TeX File #6



\documentstyle[A4]{article}
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\newcommand{\co}{\mbox{C\hspace{-.455em}\rule{0.035em}{0.67em}\hspace{.5em}}}             
\newcommand{\r}{\mbox{R\hspace{-.730em}\rule{0.05em}{0.68em}\hspace{.7em}}}         
\newcommand{\q}{\mbox{\bf$Q\hspace{-.51em}$\rule{0.05em}{0.49em}\hspace{.4em}}} 
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\newcommand{\desda}{\mbox{$\ \Longleftrightarrow$\ }}
\newcommand{\eindebewijs}{\begin{flushright} $\Box$ \end{flushright}}
\newcommand{\isdef}{\stackrel{\rm def}{=}}
\author{Ugh \\ e-mail: ugh@webprimitives.com}
\newtheorem{lemma}{Lemma}
\newtheorem{theorem}{Theorem}
\title{\sc Some Fancy Math Equations...}
\newtheorem{defi}{Definition}
\begin{document}
\maketitle
\begin{abstract} Some Fancy Math Stuff....

\end{abstract}
{\bf Some Fancy Math\/}


\bigskip
\bigskip

First, we have:
\[
n_{i+1} = \left\{ \begin{array}{ll}
	3n_i+1 & \mbox{if $n_i \equiv 1 \pmod{2}$}\\ 
	n_i / 2 & \mbox{if $n_i \equiv 0 \pmod{2}$}
	\end{array}
\right.
\]

Second, we have:

\[f(z)=\sum_{n\ge0}{f_nz^n},\quad \hat{f}(z)=\sum_{n\ge0}{f_n{z^n\over
n!}}.\]


Third, we have:

\begin{equation} \label{motzkin}
 M_n = -\frac{1}{2} \sum_{k=\lceil \frac{n+2}{2} \rceil}^{n+2}
	(-1)^k {1/2 \choose k} {k \choose n+2-k} 2^{2k-n-2} 3^{n+2-k}.
\end{equation}


Fourth, we have:

\begin{equation} \label{motzkin}
 M_n = -\frac{1}{2} \prod_{k=\lceil \frac{n+2}{2} \rceil}^{n+2}
	(-1)^k {1/2 \choose k} {k \choose n+2-k} 2^{2k-n-2} 3^{n+2-k}.
\end{equation}

Fifth, we have:

$M(z) = (1-z-\sqrt{1-2z-3z^2})/(2z^2)$

Sixth, we have:

$$
I(n,x) = \int_0^x \frac{dt}{1 + t^n}
$$

Seventh, we have:

\begin{equation}
\begin{array}{l}
\dot{x}_{1}\,=\, Ku - \frac{1}{t} \, (x_{1}-T)\\[2mm]
\dot{x}_{2}\,=\, x_{3}\\[2mm]
\dot{x}_{3}\,=\, g \left( \frac{W_{a}}{W} \, (1-\frac{T}{x_{1}})
-1-\frac{\mu}{W} \,x_{3} \right)
\end{array}
\end{equation}


Eighth, we have:

\begin{equation}
\begin{array}{c}
8000\,Tt\, y_{3} - 400(T +Kt\, u)\, y_{2}^{2} - 40\, \Big( \mu\, (Kt\,u +T)\,
y_{1} + 200\,Kt\, u + 10\,\mu\,Tt\Big)\, y_{2}\\[3mm]
- \mu^{2}(Kt\,u +T)\, y_{1}^{2} +400\,\mu Kt\, uy_{1}+40000(T-Kt\,u) \,=\,0
\end{array}
\end{equation}


Nineth, we have:

	The {\em discriminant} {\rm disc}$(e_1,\ldots,e_n)$  of $n$ elements
	$e_1,\ldots,e_n \in L(x,y)$ is defined as:
	\[ \left| \begin{array}{ccc} {\rm Tr}(e_1e_1) & \ldots & {\rm Tr}(e_1e_n) \\
	\vdots & & \vdots \\
	{\rm Tr}(e_ne_1) & \ldots & {\rm Tr}(e_ne_n) \end{array} \right| \]
	where {\rm Tr} stands for the trace map of the extension
	$L(x) \subset L(x,y)$



Tenth, we have:

	\[ \bigcup_{t=1}^{\infty} L(( \ (x-\alpha)^{\frac1t} )) \]
	
Eleventh, we have:

\[ \nabla h = \frac{\partial h}{\partial x}\nabla x +
              \frac{\partial h}{\partial y}\nabla y +
              \frac{\partial h}{\partial a}\nabla a +
              \frac{\partial h}{\partial b}\nabla b 
\]
\begin{eqnarray*}
 \nabla h & = & \cos(x)y\cdot\left(\! \begin{array}{c} 1 \\ 0
\end{array}\!\right)+\sin(x)\cdot\left(\!
\begin{array}{c} 0 \\ 1 \end{array}\!\right) + 2\,a\,b\cdot\left(\!
\begin{array}{c} \frac{\partial a}{\partial x} \\ 
 \frac{\partial a}{\partial y} \end{array}\!\right) + a^{2}\cdot\left(\!
\begin{array}{c} \frac{\partial b}{\partial x} \\ \frac{\partial b}{\partial y}
 \end{array}\!\right) 
\\
& = & \left(\begin{array}{c} \cos(x)y+2\,a\,b\cdot
              \mbox{{\sl da}[1]}+a^{2}\cdot\mbox{{\sl db}[1]} \\
              \sin(x)+2\,a\,b\cdot\mbox{{\sl da}[2]}
               +a^{2}\cdot\mbox{{\sl db}[2]}
            \end{array}
      \right)
\end{eqnarray*}

Twelfth, we have:

$$
\int_{\frac{1}{2}}^{\pi - e^{\alpha}} \frac{\beta - \gamma}{\frac{n}{k} !}


$$



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