Is Mandatory LaTeX Implementation on PF Beneficial for Members?

  • Context: LaTeX 
  • Thread starter Thread starter Greg Bernhardt
  • Start date Start date
Click For Summary
SUMMARY

The implementation of LaTeX on Physics Forums (PF) is nearing completion, with extensive testing scheduled before community release. The new BBCode tags [tex] and [latex] allow users to input LaTeX code, enhancing the forum's capability to display mathematical equations. Key features include pop-up windows for viewing code and the ability to learn from previously submitted equations. However, users have reported issues with image generation and code display, which are being addressed by the development team.

PREREQUISITES
  • Basic understanding of LaTeX syntax and commands
  • Familiarity with BBCode formatting
  • Experience with online forum interactions
  • Knowledge of troubleshooting web-based applications
NEXT STEPS
  • Research LaTeX commands for advanced mathematical notation
  • Explore troubleshooting techniques for BBCode in forums
  • Learn about LaTeX integration in web applications
  • Investigate user feedback mechanisms for software testing
USEFUL FOR

This discussion is beneficial for forum administrators, educators, and students involved in mathematics and science, as well as anyone interested in enhancing their online communication with LaTeX.

  • #31
Originally posted by enigma
Should we be worried if it gives the "Latex image is being generated, please reload" message?
Perhaps... I'm not sure what's hanging it up. I'm trying to figure that out now... eek.

- Warren
 
Physics news on Phys.org
  • #32
<br /> \frac{1}{2}<br />
 
  • #33
Argh... all of a sudden the server seems to be having a problem...

- Warren
 
  • #34
<br /> \frac{3}{4}<br />
 
  • #35
I just looked through the lists.

Are there commands for double and triple integral as well as surface and volume integral symbols?
 
  • #36
looks like we bogged the server down, once it's fixed we'll open up testing again.
 
  • #37
Woohoo... sorry about the mistake guys.

\gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}

- Warren
 
Last edited:
  • #38
Do your worst, fellas... let's see if I forgot to do anything else.

- Warren
 
  • #39
\alpha

\beta

\gamma
 
  • #40
I tried this one as an edit to a previous post, and it did not work there upon reloading.

<br /> \psi<br />
 
  • #41
A psi doesn't work?

\psi
 
  • #42
What I typed up there was "\psi"; instead, I see what I would have expected afger typing "\alpha".

I guess I'll just have to wait a little longer.
 
  • #43
okay, let's try
<br /> $\Delta V_{(R_{1}\rightarrow R_{2})}= V_{R_{1}}\left[\sqrt{\frac{2R_{2}}{R_{1}+R_{2}}}-1 \right]$
 
Last edited:
  • #44
Originally posted by chroot
A psi doesn't work?

\psi

Apparently.
 
  • #45
Originally posted by ahrkron
What I typed up there was "\psi"; instead, I see what I would have expected afger typing "\alpha".

I guess I'll just have to wait a little longer.
Yeah something is weird...

this

<br /> \psi<br />

is different from this

\psi

and it shouldn't be.

- Warren
 
  • #46
There we go!

B_0 \rightarrow J\psi K^0_s
 
  • #47
The server is trying to confuse us all! :smile:

Now you see it, now you don't, now you complain, now it's ok, it looked different just a minute ago...
 
  • #48
Originally posted by ahrkron
The server is trying to confuse us all! :smile:

Now you see it, now you don't, now you complain, now it's ok, it looked different just a minute ago...

The reason is looks wacky is simple:

1) Some graphics take longer than others to create; they're all in background processes. Sometimes, you view the thread at first, but your graphics are not done yet.

2) When a graphic is not present (because it is not done being generated), the stupid webserver will try to help you out by sending you another graphic with a similar filename. It's supposed to help you avoid 404 errors for spelling mistakes. I really hate this "feature," because in this case, it results in your post showing someone else's graphics temporarily until yours are done!

I'm going to fix this in a few minutes.. hold on.

- Warren
 
  • #49
Okay, let's try this. This should fix the weirdness.

<br /> \Delta V = - I_{sp} \times g_0 \times \ln{\frac{M_L + M_i}{M_L + M_i + M_P}}<br />

- Warren
 
Last edited:
  • #50
This stuff looks GREAT! Let's have a go at it

\int_{-\infty}^{\infty}


\stackrel \cdot x

\ddot{x} = -g
\int \ddot{x}dt = \int -gdt
\dot{x} = -gt + C
\dot{x}(0)= V_0 \Rightarrow C = V_0
\int \dot{x}dt = \int -gt + V_0 dt
x(t)= - \frac {gt^2} {2} + V_0t + X_0


\mbox{ x=y} \mbox{\iff |x-y| &lt; \epsilon} \forall \mbox{ \epsilon \in \Re}

\ddot{x} = \frac {d^2x} {dt^2}
 
Last edited:
  • #51
<br /> \begin{pmatrix} D_1t&amp;-a_{12}t_2&amp;\dots&amp;-a_{1n}t_n\\<br /> -a_{21}t_1&amp;D_2t&amp;\dots&amp;-a_{2n}t_n\\<br /> -a_{n1}t_1&amp;-a_{n2}t_2&amp;\dots&amp;D_nt<br /> \end{pmatrix}<br />
 
Last edited:
  • #52
\frac{a}{b}
 
  • #53
<br /> \int\!\!\!\int\!\!\!\int f(u,v,w)\,du\,dv\,dw<br />
 
Last edited:
  • #54
\mu
\nu
\xi
\phi
 
  • #55
g_{\mu\nu}

g_{\mu\nu}

g_{\mu\nu}

g_{\mu\nu}

g_{\mu\nu}

g_{\mu\nu}

g_{\mu\nu}

g_{\mu\nu}
 
  • #56
<br /> \int_{0}^{1} x dx = \left[ \frac{1}{2}x^2 \right]_{0}^{1} = \frac{1}{2}<br />

v(t) = v_0 + \frac{1}{2} a t^2

<br /> \int_{0}^{1} y dy = \left[ \frac{1}{2}y^2 \right]_{0}^{1} = \frac{1}{2}<br />

v(t) = v_0 + \frac{1}{2} a t^2

<br /> \int_{0}^{1} z dz = \left[ \frac{1}{2}z^2 \right]_{0}^{1} = \frac{1}{2}<br />

v(t) = v_0 + \frac{1}{2} a t^2
 
  • #57
Test.
 
  • #58
Originally posted by chroot
<br /> \int_{0}^{1} x dx = \left[ \frac{1}{2}x^2 \right]_{0}^{1} = \frac{1}{2}<br />

v(t) = v_0 + \frac{1}{2} a t^2

<br /> \int_{0}^{1} y dy = \left[ \frac{1}{2}y^2 \right]_{0}^{1} = \frac{1}{2}<br />

v(t) = v_0 + \frac{1}{2} a t^2

<br /> \int_{0}^{1} z dz = \left[ \frac{1}{2}z^2 \right]_{0}^{1} = \frac{1}{2}<br />

v(t) = v_0 + \frac{1}{2} a t^2
 
  • #59
Originally posted by chroot
<br /> \int_{0}^{1} u du = \left[ \frac{1}{2}u^2 \right]_{0}^{1} = \frac{1}{2}<br />

v(t) = v_0 + \frac{1}{2} a t^2

<br /> \int_{0}^{1} v dv = \left[ \frac{1}{2}v^2 \right]_{0}^{1} = \frac{1}{2}<br />

v(t) = v_0 + \frac{1}{2} a t^2

<br /> \int_{0}^{1} w dw = \left[ \frac{1}{2}w^2 \right]_{0}^{1} = \frac{1}{2}<br />

v(t) = v_0 + \frac{1}{2} a t^2
 
Last edited:
  • #60
Originally posted by chroot
g_{\mu\nu}

g_{\mu\nu}

g_{\mu\nu}

g_{\mu\nu}

g_{\mu\nu}

g_{\mu\nu}

g_{\mu\nu}

g_{\mu\nu}
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • Sticky
  • · Replies 60 ·
3
Replies
60
Views
189K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 766 ·
26
Replies
766
Views
742K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K