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Math Help - A nonlinear problem

  1. #1
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    A nonlinear problem

    I have been trying to find a solution to the following problem.

    x(4/3) - x(-2/3) + c = 0 where -1 < x > 1

    I think if I could solve,

    s(-2) - s + c = 0,

    I could use substitution to solve the first equation, but I have not had any luck with the second either.

    Any help would be appreciated.
    Last edited by markel; September 2nd 2012 at 02:03 PM. Reason: change limits, embellish, clean up notations, change sign of c
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  2. #2
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    Re: A nonlinear problem

    Hello, markel!

    Is there a typo?
    You have a positive exponent and a negative exponent.
    Even if the signs were alike, that "c" makes for a messy solution.


    x^{\frac{4}{3}} -x^{-\frac{2}{3}} - c \:=\:0\;\;\;\text{ for }0 \,\le\,x\,<\,1

    I will wait for your clarification.

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  3. #3
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    Re: A nonlinear problem

    Dear Soroban,

    Thanks for responding. I'm afraid the negative exponent is correct. The limit however should be -1 < x < 1. I have used an Excel spreadsheet to calculate some examples by trial and error and the formula seems to be working. The task of simplifying the formula, however, has eluded me.
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  4. #4
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    Re: A nonlinear problem

    Quote Originally Posted by markel View Post
    I have been trying to find a solution to the following problem.

    x(4/3) - x(-2/3) + c = 0 where -1 < x > 1

    I think if I could solve,

    s(-2) - s + c = 0,

    I could use substitution to solve the first equation, but I have not had any luck with the second either.

    Any help would be appreciated.
    \displaystyle \begin{align*} x^{\frac{4}{3}} - x^{-\frac{2}{3}} + c &= 0 \\ x^{\frac{2}{3}}\left( x^{\frac{4}{3}} - x^{-\frac{2}{3}} + c \right) &= 0x^{\frac{2}{3}} \\ x^{\frac{6}{3}} - 1 + c\,x^{\frac{2}{3}} &= 0 \\ X^3 + c\,X - 1 &= 0 \textrm{ if } X = x^{\frac{2}{3}}   \end{align*}

    While there is a solution to this cubic, it is NOT pretty. I suggest you read the attachment.
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