# [OS X TeX] corrupt pdf fille

Javier Elizondo javier_elizondo69 at yahoo.com
Mon Mar 17 16:49:46 EDT 2008

Hello everyone,

I have edited a latex file with Emacs and texshop and after a while of being working on it, I got a corrupted pdf file, most of the math content appears not readable. And the fonts (usualy roman) are also changed.

Everything gets again fine if I restart the computer, and again, after a while when I keep working on the file and running it a few times the problem comes back. It is an easy file, and I have run the file in a linux computer and everything is fine. However, it seems that if I look at the file using adobe acrobat reader everything seems fine.

The latex file is the following

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%              THIS FILE IS IN       LaTeX 2e
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\documentclass[12pt,spanish, english]{article}
\usepackage[latin1]{inputenc}
\usepackage[spanish]{babel}
%\usepackage{pdfsync}
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\usepackage{eucal}
\usepackage{amsgen}
\usepackage{amstext}
\usepackage{amsbsy}
\usepackage{amsopn}
\usepackage{amssymb}
\usepackage[all]{xy}

\AtBeginDocument{\decimalpoint}

\pagestyle{empty}
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\newtheorem*{ack}{\bf Acknowledgement}

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%
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%%%%%%%%%%%%  MISCELLANEOUS  %%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% ENUMERATE %%%%%%%%%%%%%%%%%%%%%%%%

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%%%%%%%%%%%%%%%%%%%%    Page layout   %%%%%%%%%%%%%%%%%%%%

%\markright{First Draft}

\hfuzz1.5pc % Don't bother to report overfull boxes if overage is < 1pc
%\vfuzz3pc % only for draftheadstyle

%\show \serieslogo@
%\show \makeatletter

%\makeatletter
%\renewcommand{\serieslogo@}{\begin{minipage}{8cm}
%To appear in \\ {\sc Journal of Algebraic Geometry}
%\end{minipage}}
%\makeatother

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%%%%%%%%%%%%%%%%%%%%    Top Matter (amsart style)  %%%%%%%%%%%%%%%%%%%%%

%\date{ }
%\title[Chow varietihttp://www.rae.es/es and Euler-Chow series]{Some remarks on Chow
%varieties and Euler-Chow series}
%\author{E. Javier Elizondo}
%\address{Instituto de Matem\'aticas, UNAM, Mexico}
%\thanks{The first author was supported in part by grants UNAM-DGAPA
%IN119298 and CONACYT 27969E}
%\email{javier at math.unam.mx}

%\author{V. Srinivas}
%\address{School of Mathematics, Tata Institute of Fundamental Research, Homi Bhabha
%\thanks{ Srinvias, here you have to write if you like to thanks}
%\email{srinivas at math.tifr.res.in}

\begin{document}
%\maketitle

\begin{flushright}
Name:\underline{\hspace{7cm} }\\
UIN:\underline{\hspace{7cm} }
\end{flushright}%\vspace{1.5cm}
\begin{flushleft}
MATH-172-501
\end{flushleft}
\begin{center}{\Large {\bf Answers to Second Midterm}}\end{center}
\begin{center}{\large Javier Elizondo}\end{center}\vspace{.3cm}

\begin{enumerate}
\item[I.] Find the integral\ $$\int \sin^{2}x \, \cos^{2}x\, \, dx$$
% Prob. 8.2 - 8
\\{\bf  Sol.} $\sin^2 x \cos^2 x = \left( \sin x \cos x\right) ^2 = \left(\frac{1}{2} \sin (2x)\right)^2 = \frac{1}{4} \sin^2 (2x) = \frac{1}{8} \left( 1-\cos (4x)\right)$ Now
the integral becomes $\int \frac{1}{8} \left( 1-\cos (4x)\right) \, dx$ which
is very easy to compute.

\item[II.] Find the integral\ $$\int \frac{dx}{x\sqrt{x^2 + 3}}.$$
%8.3 #15
\\ {\bf Sol.} $x= \sqrt{3}\tan \theta$, where
$\frac{-\pi}{2}<\theta<\frac{\pi}{2}$, and $dx = \sqrt{3}\sec^2 \theta \,d\theta.$ We also now that\newline  $\sqrt{3\tan^2 + 3} = \sqrt{3}\sqrt{\sec^2 x} = \sqrt{3} \sec x.$ Therefore the integral
becomes\newline  $\frac{1}{\sqrt{3}}\int \frac{\sec\theta \, d\theta}{\tan\theta} =\frac{1}{\sqrt{3}}\int \csc\theta \, d\theta.$
We use the formula written in the exam, and with the appropriate right
rectangle we return to the variable $x$.

\item[III.] Find \ $$\int \frac{x^2}{(x-3)(x+2)^2}\, \, dx$$
%prob. 8.4-30
\\ {\bf Sol.}
$\frac{x^2}{(x-3)(x+2)2} = \frac{A}{x-3} + \frac{B}{x+2}+\frac{C}{(x+2)^2} \Longrightarrow x^2 = A(x+2)^2 +B(x-3)(x+2) + C(x-3)$. If $x=3$ we get
$A=9/25$, if $x= -2$ we get $C=-4/5$. If we compare the coefficient of
$x^2$ in the right hand side and left hand side polynomials of the
equality we see that $1= A+B \Rightarrow B=16/25$. Then the integral is
equal to \newline
$\frac{9}{25} \ln|x-3| + \frac{16}{25} \ln|x+2| + \frac{4}{5(x+2)} + C$.

\item[IV.] Compute the improper integral $$\int_0^{\infty} xe^{-x}\, dx$$
%prob. 8.9-20
\\ {\bf Sol.} Using parts with $du =e^{-x} dx$ and $v = x$ we obtain
\newline $\int_0^{\infty} xe^{-x}\, dx = \lim_{t\rightarrow \infty}\left( -xe^{-x} - e^{-x}\right)_0^t = \lim_{t\rightarrow \infty}\left(1-(t+1)e^{-t}\right) =1 -\lim_{t\rightarrow\infty}\frac{t+1}{e^t}$ by L'Hôpital this last
limit is equal to $1-\lim_{t\rightarrow\infty}\frac{1}{e^t} = 1-0=1.$
Therefore the integral is convergent and is equal to $1$.

\item[V.] A tank contains 100 L of pure water. Brine that contains
$0.1$ kg of salt per liter enters the tank at a rate of 10
L/min. The solution is kept thoroughly mixed and drains from the tank
at the same rate. How much salt is in the tank after t minutes?
%9-Review-32
\\ {\bf Sol.} The initial condition is $y(0)=0$ where as usual $y(t)$
is the amount of salt at time $t$. Now,  $\frac{dy}{dt} = \text{rate in - rate out} =\left( 0.1 \, \times \, 10\right) - \left(\frac{y}{100}\, \times\, 10\right) = 1 - \frac{y}{10} = \frac{10-y}{10} \Rightarrow\newline \int\frac{dy}{10-y}=\int\frac{1}{10}dt \Rightarrow -\ln|10-y| = \frac{1}{10}t +C \Rightarrow 10 - y = Ae^{-t/10}.$ The
initial condition implies $A=10$, therefore $y = 10\left( 1-e^{-t/10}\right)$.

\item[VI.] Solve the initial-value problem
$$x^2 \frac{dy}{dx} + 2xy = \cos x, \,\,\,\, y(\pi) = 0$$
%9.2-19
\\ {\bf Sol.} $y^{\prime}=\frac{2}{x}y=\frac{\cos x}{x^2}$, and
$I(x)=e^{\int (2/x) dx}= x^2.$ Multiplying the differential equation
by $I(x)$ we obtain $x^2y^{\prime} + 2xy=\cos x \Rightarrow (x^2y)^{\prime} = \cos x \Rightarrow y = \frac{1}{x^2}\left(\int cos x dx + C\right) = \frac{1}{x^2}\left(\sin x + C\right) \Longrightarrow y = \frac{ \sin x}{x^2}$ where the last
equality follows from the fact that
the initial condition implies that $C = 0$.
\end{enumerate}

\end{document}

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