TeX Implementation

Math has provided us a method for which we can {\bf derive} through proofs and create theorems, and corollaries or even Laws.


This section describes some details about what you can do with MathJAX and LaTeX2HTML5!


$$\int\!\!\!\int_D dx\,dy$$

$$\int\int_D dx\,dy$$

$$x+y^2\over k+1 + 2$$

$$
n \choose k
$$

$$\sum\nolimits_{n=1}^m$$

$$\int\limits_0^{\pi\over2}$$

$$\int_{-\infty}^{+\infty}$$

the beauty of mathematics

You can write a proof using {\tt proof} environment. 


The platform will automatically apply the Q.E.D symbol at the end of your proof, and also make a nice bold {\bf Proof} at the beginning of the proof. In case you are curious, Q.E.D. comes from Latin, which means that that was asked to be demonstrated, or \emph{quod erat demonstrandum}. The mathematical symbol looks like this: $\qed$.


You can write a proof using this notation:

\begin{verbatim}
\begin{proof}
 % some typsetting + math goes here
\end{proof}
\end{verbatim}

Here is an example that proves that multiplication in the time domain is the product of a scalar and the convolution in the frequency domain. In other words, prove $f(t)g(t) \ftp \frac{1}{2\pi} (F * G) (\omega)$

\begin{proof}

\begin{align*}
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}(F*G)(\omega)e^{i\omega t}d\omega \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}F(W)G(\omega-W) dW e^{i\omega t}d\omega \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}F(W)\int_{-\infty}^{\infty}G(\omega-W) e^{i\omega t} d\omega dW \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}F(W)\int_{-\infty}^{\infty}G(\Omega) e^{i(\Omega+W) t} d\Omega dW \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}F(W)\lr{\int_{-\infty}^{\infty}G(\Omega) e^{i\Omega t} d\Omega }e^{iWt}dW \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}F(W)e^{iWt}dW \lr{\int_{-\infty}^{\infty}G(\Omega) e^{i\Omega t} d\Omega}\\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\lr{2\pi f(t)}\lr{2\pi g(t)} \\
\iftrans{(F*G)(\omega)} &= 2\pi f(t)g(t) \\
\iftrans{\frac{1}{2\pi}(F*G)(\omega)} &= f(t)g(t) \\
\end{align*}

Which shows that multiplication in the time domain is the product of a scalar and the convolution in the frequency domain.

\end{proof}


proof

Sometimes it is useful to include examples within text, and for this the {\tt example} environment works well.


You can add an example with the following syntax:

\begin{verbatim}
\begin{example}
 % some math + typesetting here
\end{example}
\end{verbatim}

Here is an example (no pun intended):


\begin{example}

Consider the causal signal $x(t) = e^{-3t}u(t)$. What is the Laplace Transform for $x(t+T)$? 
\begin{align*}
y(t) &= x(t+T) \\
y(t) &= e^{-3(t+T)}u(t+T) \\
\end{align*}

Note that the signal is no longer causal, although it is still right-sided. We know the transform for $x(t)$ and can utilize the shifting property:
\begin{align*}
x(t)&\ltp \hat{X}(s) = \frac{1}{3+s} \\
x(t+T) &\ltp e^{sT}\hat{X}(s) = \frac{e^{sT}}{3+s} \\
\end{align*}

The $RoC$ is the same in this case, except that it excludes $\infty$:
$$
RoC(y) : \{ s \st -3 \lt \Real{s} \lt \infty \} 
$$ 

\end{example}

example

\begin{definition}
If the function $x \in X$ is defined as $x : A \to B$, and $A$ is uncountable, then we say that the signal $x$ is a \emph{Continuous-Time Signal} (a.k.a. CT Signal).
\end{definition}


{\bf Tip:} You can use {\tt nicebox} to make it stand out if you like:

\begin{nicebox}
\begin{definition}
The function $f$ is a \emph{one-to-one correspondence}, or a \emph{bijection}, if it is both one-to-one and onto. 
\end{definition}
\end{nicebox}

definition

\begin{claim}
if a signal is real-valued, then its Fourier transform is conjugate symmetric. In other words, 

$$ x(n) \in \R \mbox{ } \forall n \Rightarrow X(\omega) = X^*(\omega) $$

\end{claim}


claim

The platform also supports corollary and theorem environments. You can define them like so:


\begin{verbatim}
\begin{corollary}
 corallary goes here!
\end{corollary}

\begin{theorem}
 theorem goes here!
\end{theorem}
\end{verbatim}

\begin{theorem}
{\bf The Chinese Remainder Theorem}
Let $m_1, m_2, \cdots , m_n \in \mathbb{Z}$ such that $ \gcd(m_i, m_j) = 1$ where $i \neq j$ Let $a_1, \cdots , a_n$ are arbitrary. This system of congruences
$$x \equiv a_1 \bmod m_1$$
$$x \equiv a_2 \bmod m_2$$
$$ \cdots $$
$$x \equiv a_n \bmod m_n$$
has a solution modulo $m= \prod_{i=1}^{n}m_i$. Basically we can find a value $x$ that is a solution by the formula
$$x=a_1M_1y_1 + a_2M_2y_2 + \cdots a_nM_ny_n$$
where $0 \leq x < m$, and all other solutions are congruent modulo $m$. 
\end{theorem}

theorems and corollaries

The platform also supports {\tt problem} and {\tt solution} environments:

\begin{verbatim}
\begin{problem}
Given the complex number $z = e^{i\frac{2\pi}{3}}$, determine $\sqrt[4]{z}$ . Be mindful of how many fourth roots $z$ has and identify each 
of them.
\end{problem}

\begin{solution}
\begin{align*}
z &= e^{i\frac{2\pi}{3}} \\
z &= e^{i(\frac{2\pi}{3} + 2\pi k) } \mbox{ for some } k \in \Z \\
\sqrt[4]{z} &= e^{i(\frac{2\pi}{3} + 2\pi k) \frac{1}{4}} \\
\sqrt[4]{z} &= e^{i(\frac{\pi}{6} + \frac{\pi}{2} k) \frac{1}{4}} 
\end{align*}
So we can determine that within the range [$0,2\pi$), we find that $\sqrt[4]{z} = e^{i\frac{\pi}{6}}, e^{i\frac{2\pi}{3}}, e^{i\frac{7\pi}{6}}, e^{i\frac{5\pi}{3}}$
\end{solution}
\end{verbatim}

\begin{problem}
Given the complex number $z = e^{i\frac{2\pi}{3}}$, determine $\sqrt[4]{z}$ . Be mindful of how many fourth roots $z$ has and identify each 
of them.
\end{problem}

\begin{solution}
\begin{align*}
z &= e^{i\frac{2\pi}{3}} \\
z &= e^{i(\frac{2\pi}{3} + 2\pi k) } \mbox{ for some } k \in \Z \\
\sqrt[4]{z} &= e^{i(\frac{2\pi}{3} + 2\pi k) \frac{1}{4}} \\
\sqrt[4]{z} &= e^{i(\frac{\pi}{6} + \frac{\pi}{2} k) \frac{1}{4}} 
\end{align*}
So we can determine that within the range [$0,2\pi$), we find that $\sqrt[4]{z} = e^{i\frac{\pi}{6}}, e^{i\frac{2\pi}{3}}, e^{i\frac{7\pi}{6}}, e^{i\frac{5\pi}{3}}$
\end{solution}


9.3 math