# Tests - Math

TeX Samples
TeX 샘플

## nowiki Test

$E=mc^2$


$E=mc^2$

<nowiki>$E=mc^2$</nowiki>


$E=mc^2$

## Inequality Sign Test

$1<2</math>  $1\lt 2$ [itex]2>1$


$2\gt 1$

$1\lt 2$


$1\lt 2$

$2\gt 1$


$2\gt 1$

## Inequality Sign Test 2

$a<b</math>  $a\lt b$ [itex]a < b</math>  $a \lt b$ [itex]a>b$


$a\gt b$

$a > b$


$a \gt b$

## UTF-8 Test

$전압 = 전류 \times 저항$


$전압 = 전류 \times 저항$

$저항 = \frac{전압}{전류}$


$저항 = \frac{전압}{전류}$

$償還までの合計利回り =\left(1+\frac{期間利率}{100}\right)^{期間}$


$償還までの合計利回り =\left(1+\frac{期間利率}{100}\right)^{期間}$

$n$개의 동전을 던져 앞면 $k$가 나올 확률 $P(E)$는?


$n$개의 동전을 던져 앞면 $k$가 나올 확률 $P(E)$는?

## The Lorenz Equations

\begin{align} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{align}


\begin{align} \dot{x} & = \sigma(y-x) \\ \dot{y} & = \rho x - y - xz \\ \dot{z} & = -\beta z + xy \end{align}

## The Cauchy-Schwarz Inequality

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


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

## A Cross Product Formula

$\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\ \frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0 \end{vmatrix}$


$\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\ \frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0 \end{vmatrix}$

## The probability of getting k heads when flipping n coins is

$P(E) = {n \choose k} p^k (1-p)^{ n-k}$


$P(E) = {n \choose k} p^k (1-p)^{ n-k}$

## An Identity of Ramanujan

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


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

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 for\,|q|<1.</math>  $1 + \frac{q^2}{(1-q)} + \frac{q^6}{(1-q)(1-q^2)} + \cdots = \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, \quad\quad for\,|q|\lt 1.$ ## Maxwell’s Equations [itex]\begin{align} \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{align}


\begin{align} \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{align}