Is there any closed form of $$\prod_{k=1}^{n}\left(\cos(kx)-1\right)?$$ I failed to find references on this problem in the internet.


$$\prod_{k=1}^{n}\left(\cos kx-1\right)=2^{-n} e^{-\frac{1}{2} i n (n+1) x} \left(e^{i x};e^{i x}\right)_n^2,$$ with $(\cdot;\cdot)_n$ the q-Pochhammer symbol.
I guess this counts as a "closed form", but of course it's just a rewriting of the product in terms of some named quantity.

Steven Stadnicki has suggested to compute the Fourier coefficients, $$\prod_{k=1}^{n}\left(\cos kx-1\right)=\tfrac{1}{2}a_{n,0}+\sum_{p=1}^{n(n+1)/2}a_{n,p}\cos px.$$ I find $$a_{n,p}=2^{1-n}T_{n,q},\;\;q=\tfrac{1}{2}n(n+1)+p,$$ with $T_{n,q}$ the coefficients in OEIS:A304080.

Mathematica code to test this:


  • $\begingroup$ Thanks, Beenakker. As you said the identity involving the q-Pochhammer symbol is just a rewrite of the product. I'm finding a closed form such like a simple expression…… $\endgroup$ – 好きな人がいません Jun 3 at 10:15
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    $\begingroup$ I don't quite understand what it is you are after; take $n=2$, then the product evaluates to $(\cos x-1)(\cos 2x-1)$. Is that a "closed-form" answer to your question? If it's not, what are you hoping to find, and how does that generalize to larger values of $n$? $\endgroup$ – Carlo Beenakker Jun 3 at 11:51
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    $\begingroup$ By trigonometric-product identities this product can at least be written as a finite series of harmonics; e.g. $(\cos x-1)(\cos 2x-1)$ $= \cos x\cos 2x -\cos 2x-\cos x+1$ $= \frac12(\cos3x+\cos x)-\cos 2x-\cos x+1$ $= \frac12\cos 3x-\cos2x-\frac12\cos x+1$. Possibly a closed form solution for the coefficients of this expansion? $\endgroup$ – Steven Stadnicki Jun 3 at 16:42
  • $\begingroup$ @StevenStadnicki -- neat idea, I have added this to my answer. $\endgroup$ – Carlo Beenakker Jun 3 at 18:50
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    $\begingroup$ the $p=0$ term is a bit anomalous, because the $p$-th Fourier coefficient of $\cos p x$ equals 1 for $p=1,2,3,\ldots$, but it equals 2 for $p=0$ [with the usual definition $a_p=(2/\pi)\int_0^\pi f(t)\cos pt \,dt$.] $\endgroup$ – Carlo Beenakker Jun 3 at 19:40

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