# How to typeset “evaluated at” vertical bar with two conditions instead of one?

I would like to typeset the differential of a multivariate function evaluated at a point where two of the variables take precise value. I typeset the "evaluated at" using a vertical bar, at the bottom of which I indicate the point at which it's evaluated.

When it's evaluated at just "one variable", no problem, I would just do:

$\left. \frac{\partial f}{\partial x} \right|_{y=0}$


But how could I evaluate the derivative at y=0, z=1? I could write it in one line but it becomes quite long. So ideally I would like to "stack" them. At the moment, I use:

$\left. \frac{\partial f}{\partial x} \right|_{\mathop{z=1}\limits^{y=0}}$


which is very kludgy to say the least. And it doesn't work very nicely: the y=0 is typeset a bit smaller than the z=0.

Any idea on how I could do this properly?

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I found out that I can achieve a somewhat acceptable result using the "mathtools" package, and the \substack command: $\left.\frac{\partial f}{\partial x} \right|_{\tiny\substack{y=0\\ z=1}}$ The \tiny is optional, and actually throws in a warning that it's not valid in math mode, so it's probably not the exact perfect solution. Also the second line is further down than the vertical bar, but that's a start. –  Peutch Sep 20 '11 at 10:16
The \substack command is actually provided by the amsmath package (in its v2.0 at least). –  Peutch Sep 20 '11 at 10:20

You can modify the definition of subarray and \substack to accept an optional argument:

\makeatletter
\renewenvironment{subarray}[2][c]{%
\if#1c\vcenter\else\vbox\fi\bgroup
\Let@ \restore@math@cr \default@tag
\baselineskip\fontdimen10 \scriptfont\tw@
\lineskip\thr@@\fontdimen8 \scriptfont\thr@@
\lineskiplimit\lineskip
\ialign\bgroup\ifx c#2\hfil\fi
$\m@th\scriptstyle##$\hfil\crcr
}{%
\crcr\egroup\egroup
}
\makeatother
\renewcommand{\substack}[2][c]{\subarray[#1]{c}#2\endsubarray}


Then

$\left.\frac{\partial f}{\partial x}\right|_{\substack[b]{y=0\\z=1}}$


will do. The optional argument default is c, that is, the usual placement. It can be also b, to get "bottom alignment".

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That works perfectly, thanks! –  Peutch Sep 20 '11 at 12:09

This code uses the \shortstack command in the subscript. I also use \left. before the matrix and \right| after the matrix to fit the vertical bar.

\begin{align}
\left.
\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0
\end{bmatrix}
\right|_{\shortstack{\tiny $a=0$ \\ \tiny $b=0$ \\ \tiny $c=0$ \\ \tiny $d=0$}}
\end{align}

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