# closed form for $\prod_{k=0}^{n}\left(1-2\alpha \cos\frac{2\pi k}{n}\right)$

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?

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 $$k=n$$).
• 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? May 23 '20 at 9:46
• @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. May 23 '20 at 10:00
• 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? May 23 '20 at 10:07
• @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: $\zeta^{-k}=\zeta^{n-k}$). May 23 '20 at 10:13