3

In the following excerpt, I was wondering how to center the expression J'_{+}(x^*)(y-x^*), located in the second column of the first line, between the \leq and = signs. I tried to insert \hfill with no effect.

Thanks.

enter image description here

\documentclass{extbook}
\usepackage{amsmath}

\begin{document}    
\begin{alignat*}{2}
    0 & \leq J'_{+}(x^*)(y-x^*) && = \lim_{t \downarrow 0} \frac{J \big[ x^* + t(y-x^*) \big] - J(x^*)}{t} \\
    h & \leq \lim_{t \downarrow 0} \frac{t J(y) + (1-t) J(x^*) - J(x^*)}{t} && = J(y)-J(x^*).
\end{alignat*}
\end{document}

2 Answers 2

4

You'd have to measure the width of the larger expression, which would allow you to place thinner objects within a larger box and adjust the alignment, if needed. This can be automated using eqparbox:

enter image description here

\documentclass{extbook}

\usepackage{amsmath,eqparbox}

% https://tex.stackexchange.com/a/34412/5764
\makeatletter
\NewDocumentCommand{\eqmathbox}{o O{c} m}{%
  \IfValueTF{#1}
    {\def\eqmathbox@##1##2{\eqmakebox[#1][#2]{$##1##2$}}}
    {\def\eqmathbox@##1##2{\eqmakebox{$##1##2$}}}
  \mathpalette\eqmathbox@{#3}
}
\makeatother

\begin{document}    

\begin{alignat*}{2}
  0 & \leq \eqmathbox[JJ]{                     J'_{+} (x^*) (y - x^*)                     } && = \lim_{t \downarrow 0} \frac{J \big[ x^* + t (y - x^*) \big] - J(x^*)}{t} \\
  h & \leq \eqmathbox[JJ]{\lim_{t \downarrow 0} \frac{t J(y) + (1 - t) J(x^*) - J(x^*)}{t}} && = J(y) - J(x^*).
\end{alignat*}

\end{document}

\eqmathbox[<tag>][<align>]{<stuff>} figures out the maximum width needed for <stuff> across the same <tag>. Individual <align>ment can be changed for <stuff> that isn't the widest across the same <tag> (default is centred).

If you don't enjoy this automation, then use this:

\newsavebox{\bigmath}
\savebox{\bigmath}{$\displaystyle \lim_{t \downarrow 0} \frac{t J(y) + (1 - t) J(x^*) - J(x^*)}{t}$}% Capture large math object

\begin{alignat*}{2}
  0 & \leq \makebox[\wd\bigmath]{$J'_{+} (x^*) (y - x^*)$} && = \lim_{t \downarrow 0} \frac{J \big[ x^* + t (y - x^*) \big] - J(x^*)}{t} \\
  h & \leq \lim_{t \downarrow 0} \frac{t J(y) + (1 - t) J(x^*) - J(x^*)}{t} && = J(y) - J(x^*).
2
  • Thank you, @Werner, for the detailed answer.
    – AEW
    Commented Jul 12, 2023 at 21:47
  • I’m not sure that the label should be an optional argument, because each such measurement needs a specific label.
    – egreg
    Commented Jul 13, 2023 at 11:27
5

Like this?

enter image description here

\documentclass{article}
\usepackage{amsmath}
\usepackage{IEEEtrantools}
\begin{document}

\begin{IEEEeqnarray}{cCcCl}
  0 & \leq & J'_{+}(x^*)(y-x^*) & = & \lim_{t \downarrow 0} \frac{J \big[ x^* + t(y-x^*) \big] - J(x^*)}{t} \\
  h & \leq & \lim_{t \downarrow 0} \frac{t J(y) + (1-t) J(x^*) - J(x^*)}{t} & = &  J(y)-J(x^*).
\end{IEEEeqnarray}

\end{document}
1
  • Yes, @Stephen, thank you. Is there a solution with the standard math environments?
    – AEW
    Commented Jul 12, 2023 at 21:18

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