LaTeX Is Mandatory LaTeX Implementation on PF Beneficial for Members?

  • Thread starter Thread starter Greg Bernhardt
  • Start date Start date
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The implementation of LaTeX on Physics Forums (PF) is nearing completion and is set for extensive testing, with a new bbcode format introduced for users. Mentors are encouraged to familiarize themselves with basic LaTeX to assist members as they navigate this new feature. Users can input LaTeX code between [tex] and [/tex] tags, with a popup feature allowing them to learn from previous equations. However, there are some initial issues, such as images not appearing in previews and problems with LaTeX compilation that need addressing. Overall, this enhancement is expected to significantly benefit the community by improving the presentation of mathematical expressions.
  • #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
 
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  • #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 /> $\Delta V_{(R_{1}\rightarrow R_{2})}= V_{R_{1}}\left[\sqrt{\frac{2R_{2}}{R_{1}+R_{2}}}-1 \right]$
 
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  • #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}
 

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