Daily Graph 18-8-09

f(x)=[x²]-[x]²

0≤x<1
=>[x]=0
and 0≤[x²]<1
=> [x²]-[x]²=0

1≤x<2
=>[x]=1
and 1≤[x²]<4
=> 0≤[x²]-[x]²<3

Similarly ohter intervals will eb evaluated

corrected

also i think we will be more precise to take intervals as:

1≤x<√2

[x²]-[x]² =0

√2<= x <√3

[x²]-[x]² =1

in [√3,2)

[x²]-[x]² =2

in [2,√5)

[x²]-[x]² =2