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


by Kreizhn
Tags: evaluation, latex, represent, vertical
Kreizhn
Kreizhn is offline
#1
Apr16-08, 06:58 PM
P: 743
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

[tex] \displaystyle \frac{d}{dt} |_{t=0} f(t) [/tex]

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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alphysicist
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#2
Apr16-08, 08:50 PM
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Hi Kreizhn,

Is this it?

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

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.
Kreizhn
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#3
Apr16-08, 08:56 PM
P: 743
Excellent, thank you

robphy
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#4
Apr16-08, 11:32 PM
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how I can use a vertical bar to represent evaluation in LaTeX


\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.

[tex]\frac{d}{dt} \bigg|_{t=0} f(t)[/tex]
PiRho31416
PiRho31416 is offline
#5
May5-11, 12:35 PM
P: 17
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.
Kreizhn
Kreizhn is offline
#6
May5-11, 12:42 PM
P: 743
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}
radiator
radiator is offline
#7
Feb19-12, 12:15 AM
P: 23
\mathbf{M \bigg|_{x^k} \Delta x^k = -f(x^k)}
[tex]
\mathbf{M \bigg|_{x^k} \Delta x^k = -f(x^k)}[/tex]


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