1. ## 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?

2. Work backwards.

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

$=\ f(1003)$

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

$=\ f(501)$

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

$=\ f(250)$

$=\ f(2(125))$

$=\ 1-f(125)$

$\vdots$

etc.

3. i got all the way till f(1)... then?

4. That’s it. The answer is $f(1)$ (which should be given to you).

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