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I am currently working on plotting a function and figuring out if its periodic or not. The function is as follows:

$$x_n = betacdot x_n-1+alpha gammacdot mathrmsgn(x_n-3)+alpha(1-gamma)cdot mathrmsgn(x_n-2)$$

I know $beta$ is between $0$ and $1$, $alpha>0$, $gamma>0$. Note that $mathrm sgn$ is the sign function.

I am having a hard time knowing where to start. For example, if I take
$beta=0.5$, $alpha=10$, and $gamma=3$, how would I find my $x_1$, $x_2$, and $x_3$?

For example:

$x_1$$=$$1$

$x_2$$=$$2$

$x_3$$=$$3$

$x_4$ $ = 0.5(3)+30+10(-2)=11.5$

Also I am saying that $sgn(x_1)=1$ since $x_1=1$
and $sgn(x_2)=1$ since $x_2=2$

Would this be the correct way to start plotting?

Now let's say for the three initial values that function is not periodic, would some other three different initial values make it periodic? and if so how would one go about finding those initial values to make it period? (keep guess values)

Lastly, I promise, why would someone come up with this specific function to figure out periodic or non-periodic?

New Edit:

I have been thinking what would happen if we take $0<gamma<1$?
Would it be the same or would it change?
Also, what about taking $alpha<0$?

Should I ask another question or just leave the new Edit.

With proper values for $alpha,beta,gamma, x_1, x_2$ ($x_1,x_2$ are assumed to be positive ) ---(*)

$$
-alpha(1-gamma)-alphagamma le alpha(1-gamma)mboxsign(x_n-2)+alphagamma mboxsign(x_n-3)le alpha(1-gamma)+alphagamma
$$

and the recurrences

$$
x_n^i = beta x_n-1^i-alpha(1-gamma)-alphagamma\
x_n^s = beta x_n-1^s+alpha(1-gamma)+alphagamma\
$$

have the solutions

$$
x_n^i = C_0 beta^n-1-fracalpha(1-beta^n)beta-1\
x_n^s = C_0 beta^n-1+fracalpha(1-beta^n)beta-1\
$$

and

$$
x_n^i le x_nle x_n^s
$$

Attached the plot for $x_n^i, x_n^s$ in red and $x_n$ in blue. Note that the recurrences $x_n^i, x_n^s$ should begin at $x_3$.

NOTE

Due to the structure

$$
x_n = beta x_n-1+cdots
$$

there is an exponential component all along $ntoinfty$ which eliminates the periodic behavior possibility.

To solve

$$
z_n = beta z_n-1+alpha(1-gamma)+alphagamma
$$

which is a linear recurrence we use the fact

$$
z_n = z_n^h+z_n^p\
z_n^h -beta z_n-1^h = 0\
z_n^p -beta z_n-1^p=alpha(1-gamma)+alphagamma
$$

then easily we find $z_n^h = Cbeta^n-1$. Now considering $z_n^p = C_nbeta^n-1$ and substituting into the particular we can find also the recurrence for $C_n$ which is

$$
C_n-C_n-1 = alphabeta^1-n
$$

with solution

$$
C_n = fracalpha beta left(1-beta ^-nright)beta -1
$$

and finally

$$
z_n = Cbeta^n-1+ fracalpha beta left(1-beta ^-nright)beta -1beta^n-1=C beta^n-1+fracalpha(1-beta^n)beta-1
$$

The general solution can be computed with the help of the following MATHEMATICA script

x2 = 2; x3 = 3; x1 = 1;
path = x1, x2, x3;
beta = 0.5; alpha = 10; gamma = 3;
For[i = 1, i <= 30, i++, x4 = beta x3 + alpha (1 - gamma) Sign[x2] + alpha gamma Sign[x1]; 
AppendTo[path, x4]; x1 = x2; x2 = x3; x3 = x4]

ListPlot[path, PlotRange -> All]

Regarding --- (*) the most general boundaries can be calculated with the optimization problem

$$
min_u,v(max_u,v)alpha(1-gamma)sin u+alphagamma sin v
$$

or

$$
min(max)-alpha (1-gamma )-alpha gamma ,alpha gamma -alpha (1-gamma ),alpha (1-gamma )-alpha gamma ,alpha (1-gamma )+alpha
gamma
$$