Is there a short form for this equation?

[TheIsingGuy]

I am trying to write down a compact form of[STRIKE] S = A+L/(B+C)+L/(B-C), does it make sense in general to write down S = A+L/(B[itex]\pm[/itex]C)?[/STRIKE]

Update:

I have simplified it too much in the minimal working example, the denominators is actually more complicated, let me elaborate:

[itex]S=A+\frac{L}{D+(B+C)^{2}}+\frac{L}{D+(B-C)^{2}}[/itex]

Any ideas?

 

[Orodruin]

No, it does not make sense to write it that way.

What was wrong with S = A + 2 L B/(B^2 - C^2)?

 

[adjacent]

I am trying to write down a compact form of S = A+L/(B+C)+L/(B-C), does it make sense in general to write down S = A+L/(B[itex]\pm[/itex]C)?

If you write like that, it means the answer is:
##S=A+\frac{L}{B+C} or S=A+\frac{L}{B-C}##
Which is not correct.

 

[TheIsingGuy]

I have simplified it too much in the minimal working example, the denominators is actually more complicated, let me elaborate:

[itex]S=A+\frac{L}{D+(B+C)^{2}}+\frac{L}{D+(B-C)^{2}}[/itex]

Any ideas?

 

[adjacent]

I have simplified it too much in the minimal working example, the denominators is actually more complicated, let me elaborate:

[itex]S=A+\frac{L}{D+(B+C)^{2}}+\frac{L}{D+(B-C)^{2}}[/itex]

Any ideas?

From where did this ##D## come from?

 

[TheIsingGuy]

D is supposed to be there, I had ignored the square in the first post, in which case D is not neccessary. Now it is. Any ideas how to simplify it now?

 

[adjacent]

D is supposed to be there, I had ignored the square in the first post, in which case D is not neccessary. Now it is. Any ideas how to simplify it now?

Show us your attempt

 

[TheIsingGuy]

Show us your attempt

This is not homework, I believe it is the simplest form, some people might even recognise the equation (it's a famous one), I just wanted it to fit in 1 line on a double column publication