How does one prove the following inequality?
$$\left|\frac{\sin(\pi \alpha N)}{\sin(\pi \alpha)}\right| \leq \frac{1}{2 \| \alpha \|}$$
Here $\| \alpha \|$ denotes the distance to the nearest integer.
I'm not even sure where to get started.
Thanks.
How does one prove the following inequality?
$$\left|\frac{\sin(\pi \alpha N)}{\sin(\pi \alpha)}\right| \leq \frac{1}{2 \| \alpha \|}$$
Here $\| \alpha \|$ denotes the distance to the nearest integer.
I'm not even sure where to get started.
Thanks.
By looking at the graph of the $\sin(x)$ function around $0$, we see that $$|\sin( x)|\geq \frac{2}{\pi} x$$ when $-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}$ and also $$\sin(x)\leq 1.$$ Since $|\sin(\pi\alpha)|= |\sin(\pi \|\alpha\|)|$, and $\|\alpha\|\leq \frac{1}{2}$, we may apply the above inequalities to find $$\left| \frac{\sin(N\pi\alpha)}{\sin(\pi\alpha)}\right|\leq \left| \frac{1}{\sin(\pi\|\alpha\|)}\right|\leq \frac{1}{2\|\alpha\|}.$$