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I am trying to solve the following problem however I am stuck. I tried to apply trigonometric identity but I am not finding anything useful.

Kbiir
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2 Answers2

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I guess that $x_n=\sin^2(2^n\theta)$.. Just came into my mind. I used trigonometrical identities as you suggested, and came up with $\sin 2\theta=2\sin\theta\cos\theta$.

WishofStar
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Rewrite the equation $x_{n+1}$ = $4x_n(1-x_n)$ as

$$(2x_n-1)^2=1-x_{n+1}\tag 1$$

and substitute $x_n = \sin^2\theta_n$ into (1) to get

$$(2\sin^2\theta_n-1)^2= 1-\sin^2\theta_{n+1}$$

Simplify to obtain

$$\cos^2 2\theta_n =\cos^2\theta_{n+1}\tag 2$$

which leads to one simple solution,

$$2\theta_n=\theta_{n+1}\implies \theta_n=2^n\alpha$$

Thus, $x_n=\sin^2(2^n\alpha)$.

Edit: As commented by @Andrei, the general solutions to (2) is quite involved, which can be obtained by factorizing (2)

$$\sin(\theta_{n+1}-2\theta_n)\sin(\theta_{n+1}+2\theta_n)=0$$

Then, $\sin(\theta_{n+1}-2\theta_n)=0$ leads to $\theta_{n+1}=2\theta_n+k\pi$ and $\sin(\theta_{n+1}+2\theta_n)=0$ leads to $\theta_{n+1}=-2\theta_n+k\pi$.

Quanto
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  • This is not true. $2\theta_n=-\theta_{n+1}$ is also a solution. Or, more generally $$\theta_{n+1}=\pm2\theta_n+k\pi$$ – Andrei Jan 14 '20 at 19:10
  • @Andrei - Right. I only listed one possibility. The solutions to $\cos^2 2\theta_n =\cos^2\theta_{n+1}$ is quite rich. – Quanto Jan 14 '20 at 19:20