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I Solving Qx+ x^P=R

  1. Jun 20, 2017 #1
    I am having difficulty solving Q*y - y^P = Q - 1 for y.
    If P = 2 or 0.5 or 0, I can solve it quadratically. However, I'd like to find a general solution for y. Any tips?
     
  2. jcsd
  3. Jun 20, 2017 #2

    mfb

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    There is no general solution. If P is an integer from -3 to 4, or a few special fractions, the equation has closed forms for its solutions, but for arbitrary P this is no longer true. Abel-Ruffini theorem.
     
  4. Jun 20, 2017 #3
    Where did you got this question from ?
     
  5. Jun 21, 2017 #4
    For the particular problem I am trying to solve, P ranges from 0 to 1. Is there a closed form for 0<P<1 ?
     
  6. Jun 21, 2017 #5

    mfb

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    y=1 is always a trivial solution, of course.
    Are there more real solutions in some cases? Are you interested in complex solutions?

    For 1/4, 1/3, 1/2, 2/3, 3/4 you can convert it to a quadratic, cubic or quartic equation and solve it. Otherwise there is no closed form for a solution in the general case. There can still be values of Q were you can find all solutions, but it won't work for all Q any more, unless y=1 is the only solution.
     
  7. Jun 22, 2017 #6

    WWGD

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    Isn't the result that a solution by radicals exists iff the Galois group of the equation is solvable? EDIT: I mean, of course for this equation; we can construct equations whose group is not solvable, e.g., ##A_5 ##
     
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