MARKDPWN-EXAMPLE

$$\Bigg \langle 3x+7 \bigg \rangle$$

$$F = G \left( \frac{m_1 m_2}{r^2} \right)$$

$$\left[ \frac{ N } { \left( \frac{L}{p} \right) - (m+n) } \right]$$

$\begin{matrix} 2 & 2 & 3\\ 3 & 2 & 4\\ \end{matrix}$

$\begin{pmatrix} 2 & 2 & 3\\ 3 & 2 & 4\\ \end{pmatrix}$

$\begin{vmatrix} 2 & 2 & 3\\ 3 & 2 & 4\\ \end{vmatrix}$

$\begin{Vmatrix} 2 & 2 & 3\\ 3 & 2 & 4\\ \end{Vmatrix}$

$\begin{bmatrix} 2 & 2 & 3\\ 3 & 2 & 4\\ \end{bmatrix}$

$\begin{Bmatrix} 2 & 2 & 3\\ 3 & 2 & 4\\ \end{Bmatrix}$

$\begin{cases} x^2 + 2x +2 = 0\\ x^3 + 3x^2 +5 = 0\\ \end{cases}$

$$\begin{array}{c} a+b\\ [f,g]\\ m+n \end{array}$$

$|\vec{A}|=\sqrt{A_x^2 + A_y^2 + A_z^2}$

$\vec{A}=ABC+DEF$

$\sum_{i=1}^{10} t_i$

$\int_0^\infty$

$\int_0^\infty \mathrm{e}^{-x},\mathrm{d}x$

Let $( \mathcal{T} )$ be a topological space, a basis is defined as

$$\mathcal{B} = \{B_{\alpha} \in \mathcal{T}\, |\, U = \bigcup B_{\alpha} \forall U \in \mathcal{T} \}$$

$$\begin{align*} 3x^2 \in R \subset Q \\ \mathnormal{3x^2 \in R \subset Q} \\ \mathrm{3x^2 \in R \subset Q} \\ \mathit{3x^2 \in R \subset Q} \\ \mathbf{3x^2 \in R \subset Q} \\ \mathsf{3x^2 \in R \subset Q} \\ \mathtt{3x^2 \in R \subset Q} \end{align*}$$

\displaystyle \bigg|\bigg| + \big|x - |1 - a| + \big| = 0

a_{n_i}	{$displaystyle a_{n_{i}}$}
\int_{i=1}^n	{$displaystyle \int _{i=1}^{n}$}
\sum_{i=1}^{\infty}	{$displaystyle \sum _{i=1}^{\infty }$}
\prod_{i=1}^n	{$displaystyle \prod _{i=1}^{n}$}
\cup_{i=1}^n	{$displaystyle \cup _{i=1}^{n}$}
\cap_{i=1}^n	{$displaystyle \cap _{i=1}^{n}$}
\oint_{i=1}^n	{$displaystyle \oint _{i=1}^{n}$}
\coprod_{i=1}^n	{$displaystyle \coprod _{i=1}^{n}$}

$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$

$f(x)=\frac{P(x)}{Q(x)} \ \ \textrm{and} \ \ f(x)=\textstyle\frac{P(x)}{Q(x)}$

$\frac{1+\frac{a}{b}}{1+\frac{1}{1+\frac{1}{a}}}$

Now a wild example

$$a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cdots}}}$$

$$\begin{align*} 2x - 5y &= 8 \\ 3x + 9y &= -12 \end{align*}$$

$$\begin{align*} x&=y & w &=z & a&=b+c\\ 2x&=-y & 3w&=\frac{1}{2}z & a&=b\\ -4 + 5x&=2+y & w+2&=-1+w & ab&=cb \end{align*}$$

$$\begin{gather*} 2x - 5y = 8 \\ 3x^2 + 9y = 3a + c \end{gather*}$$

Testing notation for limits

$$\lim_{h \to 0 } \frac{f(x+h)-f(x)}{h} .$$

This operator changes when used alongside text ( \lim_{h \to 0} (x-h) ).

$$S = \{ z \in \mathbb{C}\, |\, |z| < 1 \} \quad \textrm{and} \quad S_2=\partial{S}$$

$$\begin{align*} f(x) &= x^2\! +3x\! +2 \\ f(x) &= x^2+3x+2 \\ f(x) &= x^2\, +3x\, +2 \\ f(x) &= x^2\: +3x\: +2 \\ f(x) &= x^2\; +3x\; +2 \\ f(x) &= x^2\ +3x\ +2 \\ f(x) &= x^2\quad +3x\quad +2 \\ f(x) &= x^2\qquad +3x\qquad +2 \end{align*}$$

$\sum_{n=1}^{\infty} 2^{-n} = 1$

$\sum_{n=1}^{\infty} 2^{-n} = 1$

$\lim_{x\to\infty} f(x)$

$\lim_{x\to\infty} f(x)$

Depending on the value of $x$ the equation ( f(x) = \sum_{i=0}^{n} \frac{a_i}{1+x} ) may diverge or converge.

$f(x) = \sum_{i=0}^{n} \frac{a_i}{1+x}$

In-line maths elements can be set with a different style: (f(x) = \displaystyle \frac{1}{1+x}). The same is true the other way around:

Let $( \mathcal{T} )$ be a topological space, a basis is defined as

$$\mathcal{B} = \{B_{\alpha} \in \mathcal{T}\, |\, U = \bigcup B_{\alpha} \forall U \in \mathcal{T} \}$$