Cauchy Shwartz Inequality
\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)	$${\left(\sum _{k=1}^n{a}_k{b}_k\right)}^2\le \left(\sum _{k=1}^n{a}_k^2\right)\left(\sum _{k=1}^n{b}_k^2\right)$$
Cross Product
\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}	$${𝐕}_1\times {𝐕}_2=\left|\begin{array}{ccc}𝐢& 𝐣& 𝐤\\ \frac{\partial X}{\partial u}& \frac{\partial Y}{\partial u}& 0\\ \frac{\partial X}{\partial v}& \frac{\partial Y}{\partial v}& 0\end{array}\right|$$
Lorenz Eqautions
\begin{aligned} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{aligned}	$$\begin{array}{cc}\dot{x}& =\sigma (y-x)\\ \dot{y}& =\rho x-y-xz\\ \dot{z}& =-\beta z+xy\end{array}$$
Maxwell Equations
\begin{aligned} \nabla \times \vec{\mathbf{B}} - \frac1c\, \frac{\partial\vec{ \mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{ \mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{aligned}	$$\begin{array}{cc}\nabla \times \overrightarrow{𝐁}-\frac{1}{c}\frac{\partial \overrightarrow{𝐄}}{\partial t}& =\frac{4\pi }{c}\overrightarrow{𝐣}\\ \nabla \cdot \overrightarrow{𝐄}& =4\pi \rho \\ \nabla \times \overrightarrow{𝐄}+\frac{1}{c}\frac{\partial \overrightarrow{𝐁}}{\partial t}& =\overrightarrow{\mathrm{𝟎}}\\ \nabla \cdot \overrightarrow{𝐁}& =0\end{array}$$
N Choose K
P(E) = \binom{n}{k} p^k (1-p)^{ n-k} \	$$P(E)=\left(\genfrac{}{}{0em}{}{n}{k}\right)p^k(1-p{)}^{n-k}$$
Ramanujan Identity
\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} } } }	$$\frac{1}{(\sqrt{\varphi \sqrt{5}}-\varphi )e^{\frac{25}{\pi }}}=1+\frac{e^{-2\pi }}{1+\frac{e^{-4\pi }}{1+\frac{e^{-6\pi }}{1+\frac{e^{-8\pi }}{1+\dots }}}}$$
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.	$$1+\frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots =\prod _{j=0}^{\infty }\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},\text{for}|q|<1.$$