LaTeX How I can use a vertical bar to represent evaluation in LaTeX

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SUMMARY

The discussion focuses on using a vertical bar to represent evaluation in LaTeX, specifically in the context of derivatives. Users suggest various methods, including the use of \left. and \right| for automatic sizing, as well as \bigg| for manual sizing. A notable solution involves defining an "invisible tower" using the \xstrut command to control the height of the vertical bar. The final recommendation is to adjust the height of the box by modifying the \myVSpace length.

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Kreizhn
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Hey guys. This might not be the right place for this, but any ideas on how I can use a vertical bar to represent evaluation in LaTeX?

Example

\displaystyle \frac{d}{dt} |_{t=0} f(t)

I would like the vertical bar to be the size of the differential. I've tried using \left|, \right|, \vert, etc. but nothing seems to work. Thanks a lot in advance.
 
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Hi Kreizhn,

Is this it?

\left. \frac{d}{dt} \right|_{t=0} f(t)

which is given by

\left. \frac{d}{dt} \right|_{t=0} f(t)

Using either \left or \right on a period means the automatic delimiter sizing takes place, but only one delimiter is shown.
 
Excellent, thank you
 
\frac{d}{dt} \bigg|_{t=0} f(t) achieves a similar effect
with manual sizing (by using two g's) and the unneeded \left. tag.

\frac{d}{dt} \bigg|_{t=0} f(t)
 


I'm using lyx and I'm having difficulty trying to find the code for this vertical line. Any suggestions? I use the | but it's extremely small.
 


Maybe try \|?

The height of the vertical line will be determined by the {stuff} in between the \left. {stuff} \right|. If it's not naturally big, you may want to add an "invisible tower." My invisible tower is called \xstrut, and is defined as follows

\newlength{\myVSpace}% the height of the box
\setlength{\myVSpace}{3ex}% the default,
\newcommand\xstrut{\raisebox{-.5\myVSpace}% symmetric behaviour,
{\rule{0pt}{\myVSpace}}%
}
To change the height of the box, change

\setlength{\myVSpace}{Your number here}
 
\mathbf{M \bigg|_{x^k} \Delta x^k = -f(x^k)}
<br /> \mathbf{M \bigg|_{x^k} \Delta x^k = -f(x^k)}
 

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