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Does anybody know a closed form for this multiplication?

$$\prod_{k=0}^{n}\left(1-2\alpha \cos\frac{2\pi k}{n}\right)$$ where $\alpha$ is a real number

or maybe even a potential method to use to evaluate it?

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With $\zeta=e^{2\pi\mathrm{i}/n}$, we have $z^n-1=\prod\limits_{k=0}^{n-1}(z-\zeta^k)=\prod\limits_{k=0}^{n-1}(z-\zeta^{-k})$. Multiplying these, we get $$(z^n-1)^2=\prod_{k=0}^{n-1}\left(z^2-2z\cos\frac{2\pi k}{n}+1\right)=(z^2+1)^n\prod_{k=0}^{n-1}\left(1-\frac{2z}{z^2+1}\cos\frac{2\pi k}{n}\right).$$ Thus, to get your product, you have to solve $z/(z^2+1)=\alpha$ and add an extra term (with $k=n$).

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  • $\begingroup$ but this seems a bit odd to me since to have a real result for this multiplication, what you wrote confines the value of $\alpha$ in a region which is where that equation $\frac{z}{z^2+1}=\alpha$ has real answers otherwise it gives us a complex result, but we know that as long as we have $\alpha$ to be between [0,1] that multiplication should always give a real result. or i think for any real value of $alpha$ it should be real , so whats going on here? $\endgroup$
    – Jason
    May 23 '20 at 9:46
  • $\begingroup$ @Jason: If $z+1/z=1/\alpha$ is real, then $z^n+z^{-n}$ is real (even if $z$ happens to be complex). And then $(z^n-1)^2/(z^2+1)^n=\alpha^n(z^n+z^{-n}-2)$ is real too. $\endgroup$
    – metamorphy
    May 23 '20 at 10:00
  • $\begingroup$ So I have two question id be appreciated if you can help me, first how can I prove the first equality you used in your answer? (the one with $z^{n}-1=\Pi(z-\zeta^k)$ and second, in your comment I could follow exactly how you decompose that division to RHS? $\endgroup$
    – Jason
    May 23 '20 at 10:07
  • $\begingroup$ @Jason: the first one is basic algebra ($z=\zeta^k$ are distinct roots of $z^n-1=0$); going from $k$ to $-k$ is easy (it amounts just to a change of order of the terms: $\zeta^{-k}=\zeta^{n-k}$). $\endgroup$
    – metamorphy
    May 23 '20 at 10:13
  • $\begingroup$ yes, thanks, and for second question, in your comment I could not follow exactly how you decompose that division to RHS?(in my comment I accidentally wrote" I Could follow") $\endgroup$
    – Jason
    May 23 '20 at 10:18

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