Displaying equations using Tex

[ArafuraOpal]

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

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

 

[ArafuraOpal]

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

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

 

[ArafuraOpal]

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

 

[icarus2]

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

 

[Beer w/Straw]

fgfggh

 

[James R]

fgfggh indeed.

 

[Confused2]

Fail

 

[Confused2]

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

 

[Amadeusboy]

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?

 

[Neddy Bate]

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?

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.

 

[Amadeusboy]

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

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.

 

[Confused2]

There's a simple Tex editor here:-
https://www.hostmath.com/

 

[foghorn]

$\sqrt{ab}$

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

 

[foghorn]

Hmmm... I'm not getting the hang of this.

 

[foghorn]

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

 

[ralfcis]

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

 

[ralfcis]

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

 

[ralfcis]

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

 

[ralfcis]

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

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

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

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

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

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

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

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

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

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

$$v_h=Yv/(Y+1)$$

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

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

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

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

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

$$Yv =x/t'$$

$$Yu/Yw=DSR_v$$

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

$$tpad=t′−x/Yv$$

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

$$t' =xDSR_o/Yv$$

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

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

$$Y_ww = (v+u)YuYv$$

$$v_h_t=cDSR$$

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

$$DSR_v= Yu/Yw$$

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

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

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

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

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

$$t'(DSR - 1)$$

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

$$A = (v+u)$$

$$B= (1+vu/c2)$$

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

 

[ralfcis]

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