1
$\begingroup$

If $f(x) = \lfloor x^2 \rfloor - \lfloor x \rfloor^2$, and $x \in [0,n]$ where $n \in \mathbb{Z} $ then the number of elements in the range of $f(x)$ is:-

$$1.) \ 2n+1$$ $$2.)\ 4n-3$$ $$3.)\ 3n-3$$ $$4.)\ 2n-1$$

I tried it like this.

Let $x=I+f$, where $I$ is $\lfloor x \rfloor$ and $f$ is fractional part of $x$. So,

$$f(x) = \lfloor (I+f)^2 \rfloor - I^2$$ $$f(x) = \require{cancel}\cancel{I^2} +\lfloor 2If+f^2 \rfloor - \cancel{I^2}$$ $$f(x) =\lfloor 2If+f^2 \rfloor $$

But I am stuck on this step, please help me out. Thanks in advance.

$\endgroup$
2
  • $\begingroup$ $0\leq f(x) < 2(n-1) +1 +1 =2n $ $\endgroup$
    – user6
    Apr 15, 2019 at 15:36
  • $\begingroup$ The answer is 2n-1 $\endgroup$
    – user585765
    Apr 15, 2019 at 15:37

1 Answer 1

3
$\begingroup$

It is less helpful to try to solve for $f(x)$ explicitly though, as you can already figure out an upper and lower bound for $f(x)$; $l \le x < l+1$ without having to do that.

Let us write $l = \lfloor x \rfloor$. Then $f(x) = \lfloor x^2 \rfloor - l^2$ is always an integer, and can be as large as $(l+1)^2-1-l^2 = 2l$ [for $x$ close to $l+1$] but no larger; and of course can be as small as 0, if $x-l^2$ is close to 0 [make sure you see why]. So the number of elements in the range of $\lfloor x^2 \rfloor - l^2$ $=$ $\lfloor x^2 \rfloor - \lfloor x \rfloor ^2$ $=f(x)$; where $x$ varies $l \le x < l+1$ can be precisely as large as $2l+1$. [make sure you see why]. So the number of elements in the range of $\lfloor x^2 \rfloor - l^2$ $=$ $\lfloor x^2 \rfloor - \lfloor x \rfloor ^2$ $=f(x)$; where $x$ varies $l \le x \le l+1$ can be precisely as large as $2l+1$. [make sure you see why]

So can you use this to conclude that the number of elements in the range of $\lfloor x^2 \rfloor - \lfloor x \rfloor ^2$; $x \in [0,n]$ can be as large as $2(n-1)+1$ $=2n-1$ but no larger.

$\endgroup$
1
  • $\begingroup$ @RossMillikan Yes. Good catch! $\endgroup$
    – Mike
    Apr 15, 2019 at 15:54

You must log in to answer this question.