From bdab14700bea11104081218503d424df9bb42884 Mon Sep 17 00:00:00 2001 From: Eric Date: Mon, 12 Jan 2026 15:39:06 -0500 Subject: [PATCH] Add math notes --- math handout/Math_Handout_CHEG231.tex | 171 ++++++++++++++++++++++++++ 1 file changed, 171 insertions(+) create mode 100644 math handout/Math_Handout_CHEG231.tex diff --git a/math handout/Math_Handout_CHEG231.tex b/math handout/Math_Handout_CHEG231.tex new file mode 100644 index 0000000..95d4338 --- /dev/null +++ b/math handout/Math_Handout_CHEG231.tex @@ -0,0 +1,171 @@ +\documentclass[11pt]{article} + +\usepackage[margin=1in]{geometry} + +\usepackage[bookmarks=true]{hyperref} +\hypersetup{ + pdftitle={}, + pdfauthor={}, + pdfkeywords={}, + bookmarksnumbered, + breaklinks=true, + urlcolor=blue, + citecolor=black, + colorlinks=true, + linkcolor=black, +} + +\usepackage{amsmath} +\usepackage{amssymb} + +\usepackage{booktabs} +\usepackage{caption} +\captionsetup[table]{skip=1ex} + +\usepackage{graphicx} +%\graphicspath{{figures/}} + +\usepackage{parskip} +%\usepackage{indentfirst} +%\setlength{\parindent}{2em} + +\pagestyle{plain} + +\begin{document} + +% Header text +\hspace{-0.025\textwidth}\parbox[b]{1.05\textwidth}{\centering \bfseries +\vspace{-2\baselineskip} % I move it up a little bit above the 1" margins +{\Large \uppercase{CHEG 231 Chemical Engineering Thermodynamics}}\\ +{\large {Department of Chemical and Biomolecular Engineering}}\\ +{\large {University of Delaware}} +} + +\vskip 2\baselineskip + +\begin{center} + +{\uppercase{\Large Math reminders}\par} + +%{\itshape Eric\ M.\ Furst\par} +{\itshape \today \par} + +\vskip \baselineskip + +%{\large Nome: \hrulefill\hspace{1em}{\small RA}: \rule{25mm}{0.4pt}} + +\vskip \baselineskip + +%\begin{minipage}{10cm} +%\itshape\small\emph{Instruções:} Resolva todas as questões. +%Mostre todo o seu trabalho e considerações feitas. +%Inclua unidades apropriadas para todas as respostas e direções para grandezas vetoriais. +%Integridade acadêmica é esperada de todos. +%\end{minipage} + +\end{center} + +%Grupo (nome e {\small RA}):\\\answerbox{\linewidth}{25mm} + +%f(x) = \frac{1}{x} +\section{A few common integrals} +Basic integration and integrals involving functions like $\frac{1}{x}$ show up frequently in thermodynamics. +\begin{center} +\renewcommand{\arraystretch}{1.5} +\setlength{\tabcolsep}{1em} +\begin{tabular}{cc}\\ +%\multicolumn{4}{c}{Fourier transforms} \\ +$f(x)$ & $F(x) = \int f(x) dx$ \\ +\hline +$x$ & $\frac{1}{2}x^2$ \\ +$\frac{1}{x}$ & $\ln x$ \\ +$e^x$ & $e^x$ \\ +%\multicolumn{4}{l}{\footnotesize $^1${$H(t)$ is the Heaviside step function.}} \\ +%\multicolumn{4}{l}{\footnotesize $^2${$\Gamma(x) = \int^\infty_0 e^r r^{x-1}dr$ is the Gamma function. If $x$ is a positive integer, then $\Gamma(x+1) = x!$}} \\ +\end{tabular} +\end{center} + +\vspace{2\baselineskip} +Remember that the indefinite integral of a function includes a constant term, $\int f(x) dx = F(x) + c$ were $f(x) = \frac{d}{dx}F(x)$. +The definite integral is written: +\begin{equation} +\int_a^b f(x) dx= F(b) - F(a) \nonumber. +\end{equation} + +Finally, we may see integration by parts in derivations. Recall that +\begin{equation} +\int u\,dv = uv + \int v\,du \nonumber. +\end{equation} + +\section{Differentiation, too} +We will review plenty of differentiation when we discuss partial derivatives. We also write equations in a differential form and will solve simple differential equations. Be sure to review and practice the product rule and chain rule. A common mistake is to accidentally treat variables as constants and {\it vice versa}. +\begin{center} +\renewcommand{\arraystretch}{1.5} +\setlength{\tabcolsep}{1em} +\begin{tabular}{c}\\ +$\frac{d}{dx}(cu) = c\frac{du}{dx}$, where $c$ is a constant\\ +\end{tabular} +\end{center} +Contrast that to the case below when $u$ and $v$ are both functions of $x$. We must apply the \emph{product rule}: +\begin{equation} +\frac{d}{dx}(uv) = v\frac{du}{dx} + u\frac{dv}{dx} \nonumber +\end{equation} + +Finally, review the other rules of differentiation, including the product rule and the chain rule, +\begin{equation} +\frac{d}{dx}(u + v) = \frac{du}{dx} + \frac{dv}{dx} \nonumber +\end{equation} +\begin{equation} +\frac{d}{dx}\left ( \frac{u}{v} \right ) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \nonumber +\end{equation} +\begin{equation} +\frac{d}{dx}u(v) = \frac{du}{dv}\frac{dv}{dx} \nonumber +\end{equation} +\begin{equation} +\frac{1}{y}\frac{dy}{dx} = \frac{d \ln y}{dx} \nonumber +\end{equation} +\begin{equation} +d\left ( \frac{1}{x} \right ) = - \frac{1}{x^2} dx \nonumber +\end{equation} + + + +\section{Logarithmic functions} +Here, $\ln x$ is the natural logarithm ($\log_e x$). Most of the time when we see and use the function $\log$, we are referring to the logarithm with base 10, or $\log_{10}$. +\begin{center} +\renewcommand{\arraystretch}{1.5} +\setlength{\tabcolsep}{1em} +\begin{tabular}{c}\\ +$\ln A + \ln B =$ $\ln (AB)$ \\ +$\ln A - \ln B =$ $\ln \frac{A}{B}$ \\ +$\ln x^a = a\ln x$\\ +%\end{tabular} +%\begin{tabular}{c}\\ +$\ln e^x = x\ln e = x$ \\ +$\ln 1 = \ln e^0 = 0\times \ln e = 0$\\ +$\ln e = 1$\\ +\end{tabular} +\end{center} + +\section{Exponential functions} +An exponential function is Euler's number $e$ raised to the $x$ power, $f(x) = e^x$. We've noted the relationship between the natural logarithm and $e$, which actually serves as a definition of $\ln x$: $\ln x = y \textrm{\ if and only if\ } e^y = x$. +\begin{center} +\renewcommand{\arraystretch}{1.5} +\setlength{\tabcolsep}{1em} +\begin{tabular}{c}\\ +$e^0 = 1$\\ +$e^ae^b = e^{a+b}$\\ +$\frac{d}{dx}e^x = e^x$\\ +\end{tabular} +\end{center} + +\section{Taylor series expansion} +The Taylor series is often used to linearize a function. Given the function $f(x)$, we can write +\begin{equation} +f(x) \approx f(x_0) + (x-x_0)f'(x_0) + ... \nonumber +\end{equation} + +%\usepackage[style=unsrt, citestyle=unsrt]{biblatex} +%\bibliography{references} + +\end{document} \ No newline at end of file