Question

help my exam ### A BIG NEED OF HELP

Answer 1

To find the value of y where

y = [x]+[x + 1/n] + [x + 2/n] + [x + 3/n] +…… + [ x + (n-1)/n] - [ n x ]

[ x ] is the greatest integer function and { x } is the fractional part of x.

So x = [ x ] + { x }

We know:

 [ n x ] = n [ x ] , { x } ϵ [0, 1/n)

      = n [ x ] + 1 , { x } ϵ [1/n, 2/n)

       ….

     = n [ x ] + n-1 , { x } ϵ [(n-1)/n, 1)

Alternately, it is same as:

   [ n x ] = n [ x ] + r ,   for { x } ϵ [r/n, (r+1)/n),   for 0 <= r <= n-1

 

Let { x } ϵ [ p/n, (p+1)/n )  for some p,  where  0 <= p <= n-1

Then, [ n x ] = n [ x] + p

 

As x = [ x ] + { x },  

    [ x + r /n ] = [  [x] + {x} + r/n ] = [x] +  [  (p+r)/n  ]

 

Given

$y = \Sigma_{r=0}^{n-1} \{ [ x + \frac{r}{n} ] \} - [ n x ]\\\\ =\Sigma_{r=0}^{n-1} \{ [x] + [ \frac{p+r}{n} ] \} - [ n x ] \\\\ =n [x] + \Sigma_{r=0}^{n-1} \{ [ \frac{p+r}{n} ] \} - n [ x ] - p \\\\= -p + \Sigma_{r=0}^{n-1} [ \frac{p+r}{n} ] \\\\ = -p + p\\\\ = 0$