1

EDIT: Consider $m,n\in\mathbb{Z}$ with $\frac{m}{n}\in \mathbb{R} -\mathbb{Z}$ and $n>1$.

Given integers $m,n$ I want to prove that $$\bigg\lfloor \frac{m}{n} \bigg\rfloor=\frac{m}{n}-\frac{1}{2}+\frac{1}{2n}\sum_{k=1}^{n-1} \sin \bigg(\frac{2\pi km}{n}\bigg) \cot \bigg(\frac{\pi k}{n}\bigg)$$

Using a Fourier expansion I obtain $$ \bigg\lfloor \frac{m}{n} \bigg\rfloor=\frac{m}{n}-\frac{1}{2}+\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\sin (2\pi k \frac{m}{n})} {k}$$

However, I do not see how $$\frac{1}{2n}\sum_{k=1}^{n-1}\sin\bigg( \frac{2\pi km}{n} \bigg)\cot \bigg(\frac{\pi k}{n} \bigg) =\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\sin (2\pi k \frac{m}{n})} {k}$$

EDIT This link http://functions.wolfram.com/IntegerFunctions/Floor/06/ShowAll.html also claims that they are equal.

cap
  • 2,443
  • As written, the sum on the right does not converge. – Ron Gordon Mar 16 '14 at 20:55
  • Sorry, I will edit: $m/n \in \mathbb{R} - \mathbb{Z} $ with $m,n \in \mathbb{Z}$ and $n>1$ – cap Mar 16 '14 at 21:00
  • You're still not there. Look carefully at your link. – Ron Gordon Mar 16 '14 at 21:11
  • @Gordon The infinite series in question may very well converge, just not absolutely. This is a very curious sum indeed. I would think an arithmetic progression formula for sine or cosine is needed. Maybe a variation of http://math.stackexchange.com/questions/17966 – Bobby Ocean Mar 25 '14 at 10:13

1 Answers1

2

For given $n\in{\mathbb N}_{\geq1}$ the function $$f:\>{\mathbb Z}\to{\mathbb R},\quad k\mapsto\left\lfloor{k\over n}\right\rfloor -{k\over n}$$ is periodic with period $n$. Now do a discrete Fourier transform on $f$, and you will obtain a finite trigonometric sum representing $f$.

Christian Blatter
  • 226,825