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  1. #1
    rcs
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    help me to simplify this ...

    help me to simplify this ...-mmmmmmm.jpg
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  2. #2
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    let $\displaystyle x_1 = \sqrt{2}$
    and $\displaystyle x_{n+1} = \sqrt{2x_n}$

    Then $\displaystyle x_{n+1}^2=2x_n$

    If the limiting value of $\displaystyle x_n$ is $\displaystyle x$, then the limiting value of $\displaystyle x_{n+1}$ is also $\displaystyle x$.

    So $\displaystyle x^2=2x$

    $\displaystyle x^2-2x=0$
    $\displaystyle x(x-2)=0$

    So we see that $\displaystyle x=2$

    Note that the given sequence is increasing and bounded above (you can show easily by induction on n that each member of the sequence is bounded by 2). Thus the sequence is guaranteed to converge.
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  3. #3
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    Hello, rcs!

    $\displaystyle \text{Evaluate: }\;2\sqrt{2\sqrt{2\sqrt{2\hdots}}} $

    Let: .$\displaystyle x \;=\;2\sqrt{2\sqrt{2\sqrt{2\hdots}}} $

    $\displaystyle \text{Square: }\;x^2 \;=\;2\cdot\underbrace{2\sqrt{2\sqrt{2\sqrt{2\hdot s }}}}_{\text{This is }x} $

    We have: .$\displaystyle x^2 \:=\:2x \quad\Rightarrow\quad x^2-2x \:=\:0 \quad\Rightarrow\quad x(x-2) \:=\:0 $


    Therefore: .$\displaystyle x \:=\:2$

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    Soroban - I found a little mistake in each of our arguments. The answer is actually 4.

    In my argument, I misread the problem and left off the constant 2 in front. Since constants can be pulled outside limits this is no big deal, and you can just multiply my answer of 2 by 2 to get 4.

    In your argument, the right hand side of where you wrote "Square" should actually start with a 4 (since 2^2 = 4).
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  5. #5
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    Here is my corrected solution:

    Let $\displaystyle x_1 = 2\sqrt{2}$
    and $\displaystyle x_{n+1} = 2\sqrt{x_n}$

    Then $\displaystyle x_{n+1}^2=4x_n$

    If the limiting value of $\displaystyle x_n$ is $\displaystyle x$, then the limiting value of $\displaystyle x_{n+1}$ is also $\displaystyle x$.

    So $\displaystyle x^2=4x$

    $\displaystyle x^2-4x=0$
    $\displaystyle x(x-4)=0$

    So we see that $\displaystyle x=4$

    Note that the given sequence is increasing and bounded above (you can show easily by induction on n that each member of the sequence is bounded by 4). Thus the sequence is guaranteed to converge.
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  6. #6
    rcs
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    thank you so so much ... u have been so good guys. i was actually trying to solve it add i was stuck i couldn't finish it. thanks a lot
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