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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.

  • 0
    note that for irrational $\alpha$ this is a sharp inequality, since $\{\alpha N \mod 1\} $ is dense in $[0,1]$.2012-04-18

1 Answers 1

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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 $\left| \frac{\sin(N\pi\alpha)}{\sin(\pi\alpha)}\right|\leq \left| \frac{1}{\sin(\pi\|\alpha\|)}\right|\leq \frac{1}{2\|\alpha\|}.$