HandWiki:HelpMathEquations

HandWiki uses MathJax for rendering equations. You can show equations as this [math]\displaystyle{ F_2=\frac{H_1}{H^2} }[/math] 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 [math]\displaystyle{ F_2=\frac{H_1}{H^2} }[/math] 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>

[math]\displaystyle{ E=mc^2 }[/math]

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

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

Inequality Sign Test

<math>1<2</math>

[math]\displaystyle{ 1\lt 2 }[/math]

<math>2>1</math>

[math]\displaystyle{ 2\gt 1 }[/math]

<math>1\lt 2</math>

[math]\displaystyle{ 1\lt 2 }[/math]

<math>2\gt 1</math>

[math]\displaystyle{ 2\gt 1 }[/math]

Inequality Sign Test 2

<math>a<b</math>

[math]\displaystyle{ a\lt b }[/math]

<math>a < b</math>

[math]\displaystyle{ a \lt b }[/math]

<math>a>b</math>

[math]\displaystyle{ a\gt b }[/math]

<math>a > b</math>

[math]\displaystyle{ a \gt b }[/math]

[math]\displaystyle{ f(x)=1 }[/math]

UTF-8 Test

<math>전압 = 전류 \times 저항</math>

[math]\displaystyle{ 전압 = 전류 \times 저항 }[/math]

<math>저항 = \frac{전압}{전류}</math>

[math]\displaystyle{ 저항 = \frac{전압}{전류} }[/math]

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>

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

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>

[math]\displaystyle{ \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]

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>

[math]\displaystyle{ \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]

The probability of getting k heads when flipping n coins is

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

[math]\displaystyle{ P(E) = {n \choose k} p^k (1-p)^{ n-k} }[/math]

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>

[math]\displaystyle{ \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]

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>

[math]\displaystyle{ 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. }[/math]

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>

[math]\displaystyle{ \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]