\documentclass{beamer}
\usepackage[latin1]{inputenc}
\usetheme{Warsaw}
\title[3D Desktop Scanner]{3D Desktop Scanner}
\institute{Begeleider: Rein vd Boomgaard}
\author{Remco Gubbels}
\begin{document}
\begin{frame}
\titlepage
\end{frame}
\section{Introduction}
\subsection{Doel}
\begin{frame}{Doel}
\begin{itemize}
\item Fysiek model
\pause
\item Laserscan
\pause
\item Virtueel model
\end{itemize}
\end{frame}
\begin{frame}{Doel}
\begin{figure}
\includegraphics[scale=0.5]{laser_scanner.jpg}
\end{figure}
\end{frame}
\subsection{Voor de scriptie}
\begin{frame}{Informatie}
Wat was er al
\begin{itemize}
\item Het concept
\item Wat oppervlakkige informatie voor de aparte onderdelen
\item Oude scripties
\end{itemize}
\end{frame}
\subsection{Verschillende onderdelen}
\begin{frame}{Verschillende onderdelen}
\begin{enumerate}
\item Camera kalibratie
\item Beeldbewerking
\item Punt berekening
\item Grafische bewerking
\end{enumerate}
\end{frame}
\section{Kalibratie}
\begin{frame}{Camera kalibratie}
\begin{itemize}
\item Vormen van de 3D omgeving
\pause
\item Kalibratie technieken:
\begin{itemize}
\item Zhengyou Zhang's "A flexible new technique for camera calibration" (1998)
\item Direct Linear Transformation
\end{itemize}
\end{itemize}
\pause
\begin{figure}
\includegraphics[scale=0.3]{figure-1.png}
\end{figure}
\end{frame}
\begin{frame}{Zhang's Kalibratie Techniek}
Gebruikt voor de interne kalibratie matrix
\begin{itemize}
\item Maakt gebruik van paar foto's
\item Schaakbord patroon
\end{itemize}
\end{frame}
\begin{frame}{Direct Linear Transformation}
Gebruikt voor de externe kalibratie matrix
\begin{itemize}
\item Heeft bekende 2D met matchende 3D punten nodig
\end{itemize}
\end{frame}
\section{Beeldbewerking}
\begin{frame}{Beeldbewerking}
Probleem: Dedecteren van de laser
\begin{itemize}
\item Experimenteel onderzoek
\item Oplossing: RBG waarden
\end{itemize}
\end{frame}
\begin{frame}{Experimenteel onderzoek}
\begin{itemize}
\item Kleur in RGB
\item Laserlijn is vel rood
\end{itemize}
Plaatjes
\end{frame}
\section{Punt berekening}
\begin{frame}{Punt berekening}
\begin{itemize}
\item Laser plane berekenen
\item Punt op het laser plane bepalen
\item omzetten naar x, y, z waarden
\end{itemize}
\pause
\begin{figure}
\includegraphics[scale=0.2]{figure-2.png}
\end{figure}
\end{frame}
\section{Grafische bewerking}
\subsection{Bedoeling}
\begin{frame}{Bedoeling}
\begin{itemize}
\item Points cloud
\item Omzetten naar mesh
\end{itemize}
\begin{figure}
\includegraphics[scale=0.5]{figure-3.png}
\end{figure}
\end{frame}
\subsection{Delaunay}
\begin{frame}{Delaunay}
\begin{itemize}
\item Voorbereiding
\begin{itemize}
\item Grote cellen berekenen
\item Punten in cellen verdelen
\end{itemize}
\item Begin
\begin{itemize}
\item Pak een willekeurig begin punt
\item Zoek het tweede punt erbij
\end{itemize}
\item En dan?
\begin{itemize}
\item Bounding Box
\end{itemize}
\item Range Searching
\item Shelling
\end{itemize}
\end{frame}
\section{Resultaten}
\begin{frame}
Plaatjes hoofd
Plaatjes aparte onderdelen
\end{frame}
\subsection{Conclusie}
\begin{frame}
\begin{itemize}
\item Kalibratie werkt met een goede nauwkeurigheid
\item Detectie laserlijn werkt ook redelijk, maar met een paar foute metingen
\item Omrekenen punt van 2D naar 3D werkt
\item Delaunay werkt, alleen resultaat wordt beinvloed door de foute metingen
\end{itemize}
\end{frame}
\section{Wat kan er nog aan gedaan worden}
\begin{frame}
\begin{itemize}
\item Verbeteringen:
\begin{itemize}
\item Verbetering detectie laserlijn
\item Filter systeem
\end{itemize}
\pause
\item Extra's:
\begin{itemize}
\item Combineren meerdere meshes
\item Real-time
\end{itemize}
\end{itemize}
\end{frame}
\end{document}
I can compile this file without problem, there's no error. Could you please post your .log file as attachment to a post? Perhaps we would find further useful information there.
And please ensure that the posted example is really the same as the one which gave the error.
try to post a minimal example, which produces that error, so we could check and help to fix it. You could create such minimal example by making a copy of your document and reducing that.
\documentclass{beamer}
\usetheme{Singapore}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage[retainorgcmds]{IEEEtrantools}
\usepackage{graphicx}
\usepackage{multicol}
\usepackage[english]{babel}
\usepackage{color}
\title{Macroeconomics I \\ Optimization Techniques II}
\author{Andrew Keinsley}
\institute{University of Kansas}
\date{January 27, 2012}
\begin{document}
\begin{frame}
\titlepage
\end{frame}
\begin{frame}
\frametitle{Agenda}
\tableofcontents
\end{frame}
\subsection{Basic Differential Equations}
\begin{frame}
\frametitle{Basic Differential Equations}
A differential equation is an equation that involves the derivatives of a function as well as the function itself.
\vspace{.5cm}
These are the backbone of macroeconomic models. We constantly model how variable change over time, which involves the derivative of the variable with respect to time as well as the variable itself.
\end{frame}
\begin{frame}
Consider a basic differential equation:
$$\dot{y}(t) = -\alpha y(t) - x(t)$$
\vspace{.5cm}
Here we want to solve for $y(t)$. To do so, let's first gather every $y(t)$ on the left side.
\begin{equation}
\dot{y}(t) + \alpha y(t) = -x(t) \label{difeq1}
\end{equation}
\end{frame}
\begin{frame}
\frametitle{Integrating Factor}
Now we will multiply both sides of (\ref{difeq1}) by what's called the integrating factor. Here, our integrating factor is
$$e^{\int \!\! \alpha \mathrm{d}t} = e^{\alpha t}$$
\vspace{.5cm}
Thus, our differential equation becomes:
\begin{equation}
e^{\alpha t} \left[\dot{y}(t) + \alpha y(t)\right] = -e^{\alpha t}x(t) \label{difeq2}
\end{equation}
\end{frame}
\begin{frame}
Notice that the left side of (\ref{difeq2}) can be rewritten as a derivative with respect to time.
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}t} \left[ e^{\alpha t} y(t) \right] = -e^{\alpha t}x(t) \label{difeq3}
\end{equation}
\vspace{.5cm}
Slightly abusing the notation, we'll swing the $\mathrm{d}t$ in (\ref{difeq3}) to the right side.
\begin{equation}
\mathrm{d}\left[ e^{\alpha t} y(t) \right] = -e^{\alpha t}x(t) \mathrm{d}t \label{difeq4}
\end{equation}
\end{frame}
\begin{frame}
Integrating both sides of (\ref{difeq4})
\begin{equation}
\int \mathrm{d}\left[ e^{\alpha t} y(t) \right] = \int -e^{\alpha t}x(t) \mathrm{d}t \label{difeq5}
\end{equation}
\vspace{.5cm}
Which yields
\begin{equation}
e^{\alpha t} y(t) = - \int e^{\alpha t}x(t) \mathrm{d}t + C \label{difeq6}
\end{equation}
Where $C$ is the constant term.
\end{frame}
\begin{frame}
Thus,
$$y(t) = e^{-\alpha t} \left[ - \int e^{\alpha t}x(t) \mathrm{d}t + C \right]$$
\vspace{.5cm}
Most of the differential equations you'll see will solve to a much nicer form, but this is just a simple example.
\end{frame}
\begin{frame}
\frametitle{Separation of Variables}
In the Solow Model, the growth rate of labor is fixed at $n$. This differential equation can be solved using \textit{separation of variables}.
$$\frac{\dot{L}(t)}{L(t)} = n$$
\vspace{.5cm}
Rewriting this will give us a form we can handle.
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}t}L(t) = L(t)n \label{labor1}
\end{equation}
\end{frame}
\begin{frame}
Abusing the notation in (\ref{labor1}) again and rearranging some things,
\begin{equation}
\frac{\mathrm{d}L(t)}{L(t)} = n\mathrm{d}t \label{labor2}
\end{equation}
\vspace{.5cm}
Integrating both sides of (\ref{labor2}) gives us
\begin{equation}
\ln{L(t)} = nt + c \nonumber
\end{equation}
\vspace{.5cm}
Thus,
\begin{equation}
L(t) = e^{nt + c} \label{labor3}
\end{equation}
\end{frame}
\begin{frame}
We can now solve for the constant term in (\ref{labor3}).
\begin{equation}
L(t) = Ce^{nt} \label{labor4} \\
\;\;\; \mathrm{where} \; C = e^c
\end{equation}
\end{frame}
\end{document}
Thanks for the help. Let me know if you need anything else.
Alright, I copied and pasted everything except for the part I posted earlier into an new file and it worked fine. I then started re-placing each frame of the part that didn't work to see where I went wrong.
I found that every frame in the section I posted worked fine until the last one. At that point I got some sort of "there's nothing to end" error. So I eliminated the frame and typeset it again, where then I received the error term I originally posted.
I'm not sure if that helps at all, but I think it's worth a shot.
