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.
$$\prod_{k=1}^{n}\left(\cos kx1\right)=2^{n} e^{\frac{1}{2} i n (n+1) x} \left(e^{i x};e^{i x}\right)_n^2,$$
with $(\cdot;\cdot)_n$ the qPochhammer 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, $$\prod_{k=1}^{n}\left(\cos kx1\right)=\tfrac{1}{2}a_{n,0}+\sum_{p=1}^{n(n+1)/2}a_{n,p}\cos px.$$ I find $$a_{n,p}=2^{1n}T_{n,q},\;\;q=\tfrac{1}{2}n(n+1)+p,$$ with $T_{n,q}$ the coefficients in OEIS:A304080.
_{ Mathematica code to test this:}
a[n_,p_]:=2^(1n)*CoefficientList[Expand[Product[(1x^j)^2,{j,1,n}]],x][[n*(n+1)/2+p+1]]
Product[Cos[k*x]1,{k,1,n}]a[n,0]/2Sum[a[n,p]*Cos[p*x],{p,1,n*(n+1)/2}]//FullSimplify

$\begingroup$ Thanks, Beenakker. As you said the identity involving the qPochhammer 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

5$\begingroup$ I don't quite understand what it is you are after; take $n=2$, then the product evaluates to $(\cos x1)(\cos 2x1)$. Is that a "closedform" 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

3$\begingroup$ By trigonometricproduct identities this product can at least be written as a finite series of harmonics; e.g. $(\cos x1)(\cos 2x1)$ $= \cos x\cos 2x \cos 2x\cos x+1$ $= \frac12(\cos3x+\cos x)\cos 2x\cos x+1$ $= \frac12\cos 3x\cos2x\frac12\cos x+1$. 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

1$\begingroup$ 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