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Math Help - How to solve for x: (x^a)((1-x)^b)-2c = 0

  1. #1
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    How to solve for x: (x^a)((1-x)^b)-2c = 0

    Hello helpful math whizzes,

    I'm working through a game theory problem and am trying to solve this equation for x:

    (x^a)((1-x)^b)-2c = 0

    I've tried using natural logs to manage the exponents, but I keep running into dead ends. Is it possible to reduce this equation down such that x is all by its lonesome on one side?

    Also, this is my first post here, so I apologize if I've placed this in the wrong subforum.
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  2. #2
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    Re: How to solve for x: (x^a)((1-x)^b)-2c = 0

    Possibly. Do you have any additional conditions on x? For instance, is x a probability? Is a the number of successes and b the number of failures (or replace the words successes and failures with whatever game terms make sense)? Knowing a little more of the specifics might help.

    Anyway, we can start with the Binomial Theorem.

    \begin{align*}x^a(1-x)^b & = 2c \\ x^a\sum_{i=0}^b \left( (-1)^i \binom{b}{i} x^i \right) & = 2c\end{align*}

    This gives you a polynomial of degree a+b. There are many techniques for finding real solutions of polynomials. Knowing if x is a probability would help generate bounds when looking for solutions. Additionally, knowing more about the problem may help generate another formula that could lead to the same solution (potentially).

    Please let me know if I used any notation you don't understand.
    Thanks from topsquark
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  3. #3
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    Re: How to solve for x: (x^a)((1-x)^b)-2c = 0

    Oh, and if you don't mind rational exponents, x^{\tfrac{a+b}{b}} - x^{\tfrac{a}{b}} + (2c)^{\tfrac{1}{b}} = 0
    Last edited by SlipEternal; October 3rd 2013 at 03:52 PM.
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  4. #4
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    Re: How to solve for x: (x^a)((1-x)^b)-2c = 0

    Last bit of advice. If a+b>4 then there is no general formula. The best you can do is attempt to estimate the roots of the polynomial. Here is a link that offers a lot of information about estimating eigenvalues of a matrix: Gershgorin’s Theorem for Estimating Eigenvalues. And this gives an idea of how to set up the matrix to evaluate: Wolfram.com.
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