A Fractional Periodicity

Algebra Level 3

f ( x ) = x + 2 x + 3 x + + n x n ( n + 1 ) 2 x \large f(x) = \lfloor x\rfloor + \lfloor 2x\rfloor + \lfloor 3x\rfloor + \cdots + \lfloor nx\rfloor - \dfrac{n(n+1)}2 x

Find the fundamental period of the function above.

Notation : \lfloor \cdot \rfloor denotes the floor function .

None of these choices n n 1 1 1 n \frac1n

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1 solution

Young Wolf
Jun 7, 2016

f ( x ) f(x) = = [ x ] + [ 2 x ] + [ 3 x ] + . . . . + [ n x ] [x]+[2x]+[3x]+....+[nx] - ( x + 2 x + 3 x + . . . . . + n x ) (x+2x+3x+.....+nx)

= = - ( { x } + { 2 x } + { 3 x } + . . . . . { n x } ) ({\displaystyle \{x\}} +{\displaystyle \{2x\}} +{\displaystyle \{3x\}} +.....{\displaystyle \{nx\}} )

Period of f ( x ) f(x) is LCM ( 1 , 1 2 (1,\frac{1}{2} , 1 3 \frac{1}{3} ,.... 1 n ) \frac{1}{n})

= = 1 1

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