Difference between revisions of "User:Mingli/Wikitext"

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}
 
</graphviz>
 
</graphviz>
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 +
==Math example==
 +
=== The Lorenz Equations ===
 +
<source lang='html5'>
 +
<math>\begin{align}
 +
\dot{x} & = \sigma(y-x) \\
 +
\dot{y} & = \rho x - y - xz \\
 +
\dot{z} & = -\beta z + xy
 +
\end{align}</math>
 +
</source>
 +
 +
<math>\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 ===
 +
<source lang='html5'>
 +
<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>
 +
</source>
 +
<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>
 +
 +
=== A Cross Product Formula ===
 +
<source lang='html5'>
 +
<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>
 +
</source>
 +
 +
<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>
 +
 +
=== The probability of getting k heads when flipping n coins is ===
 +
<source lang='html5'>
 +
<math>P(E)  = {n \choose k} p^k (1-p)^{ n-k}</math>
 +
</source>
 +
 +
<math>P(E)  = {n \choose k} p^k (1-p)^{ n-k}</math>
 +
 +
=== An Identity of Ramanujan ===
 +
<source lang='html5'>
 +
<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>
 +
</source>
 +
 +
<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>
 +
 +
=== A Rogers-Ramanujan Identity ===
 +
<source lang='html5'>
 +
<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>
 +
</source>
 +
 +
<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>
 +
 +
=== Maxwell’s Equations ===
 +
<source lang='html5'>
 +
<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>
 +
</source>
 +
 +
<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>

Revision as of 03:53, 29 August 2018

Graphviz example

Graph for example no. 2

Math example

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]\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]\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]\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]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]\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]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]\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]