1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Converting dependable variable to independant and vice versa

  1. Absolutely Yes.

    0 vote(s)
  2. nope, only possible numerically.

    0 vote(s)
  3. May be

    0 vote(s)
  1. Jun 6, 2012 #1
    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));
    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.

  2. jcsd
  3. Jun 6, 2012 #2
    can you mutliply them before asking computer to solve for x?
  4. Jun 6, 2012 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
  5. Jun 8, 2012 #4
    Yup. That works, but gives a very huge solution which can be used to solve my system.

    Thank you very much.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook