Let $f(x) = |x+1|-|x-1|$, find $f \circ f\circ f\circ f ... \circ f(x)$ (n times).

Question

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?

Answer 1

Here's a graph of the first few iterates:

(Desmos link)

Answer 2

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

$$f^{n+2}x = f\circ f^{n+1} (x)= \begin{cases} f(-2), & x < -\frac1{2^n}\\ f(2^{n+1}x), &x\in [-\frac1{2^n}, \frac1{2^n}] \\ f(2), & x>\frac1{2^n}\\ \end{cases}\\ =\begin{cases} -2, & x < -\frac1{2^n}\\ -2, &x\in [-\frac1{2^n}, -\frac1{2^{n+1}}) \\ f(2^{n+1}x), &x\in [-\frac1{2^{n+1}}, \frac1{2^{n+1}}] \\ 2, &x\in (\frac1{2^{n+1}}, \frac1{2^n}] \\ 2, & x>\frac1{2^n}\\ \end{cases}$$ $$=\begin{cases} -2, & x < -\frac1{2^{n+1}} \\ 2^{n+2}x, &x\in [-\frac1{2^{n+1}}, \frac1{2^{n+1}}] \\ 2, &x > \frac1{2^{n+1}}\\ \end{cases}\\ = |2^{n+1}x+1| - |2^{n+1}x-1|$$

Answer 3

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 $\ge 1$.