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Math Help - Show that if x_n > 3^{0.5} then sequence x_{n+1} > 3^{0.5}

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    Show that if x_n > 3^{0.5} then sequence x_{n+1} > 3^{0.5}

    Hi, I have the following problem which I don't know how to solve... I know how to do part C, but I don't know parts A and B. I will have enough to guide myself if you help me with part A though.

    Thanks a lot!

    The whole problem is:

    "Let f(x)=(x^2+3)/(2x) for x \not= 0. Define a sequence of real numbers x_n by

    x_{n+1} =f(x_n) for n \geq 1, x_1=2.

    A) Show that if x> \sqrt 3, then f(x)> \sqrt 3.
    B) Show that if x> \sqrt 3, then x>f(x). (Hint: show x-f(x)>0.)
    C) Conclude that the sequence x_n converges."
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Show that if x_n > 3^{0.5} then sequence x_{n+1} > 3^{0.5}

    Quote Originally Posted by buenogilabert View Post
    Hi, I have the following problem which I don't know how to solve... I know how to do part C, but I don't know parts A and B. I will have enough to guide myself if you help me with part A though.

    Thanks a lot!

    The whole problem is:

    "Let f(x)=(x^2+3)/(2x) for x \not= 0. Define a sequence of real numbers x_n by

    x_{n+1} =f(x_n) for n \geq 1, x_1=2.

    A) Show that if x> \sqrt 3, then f(x)> \sqrt 3.
    B) Show that if x> \sqrt 3, then x>f(x). (Hint: show x-f(x)>0.)
    C) Conclude that the sequence x_n converges."
    A) f(x)=\frac{x^2+3}{2x}

    =\frac{x}{2}+\frac{3}{2x}

    \ge 2\sqrt{\frac{x}{2}\cdot \frac{3}{2x}} (AM-GM inequality, which holds for x>0)

    =\sqrt{3}

    So, if x>0, then f(x)\ge \sqrt{3}.

    Equality holds only when

    \frac{x^2+3}{2x}=\sqrt{3}

    x^2+3=2\sqrt{3}x

    x^2-2\sqrt{3}x+3=0

    (x-\sqrt{3})^2=0

    x=\sqrt{3}

    So, if x>0 and x\neq\sqrt{3}, then f(x)>\sqrt{3}.

    In particular, if x>\sqrt{3}, then f(x)>\sqrt{3}.
    Last edited by alexmahone; December 16th 2011 at 05:36 AM.
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