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I have a system of equations with an arbitrary number of equations (k). I'd like to use \vdots to compactly describe the system, like so:

  R(-1) &= \sum_{i=1}^m A(i)R(i-1) \\
  R(-2) &= \sum_{i=1}^m A(i)R(i-2) \\
        &\vdots                    \\
  R(-k) &= \sum_{i=1}^m A(i)R(i-k)

LaTeX output of above code with only minimal.cls and amsmath.sty

I'd like the dots, however, to be centered with = or the entire equation. Is there an elegant way of centering a column or row with AMSMath? I'm currently using an unholy combination of whitespace operators (\; \, etc.) to get the job done.

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up vote 35 down vote accepted

  R(-1) &= \sum_{i=1}^m A(i)R(i-1) \\
  R(-2) &= \sum_{i=1}^m A(i)R(i-2) \\
        &\mathrel{\makebox[\widthof{=}]{\vdots}} \\
  R(-k) &= \sum_{i=1}^m A(i)R(i-k)

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You beat me to it … anyhow, in case there is a wish to avoid using the calc package, here is my version: \setbox0\hbox{=}\mathrel{\makebox[\wd0]{\hfil\vdots\hfil}}. – Harald Hanche-Olsen Dec 23 '10 at 15:18
I like it! I had been looking at this answer, but I didn't like the need for all the temporary width definitions. \widthof is a great new tool for me. I'll leave the question open for a bit longer, but I think this is the answer. – Matt B. Dec 23 '10 at 15:18
@Harald the \hfil are not needed – Herbert Dec 23 '10 at 15:49
Thanks, Herbert! – Matt B. Dec 23 '10 at 20:19
You're right; instead of \makebox[\wd0] I had \hbox to \wd0 and then I forgot to remove the \hfil after I changed it. – Harald Hanche-Olsen Dec 23 '10 at 20:39

Consider the package mathtools, which provides several corrections for and additions to amsmath.


It also provides a comfortable solution for your problem. You can even choose between a normal (\vdotswithin) and a short (\shortvdotswithin) distance.

  a &= b \\
  & \vdotswithin{=} \\
  & = c \\
  & = d

The result convinces.

Example showing vdotswithin and shortvdotswithin

More details can be found in the documentation of the package, section "Centered \vdots", where also the example above is taken from.

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This solution worked for me while Herbert's led to a misalignement, I'm not sure why. – anderstood Oct 20 '15 at 19:04

Here is another option (not very popular among TeXans):

  R(-1) &=& \sum_{i=1}^m A(i)R(i-1) \\
  R(-2) &=& \sum_{i=1}^m A(i)R(i-2) \\
        &\vdots& \\
  R(-k) &=& \sum_{i=1}^m A(i)R(i-k)
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the spacing is lousy in an eqnarray environment. You can it improve wit \arraycolsep=1.4pt – Herbert Dec 25 '10 at 9:30
Yes, I know... this is why I said it is not a popular option... – user1999 Dec 25 '10 at 13:55

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