# Displaying equations using Tex

Discussion in 'Physics & Math' started by Pete, Dec 20, 2006.

1. ### ArafuraOpalRegistered Member

Messages:
20
$\sum_{n=1}^{40} \left( \sum_{m=1}^n \left( \frac {a \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + l(\frac {2 \pi}{3}) \right) }{3^{(m+(n^2-n)/2)}} \right) \right) , l=-1,0,1$

$\sum_{n=1}^8 \left( \sum_{m=1}^n \left( \frac {a \left( \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + k(\frac { \pi}{3} ) + \cos \left( {3^{( m+(n^2-n)/2)}} \theta / b + k(\frac { \pi}{3} ) \right) \right) }{2 \times 3^{(m+(n^2-n)/2)}} \right) \right) , k=-1,1$

3. ### ArafuraOpalRegistered Member

Messages:
20
$\sum_{n=1}^{40} \left( \sum_{m=1}^n \left( \frac {a \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + l(\frac {2 \pi}{3}) \right) }{3^{(m+(n^2-n)/2)}} \right) \right) , \ l=-1,0,1$

$\sum_{n=1}^8 \left( \sum_{m=1}^n \left( \frac {a \left( \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + k(\frac { \pi}{3} ) + \cos \left( {3^{( m+(n^2-n)/2)}} \theta / b + k(\frac { \pi}{3} ) \right) \right) }{2 \times 3^{(m+(n^2-n)/2)}} \right) \right) , \ k=-1,1$

5. ### ArafuraOpalRegistered Member

Messages:
20
$\displaystyle \sum_{n=1}^8 \left( \sum_{m=1}^n \left( \frac {a \left( \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + k \times (\frac {\pi}{3} ) \right) + \cos \left( {3^{( m+(n^2-n)/2)}} \theta / b + k \times (\frac { \pi}{3} ) \right) \right) }{2 \times 3^{(m+(n^2-n)/2)}} \right) \right) , \ k=-1,1$

7. ### icarus2Registered Senior Member

Messages:
132
${U_{es - shell}} = \frac{1}{2}(\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q^2}}}{{{R_{es}}}})$

Last edited: Apr 29, 2018

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6,549
9. ### James RJust this guy, you know?Staff Member

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39,259
fgfggh indeed.

10. ### Confused2Registered Senior Member

Messages:
609
Last edited: Jun 11, 2020
11. ### Confused2Registered Senior Member

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609
$s^{2}={v'_{x}}^{2}t'+ {v'_{y}}^{2}t' -c^{2}t'^{2}$

Last edited: Jun 11, 2020

Messages:
22
If anyone can help it would be appreciated. I have been trying to display an equation with Tex with no luck. I have refreshed the page and still get nothing. Do I have to create the thread to actually see if the equation is correctly expressed?

13. ### Neddy BateValued Senior Member

Messages:
2,546
In my experience, the Tex equations do not display correctly in the "preview" pane, but they do display correctly after they have been posted to a thread. You can use this thread to check your Tex equations for errors, before posting or creating a whole new thread. Sorry, but this forum software still has some bugs in it.

Messages:
22
$\Large \frac{1}{1-\frac{v^2}{c^2}}$

Thanks for your help Neddy Bate. I see that you have to post to see the equation displayed. The preview does not display the equation.

Last edited: Jun 28, 2020

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609
16. ### foghornValued Senior Member

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1,434
$\sqrt{ab}$

$c_a^b\sqrt[n]{ab}$

Last edited: Jun 28, 2020

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1,434
18. ### foghornValued Senior Member

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1,434
$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

19. ### ralfcisRegistered Senior Member

Messages:
421
How do I make "$h_2^$" appear as plain old text?

20. ### ralfcisRegistered Senior Member

Messages:
421
tex] h_2^0[/tex]

I've read this thread, I've googled it, this is the closest I've gotten.Just remove the 1st sq bracket

Last edited: Jan 30, 2021
21. ### ralfcisRegistered Senior Member

Messages:
421
This is my equation sheet to cut and paste equations:
I need to ferret out embedded script here



$c^2 = v^2 +v_t^2$
$Y = \frac{c}{ \sqrt{(c^2-v^2)} }$

$v_t= c/Y$
$v = c/Y_t$

$w =(v+ u) / (1 + vu/c^2)$

$c^2= ((v+u)^2+ v_t^2u_t^2) / ( 1 + vu/c^2)^2$

$(ct')^2 = (ct)^2 - x^2$

$(ct)^2= (ct')^2+ x^2$

$v/(c-v) = 2cv_h / (c - v_h)^2$

$Y = 2Y_h^2 - 1 = (c^2 + v_h^2)/(c^2 - v_h^2) = c/v_t = (1/(1-v_h/v)) -1 = 2c^2v_hv/(c^2 - v_h^2) = 2Y_h^2 v_h /v$

$v_h=Yv/(Y+1)$

$(Yv)^2 = v^2/(1-v^2)$

$v= c sqrt(Y^2-1)/Y$

$t'=x(v_h +c)/c$

$v' = Yv$
$t = Yt'$
$x'=Yx$

$Yv =x/t'$

$Yu/Yw=DSR_v$

$t'(1-Y)$
$t'=x/Yv$
$vx/c^2$

$tpad=t′−x/Yv$

$t'=sqrt(t^2- x^2)$

$t' =xDSR_o/Yv$

$wt= c / Yw = c / (YvYu(1 + vu/c^2))$

$Y_ww = (v+u)YuYv$

$cDSR = sqrt((c-v)/(c+v))$

$DSR_v= Yu/Yw$

$Y(c-v) = c/DSR$

$DSR_w= DSR_v* DSR_u$
$DSR_w^2= (c-w)/(c+w)$

$w = c(1-DSR_v^2 DSR_u^2 )/(1+DSR_v^2 DSR_u^2 )$

$DSR = Y(1-v/c)$

$t' = X(c+v_h)/c$

$t'(DSR - 1)$

$t' = X(-2v_h)/(c+v_h)$

$A = (v+u)$

$B= (1+vu/c2)$

$sqrt(c^2B^2-A^2) = c/YvYu =v_tu_t$

v Y vh" role="presentation" style="display: inline; line-height: normal; font-size: 14.6667px; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">vhvhvt" role="presentation" style="display: inline; line-height: normal; font-size: 14.6667px; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">vtvtv_h_t" role="presentation" style="display: inline; line-height: normal; font-size: 14.6667px; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">v_h_tv_h_t DSR t'

1 1 1 0 0 0

3280/3281 3281/81 40/41 81/3281 1/81 81

40/41 41/9 4/5 9/41 4/36 9

255/257 65/16 15/17 16/65 1/8 8

24/25 25/7 3/4 7/25 1/7 7

35/37 37/12 5/7 12/37 6/36 6

12/13 13/5 2/3 5/13 1/5 5

77/85 85/36 77/121 36/85 8/36 9/2

15/17 17/8 3/5 8/17 9/36 4

4/5 5/3 1/2 3/5 12/36 3

3/5 5/4 1/3 4/5 18/36 2

1/2 1.155 .268 .866 .577 1.73

8/17 17/15 1/4 15/17 3/5 5/3

5/13 13/12 1/5 12/13 24/36 3/2

1/3 1.06 .17 .943 .707 1.41

12/37 37/35 1/6 35/37 5/7 7/5

7/25 1/7 24/25 27/36 4/3

16/65 1/8 255/257 15/17 17/15

9/41 1/9 40/41 4/5

0 0 1 1

21523360/21523361 21523360/21529922 6561/21523361 1/6561

Last edited: Jan 30, 2021
22. ### ralfcisRegistered Senior Member

Messages:
421


$c^2 = v^2 +v_t^2$
$Y = \frac{c}{ \sqrt{(c^2-v^2)} }$

$v_t= c/Y$
$v = c/Y_t$

$w =(v+ u) / (1 + vu/c^2)$

$c^2= ((v+u)^2+ v_t^2u_t^2) / ( 1 + vu/c^2)^2$

$(ct')^2 = (ct)^2 - x^2$

$(ct)^2= (ct')^2+ x^2$

$$$v = v_h(v_t/c +1)$$$
$v_t = v_h_t(v/c +1)$
$$$v = 2c^2v_h / (c^2 + v_h^2)$$$
$v_t = 2c^2v_h_t / (c^2 + v_h_t^2)$

$v_h = c(c-v_h_t ) / (c+v_h_t )$
$v_h_t = c(c-v_h ) / (c+v_h )$

$v/(c-v) = 2cv_h / (c - v_h)^2$

$Y = 2Y_h^2 - 1 = (c^2 + v_h^2)/(c^2 - v_h^2) = c/v_t = (1/(1-v_h/v)) -1 = 2c^2v_hv/(c^2 - v_h^2) = 2Y_h^2 v_h /v$

$v_h=Yv/(Y+1)$

$(Yv)^2 = v^2/(1-v^2)$

$v= c sqrt(Y^2-1)/Y$

$v_t = c/Y = DSR(c+v) = v_h_t(1 +v/c)$

$t'=x(v_h +c)/c$

$v' = Yv$
$t = Yt'$
$x'=Yx$

$Yv =x/t'$

$Yu/Yw=DSR_v$

$t'(1-Y)$
$t'=x/Yv$
$vx/c^2$

$tpad=t′−x/Yv$

$t'=sqrt(t^2- x^2)$

$t' =xDSR_o/Yv$

$t_p_s= xv_p_s = xYv/(1+Y)$

$wt= c / Yw = c / (YvYu(1 + vu/c^2))$

$Y_ww = (v+u)YuYv$

$v_h_t=cDSR$

$cDSR = sqrt((c-v)/(c+v))$

$DSR_v= Yu/Yw$

$Y(c-v) = c/DSR$

$DSR_w= DSR_v* DSR_u$
$DSR_w^2= (c-w)/(c+w)$

$w = c(1-DSR_v^2 DSR_u^2 )/(1+DSR_v^2 DSR_u^2 )$

$DSR = Y(1-v/c)$

$t' = X(c+v_h)/c$

$t'(DSR - 1)$

$t' = X(-2v_h)/(c+v_h)$

$A = (v+u)$

$B= (1+vu/c2)$

$sqrt(c^2B^2-A^2) = c/YvYu =v_tu_t$

Last edited: Jan 30, 2021
23. ### ralfcisRegistered Senior Member

Messages:
421
$v=vh(vt/c+1)" role="presentation" style="display: inline; line-height: normal; font-size: 14.6667px; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">tex]v=vh(vt/c+1)[tex]v=vh(vt/c+1)$