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The summation of cosine $\sum_{k=1}^N \cos (k x)$ is well known (for example, see the previous question here) and is called Lagrange's trigonometric identity.

Is it possible to construct a similar solution for the absolute value of cosine? Namely, is there a closed form solution for the summation

$S(k, N) = \sum_{k=1}^N |\cos (k x)|$

I am hoping there is a trick I am missing...

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Nigel1
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    Even though I have no proof, I strongly believe this to be impossible. – Alex M. Nov 22 '15 at 22:12
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    I can provide a closed form solution for $S(k,N)$ that only involves a summation of $N$ terms. haha! – Michael Nov 24 '15 at 04:51
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    The reason $\sum \cos(kx)$ is easy to compute is that $2\cos(kx) = e^{ikx} + e^{-ikx}$ so the sum is just two simple geometrical series which we know how to handle. Taking the absolute value then this simplifying property is lost and it becomes much much harder. You can see a plot of $S(7,x)$ here, it does not look like very nice so it's unlikely that a simple formula exists here. – Winther Nov 27 '15 at 15:50

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