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Thread: functions n

  1. #1
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    functions n

    let 'f' be a function defined on non-negative integers satisfying the following conditions: f(2n+1) = f( n ) & f(2n) = 1 - f( n )
    find f(2007.

    need a little bit of help here.. how do i start?
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Work backwards.

    $\displaystyle f(2007)\ =\ f(2(1003)+1)$

    $\displaystyle =\ f(1003)$

    $\displaystyle =\ f(2(501)+1)$

    $\displaystyle =\ f(501)$

    $\displaystyle =\ f(2(250)+1)$

    $\displaystyle =\ f(250)$

    $\displaystyle =\ f(2(125))$

    $\displaystyle =\ 1-f(125)$

    $\displaystyle \vdots$

    etc.
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  3. #3
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    i got all the way till f(1)... then?
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    Thatís it. The answer is $\displaystyle f(1)$ (which should be given to you).

    In order to define the series, the first two values, namely $\displaystyle f(0)$ and $\displaystyle f(1),$ must be given.
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