MARKDPWN-EXAMPLE

3x+7\Bigg \langle 3x+7 \bigg \rangle
y = 1 + & \left( \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \ldots \right. \\ & \quad \left. + \frac{1}{x^{n-1}} + \frac{1}{x^n} \right)
F=G(m1m2r2)F = G \left( \frac{m_1 m_2}{r^2} \right)
[N(Lp)(m+n)]\left[ \frac{ N } { \left( \frac{L}{p} \right) - (m+n) } \right]

223324\begin{matrix} 2 & 2 & 3\\ 3 & 2 & 4\\ \end{matrix}

(223324)\begin{pmatrix} 2 & 2 & 3\\ 3 & 2 & 4\\ \end{pmatrix}

223324\begin{vmatrix} 2 & 2 & 3\\ 3 & 2 & 4\\ \end{vmatrix}

223324\begin{Vmatrix} 2 & 2 & 3\\ 3 & 2 & 4\\ \end{Vmatrix}

[223324]\begin{bmatrix} 2 & 2 & 3\\ 3 & 2 & 4\\ \end{bmatrix}

{223324}\begin{Bmatrix} 2 & 2 & 3\\ 3 & 2 & 4\\ \end{Bmatrix}

{x2+2x+2=0x3+3x2+5=0\begin{cases} x^2 + 2x +2 = 0\\ x^3 + 3x^2 +5 = 0\\ \end{cases}

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

A=Ax2+Ay2+Az2|\vec{A}|=\sqrt{A_x^2 + A_y^2 + A_z^2}

A=ABC+DEF\vec{A}=ABC+DEF

i=110ti\sum_{i=1}^{10} t_i

0\int_0^\infty

0ex,dx\int_0^\infty \mathrm{e}^{-x},\mathrm{d}x

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

B={BαTU=BαUT}\mathcal{B} = \{B_{\alpha} \in \mathcal{T}\, |\, U = \bigcup B_{\alpha} \forall U \in \mathcal{T} \}
$$ \begin{align*} RQSZ \\ \mathcal{RQSZ} \\ \mathfrak{RQSZ} \\ \mathbb{RQSZ} \end{align*}
3x2RQ3x2RQ3x2RQ3x2RQ3x2RQ3x2RQ3x2RQ\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*}

\begin{equation*} \sum_{i=1}^n i = \left(\sum_{i=1}^{n-1} i\right) + n = \frac{(n-1)(n)}{2} + n = \frac{n(n+1)}{2} \end{equation*}

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

a_{n_i}

\int_{i=1}^n

\sum_{i=1}^{\infty}

\prod_{i=1}^n

\cup_{i=1}^n

\cap_{i=1}^n

\oint_{i=1}^n

\coprod_{i=1}^n

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

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

1+ab1+11+1a\frac{1+\frac{a}{b}}{1+\frac{1}{1+\frac{1}{a}}}

Now a wild example

a0+1a1+1a2+1a3+a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cdots}}}
\newcommand*{\contfrac}[2]{% { \rlap{$\dfrac{1}{\phantom{#1}}$}% \genfrac{}{}{0pt}{0}{}{#1+#2}% } }
a_0 + \contfrac{a_1}{ \contfrac{a_2}{ \contfrac{a_3}{ \genfrac{}{}{0pt}{0}{}{\ddots} }}}
\begin{equation} \label{eq1} \begin{split} A & = \frac{\pi r^2}{2} \\ & = \frac{1}{2} \pi r^2 \end{split} \end{equation}
\begin{multline*} p(x) = 3x^6 + 14x^5y + 590x^4y^2 + 19x^3y^3\\ - 12x^2y^4 - 12xy^5 + 2y^6 - a^3b^3 \end{multline*}
2x5y=83x+9y=12\begin{align*} 2x - 5y &= 8 \\ 3x + 9y &= -12 \end{align*}
x=yw=za=b+c2x=y3w=12za=b4+5x=2+yw+2=1+wab=cb\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*}
2x5y=83x2+9y=3a+c\begin{gather*} 2x - 5y = 8 \\ 3x^2 + 9y = 3a + c \end{gather*}

Testing notation for limits

limh0f(x+h)f(x)h.\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={zCz<1}andS2=SS = \{ z \in \mathbb{C}\, |\, |z| < 1 \} \quad \textrm{and} \quad S_2=\partial{S}
f(x)=x2 ⁣+3x ⁣+2f(x)=x2+3x+2f(x)=x2+3x+2f(x)=x2+3x+2f(x)=x2  +3x  +2f(x)=x2 +3x +2f(x)=x2+3x+2f(x)=x2+3x+2\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*}

n=12n=1\sum_{n=1}^{\infty} 2^{-n} = 1

n=12n=1\sum_{n=1}^{\infty} 2^{-n} = 1

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

limxf(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)=i=0nai1+xf(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:

\begin{eqnarray*} f(x) = \sum_{i=0}^{n} \frac{a_i}{1+x} \\ \textstyle f(x) = \textstyle \sum_{i=0}^{n} \frac{a_i}{1+x} \\ \scriptstyle f(x) = \scriptstyle \sum_{i=0}^{n} \frac{a_i}{1+x} \\ \scriptscriptstyle f(x) = \scriptscriptstyle \sum_{i=0}^{n} \frac{a_i}{1+x} \end{eqnarray*}

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

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

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