Underbrace in equation makes expression a bit too big

Question

I have an expression with an \underbrace, where the parenthesis and a second partial devirative gets a little to big.

Here is the code:

\begin{fleqn}
\begin{align*}
    \frac{1}{v_G} =  \frac{1}{v_{ph}} \cdot \left(\frac{1 - \frac{\lambda}{n} \frac{\partial n}{\partial \lambda}}{1 - \left(\frac{\lambda}{n} \right)^2 \underbrace{\frac{\partial^2 n}{\partial \lambda^2}}_{=0}} \right) = \frac{1}{v_{ph}} \cdot \left( 1 - \frac{\lambda}{n} \frac{\partial n}{\partial \lambda} \right) 
\end{align*}
\end{fleqn}

and here the result:

How can I adjust the code so that the second order partial derivative is as big as the term over the fraction line?

Welcome to TeX.SX! The simplest would be removing the big parentheses which are redundant. But also the \underbrace needs a fix. — egreg, Nov 04 '18 at 14:31

score 5 · Answer 1 · answered Nov 04 '18 at 14:48

I would simply omit those redundant parentheses. And most certainly fix the size of the \underbrace object.

It's also possible (but requires a phantom) to have the right size for the parentheses:

\documentclass{article}
\usepackage{amsmath}

\makeatletter
\newcommand{\fixedunderbrace}[2]{{\mathpalette\fixed@underbrace{{#1}{#2}}}}
\newcommand{\fixed@underbrace}[2]{\fixed@@underbrace#1#2}
\newcommand{\fixed@@underbrace}[3]{\underbrace{#1#2}_{#3}}
\makeatother

\begin{document}

\begin{equation*}
\frac{1}{v_G} =
\frac{1}{v_{ph}} \cdot 
  \frac{1 - \frac{\lambda}{n} \frac{\partial n}{\partial \lambda}}
       {1 - \left(\frac{\lambda}{n} \right)^2
        \fixedunderbrace{\frac{\partial^2 n}{\partial \lambda^2}}{=0}}
= \frac{1}{v_{ph}} \cdot 
  \left( 1 - \frac{\lambda}{n} \frac{\partial n}{\partial \lambda} \right) 
\end{equation*}

\begin{equation*}
\frac{1}{v_G} =
\frac{1}{v_{ph}} \cdot 
  \left(
  \frac{1 - \frac{\lambda}{n} \frac{\partial n}{\partial \lambda}}
       {1 - \left(\frac{\lambda}{n} \right)^2
        \smash[b]{\fixedunderbrace{\frac{\partial^2 n}{\partial \lambda^2}}{=0}}}
  \right)
  \vphantom{
    \frac{1 - \frac{\lambda}{n} \frac{\partial n}{\partial \lambda}}
       {1 - \left(\frac{\lambda}{n} \right)^2
        \fixedunderbrace{\frac{\partial^2 n}{\partial \lambda^2}}{=0}}
  }
= \frac{1}{v_{ph}} \cdot 
  \left( 1 - \frac{\lambda}{n} \frac{\partial n}{\partial \lambda} \right) 
\end{equation*}

\end{document}

With mathlap or something like that command is also possible? — manooooh, Nov 04 '18 at 14:51
@manooooh No, \math[r|c|l]lap is for hiding horizontal lengths; for hiding a vertical length one needs \smash; however, the real height has to be reinforced later or the vertical spacing would break (the line below the display would be too near, in this case). — egreg, Nov 04 '18 at 14:57

score 3 · Answer 2 · answered Nov 04 '18 at 14:52

Some suggestions (the first two of which have already been made by @egreg):

Omit the large parentheses entirely -- they seem to satisfy no useful purpose.
Use \tfrac (or \textstyle\frac) instead of \frac in the first argument of \underbrace. (By default, the first argument of \underbrace is processed in display-style math mode. However, the remainder of the denominator is processed in text style, not display style. That's why it's necessary to write \tfrac instead of just \frac.)
Omit all \cdot directives. They achieve little but clutter up the appearance of the equation.
For better horizontal spacing as well as for more typographically-appropriate sizing of the parentheses in the final part of the equation, use \Bigl( and \Bigr) instead of \left( and \right).

\documentclass{article}
\usepackage{amsmath} % for "\tfrac" macro
\begin{document}
\[
\frac{1}{v_G} 
= \frac{1}{v_{ph}}  
\frac{1 - \frac{\lambda}{n} \frac{\partial n}{\partial \lambda}}{%
      1 - \bigl(\frac{\lambda}{n} \bigr)^{\!2} \!
\underbrace{\tfrac{\partial^2 n}{\partial\lambda^2}}_{=0}}
= \frac{1}{v_{ph}} \Bigl( 1 - \frac{\lambda}{n} 
    \frac{\partial n}{\partial \lambda} \Bigr)
\]
\end{document}

score 2 · Answer 3 · answered Nov 04 '18 at 15:21

You can either use the \cfrac command to have all fractions in display style, or \tfrac for the underbraced partial derivative. I don't like such a discrepancy between these two sizes in the same formula, so I suggest using the medium size commands from nccmath (~80% of \displaystyle). I give an example of both, with some spacing improvements, and in the second example, the use ofthe esdiff package to simpify typing of partial derivatives:

\documentclass{article}

\usepackage{amsmath, nccmath}
\usepackage{esdiff}
\usepackage{bigstrut}

\begin{document}

\begin{fleqn}
\begin{align*}
    \frac{1}{v_G} = \frac{1}{v_{ph}} \cdot \left(\frac{1 - \cfrac{\lambda}{n\bigstrut[b]} \cfrac{\partial n}{\partial \lambda}}{1 - \biggl(\cfrac{\lambda}{n} \biggr)^{\mkern-5mu 2} \smash[b]{\underbrace{\cfrac{\partial^2 n}{\partial \lambda^2}}_{=0}}} \right) = \frac{1}{v_{ph}} \cdot \left( 1 - \frac{\lambda}{n} \frac{\partial n}{\partial \lambda} \right)
\end{align*}
\bigskip

\begin{align*}
    \frac{1}{v_G} = \frac{1}{v_{ph}} \cdot \left(\frac{1 - \medmath{\frac{\lambda}{n\bigstrut[b]} \diffp{n}{\lambda}}}{1 - \medmath{ \left(\frac{\lambda}{n} \right)^{\mkern-5mu 2} \smash[b]{\underbrace{\diffp[2]{n}{\lambda}}_{=0}}}} \right) = \frac{1}{v_{ph}} \cdot \left( 1 - \frac{\lambda}{n} \diffp{n}{\lambda} \right)
\end{align*}
\end{fleqn}

\end{document}

@Bernhard Thank you very much! Your solution seems to be very compact. In future, I will use the \usepackage{esdiff} :) — astronerd, Nov 04 '18 at 15:28

Underbrace in equation makes expression a bit too big

3 Answers3