Let $f(x) = |x+1|-|x-1|$, find $f \circ f\circ f\circ f ... \circ f(x)$ (n times). Let $f(x) = |x+1|-|x-1|$, find $f \circ f\circ f\circ f ... \circ f(x)$  (n times). I don't know where to start... Should I use mathematical induction? But what should be my hypothesis? Should I calculate for n = 1, n = 2?
 A: First note
$$f(x) = |x+1| - |x-1|\\
= \begin{cases}
-2, & x< -1\\
2x, &  x\in [-1, 1] \\
2, &  x>1\\
\end{cases}$$
Proposition: $$\color{red}{f^{n+1}(x) = \left|2^nx+1\right| - \left| 2^nx-1\right|}$$.
Note: this is exactly equivalent to
$$f^{n+1}(x) = \begin{cases}
-2, & x< -\frac1{2^n}\\
2^{n+1}x, &  x\in [-\frac1{2^n}, \frac1{2^n}] \\
2, &  x>\frac1{2^n}\\
\end{cases}$$
Clearly true for $n=0$.  Assuming it holds for $n$, the inductive step is
$$f^{n+2}x = f\circ f^{n+1} (x)= \begin{cases}
f(-2), & x < -\frac1{2^n}\\
f(2^{n+1}x), &x\in [-\frac1{2^n}, \frac1{2^n}] \\
f(2), &  x>\frac1{2^n}\\  
\end{cases}\\
=\begin{cases}
-2, & x < -\frac1{2^n}\\
-2, &x\in [-\frac1{2^n}, -\frac1{2^{n+1}}) \\
f(2^{n+1}x), &x\in [-\frac1{2^{n+1}}, \frac1{2^{n+1}}] \\
2, &x\in (\frac1{2^{n+1}}, \frac1{2^n}] \\
2, &  x>\frac1{2^n}\\  
\end{cases}$$
$$=\begin{cases}
-2, & x < -\frac1{2^{n+1}} \\
2^{n+2}x, &x\in [-\frac1{2^{n+1}}, \frac1{2^{n+1}}] \\
2, &x > \frac1{2^{n+1}}\\  
\end{cases}\\
= |2^{n+1}x+1| - |2^{n+1}x-1|$$
A: Here is a slightly more general solution, introducing a "one-parameter" group of functions (see relationship (3)).
As function $f$ is odd, and the composition of odd functions is odd, we can restrict our study to positive values of $x$.
In this domain, we can express $f$ in the following way:
$$f(x)=\min(2x,2)$$
Let us introduce, for any $a>0$, the following notation:
$$f_a(x):=\min(2ax,2)\tag{1}$$
giving in particular $f_1=f \tag{2}.$
We have the identity (assuming $a,b \ge 1$):
$$f_{2ab}(x)=f_a(f_b(x))\tag{3}$$ from which, by an immediate recurrence, taking $a=b=1$, we get the desired answer:
$$f_{2^n}(x)=\underbrace{(f \circ f \circ \cdots \circ f)}_{(n+1) \ \text{times "f"}}(x)$$
Proof of (3): (3) is equivalent to:
$$\begin{align}
&&\forall x>0, \ \ &\min(4abx,2)=\min(2a.f_b(x),2)\\
&\iff &&\min(4abx,2)=\min([2a.\min(2bx,2)],2)\\
&\iff &&\min(4abx,2)=\min(4abx,4a,2)\end{align}$$
(by associativity of "min" operator)
proving (1) because we deal with values of $a$ which are $\ge 1$.
A: Here's a graph of the first few iterates:

(Desmos link)
