Iterates of $\frac{\sqrt{2}x}{\sqrt{x^2 +1}}$ converge to $\text{sign}(x)$.

Question

In this post, a comment states that if $f(x):= \dfrac{\sqrt{2}x}{\sqrt{x^2 +1}}$ and $F_n:=\underbrace{f\circ \dots\circ f}_{n\text{ times}}$, then the pointwise limit $\lim\limits_{n \to \infty} F_n$ is equal to the sign function $$ \text{sign}(x):=\begin{cases} 1 & : x>0\\ 0 & : x=0\\ -1 & : x<0 \end{cases}. $$

• When $x=0$, the limit is clear since $f(0)=0$ is a fixed point.
• When $x>0$, I have the bound $0<F_n(x)<f(x)^{-2^n}\to 1$.

I guess the way to go about showing that the (pointwise) limit holds is by establishing a lower limit for $x>0$ (and arguing similarly for $x<0$). However, I can't manage to figure what that lower limit should look like...

Answer 1

You can find the $n^{th}$ iterate of $f$ explicitly.

$$f(x)=\dfrac{\sqrt2x}{\sqrt{x^2+1}},$$

$$f(f(x))=\frac{\sqrt2\dfrac{\sqrt2x}{\sqrt{x^2+1}}}{\sqrt{\dfrac{2x^2}{x^2+1}+1}}=\frac{2x}{\sqrt{3x^2+1}},$$

$$f(f(f(x)))=\frac{\sqrt2\dfrac{2x}{\sqrt{x^2+1}}}{\sqrt{\dfrac{4x^2}{3x^2+1}+1}}=\frac{2\sqrt2x}{\sqrt{7x^2+1}}$$

and more generally,

$$f^{(n)}(x)=\frac{2^{n/2}x}{\sqrt{(2^n-1)x^2+1}}=\frac x{\sqrt{(1-2^{-n})x^2+2^{-n/2}}}.$$

Then as $n$ tends to infinity,

$$f^{(\infty)}(x)=\frac x{|x|}.$$

Answer 2

You may argue as follows:

• $f(x) \leq 1$ for any $x \in (0, \infty)$;

• $f(x) \geq x$ for any $x\in(0, 1]$, with equality if and only if $x = 1$.

Therefore for fixed $x> 0$, the sequence $(F_n(x))_{n\geq 1}$ is increasing, hence the limit $\lim_{n\rightarrow \infty} F_n(x)$ exists and must be equal to $1$.