebecca B. Turner.
+%
+% This work consists of the files:
+% README.md
+% rbt-mathnotes.tex
+% rbt-mathnotes.sty
+% rbt-mathnotes.cls
+% rbt-mathnotes-util.sty
+% rbt-mathnotes-messages.sty
+% rbt-mathnotes-hw.cls
+% rbt-mathnotes-formula-sheet.cls
+% examples/cheat-sheet.tex
+% examples/multivar.tex
+% examples/topology-hw-1.tex
+% and the derived files:
+% rbt-mathnotes.pdf
+% examples/cheat-sheet.pdf
+% examples/multivar.pdf
+% examples/topology-hw-1.pdf
+
+\documentclass[twocolumn, noxcolor, maketitle]{rbt-mathnotes-hw}
+\mathnotes{
+ instructor = Prof.~Ruth Charney ,
+ name = Rebecca Turner ,
+ email = rebeccaturner@brandeis.edu ,
+ course = \textsc{math} 104a (Intro to Topology) ,
+ institution = Brandeis University ,
+ semester = Spring 2020 ,
+}
+\title{Homework 1}
+\date{2020-01-18}
+\def\T{\mathcal{T}}
+\def\basis{\mathcal{B}}
+\usepackage[
+ letterpaper,
+ margin = 1in,
+]{geometry}
+\raggedbottom
+\begin{document}
+\maketitle
+
+\section{Topological Spaces}
+\subsection{Open Sets and the Definition of a Topology}
+\begin{problem}[1.7]
+ Define a topology on $\R$ (by listing the open sets within it) that contains
+ the open sets $(0,2)$ and $(1,3)$ that contains as few open sets as possible.
+\end{problem}
+$\T = \{ \emptyset, (0,2), (1,3), (1,2), (0,3), \R \}$.
+
+\subsection{Basis for a Topology}
+\begin{problem}[1.10]
+ Show that $\mathcal{B} = \{[a,b) \subset \R : a < b\}$ is a basis for a
+ topology on $\R$.
+\end{problem}
+\begin{enumerate}
+ \item \textbf{$\emptyset \in \T$, $\R \in \T$.} $\emptyset \in \T$ (by the
+ definition of the completion of a basis to a topology).
+
+ Next, we show $\R \in \T$. For all $n \in \Z_{\ge 0}$, $[n-1, n) \in \basis$
+ and $[-n + 1, -n) \in \basis$. We know that if $b_1, b_2 \in \basis$, $b_1
+ \union b_2 \in \T$, so these short intervals can be gathered together (``unionized'')
+ to produce $\R$:
+ \[ \bigunion_{n=1}^{\infty} \left( [n-1, n) \union [-n + 1, -n] \right) = \R, \]
+ so $\R \in \T$.
+
+ \item \textbf{$\T$ contains all finite intersections of elements of $\T$.}
+ Suppose we have two intervals $[a, b)$ and $[c, d)$. Then, we define
+ \begin{alignat*}{1}
+ a' &= \max(a, c) \\
+ b' &= \min(b, d).
+ \end{alignat*}
+ If $a' > b'$, the intersection $[a, b) \inter [c, d) = \emptyset$, which is
+ in $\T$. Otherwise, the intersection is $[a', b')$, which is an element of
+ $\basis$. All elements of the basis are in $\T$, so the intersection of two
+ elements is in the topology.
+
+ Thankfully, the intersection is itself always a basis element, so we can
+ use the same process to show that finite intersections are in $\T$ by
+ induction.
+
+ \item \textbf{Unions of elements of $\T$ are in $\T$.} By the definition of
+ the completion of a basis to a topology, this is true (all unions of basis
+ elements are included in $\T$).
+\end{enumerate}
+
+\begin{problem}[1.12]
+ % See example 1.9 for defn of \R_l
+ Determine which of the following are open sets in $\R_l$. In each case, prove
+ your assertion.
+ \[ A = [4,5)
+ \quad B = \{3\}
+ \quad C = [1,2]
+ \quad D = (7,8) \]
+\end{problem}
+\begin{enumerate}
+ \item $A$ is open in $\R_l$; $[4,5) \in \basis$.
+ \item $B$ is not an open set in $\R_l$; there is no $[a, b) \subset \R$ where
+ both $b > a$ and $|[a, b)| = 1$ (because $[0,1) \cong \R$, i.e.~all
+ intervals contain infinitely many points).
+
+ (Where $\cong$ means ``is isomorphic to.'')
+ \item $C$ is not open in $\R_l$ because the upper bound of an open set in
+ $\R_l$ is never inclusive. There is no set of intervals $[a_1, b_1), \dots$
+ where the union or intersection of the intervals has an inclusive upper bound.
+ \item $D$ is open because we can take
+ \[ D = \lim_{n\to\infty} \left[7 + \frac{1}{n}, 8\right), \]
+ where $[7 + 1/n, 8) \in \basis$ for any $n \in \R$ with $n \ne 0$.
+\end{enumerate}
+
+\pagebreak
+\begin{problem}[1.15]
+ An arithmetic progression in $\Z$ is a set
+ \[ A_{a,b} = \{\dots, a - 2b, a - b, a, a + b, a + 2b, \dots\} \]
+ with $a,b \in \Z$ and $b \ne 0$. Prove that the collection of arithmetic
+ progressions
+ \[ \mathcal{A} = \{ A_{a,b} : a, b \in \Z \text{ and } b \ne 0 \} \]
+ is a basis for a topology on $\Z$. The resulting topology is called the
+ arithmetic progression topology on $\Z$.
+\end{problem}
+
+\begin{proof}
+ \def\Ar#1#2{A_{#1,#2}}
+ \def\Ars{\mathcal{A}}
+ Let us describe the \emph{minimal form} of an arithmetic progression $\Ar ab$
+ to be the progression $\Ar{a'}{b'} = \Ar ab$ with $a', b' > 0$ and the
+ smallest possible $a'$; in particular, that $a' < b'$.
+
+ We can obtain the minimal form of the progression like so:
+ \begin{alignat*}{1}
+ a' &= a \bmod b \\
+ b' &= |b|, \\
+ \Ar{a'}{b'} &= \Ar ab.
+ \end{alignat*}
+
+ \begin{remark}
+ Two arithmetic progressions have the same elements if their minimal forms are
+ the same; this give an equivalence relation on $\Ars$.
+ \end{remark}
+
+ Now, suppose we have two arithmetic progressions $\Ar ab$ and $\Ar cd$. We
+ assume that the progressions are in minimal form without loss of generality. We
+ also assume that $b \le d$ (by swapping $(a,b)$ with $(c,d)$ if necessary),
+ again without loss of generality.
+
+ If $b \mid d$ and $a = c$, we have $\Ar ab \subset \Ar cd$. In particular, $\Ar
+ ab \inter \Ar cd = \Ar cd$.
+
+ If $b \mid d$ and $a \ne c$, we have $\Ar ab \inter \Ar cd = \emptyset$.
+
+ If $b \nmid d$, we have a different progression. An intersection is generated by
+ an index $(n_1, n_2)$, where
+ \begin{alignat*}{1}
+ a + b n_1 &= c + d n_2. \\
+ \intertext{We can then solve for $n_1$:}
+ t(n) &= c - a + dn \\
+ n_1 &= \frac{t(n_2)}{b}. \\
+ \intertext{Next, we have an infinite \emph{set} of possibilities for $n_2$:}
+ n_2 &\in \left\{n \in \Z : t(n) \mid b \right\}.
+ \intertext{Sorting the possible values of $n_2$ by absolute value, let us call
+ the smallest two values $i_1$ and $i_2$. Then, the difference between
+ adjacent elements in the intersection progression $\Ar ab \inter \Ar cd$
+ is $i_2 - i_1$.
+ \endgraf
+ Let}
+ a' &= a + bi_1 \\
+ b' &= i_2 - i_1 \\
+ \Ar ab \inter \Ar cd &= \Ar{a'}{b'}.
+ \end{alignat*}
+ This isn't super rigorous, admittedly (we're missing some inductive reasoning
+ about the integers to prove that there are an infinite set of valid values of
+ $n_2$, in particular), but I have some fairly convincing Haskell code. And the
+ missing steps are mostly boilerplate, and it's late at night already\dots
+
+ In all cases, the intersection of two arithmetic progressions is either empty or
+ another arithmetic progression (i.e.~either the empty set or another basis
+ element), so the same argument given above for $\R_l$ holds (namely that we can
+ extend this to all finite intersections of elements of $\Ars$ inductively).
+
+ Therefore, finite intersections are in the basis. Unions are in the completion
+ of the basis (again by definition). The special element $\emptyset$ is in the
+ completion (by definition), and $\Z = \Ar01$, so $\Z \in \basis$. Therefore,
+ $\Ars$ forms the basis of a topology on $\Z$.
+\end{proof}
+
+\pagebreak
+\subsection{Closed Sets}
+\begin{problem}[1.27(a)]
+ The infinite comb $C$ is the subset of the plane illustrated in Figure~1.17
+ and defined by
+ \begin{multline*}
+ C = \{(x,0) : 0 \le x \le 1\} \;\union \\
+ \bigg\{ \left( \frac{1}{2^n}, y \right) : n = 0,1,2, \dots \\
+ \text{ and } 0 \le y \le 1 \bigg\}.
+ \end{multline*}
+ Prove that $C$ is not closed in the standard topology on $\R^2$.
+\end{problem}
+\begin{proof}
+ Suppose $C$ is closed in the standard topology on $\R^2$. Then, its complement
+ $C^c = \R^2 \setminus C$ must be an open set.
+
+ The point $(0, 1)$ is not in $C$, so $(0, 1) \in C^c$. Every open ball in $\R^2$
+ containing $(0, 1)$ also contains a smaller open ball centered about $(0, 1)$.
+ (For example, the open ball about $(-1, 1)$ of radius $1.1$ contains the open
+ ball centered about $(0, 1)$ of radius $0.1$.)
+
+ However, every open ball centered about $(0, 1)$ contains infinitely many
+ points of $C$; if the ball has radius $r$, all the comb's ``tines'' at $x =
+ 1/2^n$ for $n > - \log_2 r$ intersect with the ball.
+
+ Therefore, every open ball containing $(0, 1)$ also contains points in $C$. As
+ a result, $C^c$ is not open, which contradicts our assumption. Therefore, $C$
+ is not closed.
+\end{proof}
+
+\begin{problem}[1.32]
+ Prove that intervals of the form $[a, b)$ are closed in the lower limit
+ topology on $\R$.
+\end{problem}
+\begin{proof}
+ Take some interval $[a, b)$. Its complement is given by $(-\infty, a) \union
+ [b, \infty)$. Given that
+ \begin{alignat*}{1}
+ (-\infty, a) &= \bigunion_{n=1}^\infty [a-n, a) \\
+ [b, \infty) &= \bigunion_{n=1}^\infty [b, b+n),
+ \end{alignat*}
+ the complement of $[a, b)$ is the union of a number of lower-limit intervals
+ in $\R$, i.e.~the basis elements. The basis elements and its unions are open
+ sets, so the complement of $[a, b)$ is an open set. Then, by the definition of
+ a closed set, $[a, b)$ is closed in $\R_l$.
+\end{proof}
+
+
+\end{document}