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

  1. #1
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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 $\displaystyle f(x)=(x^2+3)/(2x)$ for $\displaystyle x \not= 0$. Define a sequence of real numbers $\displaystyle x_n$ by

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

    A) Show that if $\displaystyle x> \sqrt 3$, then $\displaystyle f(x)> \sqrt 3$.
    B) Show that if $\displaystyle x> \sqrt 3$, then $\displaystyle x>f(x)$. (Hint: show $\displaystyle x-f(x)>0$.)
    C) Conclude that the sequence $\displaystyle x_n$ 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 $\displaystyle f(x)=(x^2+3)/(2x)$ for $\displaystyle x \not= 0$. Define a sequence of real numbers $\displaystyle x_n$ by

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

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

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

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

    $\displaystyle =\sqrt{3}$

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

    Equality holds only when

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

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

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

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

    $\displaystyle x=\sqrt{3}$

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

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