# Tests - Math

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## nowiki Test

<math>E=mc^2</math>


$E=mc^2$

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


$E=mc^2$

## Inequality Sign Test

<math>1<2</math>


$1\lt 2$

<math>2>1</math>


$2\gt 1$

<math>1\lt 2</math>


$1\lt 2$

<math>2\gt 1</math>


$2\gt 1$

## Inequality Sign Test 2

<math>a<b</math>


$a\lt b$

<math>a < b</math>


$a \lt b$

<math>a>b</math>


$a\gt b$

<math>a > b</math>


$a \gt b$

## UTF-8 Test

<math>전압 = 전류 \times 저항</math>


$전압 = 전류 \times 저항$

<math>저항 = \frac{전압}{전류}</math>


$저항 = \frac{전압}{전류}$

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


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

<math>n</math>개의 동전을 던져 앞면 <math>k</math>가 나올 확률 <math>P(E)</math>는?


$n$개의 동전을 던져 앞면 $k$가 나올 확률 $P(E)$는?

## The Lorenz Equations

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


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

## The Cauchy-Schwarz Inequality

<math>\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>


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

## A Cross Product Formula

<math>\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}</math>


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

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

<math>P(E)   = {n \choose k} p^k (1-p)^{ n-k}</math>


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

## An Identity of Ramanujan

<math>\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} } } }</math>


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

## A Rogers-Ramanujan Identity

<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})},


$1 + \frac{q^2}{(1-q)} + \frac{q^6}{(1-q)(1-q^2)} + \cdots = \prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})}, \quad\quad for\,|q|\lt 1.$

## Maxwell’s Equations

<math>\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}</math>


\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}