Hermite's identity

In mathematics, Hermite's identity, named after Charles Hermite, gives the value of a summation involving the floor function. It states that for every real number x and for every positive integer n the following identity holds:


 * $$\sum_{k=0}^{n-1}\left\lfloor x+\frac{k}{n}\right\rfloor=\lfloor nx\rfloor .$$

Proof by algebraic manipulation
Split $$x$$ into its integer part and fractional part, $$x=\lfloor x\rfloor+\{x\}$$. There is exactly one $$k'\in\{1,\ldots,n\}$$ with
 * $$\lfloor x\rfloor=\left\lfloor x+\frac{k'-1}{n}\right\rfloor\le x<\left\lfloor x+\frac{k'}{n}\right\rfloor=\lfloor x\rfloor+1.$$

By subtracting the same integer $$\lfloor x\rfloor$$ from inside the floor operations on the left and right sides of this inequality, it may be rewritten as


 * $$0=\left\lfloor \{x\}+\frac{k'-1}{n}\right\rfloor\le \{x\}<\left\lfloor \{x\}+\frac{k'}{n}\right\rfloor=1.$$

Therefore,


 * $$1-\frac{k'}{n}\le \{x\}<1-\frac{k'-1}{n} ,$$

and multiplying both sides by $$n$$ gives


 * $$n-k'\le n\, \{x\}<n-k'+1.$$

Now if the summation from Hermite's identity is split into two parts at index $$k'$$, it becomes

\begin{align}

\sum_{k=0}^{n-1}\left\lfloor x+\frac{k}{n}\right\rfloor

& =\sum_{k=0}^{k'-1} \lfloor x\rfloor+\sum_{k=k'}^{n-1} (\lfloor x\rfloor+1)=n\, \lfloor x\rfloor+n-k' \\[8pt]

& =n\, \lfloor x\rfloor+\lfloor n\,\{x\}\rfloor=\left\lfloor n\, \lfloor x\rfloor+n\, \{x\} \right\rfloor=\lfloor nx\rfloor.

\end{align}

$$

Proof using functions
Consider the function


 * $$f(x) = \lfloor x \rfloor + \left\lfloor x + \frac{1}{n} \right\rfloor + \ldots + \left\lfloor x + \frac{n - 1}{n} \right\rfloor - \lfloor nx \rfloor $$

Then the identity is clearly equivalent to the statement $$f(x) = 0$$ for all real $$x$$. But then we find,


 * $$ f\left(x + \frac{1}{n} \right) = \left\lfloor x + \frac{1}{n} \right\rfloor + \left\lfloor x + \frac{2}{n} \right\rfloor + \ldots + \left\lfloor x + 1 \right\rfloor - \lfloor nx + 1 \rfloor = f(x) $$

Where in the last equality we use the fact that $$\lfloor x + p \rfloor = \lfloor x \rfloor + p$$ for all integers $$p$$. But then $$f$$ has period $$1/n$$. It then suffices to prove that $$f(x) = 0$$ for all $$x \in [0, 1/n)$$. But in this case, the integral part of each summand in $$f$$ is equal to 0. We deduce that the function is indeed 0 for all real inputs $$x$$.