Difference between revisions of "User:Mingli/Wikitext"
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| + | ==Gallery example== | ||
| + | |||
| + | <gallery> | ||
| + | File:Gfs 1 2018-10-23.png|北美 | ||
| + | File:Gfs 2 2018-10-23.png|欧非 | ||
| + | File:Gfs 3 2018-10-23.png|亚洲 | ||
| + | File:Gfs 4 2018-10-23.png|东太平洋 | ||
| + | File:Gfs 5 2018-10-23.png|南美 | ||
| + | File:Gfs 6 2018-10-23.png|南大洋 | ||
| + | File:Gfs 7 2018-10-23.png|北极 | ||
| + | File:Gfs 8 2018-10-23.png|南极 | ||
| + | </gallery> | ||
| + | |||
| + | ==Graphviz example== | ||
| + | |||
<graphviz renderer="neato" caption="Graph for example no. 2"> | <graphviz renderer="neato" caption="Graph for example no. 2"> | ||
graph example2 { | graph example2 { | ||
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} | } | ||
</graphviz> | </graphviz> | ||
| + | |||
| + | ==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> | ||
| + | |||
| + | === 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> | ||
| + | |||
| + | === 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> | ||
| + | |||
| + | === The probability of getting k heads when flipping n coins is === | ||
| + | |||
| + | <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> | ||
| + | |||
| + | === 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> | ||
| + | |||
| + | === 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> | ||
Latest revision as of 15:30, 23 October 2018
Contents
Gallery example
北美
欧非
亚洲
东太平洋
南美
南大洋
北极
南极
Graphviz example
Graph image source changed. Reload page to display updated graph image.
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]
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]
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]
The probability of getting k heads when flipping n coins is
[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]
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|\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]
