流程图

代码

```flow
st=>start: User login
op=>operation: Operation
cond=>condition: Successful Yes or No?
e=>end: Into admin
st->op->cond
cond(yes)->e
cond(no)->op
```

结果

时序图

代码

```seq
.........
```
# or
```sequence
.........
```

结果

数学公式

行内的公式 Inline

代码

$$E=mc^2$$
Inline 行内的公式 $$E=mc^2$$ 行内的公式，行内的$$E=mc^2$$公式。
$$c = \\pm\\sqrt{a^2 + b^2}$$
$$x > y$$
$$f(x) = x^2$$
$$\alpha = \sqrt{1-e^2}$$
$$\(\sqrt{3x-1}+(1+x)^2\)$$
$$\sin(\alpha)^{\theta}=\sum_{i=0}^{n}(x^i + \cos(f))$$
$$\\dfrac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$$
$$f(x) = \int_{-\infty}^\infty\hat f(\xi)\,e^{2 \pi i \xi x}\,d\xi$$
$$\displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }$$
$$\displaystyle \left( \sum\_{k=1}^n a\_k b\_k \right)^2 \leq \left( \sum\_{k=1}^n a\_k^2 \right) \left( \sum\_{k=1}^n b\_k^2 \right)$$
$$a^2$$
$$a^{2+2}$$
$$a_2$$
$${x_2}^3$$
$$x_2^3$$
$$10^{10^{8}}$$
$$a_{i,j}$$
$$_nP_k$$
$$c = \pm\sqrt{a^2 + b^2}$$
$$\frac{1}{2}=0.5$$
$$\dfrac{k}{k-1} = 0.5$$
$$\dbinom{n}{k} \binom{n}{k}$$
$$\oint_C x^3\, dx + 4y^2\, dy$$
$$\bigcap_1^n p   \bigcup_1^k p$$
$$e^{i \pi} + 1 = 0$$
$$\left ( \frac{1}{2} \right )$$
$$x_{1,2}=\frac{-b\pm\sqrt{\color{Red}b^2-4ac}}{2a}$$
$${\color{Blue}x^2}+{\color{YellowOrange}2x}-{\color{OliveGreen}1}$$
$$\textstyle \sum_{k=1}^N k^2$$
$$\dfrac{ \tfrac{1}{2}[1-(\tfrac{1}{2})^n] }{ 1-\tfrac{1}{2} } = s_n$$
$$\binom{n}{k}$$
$$0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+\cdots$$
$$\sum_{k=1}^N k^2$$
$$\textstyle \sum_{k=1}^N k^2$$
$$\prod_{i=1}^N x_i$$
$$\textstyle \prod_{i=1}^N x_i$$
$$\coprod_{i=1}^N x_i$$
$$\textstyle \coprod_{i=1}^N x_i$$
$$\int_{1}^{3}\frac{e^3/x}{x^2}\, dx$$
$$\int_C x^3\, dx + 4y^2\, dy$$
$${}_1^2\!\Omega_3^4$$

结果

E=mc^2

Inline 行内的公式 E=mc^2 行内的公式，行内的E=mc^2公式。

c = \pm\sqrt{a^2 + b^2}

x > y

f(x) = x^2

\alpha = \sqrt{1-e^2}

(\sqrt{3x-1}+(1+x)^2)

\sin(\alpha)^{\theta}=\sum_{i=0}^{n}(x^i + \cos(f))

\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

f(x) = \int_{-\infty}^\infty\hat f(\xi)\,e^{2 \pi i \xi x}\,d\xi

\displaystyle \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }

\displaystyle \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)

a^2

a^{2+2}

a_2

{x_2}^3

x_2^3

10^{10^{8}}

a_{i,j}

_nP_k

c = \pm\sqrt{a^2 + b^2}

\frac{1}{2}=0.5

\dfrac{k}{k-1} = 0.5

\dbinom{n}{k} \binom{n}{k}

\oint_C x^3\, dx + 4y^2\, dy

\bigcap_1^n p \bigcup_1^k p

e^{i \pi} + 1 = 0

\left ( \frac{1}{2} \right )

x_{1,2}=\frac{-b\pm\sqrt{\color{Red}b^2-4ac}}{2a}

{\color{Blue}x^2}+{\color{YellowOrange}2x}-{\color{OliveGreen}1}

\textstyle \sum_{k=1}^N k^2

\dfrac{ \tfrac{1}{2}[1-(\tfrac{1}{2})^n] }{ 1-\tfrac{1}{2} } = s_n

\binom{n}{k}

0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+\cdots

\sum_{k=1}^N k^2

\textstyle \sum_{k=1}^N k^2

\prod_{i=1}^N x_i

\textstyle \prod_{i=1}^N x_i

\coprod_{i=1}^N x_i

\textstyle \coprod_{i=1}^N x_i

\int_{1}^{3}\frac{e^3/x}{x^2}\, dx

\int_C x^3\, dx + 4y^2\, dy

{}_1^2!\Omega_3^4

多行公式 Multi line

代码

```math or ```latex or ```katex

```math
f(x) = \int_{-\infty}^\infty
    \hat f(\xi)\,e^{2 \pi i \xi x}
    \,d\xi
```
```math
\displaystyle
\left( \sum\_{k=1}^n a\_k b\_k \right)^2
\leq
\left( \sum\_{k=1}^n a\_k^2 \right)
\left( \sum\_{k=1}^n b\_k^2 \right)
```
```math
\dfrac{ 
    \tfrac{1}{2}[1-(\tfrac{1}{2})^n] }
    { 1-\tfrac{1}{2} } = s_n
```
```katex
\displaystyle 
    \frac{1}{
        \Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{
        \frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {
        1+\frac{e^{-6\pi}}
        {1+\frac{e^{-8\pi}}
         {1+\cdots} }
        } 
    }
```
```latex
f(x) = \int_{-\infty}^\infty
    \hat f(\xi)\,e^{2 \pi i \xi x}
    \,d\xi
```

结果

f(x) = \int_{-\infty}^\infty \hat f(\xi)\,e^{2 \pi i \xi x} \,d\xi

\displaystyle \left( \sum\_{k=1}^n a\_k b\_k \right)^2 \leq \left( \sum\_{k=1}^n a\_k^2 \right) \left( \sum\_{k=1}^n b\_k^2 \right)

\dfrac{ \tfrac{1}{2}[1-(\tfrac{1}{2})^n] } { 1-\tfrac{1}{2} } = s_n

\displaystyle \frac{1}{ \Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{ \frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} { 1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }

f(x) = \int_{-\infty}^\infty \hat f(\xi)\,e^{2 \pi i \xi x} \,d\xi

KaTeX vs MathJax

https://jsperf.com/katex-vs-mathjax