Math in Markdown¶

MyST-Parser offers full support for mathematical expressions. Check out the Math and equations page in the MyST-Parser documentation.

This page reproduces the examples found in the official documentation, with the purpose to show how math expressions are displayed when using Befertiti for Sphinx.

Math role¶

Using the {math} role it is possible to insert mathematical expressions in the flow of a paragraph.

Since Pythagoras, we know that {math}`a^2 + b^2 = c^2` represents the
relationship among three sides of a right triangle.

Since Pythagoras, we know that \(a^2 + b^2 = c^2\) represents the relationship among three sides of a right triangle.

Math directive¶

The math directive allows you to display the mathematical expression as a block, and to refer to it from other places of your documentation.

(1)¶\[ \begin{align}\begin{aligned}(a + b)^2 = a^2 + 2ab + b^2\\\begin{split}(a + b)^2 &= (a + b)(a + b) \\ &= a^2 + 2ab + b^2\end{split}\end{aligned}\end{align} \]

The equation (1) is a quadratic equation.

Dollar delimited math¶

MyST-Parser has support for dollar delimited math expressions, which can be used to write inline or block expressions. Read the official documentation.

$$
(a + b)^2  &=  (a + b)(a + b) \\
           &=  a^2 + 2ab + b^2
$$ (mymath2)

The equation {eq}`mymath2` is also a quadratic equation.

(2)¶\[\begin{split} (a + b)^2 &= (a + b)(a + b) \\ &= a^2 + 2ab + b^2 \end{split}\]

The equation (2) is also a quadratic equation.

Direct LaTeX Math¶

It is possible to use LaTeX by adding the amsmath extension in myst_enable_extensions:

\begin{equation}\label{e:barwq}\begin{split}
H_c&=\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2}
\sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\\
&\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot
\Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr].
\end{split}\end{equation}

(3)¶\[\begin{equation}\label{e:barwq}\begin{split} H_c&=\frac{1}{2n} \sum^n_{l=0}(-1)^{l}(n-{l})^{p-2} \sum_{l _1+\dots+ l _p=l}\prod^p_{i=1} \binom{n_i}{l _i}\\ &\quad\cdot[(n-l )-(n_i-l _i)]^{n_i-l _i}\cdot \Bigl[(n-l )^2-\sum^p_{j=1}(n_i-l _i)^2\Bigr]. \end{split}\end{equation}\]