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jonk
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24 Feb 2008, 12:07 am

Okay. Now I'm having fun.

Image

which is done with:

Code:
\end{equation*}
\newcommand{\dudu}{du}
\begin{eqnarray}
  y(t) & = & \left\{
  \begin{array}{ll}
    \int_{-\infty}^{y} x(u) \dudu{} \quad \mathsf{if} \quad t<0 \\
    \\
    \int_{t}^{+\infty} \sqrt{x(u)} \dudu{} \quad \mathsf{else}
  \end{array}
  \right.
\end{eqnarray}
\begin{equation*}

Jon


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SqrachMasda
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29 Feb 2008, 3:38 pm

haha
oh
that is very cool
:D

that's one of those things i've just been too lazy to try
but seeing how you asked
and then ended up doing it
just looks like i have to try too........
....
...
.



SqrachMasda
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29 Feb 2008, 3:51 pm

i remember seeing this a long time ago
i think i tried for 10sec
and gave up
like i just did

maybe it's worth it though
http://www.physicsforums.com/showthread.php?t=8997



QuantumCowboy
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06 Apr 2008, 9:31 am

Alright, I may be missing the point here. I am unfamiliar with using Latex with the web.

However, are you saying that if I enter Latex equation code into a message body, it would be rendered?


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lau
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06 Apr 2008, 10:23 am

QuantumCowboy wrote:
Alright, I may be missing the point here. I am unfamiliar with using Latex with the web.

However, are you saying that if I enter Latex equation code into a message body, it would be rendered?

Yes, if you were on "Physics Forums" and you put the "tex" BBCode tags around it.

Here, the best we have come up with is the Laeqed approach. The Latex code is embedded in the PNG images posted above. If you want to edit one of those equations, you would need to save the PNG locally, run Laeqed on it to make your changes to the Latex code in it, save it, upload it somewhere as an image, and finally make a post with that image in it.

As in, slightly correcting Jonk's last formula - giving:

Image

He had the upper bound of the first integral as "y" instead of "t", which cannot have been what he intended.


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dark_mage
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06 Apr 2008, 11:09 am

Nice now let's see what would happen if we did this:

taking that original y(t) and define x(u) = 1/u
at that point the limits of integration would have to be the limit as t approaches 0 from -inf for the top function (I might have this mixed up but let's see what happens when actually try this)



lau
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06 Apr 2008, 2:50 pm

dark_mage wrote:
Nice now let's see what would happen if we did this:

taking that original y(t) and define x(u) = 1/u
at that point the limits of integration would have to be the limit as t approaches 0 from -inf for the top function (I might have this mixed up but let's see what happens when actually try this)

Erm... jonk's equation pair was just an example of layout. It happens to produce a function that isn't exactly very interesting. With your example function x(u)=1/u, both definite integrals are easily simplified. It produces (a positive) infinity for y(t) at t=0. All negative values of t produce negative results, tending to negative infinity near t=0. I.e. y(t) is not continuous at t=0.


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dark_mage
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11 Apr 2008, 12:33 pm

Sorry I understood what he was trying to do and was just thinking of an example to try that particular equation out. :D