허근 ω의 실수부분

$x ^{3} + 1 = ( x + 1 ) ( x ^{2} - x + 1 ) = 0$에서 한 허근이 $\omega$이므로 $\omega ^{2} - \omega + 1 = 0$ ∴ $\displaystyle \omega = \frac{1 \pm \sqrt{3} i}{2}$ 이 때, $f ( n )$은 $\omega ^{n}$의 실수 부분이고, $\omega ^{3} = - 1$, $\omega ^{2} = \omega - 1$ 이므로 $f ( n )$을 차례로 구하면 $\displaystyle \omega = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$이므로 $\displaystyle f ( 1 ) = \frac{1}{2}$ $\omega ^{2} = \omega - 1$이므로 $\displaystyle f ( 2 ) = - \frac{1}{2}$ $\omega ^{3} = - 1$이므로 $f ( 3 ) = - 1$ $\omega ^{4} = \omega ^{3} \omega = - \omega$이므로 $\displaystyle f ( 4 ) = - f ( 1 ) = - \frac{1}{2}$ $\omega ^{5} = \omega ^{3} \omega ^{2} = - \omega ^{2}$이므로 $\displaystyle f ( 5 ) = - f ( 2 ) = \frac{1}{2}$ $\omega ^{6} = ( \omega ^{3} ) ^{2} = 1$이므로 $f ( 6 ) = - f ( 3 ) = 1$ $\omega ^{7} = ( \omega ^{3} ) ^{2} \omega = \omega$이므로 $f ( 7 ) = f ( 1 )$ ⋮ 따라서, $f ( n )$은 주기가 6인 함수이고, $f ( 1 ) + f ( 2 ) + \cdots + f ( 6 ) = 0$ ∴ $\displaystyle \sum\limits _{{k} = 1} ^{999} \left\{ f ( k ) + \frac{1}{3} \right\} = \sum\limits _{{k} = 1} ^{999} f ( k ) + \sum\limits _{{k} = 1} ^{999} \frac{1}{3}$ $\displaystyle = 166 \sum\limits _{{k} = 1} ^{6} f ( k ) + f ( 997 ) + f ( 998 ) + f ( 999 )$$\displaystyle + 999 \cdot \frac{1}{3}$ $= 16 \cdot 0 + f ( 1 ) + f ( 2 ) + f ( 3 ) + 333$ $\displaystyle = 0 + \left( \frac{1}{2} - \frac{1}{2} - 1 \right) + 333$ $= 332$