Proving $\sum\limits_{k=0}^{n-1} \Bigl[x + \frac{k}{n}\Bigr] = [nx]$ [duplicate]

I'm doing some exercises on Apostol's calculus, on the floor function. Now, he doesn't give an explicit definition of $[x]$, so I'm going with this one:

DEFINITION Given $x\in \Bbb R$, the integer part of $x$ is the unique $z\in \Bbb Z$ such that $$z\leq x < z+1$$ and we denote it by $[x]$.

Now he asks to prove some basic things about it, such as: if $n\in \Bbb Z$, then $[x+n]=[x]+n$

So I proved it like this: Let $z=[x+n]$ and $z'=[x]$. Then we have that

$$z\leq x+n<z+1$$

$$z'\leq x<z'+1$$

Then $$z'+n\leq x+n<z'+n+1$$

But since $z'$ is an integer, so is $z'+n$. Since $z$ is unique, it must be that $z'+n=z$.

However, this doesn't seem to get me anywhere to prove that $$\left[ {2x} \right] = \left[ x \right] + \left[ {x + \frac{1}{2}} \right]$$

in and in general that

$$\left[ {nx} \right] = \sum\limits_{k = 0}^{n - 1} {\left[ {x + \frac{k}{n}} \right]} $$

Obviously one could do an informal proof thinking about "the carries", but that's not the idea, let alone how tedious it would be. Maybe there is some easier or clearer characterization of $[x]$ in terms of $x$ to work this out.

Another property is $$[-x]=\begin{cases}-[x]\text{ ; if }x\in \Bbb Z \cr-[x]-1 \text{ ; otherwise}\end{cases}$$

I argue: if $x\in\Bbb Z$, it is clear $[x]=x$. Then $-[x]=-x$, and $-[x]\in \Bbb Z$ so $[-[x]]=-[x]=[-x]$. For the other, I guess one could say:

$$n \leqslant x < n + 1 \Rightarrow - n - 1 < x \leqslant -n$$

and since $x$ is not an integer, this should be the same as $$ - n - 1 \leqslant -x < -n$$

$$ - n - 1 \leqslant -x < (-n-1)+1$$

So $[-x]=-[x]-1$

Solution 1:

This is one of my favourite exercises, because of the following neat solution:

Fix $n$. Let

$$ f(x) := \sum\limits_{k=0}^{n-1} \Biggl[x + \frac{k}{n}\Biggr] - [nx] \,.$$

Then $f(x) =0 \forall x \in [0,\frac{1}{n})$ since all terms are zero, and it is easy to prove that $f(x+\frac{1}{n})=f(x)$.

It follows imediately that $f$ is identically 0.

Solution 2:

Let $n=\lfloor x\rfloor$, and let $\alpha=x-n$; clearly either $0\le\alpha<\frac12$, or $\frac12\le\alpha<1$. Then

$$\lfloor 2x\rfloor=\lfloor 2n+2\alpha\rfloor=2n+\lfloor 2\alpha\rfloor=\begin{cases} 2n,&\text{if }0\le\alpha<\frac12\\ 2n+1,&\text{if }\frac12\le\alpha<1\;, \end{cases}$$

and

$$\left\lfloor x+\frac12\right\rfloor=\left\lfloor n+\alpha+\frac12\right\rfloor=n+\left\lfloor\alpha+\frac12\right\rfloor=\begin{cases} n,&\text{if }0\le\alpha<\frac12\\ n+1&\text{if }\frac12\le\alpha<1\;; \end{cases}$$

since $\lfloor x\rfloor=n$, the first result is immediate.

The general case is handled similarly, except that there are $n$ cases; for $k=0,\dots,n-1$, case $k$ is $$\frac kn\le\alpha<\frac{k+1}n\;.$$