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Math Help - Evaluate a recursion function.

  1. #1
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    Mesnil, Mauritius
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    Question Evaluate a recursion function.

    2- \frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{...}}}} = ?

    (A) 0
    (B) 1
    (C) 2
    (D) 1/2
    (E) -1/2

    The answer booklet is separate from the book and I do not have it.
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, shailen.sobhee!

    2-\frac{1}{2-\dfrac{1}{2-\dfrac{1}{2- \hdots}}} \;=\;?

    . . (A)\;0 \qquad (B)\;1 \qquad (C)\;2 \qquad (D)\;\tfrac{1}{2} \qquad (E)\;-\tfrac{1}{2}

    \text{Let: }\:x \;=\;2 - \frac{1}{2-\dfrac{1}{2-\dfrac{1}{2 - \hdots}}}

    \text{Then: }\:x \;=\;2 - \frac{1}{x}

    Multiply by x\!:\;\;x^2 \:=\:2x - 1 \quad\Rightarrow\quad x^2-2x+1\:=\:0 \quad\Rightarrow\quad(x-1)^2 \:=\:1


    Therefore: . \boxed{x \:=\:1}

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  3. #3
    Junior Member
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    Mesnil, Mauritius
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    That's brilliant.

    There there exist a recursion in almost all expressions; take a very simple case:

    y= \frac{x}{y} = \frac{x}{\frac{x}{y}} = \frac{x}{\frac{x}{\frac{x}{y}}} = \frac{x}{\frac{x}{\frac{x}{\frac{x}{...}}}}
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