2

Consider the following MWE:

\documentclass{scrartcl}

\usepackage{mathtools} 
\usepackage[upright]{fourier} 

\begin{document}

\begin{gather*}
    \sigma_i+\sigma_n\geqslant\tau\left[\nu_{\sigma}^{Q}(\{i\})+\nu(R_i^{-1}\cup\{n\})-\sum_{k\in R_i^{-1}}\sigma_k\right]+(1-\tau)\left[\nu_{\sigma}^Q(\{n\})+\nu(R_n^{-1}\cup\{i\})-\sum_{k\in R_n^{-1}}\sigma_k\right]
\end{gather*}

\begin{gather*}
    \sigma_i+\sigma_n\geqslant\tau\Bigg[\nu_{\sigma}^{Q}(\{i\})+\nu(R_i^{-1}\cup\{n\})-\sum_{k\in R_i^{-1}}\sigma_k\Bigg]+(1-\tau)\left[\nu_{\sigma}^Q(\{n\})+\nu(R_n^{-1}\cup\{i\})-\sum_{k\in R_n^{-1}}\sigma_k\right]
\end{gather*}

\end{document}

As you can see in the screenshot below, the squared parentheses of the two parts of the expression are not of the same size. I've been playing around to get them of the same size, but it turns out that I can only do so by commenting \usepackage{fourier}, something I would personally like to avoid. Using \left[\right] is too big (case 1), whereas using \Bigg[\Bigg] (the biggest I know) is not big enough (case 2).

enter image description here

My question is thus simple: how can I get them of the same size without dropping \usepackage{fourier}?

Thank you all very much for your time and effort.

  • 1
    You could use, for instance, \renewcommand{\BIG}{\bBigg@{3}}, then use \BIG instead. – Phelype Oleinik Oct 15 '18 at 22:25
  • 1
    <<case 2 id fine for me. I don't think delimiters have to encompass the totality of the sum symbol, including the indices. – Bernard Oct 15 '18 at 22:29
  • @PhelypeOleinik Thank you for your comment. If it were an answer, I'd accept it... – Héctor Oct 15 '18 at 23:27
  • 1
    Yes, but the simplest solution consists in using something like \Biggl[ ... \Biggr] in both cases. – Bernard Oct 15 '18 at 23:31
  • 1
    That's a possibility. Personally, I very rarely use left ...\right.. The horizontal spacing is better. – Bernard Oct 15 '18 at 23:34
4

What is causing you this trouble is the i lowered a tad too much:

enter image description here

There are a few approaches to overcome this and get the delimiter size you want. To name a few:

  • You can \smash the i to make it have zero depth and height, then it won't be a problem anymore:

    \begin{equation*}
       \sigma_i+\sigma_n\geqslant
          \tau \left[\nu_{\sigma}^Q(\{i\})+\nu(R_i^{-1}\cup\{n\})-\sum_{k\in R_{\smash{i}}^{-1}}\sigma_k\right]
      +(1-\tau)\left[\nu_{\sigma}^Q(\{n\})+\nu(R_n^{-1}\cup\{i\})-\sum_{k\in R_n^{-1}}\sigma_k\right]
    \end{equation*}
    
  • You can also change TeX's \delimiterfactor and \delimitershortfall and let TeX adjust the delimiters accordingly:

    \begin{equation*}
       \delimitershortfall=7pt % Found by trial-and-error
       \delimiterfactor=810 % Move outside the `equation*` to make the effect global
       \sigma_i+\sigma_n\geqslant
          \tau \left[\nu_{\sigma}^Q(\{i\})+\nu(R_i^{-1}\cup\{n\})-\sum_{k\in R_i^{-1}}\sigma_k\right]
      +(1-\tau)\left[\nu_{\sigma}^Q(\{n\})+\nu(R_n^{-1}\cup\{i\})-\sum_{k\in R_n^{-1}}\sigma_k\right]
    \end{equation*}
    
  • Or you can define a new size with amsmath's (loaded by mathtools) \bBigg@. Note that simply using \BIG you get a different spacing around the delimiters.

    \makeatletter
    \def\BIG{\bBigg@{3}} % \big = 1; \Big = 1.5; \bigg = 2; and \Bigg = 2.5
    \def\BIGl{\mathopen\BIG}
    \def\BIGr{\mathclose\BIG}
    \makeatother
    
    \begin{equation*}
       \sigma_i+\sigma_n\geqslant
          \tau \BIGl[\nu_{\sigma}^Q(\{i\})+\nu(R_i^{-1}\cup\{n\})-\sum_{k\in R_i^{-1}}\sigma_k\BIGr]
      +(1-\tau)\BIGl[\nu_{\sigma}^Q(\{n\})+\nu(R_n^{-1}\cup\{i\})-\sum_{k\in R_n^{-1}}\sigma_k\BIGr]
    \end{equation*}
    

Full code:

\documentclass{scrartcl}

\usepackage{mathtools}
\usepackage[upright]{fourier}

\begin{document}

\begin{equation*}
   \sigma_i+\sigma_n\geqslant
      \tau \left[\nu_{\sigma}^Q(\{i\})+\nu(R_i^{-1}\cup\{n\})-\sum_{k\in R_i^{-1}}\sigma_k\right]
  +(1-\tau)\left[\nu_{\sigma}^Q(\{n\})+\nu(R_n^{-1}\cup\{i\})-\sum_{k\in R_n^{-1}}\sigma_k\right]
\end{equation*}

\begin{equation*}
   \sigma_i+\sigma_n\geqslant
      \tau \left[\nu_{\sigma}^Q(\{i\})+\nu(R_i^{-1}\cup\{n\})-\sum_{k\in R_{\smash{i}}^{-1}}\sigma_k\right]
  +(1-\tau)\left[\nu_{\sigma}^Q(\{n\})+\nu(R_n^{-1}\cup\{i\})-\sum_{k\in R_n^{-1}}\sigma_k\right]
\end{equation*}

\begin{equation*}
   \delimitershortfall=7pt
   \delimiterfactor=810
   \sigma_i+\sigma_n\geqslant
      \tau \left[\nu_{\sigma}^Q(\{i\})+\nu(R_i^{-1}\cup\{n\})-\sum_{k\in R_i^{-1}}\sigma_k\right]
  +(1-\tau)\left[\nu_{\sigma}^Q(\{n\})+\nu(R_n^{-1}\cup\{i\})-\sum_{k\in R_n^{-1}}\sigma_k\right]
\end{equation*}

\makeatletter
\def\BIG{\bBigg@{3}}
\def\BIGl{\mathopen\BIG}
\def\BIGr{\mathclose\BIG}
\makeatother

\begin{equation*}
   \sigma_i+\sigma_n\geqslant
      \tau \BIGl[\nu_{\sigma}^Q(\{i\})+\nu(R_i^{-1}\cup\{n\})-\sum_{k\in R_i^{-1}}\sigma_k\BIGr]
  +(1-\tau)\BIGl[\nu_{\sigma}^Q(\{n\})+\nu(R_n^{-1}\cup\{i\})-\sum_{k\in R_n^{-1}}\sigma_k\BIGr]
\end{equation*}

\end{document}

enter image description here

  • Great answer, since it provides a few cool alternatives. Thank you! – Héctor Oct 16 '18 at 0:35

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