Proving Hermite's Identity in a different approach

Question

To prove

$$S=\left [x \right]+\left [x+\frac{1}{n} \right]+\left [x +\frac{2}{n}\right]+\cdots+\left [x +\frac{n-1}{n}\right]=\left [nx \right]$$

using and starting with $$x-1 \lt \left [x \right]\le x \tag{1}$$ we have

$$x+\frac{1}{n}-1 \lt \left [x +\frac{1}{n}\right] \le x+\frac{1}{n} \tag{2}$$

$$x+\frac{2}{n}-1 \lt \left [x +\frac{2}{n}\right] \le x+\frac{2}{n} \tag{3}$$

$$x+\frac{3}{n}-1 \lt \left [x +\frac{3}{n}\right] \le x+\frac{3}{n} \tag{4}$$

and so on till

$$x+\frac{n-1}{n}-1 \lt \left [x +\frac{n-1}{n}\right] \le x+\frac{n-1}{n} \tag{n}$$

Adding all we get

$$nx+\frac{n(n-3)}{2} \lt S \le nx+\frac{n(n-1)}{2}$$

Now how can we prove that between $nx+\frac{n(n-3)}{2}$ and $nx+\frac{n(n-1)}{2}$ there is only on integer which is $\left [nx \right]$?

Answer 1

Here is a proof I came up with about a year ago:

For all $x$ which are real numbers, prove that $\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$

The basic idea is that you have to see how far $x$ is from $\lfloor x \rfloor$ and, if $d = x-\lfloor x \rfloor$, what $\lfloor nd \rfloor$ is. This allows you do get the sum.

Answer 2

This method doesn't work.

Adding all of them we get $$nx+\frac{\frac{n(n-1)}2}{n}-n<S\le nx+\frac{\frac{n(n-1)}2}{n}$$ or $$nx-\frac{n+1}2<S\le nx+\frac{n-1}2\implies -\frac n2<S-nx+\frac12\le\frac n2$$ and this interval gets wider and wider as $n$ gets larger.

Answer 3

Let us assume a new function

$$f(x) = \sum_{k=0}^{n-1} \left( \left[x+ \frac{k}{n} \right]\right) - \left[nx \right]$$

Now let's replace $x \rightarrow x+\frac{1}{n}$ in $f(x)$, we get

$$f\left(x+\frac{1}{n}\right) = \sum_{k=0}^{n-1} \left( \left[(x+\frac{1}{n})+ \frac{k}{n} \right] \right)- \left[n(x +\frac{1}{n})\right]$$ $$ = \sum_{k=0}^{n-1} \left( \left[(x)+ \frac{k+1}{n} \right] \right)- \left[n(x)\right]-\left[1\right]$$ $$ = \sum_{k=1}^{n} \left( \left[(x)+ \frac{k}{n} \right] \right)- \left[n(x)\right]-1$$ $$ = \sum_{k=0}^{n} \left( \left[(x)+ \frac{k}{n} \right] \right)-[x]- \left[n(x)\right]-1$$ $$ = \sum_{k=0}^{n-1} \left( \left[(x)+ \frac{k}{n} \right] \right)+[x+\frac{n}{n}]-[x]- \left[n(x)\right]-1$$ $$ = \sum_{k=0}^n \left( \left[x+ \frac{k}{n} \right]\right) - \left[nx \right]$$ $$= f(x)$$

$\therefore$ $f(x)$ is periodic function with period equal to $\frac{1}{n}$

& for all $x \in [0,\frac{1}{n})$ , $f(x) = 0$

Hence, we get $$\sum_{k=0}^{n-1} \left( \left[x+ \frac{k}{n} \right]\right) = \left[nx \right]$$

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$$S= \sum_{k=0}^{n-1} \left[x+ \frac{k}{n} \right]$$ $$=\sum_{k=0}^{n-1} \left[ \left[x\right] + \{x\} + \frac{k}{n} \right]$$ $$S= n \left[x\right] + \sum_{k=0}^{n-1} \left[\{x\}+ \frac{k}{n} \right]$$ Now, to find all values of $k$ for which $\left[\{x\} + \frac{k}{n}\right] = 1$

So let $$\left[\{x\} + \frac{k}{n}\right] = 1$$ $$\Rightarrow 1≤\{x\} + \frac{k}{n}<2$$

Since, $\{x\} + \frac{k}{n} <2 $ is always true for all $k \in [0,n-1)$

$\therefore$ $$ \{x\} + \frac{k}{n}≥1$$

$$k≥(1-\{x\})n$$ And $k\in I$, $\therefore $ we can write $$k≥\left[ (1-\{x\})n\right]$$ $\therefore$ total no of values of $k$ for which $\left[\{x\} + \frac{k}{n}\right] = 1$ are $$(n-1)-\left[ (1-\{x\})n\right]+1$$ $$= n-1 - [n] -[n\{x\}] +1$$ $$=[n\{x\}]$$ $\therefore$ $$\sum_{k=0}^{n-1} \left[\{x\}+ \frac{k}{n} \right]=[n\{x\}]$$ $\therefore$ $$S= n \left[x\right] + \sum_{k=0}^{n-1} \left[\{x\}+ \frac{k}{n} \right]=n[x] + [n\{x\}]$$ $$= [n[x]]+[n\{x\}]$$ $$= [n([x]+\{x\})]$$ $$S= \sum_{k=0}^{n-1} \left[x+ \frac{k}{n} \right]=[nx]$$

Q.E.D.