Jay Mckay

2022-01-29

How to solve $\mathrm{cos}\left({2}^{n}x\right)=\mathrm{cos}\left(x\right)$
I am given the function $F:\mathbb{R}\to \mathbb{R}$ , $x↦4x\left(1-x\right)$ and I have to find all n-cycles of the function. I have already reduced this to the problem of solving $\mathrm{cos}\left({2}^{n}x\right)=\mathrm{cos}\left(x\right)$ for $x\in \left[0,\pi \right]$. Using Mathematica and $n\in \left\{1,2,3,4\right\}$, I concluded that the solutions must be $\left\{\frac{2k\pi }{{2}^{n}-1},\frac{2k\pi }{{2}^{n}+1}\right\}\cap \left[0,\pi \right]$. Since the equation obviously has ${2}^{n}$ solutions and the solution set has ${2}^{n}$ elements, it only remains to verify that they are indeed solutions to the equation. I tried using $\mathrm{cos}\left(2x\right)=2{\mathrm{cos}}^{2}\left(x\right)-1$ and proving this by induction on n, but I was not able to.

Expert

Using the identity
$\mathrm{cos}A-\mathrm{cos}B=2\mathrm{sin}\left(\frac{B-A}{2}\right)\mathrm{sin}\left(\frac{B+A}{2}\right)$ ,
we have
$0=\mathrm{cos}\left(x\right)-\mathrm{cos}\left({2}^{n}x\right)=2\mathrm{sin}\left(\frac{{2}^{n}-1}{2}x\right)\mathrm{sin}\left(\frac{{2}^{n}+1}{2}x\right)$,
which holds iff

for some $k\in \mathbb{Z}$

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