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行內、區塊
When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$ x = {-b \pm \sqrt{b^2-4ac} \over 2a} $$
羅倫茲方程
$$ \begin{aligned} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{aligned} $$
柯西－史瓦茲不等式
$$ \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right) $$
向量外積
$$ \mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\ \frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0 \end{vmatrix} $$
投擲 n 個硬幣得到 k 個人頭的機率
$$ P(E) = {n \choose k} p^k (1-p)^{ n-k} $$
An Identity of Ramanujan
$$ \frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\ldots} } } } $$
A Rogers-Ramanujan Identity
$$ 1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots = \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, \quad\quad \text{for $|q|<1$}. $$
馬克士威方程
$$ \begin{aligned} \nabla \times \vec{\mathbf{B}} -\ \frac1c\ \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\[1.0em] \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\[0.5em] \nabla \times \vec{\mathbf{E}}\ +\ \frac1c\ \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\[1.0em] \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned} $$
Inline math: \(\varphi = \dfrac{1+\sqrt5}{2}= 1.6180339887…\)
Block math: $$ \varphi = 1+\frac{1} {1+\frac{1} {1+\frac{1} {1+\cdots} } } $$