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I need to calculate as part of a proof the maximum and minimum of this function analytically: $$f_n(\varphi) = \sum_{k=0}^{n-1}\left|\sin \left(\varphi-\frac{2\pi k}{n}\right) \right|$$ whereby $\varphi \in [0,2\pi)$.

I thought to try it with a complex approach, as it looks quite similar to root of unity. However, the absolute value makes me trouble in any simplification.

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  • $\begingroup$ With Cauchy-Schwarz I get $\sum_{k=0}^{n-1}\left|\sin \left(\varphi-\frac{2\pi k}{n}\right) \right| \ge \left| \sum_{k=0}^{n-1}\sin \left(\varphi-\frac{2\pi k}{n}\right) \right| = 0 $ which seem for me by definition already case. $\endgroup$
    – bilaljo
    Commented Feb 26, 2023 at 14:17
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    $\begingroup$ And for odd $n$ use the fact that $$f_n(\phi) = \frac{1}{2}f_{2n}(\phi)$$. $\endgroup$
    – Jkbb
    Commented Feb 26, 2023 at 15:15
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    $\begingroup$ Note that $f_n(\varphi) = f_n(\varphi + \frac{2\pi}{n})$ so we can focus on the interval $\varphi \in [0,\frac{2\pi}{n})$. On this interval, for even numbers $n$ we have: $$f_n(\varphi) = 2\sum_{k=0}^{n/2}\sin \left(\varphi-\frac{2\pi k}{n}\right) $$ Thus $f_n$ is differentiable on this interval and so the extrema are attained possibly at $0$ or the zeros of the derivative. $$f'_n(\phi) = 2\sum_{k=0}^{n/2}\cos \left(\varphi-\frac{2\pi k}{n}\right)$$ which is zero exactly at $\frac{\pi}{n}$. $\endgroup$
    – Jkbb
    Commented Feb 26, 2023 at 15:19
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    $\begingroup$ Typo: I meant to write $f'_n(\varphi)$ instead of $f'_n(\phi)$ in the last equation. $\endgroup$
    – Jkbb
    Commented Feb 26, 2023 at 17:45
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    $\begingroup$ For odd $n$ you can't drop the absolute value of the $|\sin(\varphi - \frac{2\pi(n+1)/2}{n})|$ in the sum (even on our reduced interval) and so you can't differentiate the sum. I would rather do it the way I explained in the first comment - reducing it to the even $n$ case. $\endgroup$
    – Jkbb
    Commented Mar 1, 2023 at 20:40

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