[OS X TeX] corrupt pdf fille

[Javier Elizondo]

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

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%\show \makeatletter

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%To appear in \\ {\sc Journal of Algebraic Geometry}
%\end{minipage}}
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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
%Road, Mumbai-400005, India}
%\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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