Latest geometry breaks beamer for me.

[Jeffrey Goldberg]

Beamer was working for until I updated geometry to

Package: geometry 2010/02/12 v5.0 Page Geometry

Now I get

! Undefined control sequence.
\Gm@lmargin ->\Geom@lmargin

at \begin{document}

Here are excerpts from my log
=================================
This is pdfTeX, Version 3.1415926-1.40.10 (TeX Live 2009)
(format=pdflatex 2010.2.10) 15 FEB 2010 19:37
entering extended mode
\write18 enabled.
%&-line parsing enabled.
**9.2-lesson.tex
(./9.2-lesson.tex
LaTeX2e <2009/09/24>

[...]

(/usr/local/texlive/2009/texmf-dist/tex/latex/beamer/beamer.cls
(/usr/local/texlive/2009/texmf-dist/tex/latex/beamer/beamerbasercs.sty
Package: beamerbasercs 2007/01/28 (rcs-revision 1.4)
)
Document Class: beamer 2007/03/11 cvs version 3.07 A class for
typesetting pres
entations (rcs-revision 1.70)
(/usr/local/texlive/2009/texmf-dist/tex/latex/beamer/beamerbasemodes.sty
Package: beamerbasemodes 2007/01/28 (rcs-revision 1.22)

[...]

(/usr/local/texlive/2009/texmf-dist/tex/latex/geometry/geometry.sty
Package: geometry 2010/02/12 v5.0 Page Geometry

[...]

*geometry* driver: auto-detecting
*geometry* detected driver: pdftex
! Undefined control sequence.
\Gm@lmargin ->\Geom@lmargin

l.51 \begin{document}

? x
=================================

And here is my preamble: (Yes, I know I have some redundant \usepackage
instances in there, but I want to keep this as close to what was working
earlier)

=======================
\documentclass{beamer}
%\mode<trans>

\usepackage{geometry}
\usepackage{pgfpages}
\pgfpagesuselayout{resize to}[letterpaper,border shrink=5mm,landscape]
\beamertemplatenavigationsymbolsempty

%\setbeamersize{text margin left=2cm}
\usecolortheme{dove}
\usefonttheme{structurebold}
\title{9.2: Property of Polynomials}
\date{February 16, 2010}
%\useoutertheme{infolines}
%\author{Honors Algebra I}

\usepackage{array}
\newcolumntype{C}{>{$}c<{$}}
\newcolumntype{L}{>{$}l<{$}}
\newcolumntype{R}{>{$}r<{$}}
\renewcommand{\arraystretch}{1.3}

\usepackage{amsmath,amsthm,amsfonts}
\usepackage{graphicx}

\usepackage{color}

\newcommand{\settnotecolor}[1]{\def\tnotecolor{#1}}

\settnotecolor{red}

\newcommand{\tn}[1]{{\small \color{\tnotecolor}#1}}

\newenvironment{tnote}
{\renewcommand{\baselinestretch}{1}\parskip=.5\baselineskip \small
\color{\tnotecolor}}{}

\newcommand{\superscript}[1]{\ensuremath{^{\textrm{#1}}}}
\newcommand{\subscript}[1]{\ensuremath{_{\textrm{#1}}}}

\renewcommand{\th}[0]{\superscript{th}}
\newcommand{\st}[0]{\superscript{st}}
\newcommand{\nd}[0]{\superscript{nd}}
\newcommand{\rd}[0]{\superscript{rd}}

%\newtheorem*{definition}{Definition}
\newtheorem*{remember}{Remember}

%%% BEGIN DOCUMENT
\begin{document}
===================

As always, any help would be appreciated. I'm using texlive-2009 via
MacTeX on OS X 10.6.2.

Cheers,

-j

--
Jeffrey Goldberg http://goldmark.org/jeff/
I rarely read HTML or poorly quoting posts
Reply-To address is valid

[Ulrike Fischer]

Am Mon, 15 Feb 2010 19:52:41 -0600 schrieb Jeffrey Goldberg:

> Beamer was working for until I updated geometry to
>
> Package: geometry 2010/02/12 v5.0 Page Geometry
>
> Now I get
>
> ! Undefined control sequence.
> \Gm@lmargin ->\Geom@lmargin

geometry 5..0 has removed the options: 'compat2' and this breaks
some code in beamerbasecompatibility.sty.
There is already a report in the beamer bug database.

In the meanwhile use this

\makeatletter\let\ifGm@compatii\relax\makeatother
\documentclass{beamer}
...

> And here is my preamble: (Yes, I know I have some redundant \usepackage
> instances in there, but I want to keep this as close to what was working
> earlier)

A simpler example would make life much easier for the people who try
to identify the problem.

In fact already this breaks
\documentclass{beamer}
\begin{document}
abc
\end{document}

--
Ulrike Fischer

[teejayst...@gmail.com]

I have problem with producing this slide on another system. the error occurs with begin document, begin frame and end frame... I don't know what could be the cause.


\documentclass[aspectratio=169]{beamer}
\mode<beamer>{%
\usetheme{CambridgeUS}
}
\title{\textbf{MAGNETOHYDRODYNAMIC POISEUILLE FLOW OF VISCOUS FLUID WITH VARIABLE VISCOSITY IN A POROUS MEDIUM}}
\begin{document}
\begin{frame}
\titlepage
\color{magenta}
\begin{center}Department of Mathematical Sciences,\\The Federal University of Technology, Akure, Ondo State, Nigeria.\\By\end{center}
\begin{center}
{{\textbf{AFOLABI, FATIMO OPEYEMI}}\\}
{{\textbf{MTS/09/8362}}\\}
\end{center}
\end{frame}

\begin{frame}
\frametitle{\textbf{ABSTRACT}}
This project investigates the effects of flow parameters on velocity and temperature of magnetohydrodynamic poiseuille flow of viscous fluid with variable viscosity in a porous medium. The relevant governing differential equations have been transformed to non-linear coupled equations and are solved numerically using Runge Kutta in conjunction with shooting techniques to give numerical solutions to the problem. We highlight the effects of permeability parameter, magnetic parameter, pressure gradient, Brinkman constant and temperature difference between the wall of the channel. The results obtained are displayed graphically to illustrate the effects of flow control parameters.
\end{frame}

\begin{frame}
\frametitle{\textbf{INTRODUCTION}}
A fluid is a substance that deforms continually when subjected to shear stress, no matter how small the shear stress may be.\\ Poiseuille flow is a type flow in which an applied pressure difference causes fluid motion between stationary surfaces. \\
The conditions necessary for this type of flow are:
\begin{enumerate}[1.]
\item The flow is irrotational.
\item The pressure is not constant, so the pressure gradient is not zero, i.e.$\frac{\partial {p}}{\partial{x}} \neq0$.
\item The flow has no point of inflection.
\item The viscosity is temperature dependent.
\item Velocity of the fluid for the lower plate is zero and the same thing holds for the upper plate, i.e. at y=0, U=0 and at y=h, U=0.
\end{enumerate}
\end{frame}

\begin{frame}
\begin{center}
\begin{figure}[h]
\centering
\includegraphics[width=8cm, height=3cm]{fatty}
\caption{Poiseuille flow between two plates}
\label{fig:fatty}
\end{figure}
\end{center}
\end{frame}

\begin{frame}
\frametitle{\textbf{Aim and Objectives}}
The aim of this project is to obtain a numerical solution for a magnetohydrodynamic poiseuille flow of viscous fluid with variable viscosity.\\

The objectives of this project is to:
\begin{enumerate}[i.]
\item Use numerical method to solve differential equation arising from physical model;
\item Develop algorithms using symbolic mathematical software known as MATLAB to generate numerical values(using Runge Kutta Gill and Newton Raphson for the interpolation) of the solution and plot the graphs;
\item Investigate the different qualitative effects of varying the dimensionless parameters.\\
\end{enumerate}
\end{frame}

\begin{frame}
\frametitle{\textbf{MATHEMATICAL FORMULATION OF THE PROBLEM}}
Considering a magnetohydrodynamic poiseuille flow of viscous fluid with variable viscosity in a porous medium with constant thermo-physical properties, the resulting differentials equations governing the fluid is:\\
\begin{equation}
\frac{du^{\ast}}{dy}=0 \,\,\,(Continuity \,\,\,equation)
\end{equation}
\begin{equation}
\frac{1}{\rho}\frac{d}{dy}[\mu_0\exp^{-T^{\ast}}\frac{du^{\ast}}{dy}] - \frac{1}{\rho}\frac{dp^{\ast}}{dx} - \frac{\mu_{0}\exp^{-T^{\ast}}u^{\ast}}{\rho K^{\ast}} - \frac{\sigma B_{0}^{2}u^{\ast} }{\rho}= 0\,\,\,(Momentum\,\,\,equation)
\end{equation}
\begin{equation}
\lambda\frac{d^{2}T^{\ast}}{dy^{2}} +\mu _{0} \exp^{-T^{\ast}}(\frac{du^\ast}{dy})^{2} = 0\,\,\,(Energy\,\,\, Balance\,\,\, Equation)
\end{equation}

Subjects to the conditions:\\
\begin{equation}
\left. \begin{array}{rrl}

u^{\ast} = 0,\,\,\, T^{\ast} = T _{0},\,\,\, y = 0\\
u^{\ast} = U,\,\,\, T^{\ast} = T_{w},\,\,\, y = h\\
\end{array}\right\}
\end{equation}
\end{frame}

\begin{frame}
\frametitle{\textbf{NON-DIMENSIONALIZATION}}
\textbf{Transformation of Continuity Equation}
\begin{equation}
\begin{aligned}[left]
\frac{du^{\ast}}{dy} = \frac{1}{h}\frac{d}{d\eta}uU =\frac{U}{h} \frac{du}{d\eta} = 0 \\
\Rightarrow \frac{du}{d\eta} = 0 \\
\end{aligned}
\end{equation}
\textbf{ Transformation of Momentum equation}
\begin{equation}
\frac{1}{\rho} \frac{1}{h}\frac{d}{d\eta}[\mu_{0} \exp^{-(\alpha \theta + T_{0})} \frac{U}{h}\frac{du}{d\eta}] - \frac{1}{\rho} \rho U^{2} \frac{dp}{dx} - \frac{\mu_{0} \exp^{-(\alpha \theta + T_{0})}Uu }{\rho K^{\ast}} - \frac{\sigma B_{0}^{2} Uu}{\rho} = 0
\end{equation}
\end{frame}

\begin{frame}
\begin{equation}
\frac{U\mu_{0}}{\rho h^{2}}\exp^{-T_{0}}\frac{d}{d\eta}[\exp^{-\alpha \theta}\frac{du}{d\eta}] - U^{2}\frac{dp}{dx} - \frac{\mu_{0}U\exp^{-T_{o}}}{\rho K^{\ast}} \exp^{-\alpha\theta}u - \frac{\sigma B_{0}^{2}Uu}{\rho} = 0
\end{equation}

Multiply through by $\frac{\rho h^{2}}{U\mu_{0}}\exp^{T_{o}}$\\

\begin{equation}
\frac{d}{d\eta}[\exp^{-\alpha\theta}\frac{du}{d\eta}] - \frac{U\rho h^{2}}{\mu_{0}} \exp^{T_{0}}\frac{dp}{dx} -\frac{h^{2}}{K^{\ast}}\exp^{-\alpha\theta}u - \frac{\sigma h^{2}B_{0}^{2} \exp^{T_{0}}}{\mu_{0}}u = 0
\end{equation}
We assume that\\
\begin{equation}
\left. \begin{array}{rrl}
Km = \frac{U\rho h^{2}}{\mu_{0}}\exp^{T_{0}}\frac{dp}{dx} \\\\
Kp = \frac{K^{\ast}}{h^{2}} \\\\
Mm^{2} = \frac{\sigma h^{2}B_{0}}{\mu_{0}}\exp^{T_{o}} \\
\end{array}\right\}
\end{equation}
\end{frame}

\begin{frame}
\begin{equation}
\frac{d}{d\eta}[\exp^{-\alpha\theta}\frac{du}{d\eta}] - Km -\frac{u}{Kp}\exp^{-\alpha\theta} - Mm^{2}u = 0
\end{equation}

\begin{equation}
\frac{d}{d\eta}[\exp^{-\alpha\theta}\frac{du}{d\eta}] - Km -[\frac{1}{Kp} + Mm^{2}]u = 0
\end{equation}

\textbf{Transformation of Energy Equation}
\begin{equation}
\lambda\frac{d^{2}T^{\ast}}{dy^{2}} + \mu_{0}\exp^{-T^{\ast}}(\frac{du^{\ast}}{dy})^{2} = 0
\end{equation}

\begin{equation}
\lambda\frac{d^{2}}{dy^{2}}(\alpha\theta + T_{0}) + \mu_{0} \exp^{-(\alpha\theta + T_{0})} (\frac{1}{h}\frac{Udu}{d\eta})^{2} = 0
\end{equation}

\begin{equation}
\frac{\lambda}{h^{2}}\frac{d^{2}}{d\eta^{2}}(\alpha\theta + T_{0}) + \frac{\mu_{0}U^{2}}{h^{2}}\exp^{-\alpha\theta}\exp^{-T_{0}}(\frac{du}{d\eta})^{2} = 0
\end{equation}

\begin{equation}
\frac{\alpha\lambda}{h^{2}}\frac{d^{2}\theta}{d\eta^{2}} + \frac{\mu_{0}U^{2}}{h^{2}}\exp^{-\alpha\theta}\exp^{-T_{0}}(\frac{du}{d\eta})^{2} = 0
\end{equation}
\end{frame}

\begin{frame}
Multiply through by $ \frac{h^{2}}{\alpha\lambda}$

\begin{equation}
\frac{d^{2}\theta}{d\eta^{2}} + \frac{\mu_{0}U^{2}}{\alpha\lambda} \exp^{-T_{o}}\exp^{-\alpha\theta}(\frac{du}{d\eta})^{2} = 0
\end{equation}
Let $Br_{m} = \frac{\mu_{0}}{\alpha\lambda} U^{2}\exp^{-T_{0}}$

\begin{equation}
\frac{d^{2}\theta}{d\eta^{2}} + Br_{m}\exp^{-\alpha\theta}(\frac{du}{d\eta})^{2} = 0
\end{equation}


\end{frame}

\begin{frame}
\frametitle{\textbf{DIMENSIONLESS PROBLEM}}
\begin{equation}
\frac{d}{d\eta}[\exp^{-\alpha\theta}\frac{du}{d\eta}] - Km - [\frac{1}{Kp}\exp^{-\alpha\theta} + Mm^{2}]u = 0\,\,\, (Momentum\,\,\, Equation)
\end{equation}

\begin{equation}
\frac{d^{2}u}{d\eta^{2}} + Brm\exp^{-\alpha\theta}(\frac{du}{d\eta})^{2} = 0 \,\,\,(Energy \,\,\,Equation)
\end{equation}
Boundary Conditions\\
\begin{equation}
\begin{aligned}
u(0) = 0,\,\,\, u(1) = 0\\\\
\theta(0) = 0,\,\,\, \theta(1) = 1 \\
\end{aligned}
\end{equation}
\end{frame}

\begin{frame}
\textbf{NUMERICAL SOLUTION}
\begin{table}[h!]
\centering
\caption1{Effect of Temperature difference($\alpha$) on Velocity distribution}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{6}{|c|}{u} \\
\hline
$\eta$ & $\alpha=2.5$ & $\alpha=3.0$ & $\alpha=3.5$ & $\alpha=4$ & $\alpha=4.5$ \\
\hline
0 & 0 & 0 & 0 & 0 & 0\\
\hline
0.1 & 0.07022 & 0.096312 & 0.08169 & 0.07257 & 0.06612 \\
\hline
0.2 & 0.19454 & 0 0.22103 & 0.19213& 0.24607 & 0.21710 \\
\hline
0.3 & 25532 & 0.29160 & 0.34612 & 0.33640 & 0.39175\\
\hline
0.4 & 0.37648 & 0.43479 & 0.50811 & 0.50956 & 0.57432 \\
\hline
0.5 & 0.42422 & 0.50341 & 0.58777 & 0.68097 & 0.74509\\
\hline
0.6 & 0.43285 & 0.55067 & 0.65069 & 0.74116 & 0.80164 \\
\hline
0.7 & 0.40125 & 0.54997 & 0.65396 & 0.75766 & 0.84341 \\
\hline
0.8 & 0.33746 & 0.44127 & 0.60496 & 0.65711 & 0.78887 \\
\hline
0.9 & 0.14237 & 0.22937 & 0.30330 & 0.46146 & 0.54749\\
\hline
1 & 0 & 0 & 0 & 0 & 0\\
\hline
\end{tabular}
\end{table}
\end{frame}

\begin{frame}
\begin{figure}[h!]
\centering
\caption1{Variation of velocity profile with temperature difference with Kp=6, Km=-1, Mm=1,$Br_{m}=18$}
\label{fig:latalpha}
\includegraphics[width=9cm, height=6cm]{latalpha}
\end{figure}
\end{frame}

\begin{frame}
\begin{table}[h!]
\centering
\caption2{Effect of Pressure gradient (Km) on Velocity distribution}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{6}{|c|}{u} \\
\hline
$\eta$ & Km=-1 & Km=-1.2 & Km=-1.4 & Km=-0.6 & Km=-0.8\\
\hline
0 & 0 & 0 & 0 & 0 & 0\\
\hline
0.1 & 0.096312 & 0.08333 & 0.11332 & 0.07164 & 0.06374\\
\hline
0.2 & 0.22103 & 0.26868 & 0.32188 & 0.12485 & 0.1142\\
\hline
0.3 & 0.29160 & 0.43492 & 0.51562 & 0.15321 & 0.23440\\
\hline
0.4 & 0.43479 & 0.61233 & 0.83781 & 0.17867 & 0.29375\\
\hline
0.5 & 0.50341 & 0.52188 & 0.93851 & 0.22613 & 0.34877\\
\hline
0.6 & 0.55067 & 0.75426 & 0.98578 & 0.26124 & 0.38820\\
\hline
0.7 & 0.54997 & 0.70028 & 0.94581 & 0.26862 & 0.39092\\
\hline
0.8 & 0.44127 & 0.62211 & 0.75816 & 0.18286 & 0.35511\\
\hline
0.9 & 0.22937 & 0.37194 & 0.43826 & 0.10378 & 0.20283\\
\hline
1 & 0 & 0 & 0 & 0 & 0\\
\hline
\end{tabular}
\end{table}
\end{frame}

\begin{frame}
\begin{figure}[h!]
\centering
\caption2{Variation of velocity profile with Pressure gradient (Km) with $\alpha$=3, Kp=6, Mm=1, $Br_{m}=18$}
\label{fig:latkm}
\includegraphics[width=9cm, height=6cm]{latkm}
\end{figure}
\end{frame}

\begin{frame}
\begin{table}[h!]
\centering
\caption3{Effect of Temperature difference($\alpha$) on Temperature distribution}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{6}{|c|}{$\theta$} \\
\hline
$\eta$ & $\alpha=0.5$ & $\alpha=1.0$ & $\alpha=1.5$ & $\alpha=2.0$ & $\alpha=2.5$ \\
\hline
0 & 0 & 0 & 0 & 0 & 0\\
\hline
0.1 & 0.10000 & 0.10000 & 0.10000 & 0.10000 & 0.10000 \\
\hline
0.2 & 0.20552 & 0 0.20068 & 0.20070 & 0.20142 & 0.20178\\
\hline
0.3 & 0.30011 & 0.30020 & 0.30041 & 0.30082 & 0.30192\\
\hline
0.4 & 0.40001 & 0.40125 & 0.40131 & 0.40172 & 0.40218 \\
\hline
0.5 & 0.50151 & 0.50171 & 0.50183 & 0.50192 & 0.50214\\
\hline
0.6 & 0.60201 & 0.60269 & 0.60272 & 0.60278 & 0.60338 \\
\hline
0.7 & 0.70211 & 0.70236 & 0.70246 & 0.70320 & 0.70322 \\
\hline
0.8 & 0.80238 & 0.80438 & 0.80614 & 0.80673 & 0.80712 \\
\hline
0.9 & 0.90276 & 0.90305 & 0.90348 & 0.90324 & 0.90368\\
\hline
1 & 1 & 1 & 1 & 1 & 1\\
\hline
\end{tabular}
\end{table}
\end{frame}

\begin{frame}
\begin{figure}[h!]
\centering
\caption3{Variation of Temperature profile with Temperature difference with Kp=0.2, Km=-1, Mm=1}
\label{fig:alphatemplat}
\includegraphics[width=9cm, height=6cm]{alphatemplat}
\end{figure}
\end{frame}

\begin{frame}
\begin{table}[h!]
\centering
\caption4{Effect of Pressure gradient (Km) on Temperature distribution}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{6}{|c|}{$\theta$} \\
\hline
$\eta$ & Km=-1 & Km=-0.8 & Km=-0.6 & Km=-0.4 & Km=-0.2\\
\hline
0 & 0 & 0 & 0 & 0 & 0\\
\hline
0.1 & 0.10000 & 0.10000 & 0.10000 & 0.10000 & 0.10000\\
\hline
0.2 & 0.20068 & 0.20978 & 0.22198 & 0.24077 & 0.27636\\
\hline
0.3 & 0.30099 & 0.31281 & 0.32311 & 0.34166 & 0.36984\\
\hline
0.4 & 0.40125 & 0.41945 & 0.44090 & 0.44810 & 0.46821\\
\hline
0.5 & 0.50101 & 0.52281 & 0.53261 & 0.53794 & 0.55289\\
\hline
0.6 & 0.60269 & 0.63068 & 0.66882 & 0.67840 & 0.68291\\
\hline
0.7 & 0.70326 & 0.70328 & 0.72733 & 0.74026 & 0.74261\\
\hline
0.8 & 0.80336 & 0.80568 & 0.82332 & 0.83268 & 0.8412\\
\hline
0.9 & 0.91653 & 0.92512 & 0.92693 & 0.92753 & 0.92811\\
\hline
1 & 1 & 1 & 1 & 1 & 1\\
\hline
\end{tabular}
\end{table}
\end{frame}

\begin{frame}
\begin{figure}[h!]
\centering
\caption4{Variation of Temperature profile with Pressure gradient with $\alpha$=1, Kp=0.2, Mm=1}
\label{fig:kmtemplat}
\includegraphics[width=9cm, height=6cm]{kmtemplat}
\end{figure}
\end{frame}

\begin{frame}
\frametitle{\textbf{RESULT AND DISCUSSION}}
\textbf{Effects of Temperature change$(\alpha)$ on velocity}\\
The results in Table 1 are ascertained in Figure 1 above. The effects of temperature change on the velocity of the flow field is shown in the Figure, for five values of temperature difference, the velocity is found increasing with increase in temperature change.\\
\textbf{Effects of Pressure gradient(Km) on velocity}\\
The results in Table 2 are ascertained in the Figure 2 above. The effects of pressure gradient on the velocity of the flow field is shown in the Figure, for five values of Pressure gradient, the velocity found decreasing with increase in pressure gradient.\\
\end{frame}

\begin{frame}
\textbf{Effects of Temperature change$(\alpha)$ on temperature}\\
The results in Table 3 are ascertained in Figure 3 above. The effects of temperature change on the temperature of the flow field is shown in the Figure, for five values of temperature change, the temperature found increasing with increase in temperature change.\\
\textbf{Effects of Pressure gradient(Km) on temperature}
The results in table 4 are ascertained in Figure 4. The effects of pressure gradient on the temperature of the flow field is shown in Figure, for five values of Pressure gradient, the temperature found decreasing with increase in pressure gradient.\\
\end{frame}

\begin{frame}
\frametitle{\textbf{RECOMMENDATION AND CONCLUSION}}
\textbf{Recommendation}\\
I recommended that courses on fluid and solid mechanics should be taken as compulsory courses for every student of industrial mathematics because of its wide application and it will increase their choices of specialization.\\
Today fluid mechanics has become an integral part in diverse fields, for example in medicine, meteorology, astronautics, oceanography and every engineering disciplines. Fluid can be found everywhere, our body system is composed of fluid.\\
\textbf{Conclusion}\\
The investigation reveals the following feature on the velocity and temperature of the fluid flow:
\begin{enumerate}[i)]
\item Increase in temperature difference leads to increase in velocity and temperature of the flow.
\item Increase in the pressure gradient of leads to decrease in velocity and temperature of the flow.
\end{enumerate}
\end{frame}

\begin{frame}
\frametitle{\textbf{REFERENCES}}
\begin{thebibliography}{100}
\bibitem{1} K.S. Adegbie and F.I Alao(2007), \emph{Flow of temperature-dependent viscous fluid between parallel heated walls: Exact analytical solutions in the presence of viscous dissipation}, Journal of mathematics and statistics, 3(1), 12-14.
\bibitem{6} U.Sarmal and G.C. Hazarika(2011). \emph{Effects of variable viscosity and thermal conductivity on heat and mass transfer flow along a vertical plate in the presence of magnetic field}, vol 5(1), pp.100-106.
\bibitem{11}Baoku(2010). \emph{Magnetic field and thermal radiation effects on steady hydroddynamic couette flow through a porous channel}. vol 5, pp. 215-228.
\bibitem{12} U. Sarmal and G.C. Hazarika(2011) \emph{Effects of variable viscosity and thermal conductivity on heat and mass transfer flow along a vertical plate in the presence of a magnetic field}. vol 5(1), pp.100-106
\bibitem{13} M.Modather and E.Roshy(2012). \emph{Variable viscosity effect on heat transfer over a continuous moving surface with variable internal heat generation in micropolar fluids}. vol 6(128) 6365-6379
\end{thebibliography}

\end{frame}
\begin{frame}
\color{magenta}
\fontsize{50}{20}
\begin{center}
{{\textbf{THANKS}}\\}
\end{center}
\begin{center}
{{\textbf{FOR}}\\}
\end{center}
\begin{center}
{{\textbf{LISTENING}}\\}
\end{center}
\end{frame}
\end{document}

[Karl Ratzsch]

Am 29.07.2015 um 14:29 schrieb teejayst...@gmail.com:
> I have problem with producing this slide on another system. the error occurs with begin document, begin frame and end frame... I don't know what could be the cause.

Before anyone bites off your head for alleged laziness ;-) :

- Strip your slides down until the error vanishes, then go one step
back, and report that. Minimal examples!
- What is your system, and what is the "other" system?

Best, Karl

[teejayst...@gmail.com]

the new system is TeXstudio 2.6.2 (SVN 4110M)
Using Qt Version 4.8.5, compiled with Qt 4.8.5 R......... I don't know the name of the old version.

[Herbert Voss]

Am 29.07.2015 um 14:29 schrieb teejayst...@gmail.com:

With

\PassOptionsToPackage{demo}{graphicx}

> \documentclass[aspectratio=169]{beamer}
> \mode<beamer>{%

[..]

I have no problem with your example and cannot see why
geometry should not work. I have up-to-date TL 2015

Herbert