What is the the $n$ times iteration of $f(x)=\frac{x}{\sqrt{1+x^2}}$?

We were asked to determine the composition $f \circ f \circ f \circ…\circ f $, $n$ times, where $f(x)=\dfrac{x}{\sqrt{1+x^2}}$.
Anyone has an idea?

Solutions Collecting From Web of "What is the the $n$ times iteration of $f(x)=\frac{x}{\sqrt{1+x^2}}$?"

Well, try to find some pattern:

$$\begin{align*}f^2(x)=f(f(x))&=f\left(\frac x{\sqrt{1+x^2}}\right)=\frac{\frac x{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}=\frac x{\sqrt{1+2x^2}}\\
f^3(x)=f(f^2(x))&=f\left(\frac x{\sqrt{1+2x^2}}\right)=\frac{\frac x{\sqrt{1+2x^2}}}{\sqrt{1+\frac{x^2}{1+2x^2}}}=\frac x{\sqrt{1+3x^2}}\ldots\ldots\end{align*}$$

Now a little induction could help here…

Let’s denote $f^n = f \circ … \circ f$ with $n$ times the function $f$, and suppose (which is obviously true for $n=1$), that:
$$f^n = \frac{x}{\sqrt{1+n x^2}}$$


$$f^{n+1}=f \circ f^n= \frac{\frac{x}{\sqrt{1+n x^2}}}{\sqrt{1+\left(\frac{x}{\sqrt{1+n x^2}}\right)^2}}$$
$$=\frac{\frac{x}{\sqrt{1+n x^2}}}{\sqrt{\frac{1+(n+1)x^2}{1+nx^2}}}=\frac{x}{\sqrt{1+(n+1)x^2}}$$

And you are done, by induction.

When I saw @DonAntonio’s complete answer, I realized that this is a disguised form of something that I have been aware of for a long time.

Consider the “fractional linear” function $g(x)=x/(x+1)$. If you’ve studied complex variables, you know that this is completely described by the matrix
A=\pmatrix{1&0\cr1&1}\,,\qquad A^n=\pmatrix{1&0\cr n&1}\,.
This means that the $n$-fold iteration of $g$ is $g^{(n)}(x)=x/(nx+1)$. Now “conjugate” by the squaring function, $G(x)=\left[g(x^2)\right]^{1/2}$. This $G$ is the original given function. This pseudo-conjugation operation applies also to the $n$-fold iterates, so that $G^{(n)}(x)=\left[g^{(n)}(x^2)\right]^{1/2}$, and this is exactly the function that @DonAntonio’s inductive process found.

we see that
…… continuing this way