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Thread: Evaluate a recursion function.

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

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

    (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!

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

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

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

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

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


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

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