GNU bug report logs - #66559
13.2.2; Math symbols become bold within theorem environment

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Message #17 received at 66559 <at> debbugs.gnu.org (full text, mbox):

From: Arash Esbati <arash <at> gnu.org>
To: Ikumi Keita <ikumi <at> ikumi.que.jp>
Cc: 66559 <at> debbugs.gnu.org, Jihuan Tian <jihuan_tian <at> hotmail.com>
Subject: Re: bug#66559: 13.2.2; Math symbols become bold within theorem
 environment
Date: Sun, 10 Mar 2024 11:21:28 +0100

[Message part 1 (text/plain, inline)]

Ikumi Keita <ikumi <at> ikumi.que.jp> writes:

> Ah, indeed. Now I can see the difference in the preview-latex. All
> (most?) math expressions in Theorem environnment are in bold. I'm very
> sorry, Jihuan.
>
> I guess that they inherit the boldness of the portion of the theorem
> title (in this case, "Implicit function theorem"). However, it's beyond
> my ability to work around this.

I had a second look at this.  I think this isn't a bug; one has to tell
preview how to deal with the Theorem environment by setting something
like this in the preamble:

\usepackage[displaymath,floats,graphics,footnotes,textmath]{preview}
\PreviewEnvironment*[{[]}]{Theorem}

The result looks like this and Ok for me with "emacs -Q":

[preview.png (image/png, inline)]

[Message part 3 (text/plain, inline)]

Can you reproduce this?  This is the file I used:

--8<---------------cut here---------------start------------->8---
\documentclass{article}
\usepackage{amsmath}
\usepackage[standard, framed, amsmath, hyperref, thmmarks, thref]{ntheorem}
\usepackage[displaymath,floats,graphics,footnotes,textmath]{preview}
\PreviewEnvironment*[{[]}]{Theorem}

\begin{document}
\begin{Theorem}[Implicit function theorem]
  \label{theo:implicit-func}
  Let $A$ be an open set in $\mathbb{R}^{n+r}$ and
  $f: A \rightarrow \mathbb{R}^r$ be $\mathbb{C}^r$. $f$ can be
  written as $f(x,y)$, where $x \in \mathbb{R}^n$ and
  $y \in \mathbb{R}^{r}$. Assume $(a,b) \in A$, where
  $a \in \mathbb{R}^n$, $b \in \mathbb{R}^r$ and $f(a,b) = 0$, and the
  Jacobian
  $\vert\frac{\partial f}{\partial y}\vert_{x=a, y=b} \neq 0$. Then
  $\exists$ neighborhood $B$ of $a$ in $\mathbb{R}^n$ and a unique
  $\mathbb{C}^r$ function $g: B \rightarrow \mathbb{R}^r$ such that
  $g(a) = b$ and $f(x, g(x)) = 0 \; (\forall x \in B)$, i.e.
  $y \in \mathbb{R}^r$ can be differentiably represented by
  $x \in \mathbb{R}^n$ in a neighborhood of $(a,b)$.
\end{Theorem}

  Let $A$ be an open set in $\mathbb{R}^{n+r}$ and
  $f: A \rightarrow \mathbb{R}^r$ be $\mathbb{C}^r$. $f$ can be
  written as $f(x,y)$, where $x \in \mathbb{R}^n$ and
  $y \in \mathbb{R}^{r}$. Assume $(a,b) \in A$, where
  $a \in \mathbb{R}^n$, $b \in \mathbb{R}^r$ and $f(a,b) = 0$, and the
  Jacobian
  $\vert\frac{\partial f}{\partial y}\vert_{x=a, y=b} \neq 0$. Then
  $\exists$ neighborhood $B$ of $a$ in $\mathbb{R}^n$ and a unique
  $\mathbb{C}^r$ function $g: B \rightarrow \mathbb{R}^r$ such that
  $g(a) = b$ and $f(x, g(x)) = 0 \; (\forall x \in B)$, i.e.
  $y \in \mathbb{R}^r$ can be differentiably represented by
  $x \in \mathbb{R}^n$ in a neighborhood of $(a,b)$.
\end{document}

%%% Local Variables:
%%% mode: latex
%%% TeX-master: t
%%% End:
--8<---------------cut here---------------end--------------->8---

Best, Arash

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