Converting dependable variable to independant and vice versa

  • Context: Graduate 
  • Thread starter Thread starter venki_k07
  • Start date Start date
  • Tags Tags
    Variable

Is this possible to solve this analytically?

  • Absolutely Yes.

    Votes: 0 0.0%
  • nope, only possible numerically.

    Votes: 0 0.0%
  • May be

    Votes: 0 0.0%

  • Total voters
    0
  • Poll closed .
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
venki_k07
Messages
5
Reaction score
0
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
 
Physics news on Phys.org
can you mutliply them before asking computer to solve for x?
 
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.
 
Yup. That works, but gives a very huge solution which can be used to solve my system.

Thank you very much.