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Math Help - How to find roots of the equation system?

  1. #1
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    How to find roots of the equation system?

    x_1+2/{x_1}=2x_2
    x_2+2/{x_2}=2x_3
    .
    .
    .
    x_n+2/{x_n}=2x_1

    x is real number. How can I solve this equation system?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by gabor7896 View Post
    x_1+2/{x_1}=2x_2
    x_2+2/{x_2}=2x_3
    .
    .
    .
    x_n+2/{x_n}=2x_1

    x is real number. How can I solve this equation system?
    Hi

    Let (u_n)_n the sequence defined by the value of u_1 and u_{n+1} = \frac{u_n}{2} + \frac{1}{u_n} = f\left(u_n\right) where f(x) = \frac{x}{2} + \frac{1}{x}

    I am just considering the restriction of f over ]0,+\infty[ since it is similar over ]-\infty,0[. You can show that f is decreasing up to \sqrt{2} and increasing after \sqrt{2} and f\left(\sqrt{2}\right) = \sqrt{2}

    Moreover f\left(]0,\sqrt{2}]\right) \subset [\sqrt{2},+\infty[ and f\left([\sqrt{2},+\infty[\right) \subset [\sqrt{2},+\infty[

    This means that at least from n=2 u_n \in [\sqrt{2},+\infty[

    You can show by induction that \forall n  ~ u_n > 0

    Then u_{n+1}-u_n = \frac{2-u_n^2}{2u_n} < 0

    (u_n)_n is decreasing (at least from n=2) and higher than \sqrt{2} therefore converges towards the solution of f(x) = x which is \sqrt{2}

    The only way to get a stationary sequence is x_1 = x_2 = ... = x_n = \sqrt{2}

    The same for -\sqrt{2}
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