# \bf cause part of equation undisplayed

## background

I want to show a multi-normal distribution, thus $x$ should be a vector. So I decide to use \bf or \mathbfto bold it on overleaf. However, when I bold my x， the \Sigma is undisplayed. I don't know why, maybe it is due to my loading packages or the example environment? I provide MWE below:

\documentclass[a4paper]{article}

\usepackage[english]{babel}
\usepackage{inputenc}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage[colorinlistoftodos]{todonotes}
\usepackage{ctex}
\usepackage[]{CJK}

\usepackage{amsthm}
\newtheoremstyle{break}
{\topsep}{\topsep}%
{\itshape}{}%
{\bfseries}{}%
{\newline}{}%
\theoremstyle{break}
\newtheorem{assumption}{Assumption}
\newtheorem{example}{Example}
\let\oldref\ref
\renewcommand{\ref}{(\oldref{#1})}

\begin{document}

\begin{example}
\begin{equation}
\begin{aligned}
\log{\frac{g(\bf{x})}{f(\bf{x})}} =& \frac{1}{2}\log{\frac{|\Sigma_1|}{|\Sigma_2|}} +
\frac{1}{2}(\mu_1^T \Sigma_1^{-1}\mu_1 - \mu_2^T \Sigma_2^{-1}\mu_2) \\
&+ (\mu_2^{T}\Sigma_2^{-1} - \mu_1^{T}\Sigma_1^{-1})\bf{x} +
\frac{1}{2}\bf{x}^{T}(\Sigma_1^{-1} - \Sigma_2^{-1})\bf{x}
\end{aligned}
\end{equation}
\end{example}

\end{document}



If I do not \bf{x}, the code is like this:

\begin{example}
\begin{equation}
\begin{aligned}
\log{\frac{g(\bf{x})}{f(\bf{x})}} =& \frac{1}{2}\log{\frac{|\Sigma_1|}{|\Sigma_2|}} +
\frac{1}{2}(\mu_1^T \Sigma_1^{-1}\mu_1 - \mu_2^T \Sigma_2^{-1}\mu_2) \\
&+ (\mu_2^{T}\Sigma_2^{-1} - \mu_1^{T}\Sigma_1^{-1})x + \frac{1}{2}x^{T}(\Sigma_1^{-1} - \Sigma_2^{-1})x
\end{aligned}
\end{equation}
\end{example}


Then, the display is right like this: • \bf is very different from \mathbf (say) in that it does not take an argument. It is also wrong in an equation. So use \mathbf{x} or \boldsymbol{x} but not \bf. If you want to confine the effects of \bf to x, use {\bf x}, but, as I said, you should not use it at all. – Schrödinger's cat Jan 17 at 4:39
• \let\oldref\ref \renewcommand{\ref}{(\oldref{#1})} is another stunt that is really not to be used. There is the beautiful command \eqref for that, and clever users use the cleveref package with \cref. – Schrödinger's cat Jan 17 at 4:42
• @Schrödinger'scat Thank you so much! Change all \bf to \mathbf works very well. And your clever ref advise works too! – Travis Jan 17 at 4:47
• \bf, along with most legacy two-letter font selection commands, has been deprecated for 25 years. Where did you find \bf{x} recommended to get a boldface letter in math? The bin is waiting for that guide. – egreg Jan 17 at 8:57
• Not from document, my senior classmate told me to use \bf.... @egreg – Travis Jan 17 at 13:16

With the release of LaTeX2ε in June 1994,* the commands

\bf  \it  \sl  \rm


have been deprecated. The LaTeX kernel doesn't define them; however, document classes may define them (and indeed, the standard classes do, for backward compatibility reasons).

The fact that \bf seems to work in your document is not a reason for using it.

Besides, the pre-1994 older manuals told one to say {\bf x} in order to get a boldface letter in math. Newer manuals tell you to use \mathbf{x}.

If you do \usepackage{bm} (after loading amsmath), you have the alternative of using \bm{x} that you may like more for your vectors.

Never ever use \bf again and remove it from all your documents.

As far as your code is concerned, if I compile your code with TeX Live 2018, I get The Sigmas you don't see are there, but they're boldface, which I don't think you want. There's much more unwanted boldface, too.

\documentclass[a4paper]{article}

\usepackage[english]{babel}
\usepackage{inputenc}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage[colorinlistoftodos]{todonotes}
\usepackage[fontset=windows]{ctex}
\usepackage[]{CJK}

\usepackage{amsthm}
\newtheoremstyle{break}
{\topsep}{\topsep}%
{\itshape}{}%
{\bfseries}{}%
{\newline}{}%
\theoremstyle{break}
\newtheorem{assumption}{Assumption}
\newtheorem{example}{Example}
\let\oldref\ref
\renewcommand{\ref}{(\oldref{#1})}

\begin{document}

\begin{example}
\begin{multline}
\log{\frac{g(\mathbf{x})}{f(\mathbf{x})}} = \frac{1}{2}\log{\frac{|\Sigma_1|}{|\Sigma_2|}} +
\frac{1}{2}(\mu_1^T \Sigma_1^{-1}\mu_1 - \mu_2^T \Sigma_2^{-1}\mu_2) \\
+ (\mu_2^{T}\Sigma_2^{-1} - \mu_1^{T}\Sigma_1^{-1})\mathbf{x} +
\frac{1}{2}\mathbf{x}^{T}(\Sigma_1^{-1} - \Sigma_2^{-1})\mathbf{x}
\end{multline}
\end{example}

\end{document}
` Footnote

* Yes, that's more than a quarter of a century, at the date of writing.