Prove $\prod\limits_{k=0}^{n-1} \left(x^2-2x\cos \left(\alpha+\frac{2k\pi}{n}\right)+1\right)=x^{2n}-2x^n\cos(n\alpha)+1$

Question

I have read in a paper that there is a formula as follows: $$\prod_{k=0}^{n-1} \left(x^2-2x\cos\left(\alpha+\frac{2k\pi}{n}\right)+1\right)=x^{2n}-2x^n\cos(n\alpha)+1.$$ In the paper they said that we can find such formula in ''Table of integrals, series and products'' by Gradshteyn and Ryzhik, and indeed I found it there with a reference to another book called ''Summation of Series'' by Jolley. In the latter book unfortunately I couldn't find the formula. So I do not know what to do. Can someone give me a proof of the above formula, or any other reference where the proof exists? Thanks for any help!

Answer 1

Best done by passing to the complex plane in my opinion. Note the following:

$(x-e^{i\theta})(x-e^{-i\theta})=x^2-2x \cos\theta +1$

Try using this and pushing factors around on the left hand side a bit to try get closer to the right hand side.

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We can use this to factor each expression on the left-hand side as $(x-e^{i(\alpha+2k\pi/n)})(x-e^{-i(\alpha+2k\pi/n)})$

So, our proof now relies on showing that:

$\prod_{k=0}^{n-1}(x-e^{i(\alpha+2k\pi/n)})(x-e^{-i(\alpha+2k\pi/n)})=(x^n-e^{in\alpha})(x^n-e^{-in\alpha})$

This is very close to a factorisation of $x^n-1$ involving roots of unity, $\prod_{k=0}^{n-1}(x-e^{2k\pi i/n})=x^n-1$. Let's split it into two parts:

$A(x)=\prod_{k=0}^{n-1}(x-e^{i(\alpha+2k\pi/n)})$

$B(x)=\prod_{k=0}^{n-1}(x-e^{-i(\alpha+2k\pi/n)})$

Substituting $x=te^{i\alpha}$ and $x=te^{-i\alpha}$ respectively, we quickly see:

$A(te^{i\alpha})=\prod_{k=0}^{n-1}(te^{i\alpha}-e^{i(\alpha+2k\pi/n)})=e^{in\alpha}\prod_{k=0}^{n-1}(t-e^{2k\pi i/n)})=e^{in\alpha}(t^n-1)$

$B(te^{-i\alpha})=\prod_{k=0}^{n-1}(te^{-i\alpha}-e^{-i(\alpha+2k\pi/n)})=e^{-in\alpha}\prod_{k=0}^{n-1}(t-e^{2k\pi i/n)})=e^{-in\alpha}(t^n-1)$

Moving back to the expressions in terms of $x$, this gives us $A(x)=x^n-e^{in\alpha}$ and $B(x)=x^n-e^{-in\alpha}$, and by applying the formula from the very start, we arrive at the desired identity.

Answer 2

If $x^{2n}-2x^n\cos nA+1=0, x\ne0$

$$x^n=\cos nA\pm i\sin nA=e^{\pm i(nA+2m\pi)}$$ where $m$ is any integer

$$x=e^{\pm i\left(\dfrac{2m\pi}n+A\right)}$$ where $m\equiv 0,1,\cdots,m-2,m-1\pmod m$

Now $\left(x-e^{i\left(\dfrac{2m\pi}n+A\right)}\right)\left(x-e^{-i\left(\dfrac{2m\pi}n+A\right)}\right)=x^2-2x\cos\left(\dfrac{2m\pi}n+A\right)+1$