HandWiki:HelpMathEquations

HandWiki uses MathJax for rendering equations. You can show equations as this $F_2=\frac{H_1}{H^2}$ inline. This is programmed as:

You can show equations as this <math>F_2=\frac{H_1}{H^2}</math> inline. This is programmed as:

You can show equations using the LaTeX syntax "dollar sign":

$F_2=\frac{H_1}{H^2}$

As you can see, it positions the equation at centre. This is programmed as:

$F_2=\frac{H_1}{H^2}$ 

Alternatively, you can do this formula $F_2=\frac{H_1}{H^2}$ as:

<math inline>F_2=\frac{H_1}{H^2}</math>

More examples

See more examples here (original source [1].

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

$E=m{c}^2$

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

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

Inequality Sign Test

<math>1<2</math>

$1<2$

<math>2>1</math>

$2>1$

<math>1\lt 2</math>

Failed to parse (syntax error): {\displaystyle 1\lt 2}

<math>2\gt 1</math>

Failed to parse (syntax error): {\displaystyle 2\gt 1}

Inequality Sign Test 2

<math>a<b</math>

$a<b$

<math>a < b</math>

$a<b$

<math>a>b</math>

$a>b$

<math>a > b</math>

$a>b$

$f(x)=1$

UTF-8 Test

<math>전압 = 전류 \times 저항</math>

Failed to parse (syntax error): {\displaystyle 전압 = 전류 \times 저항}

<math>저항 = \frac{전압}{전류}</math>

Failed to parse (syntax error): {\displaystyle 저항 = \frac{전압}{전류}}

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>

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

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>

${\left(\sum _{k=1}^n{a}_k{b}_k\right)}^2\le \left(\sum _{k=1}^n{a}_k^2\right)\left(\sum _{k=1}^n{b}_k^2\right)$

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>

${\mathbf{𝐕}}_1\times {\mathbf{𝐕}}_2=\left|\begin{array}{ccc}\mathbf{𝐢}& \mathbf{𝐣}& \mathbf{𝐤}\\ \frac{\partial X}{\partial u}& \frac{\partial Y}{\partial u}& 0\\ \frac{\partial X}{\partial v}& \frac{\partial Y}{\partial v}& 0\end{array}\right|$

The probability of getting k heads when flipping n coins is

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

$P(E)=(\genfrac{}{}{0ex}{}{n}{k})p^k(1-p{)}^{n-k}$

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>

$\frac{1}{(\sqrt{\varphi \sqrt{5}}-\varphi )e^{\frac{2}{5}\pi }}=1+\frac{e^{-2\pi }}{1+\frac{e^{-4\pi }}{1+\frac{e^{-6\pi }}{1+\frac{e^{-8\pi }}{1+\dots }}}}$

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})},
\quad\quad for\,|q|<1.</math>

$1+\frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots =\prod _{j=0}^{\mathrm{\infty }}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},for|q|<1.$

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>

$\begin{array}{rl}\mathrm{\nabla }\times \overrightarrow{\mathbf{𝐁}}-\frac{1}{c}\frac{\partial \overrightarrow{\mathbf{𝐄}}}{\partial t}& =\frac{4\pi }{c}\overrightarrow{\mathbf{𝐣}}\\ \mathrm{\nabla }\cdot \overrightarrow{\mathbf{𝐄}}& =4\pi \rho \\ \mathrm{\nabla }\times \overrightarrow{\mathbf{𝐄}}+\frac{1}{c}\frac{\partial \overrightarrow{\mathbf{𝐁}}}{\partial t}& =\overrightarrow{\mathbf{𝟎}}\\ \mathrm{\nabla }\cdot \overrightarrow{\mathbf{𝐁}}& =0\end{array}$