## 流程图

### 代码

flowst=>start: User loginop=>operation: Operationcond=>condition: Successful Yes or No?e=>end: Into adminst->op->condcond(yes)->econd(no)->op

## 时序图

### 代码

seq.........# orsequence.........

## 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

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