Re: [O] [odt] equation labels

[Myles English]

>> On Mon, 31 Oct 2011 03:41:18 +0530, Jambunathan K said:

  > Myles English <address@hidden> writes:
  >> I have found that Equations become labelled as Figures in the
  >> version I am using:
  >> 
  >> emacs 23.3.1 org-mode from git commit 71f1c1be (Oct 26) The test
  >> equations in latex-mathml.org in this message:
  >> 
  >> http://lists.gnu.org/archive/html/emacs-orgmode/2011-09/msg00198.html
  >> 
  >> are labelled as "Equation" in the odt files but when I export it
  >> fresh I get "Figure".

  > This was a regression. I pushed a fix few moments ago. Could you
  > please pull again?

Thanks for the push, there are three things I notice now:

1) my document won't open and causes libreoffice to crash! I get:
   "terminate called after throwing an instance of
     what():  vector::_M_default_append" on the command line

2) the first equation in latex-mathml.org is not numbered, I would
   expect this if it was using a begin{equation*} environment but not a
   begin{equation}.

3) the second equation looks a bit like this:

   x=root(b)               (1)
   Radicals

   but I would have expected something like:

   x=root(b)
   Equation 1.: Radicals

Is there a new variable that I need to set to get (e.g.) "Equation 1."?

Just to be explicit, the test file latex-mathml.org I am referring to
contains:

#+TITLE:     latex-mathml.org
#+AUTHOR:    Jambunathan K
#+EMAIL:     address@hidden
#+DATE:      2011-09-09 Fri
#+DESCRIPTION:
#+KEYWORDS:
#+LANGUAGE:  en
#+OPTIONS:   H:3 num:t toc:t \n:nil @:t ::t |:t ^:t -:t f:t *:t <:t
#+OPTIONS:   TeX:t LaTeX:t skip:nil d:nil todo:t pri:nil tags:not-in-toc

#+EXPORT_SELECT_TAGS: export
#+EXPORT_EXCLUDE_TAGS: noexport
#+LINK_UP:   
#+LINK_HOME: 
#+XSLT:

* LaTeX Fragments

** LaTeX Fragment1
#   See org-format-latex-options

    There is a equation down below.

   \begin{equation}
     e = \frac{1}{2}mv^2
   \end{equation}

** LaTeX Fragment2

#+CAPTION: Radicals
#+LABEL: Equation:1
   \begin{equation}
   x=\sqrt{b}
   \end{equation}

   If $a^2=b$ and \( b=2 \), then the solution must be either $$
   a=+\sqrt{2} $$ or \[ a=-\sqrt{2} \].


Myles