Testing: Python Markdown Extensions

The Back Story

A compilation of test for the various Python-Markdown extensions found in Pyhton's version of Markdown. To support Latex, you need to enable the latex plugin. See MathJax Many of the examples below where taken from the MathJax Demo page.

Inline Latex Symbols

Greek letters like $\alpha, \beta, \gamma, \phi, \nabla, \pi, \rho$.

Greek letters like $\alpha, \beta, \gamma, \phi, \nabla, \pi, \rho$.

Latex Math

For the eqnarray and aligned equation examples below, you'll notice that you have to use \\\ instead of \\ for the line breaks.

Time-dependent Schrödinger Equation

$$
\begin{equation}
    i\hbar\frac{\partial}{\partial t}\Psi = \hat{H}\Psi
\end{equation}
$$

$$ \begin{equation} i\hbar\frac{\partial}{\partial t}\Psi = \hat{H}\Psi \end{equation} $$

or its one-dimension version

$$
\left [ - \frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} + V \right ] \Psi
= i \hbar \frac{\partial}{\partial t} \Psi
$$

$$ \left [ - \frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} + V \right ] \Psi = i \hbar \frac{\partial}{\partial t} \Psi $$

Maxwell's Equations

$$
\begin{eqnarray}
    \nabla \cdot \vec{E} &=& 4\pi\rho \\\
    \nabla \cdot \vec{B} &=& 0 \\\
    \nabla \times \vec{E} &=& -\frac{1}{c}\frac{\partial \vec{B}}{\partial t} \\\
    \nabla \times \vec{B} &=& \frac{1}{c}\left(4\pi\vec{J} + \frac{\partial \vec{E}}{\partial t}\right)
\end{eqnarray}
$$

$$ \begin{eqnarray} \nabla \cdot \vec{E} &=& 4\pi\rho \\ \nabla \cdot \vec{B} &=& 0 \\ \nabla \times \vec{E} &=& -\frac{1}{c}\frac{\partial \vec{B}}{\partial t} \\ \nabla \times \vec{B} &=& \frac{1}{c}\left(4\pi\vec{J} + \frac{\partial \vec{E}}{\partial t}\right) \end{eqnarray} $$

Another version of Maxwell's Equations

$$
\begin{aligned}
    \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\\
    \nabla \cdot \vec{\mathbf{B}} & = 0 \\\
    \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\\
    \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}}
\end{aligned}
$$

$$ \begin{aligned} \nabla \cdot \vec{\mathbf{E}} & = 4 \pi \rho \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \\ \nabla \times \vec{\mathbf{E}}\, +\, \frac1c\, \frac{\partial\vec{\mathbf{B}}}{\partial t} & = \vec{\mathbf{0}} \\ \nabla \times \vec{\mathbf{B}} -\, \frac1c\, \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \frac{4\pi}{c}\vec{\mathbf{j}} \end{aligned} $$

Rogers-Ramanujan Identity

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

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

Identity of Ramanujan

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

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

Something more complex

$$
\begin{aligned}
    & \phi(x,y) = \phi \left(\sum_{i=1}^n x_ie_i, \sum_{j=1}^n y_je_j \right) = \\\
    & \sum_{i=1}^n \sum_{j=1}^n x_i y_j \phi(e_i, e_j) = \\\
    & (x_1, \ldots, x_n) \left( \begin{array}{ccc} \\\
        \phi(e_1, e_1) & \cdots & \phi(e_1, e_n) \\\
        \vdots & \ddots & \vdots \\\
        \phi(e_n, e_1) & \cdots & \phi(e_n, e_n)
    \end{array} \right)
    \left( \begin{array}{c}
        y_1 \\\
        \vdots \\\
        y_n
    \end{array} \right)
\end{aligned}
$$

$$ \begin{aligned} & \phi(x,y) = \phi \left(\sum_{i=1}^n x_ie_i, \sum_{j=1}^n y_je_j \right) = \\ & \sum_{i=1}^n \sum_{j=1}^n x_i y_j \phi(e_i, e_j) = \\ & (x_1, \ldots, x_n) \left( \begin{array}{ccc} \\ \phi(e_1, e_1) & \cdots & \phi(e_1, e_n) \\ \vdots & \ddots & \vdots \\ \phi(e_n, e_1) & \cdots & \phi(e_n, e_n) \end{array} \right) \left( \begin{array}{c} y_1 \\ \vdots \\ y_n \end{array} \right) \end{aligned} $$

Cross Product Formula

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

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