TITLE: Solving logistic equation $x_{n+1}= \mu x_n ( 1- x_n) $ for $\mu = 2 $ and $\mu = 4$
QUESTION [3 upvotes]: I have a question related to solving logistic equation $x_{n+1}= \mu x_n ( 1- x_n) $, for $\mu =2 $ and $\mu = 4$. This is an exercise in Elaydi's  "Discrete Chaos".

For the case $\mu=2$, there was a hint to use substitution $x_n = \frac{1}{2}(1- y_n)$. The equation then reduces to $y_{n+1}= (y_{n})^2$ and the solution is $y_n= (y_0)^{2n}$, where $y_0$ is initial condition. Now, do we have to express the final solution is terms of $x_n$ i.e. to write $y_n= 1- 2x_n$, $y_0 = 1-2 x_0$, so $ 1-2x_n= (1+2x_0)^{2n}$ i.e. $ x_n= \frac{1}{2} ( 1- (1+ 2x_0)^{2n}) $?

In case $\mu=4$, we should use substitution $x_n = (\sin \theta_n)^2$. Similarly as in the 1st case, we will get equations that depend on $\theta_n$, but do we have to express in terms of $x_n$ and would then $\theta_n $ be $\arcsin(\sqrt{x_n})$ or something else? Do we have any restrictions on $x_n$ e.g. $x_n=(\sin \theta_n)^2\geq 0$? And if we have initial condition $x_0$, what is $\theta_0$?


I couldn't find any explanation and I would appreciate any help. Thanks a lot in advance!

REPLY [0 votes]: It seems that you have answered most of the questions yourself.
Several things to point out:

For $\mu = 2$, the solution is $y_n = y_0^{2^n}$ instead of $y_n = y_0^{2n}$. You can express it back in terms of $x_n$, so that $x_n = \frac 1 2(1 - (1 + 2x_0)^{2^n}))$.

For $\mu = 4$, it is better to use the expression $x_n = \frac 1 2(1 - \cos \theta_n)$. You then get $\cos \theta_{n + 1} = \cos(2\theta_n)$, from which you deduce that $\cos \theta_n = \cos (2^n\theta_0)$.
This then can be rewritten in terms of $x_n$, i.e. $x_n = \frac 1 2(1 - \cos(2^n \theta_0))$, where $\theta_0 = \arccos(1 - 2x_0).$

For initial value $0 \leq x_0 \leq 1$, you get a real number $\theta_0$. Otherwise, there always exists a complex number $\theta_0$ such that $\cos(\theta_0) = 1 - 2x_0$. It is not unique, but any two such $\theta_0$ differ by an integral multiple of $2\pi$, thus in the final formula, the value of $\cos(2^n\theta_0)$ doesn't depend on which $\theta_0$ you choose.
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