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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] = ?
3
  • 1
    Related: Left/Right across multiline equation and \left and \right in equation across multiple lines (applies to terms in different scopes as well) Commented Nov 10, 2012 at 19:48
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    Three \lefts and four \rights by the way … :) Commented Nov 10, 2012 at 19:51
  • 1
    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'
    Commented Nov 11, 2012 at 8:41

1 Answer 1

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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] = ?
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    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.. Commented Nov 10, 2012 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'
    Commented Nov 11, 2012 at 8:43

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