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I am having trouble understanding one of the inequalities involved in the proof of the Pólya-Vinogradov inequality, more precisely

$$ \left \vert{\frac{\sqrt p} p \sum_{a \mathop = 1}^{p-1} \frac{e^{\pi i a n / p} \sin \left({\pi a n / p}\right)} {e^{\pi i a / p} \sin \left({\pi a / p}\right)} }\right \vert \le \frac{\sqrt p} p \sum_{a \mathop = 1}^{p-1} \left \vert{\frac 1 {\sin \left({\pi \left \langle {a / p}\right \rangle}\right)} }\right \vert$$

where $\left\langle{x}\right\rangle$ denotes the absolute value of the difference between $x$ and the closest integer to $x$.

I think is using the triangle inequality, and as $\left\vert \frac{e^{\pi i a n / p}}{e^{\pi i a / p}} \right \vert =1$ it only remains to justify that

$$\left\vert \frac{sin( \pi an/p)}{sin( \pi a/p)} \right \vert \le \left\vert \frac{1}{\sin \left({\pi \left \langle {a / p}\right \rangle}\right)} \right \vert $$

but I can't see where the $\left \langle {a / p}\right \rangle$ comes up.

Thanks very much, any idea will be appreciated.

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  • $\begingroup$ Just a question. What is $\langle a/p\rangle$? I would write the floor function $[a/p]$. $\endgroup$ – Dietrich Burde Nov 5 '16 at 21:53
  • $\begingroup$ Is the absolute value of the difference between $a/p$ and the closest integer to $a/p$ $\endgroup$ – JulianP Nov 5 '16 at 22:06
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    $\begingroup$ so $\langle x \rangle = |x-\lfloor x+1/2\rfloor|$ $\endgroup$ – reuns Nov 5 '16 at 22:15
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    $\begingroup$ for any integer $k$ : $|\sin(\pi x)| = |\sin(\pi\ |x-k|\ \ )|$ and hence $|\sin(\pi x)| = \sin(\pi \langle x \rangle)$ $\endgroup$ – reuns Nov 5 '16 at 22:18

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