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author | Karl Berry <karl@freefriends.org> | 2021-11-30 22:53:25 +0000 |
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committer | Karl Berry <karl@freefriends.org> | 2021-11-30 22:53:25 +0000 |
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tree | 3882071f6375e96600c8061064ede486d98d0cff /Master/texmf-dist/doc/latex/rbt-mathnotes/examples | |
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diff --git a/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/cheat-sheet.pdf b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/cheat-sheet.pdf Binary files differnew file mode 100644 index 00000000000..8f92b8337fc --- /dev/null +++ b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/cheat-sheet.pdf diff --git a/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/cheat-sheet.tex b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/cheat-sheet.tex new file mode 100644 index 00000000000..11ff82a5a90 --- /dev/null +++ b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/cheat-sheet.tex @@ -0,0 +1,227 @@ +%% 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} diff --git a/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/multivar.pdf b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/multivar.pdf Binary files differnew file mode 100644 index 00000000000..e6fdd5f7a3d --- /dev/null +++ b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/multivar.pdf diff --git a/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/multivar.tex b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/multivar.tex new file mode 100644 index 00000000000..fc026754f68 --- /dev/null +++ b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/multivar.tex @@ -0,0 +1,507 @@ +%% 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} diff --git a/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/topology-hw-1.pdf b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/topology-hw-1.pdf Binary files differnew file mode 100644 index 00000000000..3bbc7d93fbd --- /dev/null +++ b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/topology-hw-1.pdf diff --git a/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/topology-hw-1.tex b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/topology-hw-1.tex new file mode 100644 index 00000000000..406ce622ad2 --- /dev/null +++ b/Master/texmf-dist/doc/latex/rbt-mathnotes/examples/topology-hw-1.tex @@ -0,0 +1,246 @@ +%% 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} |