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Math Help - Root problem

  1. #1
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    Question Root problem

    Find x:
    x = sq.root[1 + sq.root( 1 + sq.root(1 + sq.root(1 +....
    "The dots mean that the series go till infinity.
    By attempt the answer should be (1 +-sq.root(5))/2 but that is not the answer.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Root problem

    /the answer is  \frac{1 \frac{+}{-} \sqrt(5)}{2}

    The infinite nested radicals,  \sqrt(1 + \sqrt(1 + \sqrt(1 + .... = golden ratio.

    Since golden ratio says  \frac{a+b}{a} = \frac{a}{b} = x then  1 + \frac{b}{a} = 1 + \frac{1}{x} = x so solving  x^2 - x - 1 = 0
    Last edited by jakncoke; December 21st 2012 at 11:14 PM.
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  3. #3
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    Re: Root problem

    Could you please write as to what is the answer given. The answer you have got appears to be correct.
    Root problem-series.png
    Thanks from puresoul
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  4. #4
    Senior Member jakncoke's Avatar
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    Re: Root problem

    -------
    Last edited by jakncoke; December 21st 2012 at 11:15 PM.
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  5. #5
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    Re: Root problem

    Hello, Aditya3003!

    \text{Evaluate: }\:x \:=\: \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}

    By attempt the answer should be: \frac{1\pm\sqrt{5}}{2}, but that is not the answer.

    Since x is obviously positive,
    . . you must reject the negative root: \tfrac{1-\sqrt{5}}{2}

    While solving, you squared the equation, right?
    . . This can introduce an extraneous root.
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