Is Your LaTeX Code Previewing Correctly?

[OlderDan]

I guess I'd better leave one message here so I don't keep creating new threads. I can't see any LaTeX in preview mode, so I need a place to actually post a message to see if the code is good. I will add and delete messages to this thread for that purpose.

 

[Gokul43201]

I guess I'd better leave one message here so I don't keep creating new threads. I can't see any LaTeX in preview mode, so I need a place to actually post a message to see if the code is good.

You're not alone. There's been that LaTeX problem here for a few months now...speaking of which: long time, no see! Welcome back.

 

[OlderDan]

You're not alone. There's been that LaTeX problem here for a few months now...speaking of which: long time, no see! Welcome back.

Thanks. I got real busy there for some time. I may be here in spurts. Don't know for sure.

Now I can't even get the thing to display when I post the message grrrrrrrrrrrr!

 

[jtbell]

Adding a title

Why not just leave the final equations here, as examples for other people to learn from? At least if you're testing some new technique; you can add a title or note describing it, if you get it to work.

Or can you add a title to a previously-posted message? Let's see... Aha, you can, if you "Go Advanced" when editing.

 

[OlderDan]

Why not just leave the final equations here, as examples for other people to learn from? At least if you're testing some new technique; you can add a title or note describing it, if you get it to work.

Or can you add a title to a previously-posted message? Let's see... Aha, you can, if you "Go Advanced" when editing.

I could do that. The ones that have worked have been posted in context. The stuff I was doing here was just checking the LaTeX, so it had no discussion to go with it.

Here's the one I was working on last night that I could not get to view even here. Let's see if it works

$$c \tau ' = \sqrt {l^2 \sin ^2 \theta + \left( {\frac{{l\cos \theta }}{\gamma } - v\tau '} \right)^2 }$$

$$c^2 \tau '^2 = l^2 \left( {1 - \cos ^2 \theta } \right) + \frac{{l^2 \cos ^2 \theta }}{{\gamma ^2 }} - 2\frac{{l\cos \theta }}{\gamma }v\tau ' + v^2 \tau '^2$$

$$c^2 \tau '^2 = l^2 \left( {1 - \cos ^2 \theta + \frac{{\cos ^2 \theta }}{{\gamma ^2 }}} \right) - 2\frac{{l\cos \theta }}{\gamma }v\tau ' + v^2 \tau '^2$$

$$0 = \left( {c^2 - v^2 } \right)\tau '^2 + 2\frac{{l\cos \theta }}{\gamma }v\tau ' - l^2 \left( {1 - \cos ^2 \theta \left( {1 - \frac{1}{{\gamma ^2 }}} \right)} \right)$$

$$0 = \left( {1 - \beta ^2 } \right)c^2 \tau '^2 + 2\frac{{l\cos \theta }}{\gamma }\beta c\tau ' - l^2 \left( {1 - \beta ^2 \cos ^2 \theta } \right)$$

$$0 = \frac{{c^2 \tau '^2 }}{{\gamma ^2 }} + 2l\beta \cos \theta \frac{{c\tau '}}{\gamma } - l^2 \left( {1 - \beta ^2 \cos ^2 \theta } \right)$$

$$\frac{{c\tau '}}{\gamma } = \frac{{ - 2l\beta \cos \theta \pm \sqrt {\left( {2l\beta \cos \theta } \right)^2 + 4l^2 \left( {1 - \beta ^2 \cos ^2 \theta } \right)} }}{2}$$

$$\frac{{c\tau '}}{\gamma } = - l\beta \cos \theta \pm l\sqrt {\beta ^2 \cos ^2 \theta + 1 - \beta ^2 \cos ^2 \theta }$$

$$\frac{{c\tau '}}{\gamma } = - l\beta \cos \theta \pm l = l - l\beta \cos \theta = l\left( {1 - \beta \cos \theta } \right)$$

$$\tau ' = \left( {1 - \beta \cos \theta } \right)\gamma \frac{l}{c}$$

OK I think I can make this work now.

 

[OlderDan]

$$\ddot{\psi} + \omega_0 ^2 \psi = 0$$

 

[OlderDan]

Well, the question says specifically to use the Lagrange Equation. My problem is I don't know what to integrate and what I'm integrating over. Also, I'm assuming I have to use either one of these relations:

$$\Gamma\left(p\right)=\int^{\infty}_{0}x^{p-1}e^{-x}dx$$

or

$$\Gamma\left(p+1\right) = p\Gamma\left(p\right)$$

I think there is another form that will help you. See

http://mathworld.wolfram.com/GammaFunction.html

equation (5)

or

http://numbers.computation.free.fr/Constants/Miscellaneous/gammaFunction.html

equation (1)

Your equation

$$\frac{d}{dt}m\dot{x}=-\frac{m}{2x}$$

can be integrated by separating variables

$$\frac{dv}{dt}=-\frac{1}{2x}$$

$$\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=-\frac{1}{2x}$$

$$vdv =-\frac{dx}{2x}$$

$\Gamma\left(\frac{1}{2}\right)$

 

[OlderDan]

$$ma = \frac{m}{L}g\left[ {y - \left( {L - y} \right)} \right] = mg\left( {\frac{{2y - L}}{L}} \right)$$

$$a = g\left( {\frac{{2y - L}}{L}} \right)$$

$$v\frac{{dv}}{{dy}} = g\left( {\frac{{2y - L}}{L}} \right)$$

$$vdv = g\left( {\frac{{2y}}{L} - 1} \right)dy$$

$$\frac{{v^2 }}{2} = g\left( {\frac{{y^2 }}{L} - y} \right) + C$$

$$C = g\left( {y_o - \frac{{y_o ^2 }}{L}} \right)$$ where $v$ is zero at $y_o$

$$v^2 = 2g\left( {\frac{{y^2 }}{L} - y + y_o - \frac{{y_o ^2 }}{L}} \right)$$ for any $y$ starting from rest at $y_o$

$$v^2 = 2g\left( {\frac{{y^2 }}{L} - y + \frac{L}{4}} \right)$$ if $y_o = \frac{L}{2}$

$$v_L ^2 = 2g\left( {y_o - \frac{{y_o ^2 }}{L}} \right) = 2gy_o \left( {1 - \frac{{y_o }}{L}} \right)$$ when $y = L$

$$v_{\max } ^2 = g\frac{L}{2}$$ when $y_o = \frac{L}{2}$ and $y = L$

 

[Doc Al]

You guys realize that we have an existing thread dedicated to practicing Latex? https://www.physicsforums.com/showthread.php?t=8997"

Yes, welcome back OlderDan!

 

[OlderDan]

You guys realize that we have an existing thread dedicated to practicing Latex? https://www.physicsforums.com/showthread.php?t=8997"

Yes, welcome back OlderDan!

I didn't realize that other area was available for this sort of thing. I can do things over there if that is preferred.

 

[Ouabache]

Greg has since, locked the LaTeX typesetting thread. It was recommended in the thread, to delete our work after getting our LaTeX formatted. But a lot of folks left their work posted. Actually after seeing some of the more exotic (1, 2, and 3 ) LaTeX posts, I am glad those were left. Greg recommends using this http://at.org/~cola/tex2img/index.php from now on, to practise typesetting.

 

[berkeman]

Greg has since, locked the LaTeX typesetting thread. It was recommended in the thread, to delete our work after getting our LaTeX formatted. But a lot of folks left their work posted. Actually after seeing some of the more exotic (1, 2, and 3 ) LaTeX posts, I am glad those were left. Greg recommends using this http://at.org/~cola/tex2img/index.php from now on, to practise typesetting.

Hmmm. The converter didn't work for me. Gave me an error for this:

$$\gamma^2 = \pi R^2 \Xi$$

 

[cristo]

Hmmm. The converter didn't work for me. Gave me an error for this:

$$\gamma^2 = \pi R^2 \Xi$$

Did you tick the "math mode" box at the bottom left? It didn't work for me until i noticed that!

 

[berkeman]

Did you tick the "math mode" box at the bottom left? It didn't work for me until i noticed that!

Hah! That fixed it. Thanks cristo!

$$\gamma^2 = \pi R^2 \Xi$$