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Converting dependable variable to independant and vice versa

  1. Absolutely Yes.

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  2. nope, only possible numerically.

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  3. May be

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

    haruspex

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    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.
     
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