http://ubuntuforums.org/showthread.php?t=1578465
I have the attachments posted there, but I am also posting the contents of the .tex file below.
********************** ubuntu forums post ****************
Hi, I'm making an equation sheet and I am having some minor problems with it. For some reason I am getting an extra page of output before the actual equation sheet. I don't know what is causing the problem, but I have never used the minipage environment before so it may be my inexperience with that. I am attaching the .tex file (named .txt because I cannot upload .tex files to this server) and the .pdf.
I really would appreciate someone taking a look at this and letting me know what simple mistake I'm making.
TIA.
Code: Select all
% use
% xdvi -paper usr formulae &
% and
% dvips -t landscape formulae
% to preview and make PostScript
%
\documentclass{article}
\usepackage{amssymb}
%%\usepackage{landscape}
\setlength{\textheight}{8in}
\setlength{\textwidth}{10.25in}
\setlength{\topmargin}{-0.87in}
\setlength{\oddsidemargin}{-0.5in}
\setlength{\evensidemargin}{-0.5in}
\setlength{\parindent}{0pt}
\title{Formula Sheet}
\author{}
\date{}
\renewcommand{\baselinestretch}{0.8}
\newif\iftechexplorer\techexplorerfalse
\markright{--------- Formula Sheet, Rajan Vatassery -------- \ }% \hrulefill\ }
\begin{document}
\pagestyle{headings}
%%\raggedright
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{\small
\iftechexplorer%
\maketitle %
\fi%
\begin{minipage}[t]{3.65in}
\section*{Basics}
Quadratic Equation, $ax^2+bx+c=0$~:
\[
x = { {-b \pm \sqrt{b^2-4ac}} \over {2a} }
\]
Euler Equation:
\[
e^{i x} = \cos x + i \sin x
\]
Boltzmann Equation:
\[
\frac{P_{1}}{P_{0}} = e^{\frac{-\Delta E}{kT}}
\]
{\bf Newton's Laws:}
\begin{itemize}
\item[1$^{st}$:] Law of Inertia
\item[2$^{nd}$:] $ F = ma $
\item[3$^{rd}$:] Any action has an equal and opposite reaction
\end{itemize}
{\bf Laws of Thermodynamics:}
\begin{itemize}
\item[1$^{st}$:] Conservation of energy, heat transfer
\item[2$^{nd}$:] $ S = k_{B} \ln W $ or $\Delta S$ is positive
\item[3$^{rd}$:] At 0K, S may be 0 for a given substance
\end{itemize}
\section*{Fundamental Constants}
\begin{itemize}
\addtolength{\itemsep}{-5pt}
\item[$e = $]
charge of electron $ = 1.60\times10^{-19}$ C
\item[$m_e = $]
mass of electron $ = 9.11\times10^{-31}$ kg % \\ $ = 0.511$ MeV/$c^2$
\item[$m_n = $]
mass of neutron or proton $ = 1.67\times10^{-27}$ kg % \\ $ = 938$ MeV/$c^2$
\item[$k_B = $]
Boltzmann's constant $ = 1.38 \times 10^{-23} \>$J/K
\item[$\mu_0 = $]
$ 4 \pi \times 10^{-7}$ Wb$\>$A$^{-1}$m$^{-1}$
\item[$\varepsilon_0 = $]
$ 8.85 \times 10^{-12} \>$C$^2$N$^{-1}$m$^{-2}$
\end{itemize}
\section*{Fluorescence Phenomena}
From Sharma and Shulman, 1999.
\[
\tau = 1/k
\]
Fluorescent Lifetime: Concentration decrease by 1/e
\vspace{10pt}
{\large FRET:}
\[
k_{ET} = \frac{1}{\tau_{D}}\left(\frac{R_{0}}{R}\right)^{6}
\]
where $k_{ET}$ is the rate constant for FRET, $\tau_{D}$ is the
excited state lifetime, $R$ is the mean distance between donor and
acceptor, $R_{0}$ is ``critical separation:''
\[
R_{0} = \frac{3000}{4 \tau N_{av} [A]_{1/2}}
\]
where $[A]_{1/2}$ is concentration of acceptor molecules where quantum
yield of donor fluorescence becomes 1/2 of yield without acceptor.
\vspace{10pt}
{\large Rotational Depolarization Analysis:}
\[
p = \left[\left(\frac{1}{p_{0}}-\frac{1}{3}\right)\left(1+\frac{RT}{\eta}\frac{\tau}{V}\right)+\frac{1}{3}\right]^{-1}
\]
Depolarization from FRET, in absence of rotational motion:
\[
p = \left[\left(\frac{1}{p_{0}}-\frac{1}{3}\right)\left(1+\frac{105 NR^{6}_{0}C}{a^{3}_{M}}\right)+\frac{1}{3}\right]^{-1}
\]
where $a_{M})$ is the effective molecular radius, $R_{0}$ is the
distance between dipoles at which the probability of emission is equal
to probability of ET.
\end{minipage}
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\hspace{0.05in}
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\begin{minipage}[t]{3.65in}
\section*{NMR Equations}
From Lambert, et al. 1998
Larmour Frequency Relations:
\[
\Delta E = \hbar\omega_{0} = h\nu_{0} = \gamma\hbar B_{0}
\]
Magnetic Moment:
\[
\mu = \gamma \hbar \textrm{I}
\]
Relaxation Events:
From PJ Hore 1995
\[
\frac{1}{T_{1}} =
\left(\frac{\mu_{0}}{4\pi}\right)^{2}\frac{\gamma_{I}^{2}
\gamma_{S}^{2} \hbar^{2} \tau_{c}}{r^{6}}
\]
where two dipolar-coupled nuclei, I and S, serve to relax each other.
\vspace{10pt}
Linewidth is influenced by T$_{2}$ according to:
\[
\frac{1}{T_{2}\pi} = \Delta \nu
\]
In the approximation of random fluctuating fields giving rise to
relaxation of spin polarization, T$_{1}$ and T$_{2}$ follow the
relations:
\[
\frac{1}{T_{1}} = \gamma^{2}\langle B^{2}\rangle J(\omega_{0})
\]
\[
\frac{1}{T_{2}} = \frac{1}{2}\gamma^{2}\langle B^{2}\rangle J(\omega_{0}) +
\frac{1}{2}\gamma^{2}\langle B^{2}\rangle J(0)
\]
Karplus Relation:
\[
^{3}J = A + B \cos \theta + C \cos^{2} \theta
\]
where A, B, C are empirically calculated (initial guess (2, -1, 10 Hz)).
%\end{minipage}
%
%\hspace{0.05in}
%
%\begin{minipage}[t]{3.25in}
\section*{Electron Transfer Theory}
Marcus Relation
\[
k_{et} = \frac{2\pi}{\hbar}|H_{AB}| \frac{1}{\sqrt{4\pi \lambda k_{b} T}} exp \left(-\frac{(\lambda + \Delta G^{\circ})^{2}}{4 \lambda k_{b} T}\right)
\]
Fermi's Golden Rule
\[
k_{et} = \frac{2 \pi}{\hbar} |<f|H'|i>|^{2}\rho
\]
\section*{Engineering}
Fick's First Law
\[
J = -\mathcal{D}\frac{\partial \phi}{\partial x}
\]
Fick's Second Law
\[
\frac{\partial \phi}{\partial t} = \mathcal{D} \frac{\partial^{2} \phi}{\partial x^{2}}
\]
Stokes-Einstein Formula
\[
\mathcal{D} = \frac{k_{B} T}{6 \pi \eta R_{H}}
\]
\section*{Signal Processing}
Fourier Transform
\[
f (s) = \int_{-\infty}^{\infty} e^{-2 \pi i s \xi} f (\xi) d\xi
\]
\end{minipage}
}
\end{document}
