What is the $n$ times iterate of $f(x)=\frac{x}{\sqrt{1+x^2}}$? We were asked to determine  the composition $f \circ f \circ f \circ \cdots \circ f $,  $n$ times, where $f(x)=\dfrac{x}{\sqrt{1+x^2}}$.
Has anyone an idea?
 A: 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:
$$f^n(x) = \frac{x}{\sqrt{1+n x^2}}$$
Then
$$f^{n+1}(x)=f \circ f^n(x)= \frac{\frac{x}{\sqrt{1+n x^2}}}{\sqrt{1+\left(\frac{x}{\sqrt{1+n x^2}}\right)^2}}$$
$$=\frac{\frac{x}{\sqrt{1+n x^2}}}{\sqrt{\frac{1+(n+1)x^2}{1+nx^2}}}=\frac{x}{\sqrt{1+(n+1)x^2}}$$
And you are done, by induction.
A: 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.
A: 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...
A: $$f(x)=\frac{x}{\sqrt{1+x^2}}$$
we see that
$$f^2(x)=f(f(x))=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+(\frac{x}{\sqrt{1+x^2}})^2}}=\frac{x}{\sqrt{1+2x^2}}$$
...... continuing this way
$$f^n(x)=\frac{x}{\sqrt{1+nx^2}}$$
