Ikumi Keita <ik...@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":
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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