How to solve \cos(2^n x) = \cos(x) I am given the function F:\mathbb{R}\to\mathbb{R}

Jay Mckay

Jay Mckay

Answered question

2022-01-29

How to solve cos(2nx)=cos(x)
I am given the function F:RR , x4x(1x) and I have to find all n-cycles of the function. I have already reduced this to the problem of solving cos(2nx)=cos(x) for x[0,π]. Using Mathematica and n{1,2,3,4}, I concluded that the solutions must be {2kπ2n1,2kπ2n+1}[0,π]. Since the equation obviously has 2n solutions and the solution set has 2n elements, it only remains to verify that they are indeed solutions to the equation. I tried using cos(2x)=2cos2(x)1 and proving this by induction on n, but I was not able to.

Answer & Explanation

enveradapb

enveradapb

Beginner2022-01-30Added 13 answers

Using the identity
cosAcosB=2sin(BA2)sin(B+A2) ,
we have
0=cos(x)cos(2nx)=2sin(2n12x)sin(2n+12x),
which holds iff
sin(2n12x)=0  OR  sin(2n+12x)=0
2n12x=kπ  OR  2n+12x=kπ
x=2kπ2n1  OR  x=2kπ2n+1
for some kZ

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