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Math Help - sequence

  1. #1
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    sequence

    a>1, X1>sqr(a).

    X(n+1) = (a+X(n)) / (1+X(n))

    Prove that X1>X3>X5..............
    and X2<X4<X6..............




    Thank you
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  2. #2
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    Quote Originally Posted by felixmcgrady View Post
    a>1, X1>sqr(a).

    X(n+1) = (a+X(n)) / (1+X(n))

    Prove that X1>X3>X5..............
    and X2<X4<X6..............

    Thank you

    Perhaps a little clearer: a>1\,,\,\,x_1>\sqrt{a}\,,\,\,x_{n+1}=\frac{a+x_n}{  1+x_n}.

    a) Prove first that \forall\,n\,\in\mathbb{N}\,,\,\,x_{2n-1}>\sqrt{a}>x_{2n}\, (Hint: For odd n\,,\,\sqrt{a}<x_n=\frac{a+\frac{a+x_{n-2}}{1+x_{n-2}}}{1+\frac{a+x_{n-2}}{1+x_{n-2}}} =\frac{2a+(a+1)x_{n-2}}{2x_{n-2}+(a+1)}\Longleftrightarrow

    4ax_{n-2}^2+4a(a+1)x_{n-2}+a(a+1)^2<4a^2+4a(a+1)x_{n-2}+(a+1)^2x_{n-2}^2 ... and etc. (induction, of course). For even n do something simmilar)


    b) Next...induction, again and this time let us do the one with even indexes: we want to prove that

    \forall\,\mbox{ even }n\,\in\mathbb{N}\,,\,\,x_n<x_{n+2}\Longleftrighta  rrow\,x_n<\frac{a+\frac{a+x_n}{1+x_n}}{1+\frac{a+x  _n}{1+x_n}} =\frac{2a+(a+1)x_n}{2x_n+(a+1)}\Longleftrightarrow 2x_n^2+(a+1)x_n<2a+(a+1)x_n.. and etc., and something simmilar to show x_{2n-1}>x_{2n+1}

    In fact the above's enough to show that both sequences \{x_{2n}\}\,,\,\{x_{2n-1}\} converge. Now you may want to show they both converge and to the same limit...

    Tonio
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  3. #3
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    Quote Originally Posted by felixmcgrady View Post
    a>1, X1>sqr(a).

    X(n+1) = (a+X(n)) / (1+X(n))

    Prove that X1>X3>X5..............
    and X2<X4<X6..............




    Thank you
    Just an added note. You probably could show the claim by solving the difference equation exactly.

    If you let x_n = \sqrt{a} \cdot \frac{y_n - 1}{y_n + 1}

    then your difference equation becomes

     <br />
y_{n+1} = - \frac{\sqrt{a}+1}{\sqrt{a}-1}\cdot y_n<br />

    which is linear with the solution

    y_n = c \left( -\, \frac{\sqrt{a}+1}{\sqrt{a}-1}\right)^n.
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