Periodic fixed point

Algebra Level 5

Consider the sequence: x n + 1 = 4 x n ( 1 − x n ) x_{n+1} = 4x_n(1 - x_n)

Call a point x 0 ∈ [ 0 , 1 ] x_0\in[0, 1] is r − r- periodic if x r = x 0 x_r=x_0 . For example, x 0 = 0 x_0 = 0 is always a r − r- periodic fixed point for any r r .

Let N N be the number of positive 2015 − 2015- periodic fixed points.

Find the last 3 digits of N N .


The answer is 767.

Khang Nguyen Thanh by Khang Nguyen Thanh , 27, Vietnam

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1 solution

Mietantei Conan
Oct 8, 2015

Plug x n = 0.5 − 0.5 b n x_n=0.5-0.5b_n and get b n + 1 = 2 b n 2 − 1 b_{n+1}=2b_n^2-1 . From condition 0 < = x 0 < = 1 0<=x_0<=1 we see − 1 < = b 0 < = 1 -1<=b_0<=1 . Plug b 0 = cos ⁡ θ b_0=\cos \theta . A simple induction shows b n = cos ⁡ 2 n θ b_n=\cos 2^n\theta . Therefore we need to solve cos ⁡ 2 2015 θ = cos ⁡ θ \cos 2^{2015}\theta=\cos \theta

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