Find $f(\frac{2\pi k}{2^n \pm1})$ given $f(x)=\prod_{i=1}^{n-1}[2\cos(2^{i-1}x)-1],n\geq1$.

Question

We are required to find $f(\frac{2\pi k}{2^n \pm1})$ given

$f(x)=\prod_{i=1}^{n-1}[2\cos(2^{i-1}x)-1],n\geq1$

Let's start by removing the product notation

$\Rightarrow f(x)=(2\cos x-1)(2\cos 2x-1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)$

Now multiplying and dividing by $2\cos x+1$,

$\Rightarrow f(x)=\frac{(2\cos x+1)(2\cos x-1)(2\cos 2x-1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)}{2\cos x+1}$

We know, $a^2-b^2=(a+b)(a-b)$

$\Rightarrow f(x)=\frac{(4\cos^2x-1)(2\cos 2x-1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)}{2\cos x+1}$

Now, $\cos2x=2\cos^2x-1$

$\Rightarrow f(x)=\frac{(2\cos2x+1)(2\cos 2x-1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)}{2\cos x+1}$

$\Rightarrow f(x)=\frac{(2\cos 2^2x+1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x-1)}{2\cos x+1}$

We can now repeatedly apply the difference of squares formula and our function reduces to

$\Rightarrow f(x)=\frac{4\cos^2 2^{n-1}x-1}{2\cos x+1}$

$\Rightarrow f(x)=\frac{2\cos 2^nx+1}{2\cos x+1}$

How do I proceed from here on out?

Edit- We just need to know that $\cos(2k)=\cos(2^nk)$. Thus the denominator and the numerator cancel out and we get

$f(\frac{2\pi k}{2^n \pm1})=1$ and that's the answer.

Answer 1

In fact you have solved roughly all of it. You just need to know that:

1. $\frac{2^n2\pi k}{2^n +1} +\frac{2\pi k}{2^n+1}=2\pi k$

2. $\frac{2^n2\pi k}{2^n -1} -\frac{2\pi k}{2^n-1}=2\pi k$

So in the end you get the answer $ 1 $ .