/qqmbr

Mathematical layout for web and paper

Primary LanguagePythonMIT LicenseMIT

qqmbr

Mathematical layout for web and paper

qqmbr is a publishing engine for mathematical texts that aims on different media: web, mobile and paper.

Highlights

  • Clean syntax, based on indentml markup. It's a mixture of Python-style indent-based blocks and LaTeX-style commands beginning with backslash, see code samples below.
  • Excellent equation support on web and mobile, thanks to MathJax.
  • All what you need to structure your text: different levels of headers, environments (theorem, lemma and so on).
  • References by labels.
  • Preview of equations and definitions on hover (a.k.a. snippets, see example here, look for «Контрольный вопрос»)).
  • Interactive quizes, (see example here).
  • Programmatically generated images embedding (matplotlib and plotly).

View in action

You can look at my Lecture Notes on ODE sources (see e.g. this qqmbr source and its HTML render, in Russian) or at the code sample below.

You can also play with limited subset of qqmbr features in live demo.

Inspiration

qqmbr was inspired by various projects and conceptions:

  • TeX and LaTeX for math and backslashes.
  • Python for the importance of indents (and qqmbr is written in Python!);
  • YAML for indent-based markup language;
  • DocOnce for one source to any media approach and some ideas on realization;
  • S-expressions for simplicity (why they have attributes in XML?).

TODO

The following features are on the to-do list:

  • QqLaTeXFormatter that converts qqmbr source code to LaTeX.
  • more layour features:
    • more math environments (gather, multline and so on);
    • admonitions (boxes for warning, notice, question, etc.), like in DocOnce;

You are welcome to participate with pull requests and issue-reporting.

Code sample

This is an example of qqmbr markup. See live demo.

\chapter Euler's formula
\section Complex numbers \label sec:comlex

One can introduce \em{complex numbers} by considering so-called \em{imaginary unit} $i$, such that:
\equation \label eq:i
    i^2 = -1.
\definition
    \emph{Complex number} is a number of a form $x+iy$, where $x, y \in \mathbb R$.

\section Exponent of complex number \label sec:exp
\theorem \label thm:1
    Let $x$ be real number. Then
    \equation \label eq:main
        e^{ix} = \cos x+i\sin x.
\proof
    Let us recall series for exponent, sine and cosine:
    \align
        \item \label eq:series-exp
            e^z &= 1 + z + \frac{z^2}{2} + \frac{z^3}{3!} + \ldots = \sum_{k=0}^\infty \frac{z^k}{k!}
        \item \label eq:series-sin
            \sin z &= z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \ldots
        \item \label eq:series-cos
            \cos z &= 1 - \frac{z^2}{2} +\frac{z^4}{4!} - \frac{z^6}{6!} + \ldots
    It follows from \ref{eq:i} that
    \eq
        i^k=\begin{cases}
            1 & \text{for } k = 4m\\\\
            i & \text{for } k = 4m+1\\\\
            -1 & \text{for } k = 4m+2\\\\
            -i & \text{for } k = 4m+3
            \end{cases}
    Let us put $z=ix$ into \ref{eq:series-exp}. For even $k$, $(ix)^k$ is real and for odd $k$ it is imaginary. Moreover, the sign alternates when one pass to the next term. It follows immediately that the real part of \ref{eq:series-exp} is equal to \ref{eq:series-sin} and the imaginary part is equal to \ref{eq:series-cos} with substitution $z=ix$.

    This finished the proof of \ref[Euler's formula\nonumber][thm:1].

\chapter Corollary

It follows from \ref[Theorem][thm:1] from \ref[Section][sec:exp] that
\eq
    \sin x = \frac{e^{ix}-e^{-ix}}{2i}.
Therefore,
\align
    \item \nonumber
        \sin 2x &= \frac{e^{2ix}-e^{-2ix}}{2i} = \frac{1}{2i}((e^{ix})^2-(e^{-ix})^2)=
    \item \nonumber
        &\frac{1}{2i}(e^{ix}-e^{-ix})(e^{ix}+e^{-ix})=2\sin x \cos x.

\question
    Express $\cos x$ in terms of exponents. (Click on pencil to check the correct answer.)
    \quiz
        \choice
            $(e^{ix}+e^{-ix})/(2i)$
            \comment
                No, this is complex number!
        \choice \correct
            $(e^{ix}+e^{-ix})/2$
            \comment
                Yes! That's right!
        \choice
            $(e^{ix}-e^{-ix}))/(2)$
            \comment
                No, that's $-i\sin x$.

\exercise
    Express $\cos 2x$ in terms of $\sin x$ and $\cos x$.