How find this $x_{1},x_{2},\cdots,x_{1994}$ with this following system equation $3+2x_{i+1}=3|x_{i}-1|-|x_{i}|$

Question

Find the all real numbers $x_{1},x_{2},\cdots,x_{1994}$ such $$3+2x_{i+1}=3|x_{i}-1|-|x_{i}|,i=1,2,3,\cdots,1994$$ where $x_{1995}=x_{1}$

It is clear equivalent solve following system equation real solution: $$\begin{cases} 3+2x_{2}=3|x_{1}-1|-|x_{1}|\\ 3+2x_{3}=3|x_{2}-1|-|x_{2}|\\ 3+2x_{4}=3|x_{3}-1|-|x_{3}|\\ \cdots\\ 3+2x_{1994}=3|x_{1993}-1|-|x_{1993}|\\ 3+2x_{1}=3|x_{1994}-1|-|x_{1994}| \end{cases}$$

I have some try,and can't solve it.

if we add this all equation,we have $$3\cdot 1994+2\sum_{i=1}^{1994}x_{i}=3\sum_{i=1}^{1994}(|x_{i}-1|-|x_{i}|)$$ maybe consider this inequality $$3+2x_{2}=|3x_{1}-3|-|x_{1}|\le |3x_{1}-3-x_{1}|=|2x_{1}-3|\le 2x_{1}+3?$$ I guess we can prove $$x_{1}=x_{2}=\cdots=x_{1994}?$$

Answer 1

$x_{i+1} = f(x_i) = \left\{ \begin{array}{ll} -x & \text{ if $x < 0$ }\\ -2x & \text{ if $0 \le x \le 1$ }\\ x - 3 & \text{ if $1 < x $}\end{array}\right.$ this dynamical system has three types of periodic orbit:

(a) an attracting $4$-periodic orbit $\{x, x-3, 3-x, -x\}$ for $1 \le x < 1.5$ and $1.5 < x \le 2.$

(b) an attracting $2$-periodic orbit $\{1.5, -1.5\}$

(c) a repelling $1$-periodic orbit $\{0\}.$

starting at any point, the orbit of the point eventually ends up in one of the three classes of the orbits above. therefore you only need to look for the starting points among the points in the periodic orbits listed above.

$1995 = 4*498 + 3,$ so none of the elements of the $4$-orbits can be $x_1.$ therefore the only candidates for $x_1$ such that $x_1 = x_{1995}$ are $x_1 = \pm 1.5, 0$; the laments of $1$ and $2$-periodic points.

Answer 2

$x_{i+1}=\left\{\begin{array}{}-x_i,x_i<0\\x_i-3,x_i>1\\-2x_i,0<x_i<1\end{array}\right.$
How can that sequence of functions end up where it started?
If any $x_i<-2$, then $x_{i+1}>-2$ and the cycle will never return below $-2$.
If any $x_i>2$ then $x_j$ decrease until they are less than 2, and they cycle never returns above $2$.

Answer 3

0 is stable.

Other numbers between -1 and 1 drift out to numbers with magnitude between 1 and 2.

Numbers with magnitude between 1 and 2 run a 4-length cycle above and below magnitude 1.5, except 1.5 and -1.5 which alternate.

Numbers with magnitude above 2 reduce to 2 or below.

The only three numbers which could qualify for $a_1 = a_{1995}$ are therefore {-1.5, 0, 1.5}.