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How to solve V=B+CT(V^N) for V?

  1. Sep 1, 2012 #1
    Does anyone know how to solve V=B+CT(V^N) for V?

    The answer is V=[B^(1-N)+C(N-1)T]^(1/(1-N))
    I don't understand how to get there.
    Please help me out. Thank you
  2. jcsd
  3. Sep 1, 2012 #2


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    Hey thiago.omena and welcome to the forums.

    Are these matrices or just plain old numbers?
  4. Sep 1, 2012 #3
    In the question? They are just variables. I am trying to isolate the V.
  5. Sep 2, 2012 #4


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    What makes you think that's the answer? It doesn't look right to me. In fact, I do not believe there is a solution in closed form.

    Btw, it's better to use the formatting this site provides. If you click Go Advanced you'll see a toolbar with e.g. X2 for superscript. Select the text to be superscripted and click the icon:
  6. Sep 2, 2012 #5


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    [itex]V= B+ CTV^N[/itex] is the same as the polynomial equation [itex]CTV^N- V+ B= 0[/itex]. There is NO general formula for solving a polynomial equation of degree greater than 5. In fact, it was proved about a hundred years ago that there cannot be a formula (involving only roots of numbers) for such equations.

    And it is easy to show that your [itex]V=[B^{1-N}+C(N-1)T]^(1/(1-N))[/itex] is NOT a general solution to the equation. In the case that C= T= 1, B= -2 and N= 2, the equation becomes [itex]V= -2+ V^2[/itex] which is the same as [itex]V^2- V- 2= (V- 2)(V+ 1)= 0[/itex] which has roots 2 and -1. But your formula gives [itex]V= [(-2)^{-1}+ (1)(-1)(1)]^{1/-1}= [-1/2+ -1]^{-1}= [-3/2]^{-1}= -2/3[/itex] which is NOT a solution.
  7. Sep 3, 2012 #6


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    Is this a thermodynamic equation? If it is, you're likely looking for an asymptotic solution for V >> 1 and N >> 1. As has been said, solving the equation exactly is not possible in general, but an approximate solution in the large V and/or N limits may be possible, and is perhaps the answer you claim to have.
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