KaTeX

行内公式

${x | Ax = b}$

> <span>$\{x | Ax = b\}$</span>

$\mathcal{X}_{c^*}<1, \mathcal{X}_{q^*}<1$

$E = mc^2$

$\frac{\partial}{\partial t}$

$E = mc^2$
$\frac{\partial}{\partial t}$

行间公式

$$ E = mc^2 $$

$$ E = mc^2 $$

注意,在 Markdown 文件中直接书写时,你需要多一个 \ 来转译 \。或直接使用 $$ E = mc^2 $$ 的形式。

使用 \\[ E = mc^2 \\] 而不是 \[ E = mc^2 \]

如果你有过多需要转译的字符,你可以直接使用 HTML 标签包裹它(内部的字符将不会被作为 Markdown 解析),而无需使用多个 \ 来转译。

譬如:

\[ E = mc^2 \]
<div>\[ E = mc^2 \]</div>

\[ E = mc^2 \]

\[ E = mc^2 \]

\( E = mc^2 \)

\( E = mc^2 \)

$$ \vec a=\frac{d\vec v}{dt}=\frac{d(\frac{dr}{dt}\vec e*{i}+r\frac{d\theta}{dt}\vec e*{j})}{dt}=\frac{d^2r}{dt^2}\vec e*{i}+\frac{dr}{dt}\frac{d\vec e*{i}}{dt}+\frac{dr}{dt}\frac{d\theta}{dt}\vec e*{j}+r\frac{d^2\theta}{dt^2}\vec e*{j}+r\frac{d\theta}{dt}\frac{d\vec e_{j}}{dt} $$

\[ \vec a=\frac{d\vec v}{dt}=\frac{d(\frac{dr}{dt}\vec e_{i}+r\frac{d\theta}{dt}\vec e_{j})}{dt}=\frac{d^2r}{dt^2}\vec e_{i}+\frac{dr}{dt}\frac{d\vec e_{i}}{dt}+\frac{dr}{dt}\frac{d\theta}{dt}\vec e_{j}+r\frac{d^2\theta}{dt^2}\vec e_{j}+r\frac{d\theta}{dt}\frac{d\vec e_{j}}{dt} \]

$$ m_t=g_t $$

$$ V_t=1 $$

$$ m_t=g_t $$
$$ V_t=1 $$

$$ \eta_t=lr*{\frac {m_t}{\sqrt V_t}}=lr*g_t $$

$$ w_{t+1}=w_t-\eta_t=w_t-lr*{\frac {m_t}{\sqrt V_t}}=w_t-lr*g_t $$

$$ {\Rightarrow \boxed{w_{t+1}=w_t-lr*{\frac {\partial loss}{\partial w_t}}}} $$

$$ \eta_t=lr*{\frac {m_t}{\sqrt V_t}}=lr*g_t $$
$$ w_{t+1}=w_t-\eta_t=w_t-lr*{\frac {m_t}{\sqrt V_t}}=w_t-lr*g_t $$

<font size=5 color=red>

$$ {\Rightarrow \boxed{w_{t+1}=w_t-lr*{\frac {\partial loss}{\partial w_t}}}} $$

</font>
$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$
<div>
$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$
</div>
$$ \begin{equation} \left\{ \begin{aligned} x=a\cos\theta\\ y=b\sin\theta\\ \end{aligned} \right. \end{equation} $$
<div>
$$
\begin{equation}
\left\{
\begin{aligned}
x=a\cos\theta\\
y=b\sin\theta\\
end{aligned}
\right.
\end{equation}
$$
</div>