LaTeX implementation ready for testing on Physics Forums
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looks like we bogged the server down, once it's fixed we'll open up testing again.
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okay, let's try
[itex] $\Delta V_{(R_{1}\rightarrow R_{2})}= V_{R_{1}}\left[\sqrt{\frac{2R_{2}}{R_{1}+R_{2}}}-1 \right]$[/itex]
[itex] $\Delta V_{(R_{1}\rightarrow R_{2})}= V_{R_{1}}\left[\sqrt{\frac{2R_{2}}{R_{1}+R_{2}}}-1 \right]$[/itex]
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Yeah something is weird...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.
this
[tex] \psi[/tex]
is different from this
[tex]\psi[/tex]
and it shouldn't be.
- Warren
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Originally posted by ahrkron
The server is trying to confuse us all!
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.
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This stuff looks GREAT! Let's have a go at it
[tex]\int_{-\infty}^{\infty}[/tex]
[tex]\stackrel \cdot x[/tex]
[tex]\ddot{x} = -g[/tex]
[tex]\int \ddot{x}dt = \int -gdt[/tex]
[tex]\dot{x} = -gt + C[/tex]
[tex]\dot{x}(0)= V_0 \Rightarrow C = V_0[/tex]
[tex]\int \dot{x}dt = \int -gt + V_0 dt[/tex]
[tex]x(t)= - \frac {gt^2} {2} + V_0t + X_0[/tex]
[tex]\mbox{ x=y} \mbox{\iff |x-y| < \epsilon} \forall \mbox{ \epsilon \in \Re}[/tex]
[tex]\ddot{x} = \frac {d^2x} {dt^2}[/tex]
[tex]\int_{-\infty}^{\infty}[/tex]
[tex]\stackrel \cdot x[/tex]
[tex]\ddot{x} = -g[/tex]
[tex]\int \ddot{x}dt = \int -gdt[/tex]
[tex]\dot{x} = -gt + C[/tex]
[tex]\dot{x}(0)= V_0 \Rightarrow C = V_0[/tex]
[tex]\int \dot{x}dt = \int -gt + V_0 dt[/tex]
[tex]x(t)= - \frac {gt^2} {2} + V_0t + X_0[/tex]
[tex]\mbox{ x=y} \mbox{\iff |x-y| < \epsilon} \forall \mbox{ \epsilon \in \Re}[/tex]
[tex]\ddot{x} = \frac {d^2x} {dt^2}[/tex]
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[tex]
\int_{0}^{1} x dx = \left[ \frac{1}{2}x^2 \right]_{0}^{1} = \frac{1}{2}[/tex]
[tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]
[tex] \int_{0}^{1} y dy = \left[ \frac{1}{2}y^2 \right]_{0}^{1} = \frac{1}{2}[/tex]
[tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]
[tex] \int_{0}^{1} z dz = \left[ \frac{1}{2}z^2 \right]_{0}^{1} = \frac{1}{2}[/tex]
[tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]
[tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]
[tex] \int_{0}^{1} y dy = \left[ \frac{1}{2}y^2 \right]_{0}^{1} = \frac{1}{2}[/tex]
[tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]
[tex] \int_{0}^{1} z dz = \left[ \frac{1}{2}z^2 \right]_{0}^{1} = \frac{1}{2}[/tex]
[tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]
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Originally posted by chroot
[tex] \int_{0}^{1} x dx = \left[ \frac{1}{2}x^2 \right]_{0}^{1} = \frac{1}{2}[/tex]
[tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]
[tex] \int_{0}^{1} y dy = \left[ \frac{1}{2}y^2 \right]_{0}^{1} = \frac{1}{2}[/tex]
[tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]
[tex] \int_{0}^{1} z dz = \left[ \frac{1}{2}z^2 \right]_{0}^{1} = \frac{1}{2}[/tex]
[tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]
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Originally posted by chroot
[tex] \int_{0}^{1} u du = \left[ \frac{1}{2}u^2 \right]_{0}^{1} = \frac{1}{2}[/tex]
[tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]
[tex] \int_{0}^{1} v dv = \left[ \frac{1}{2}v^2 \right]_{0}^{1} = \frac{1}{2}[/tex]
[tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]
[tex] \int_{0}^{1} w dw = \left[ \frac{1}{2}w^2 \right]_{0}^{1} = \frac{1}{2}[/tex]
[tex]v(t) = v_0 + \frac{1}{2} a t^2[/tex]
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