# Converting dependable variable to independant and vice versa

## Is this possible to solve this analytically?

Poll closed Jun 11, 2012.

0 vote(s)
0.0%

0 vote(s)
0.0%

0 vote(s)
0.0%
1. Jun 6, 2012

### venki_k07

Hello guys,
First of all, thanks for looking at this post and trying to help me.

I have function which is shown below,

F1= (ε1-√(1+(ε1 μ1-1)x^2))/(ε1+√(1+(ε1 μ1-1)x^2));
F2= (ε2-√(1+(ε2 μ2-1)x^2))/(ε2+√(1+(ε2 μ2-1)x^2));
F=F1*F2;
Here, F is a function of x --> F(x)
I need to find x(F) , x as a function of F. Is there any mathematical method for doing this.?
x(F1) and x(F2) is possible using mathematica and it is easy too. But i don't know how to find x(F).

I tried this in mathematica but it is not giving me any result.

Thanks,
Venkatesh

2. Jun 6, 2012

### algebrat

can you mutliply them before asking computer to solve for x?

3. Jun 6, 2012

### haruspex

Writing as F1 = (a-√b)/(a+√b), F2 = (c-√d)/(c+√d),
F(a+√b)(c+√d) = (a-√b)(c-√d)
(F-1)(ac+√(bd)) = - (F+1)(a√d + c√b)
(F-1)2(a2c2+bd+2ac√(bd)) = (F+1)2(a2d+c2b+2ac√(bd))
(F-1)2(a2c2+bd) - (F+1)2(a2d+c2b) = 8Fac√(bd)
Squaring again produces a quadratic in b and d. The only x terms are in the form of x2 within b and d. There's an uncancelled b2d2 term and nothing higher, therefore the polynomial is a quartic in x2.

4. Jun 8, 2012

### venki_k07

Yup. That works, but gives a very huge solution which can be used to solve my system.

Thank you very much.