A way of handling \left and \right brackets

Question

I'm looking a way of making something like this:

\underbrace{
    \left[\nabla\times
    \left[\nabla\times
    \left[\ldots\nabla\times
}_{
    \infty\text{-times taking curl operator}
}
    \mathbf{V}\right]\right]\ldots\right] = ?

Related: Left/Right across multiline equation and \left and \right in equation across multiple lines (applies to terms in different scopes as well) — Qrrbrbirlbel, Nov 10 '12 at 19:48
I don't see a reason for \left and \right here since they don't contain anything large. Wouldn't simple \underbrace{[\nabla\times[\nabla\times[\cdots\nabla\times}_{\infty\text{-times taking curl operator}}\mathbf{V}]]\cdots] work? — yo', Nov 11 '12 at 08:41

score 1 · Accepted Answer · answered Nov 10 '12 at 19:43

1

It's easy to achieve this inserting an appropriate \right and \left null delimiters:

\left.
\right.

The correspondent code would be:

\underbrace{
    \left[\nabla\times
    \left[\nabla\times
    \left[\ldots\nabla\times
    \right.\right.\right.
}_{
    \infty\text{-times taking curl operator}
}
    \mathbf{V}\left.\left.\left.\right]\right]\right]\ldots\right] = ?

answered Nov 10 '12 at 19:43

m0nhawk

9,664

3

Either setting \nulldelimiterspace=0pt inside the formula or adding \kern-3\nulldelimiterspace (or \kern-4\nulldelimiterspace) help to get rid of the additional spacing by \left. or \right.. – Heiko Oberdiek Nov 10 '12 at 19:57
As far as I see, your solution would not wrong if there were \frac{V}{W} instead of \mathbf{V}. I think that \vphantom should be used here at the proper places. – yo' Nov 11 '12 at 08:43

A way of handling \left and \right brackets

1 Answers1