I played around with the equation
$$ f_{n}(x)=\frac{x}{\sqrt{1+nx^2}} $$
which is the $n$th composition of
$$ f(x)=\frac{x}{\sqrt{1+x^2}} $$
Surprising to me, it is a good approximation for $arctan(x)$ when plugging in $n=\frac{1}{2}$.
I tried comparing their Taylor-Series-Expansion but it didn't seem fruitful to me.
Is there maybe a connection to the derivative since
$$arctan'(x)=\frac{1}{1+x^2}$$
If you have any ideas why this is the case I would be glad to hear them.
Thanks in advance!
There is something interesting in the $\frac 1 2$.
Suppose that you want the best fit of the arctangent, say between $0$ and $t$ using such a model. To find the best $a$, consider the infinite norm $$\Phi(a,t)=\int_0^t \Big( \tan ^{-1}(x)-\frac{x}{\sqrt{1+a x^2}} \Big)^2\,dx$$ which is fully explicit (even if not very nice). This is equivalent to a non linear regression based on an infinite number of data points.
Now, for a given value of $t$, solve for $a$ the equation $$\frac {\partial \Phi(a,t)}{\partial a}=0$$
A few numbers $$\left( \begin{array}{cc} t & a_{\text{opt}} \\ 5 & 0.520880 \\ 6 & 0.509581 \\ 7 & 0.500667 \\ 8 & 0.493425 \\ 9 & 0.487405 \\ 10 & 0.482305 \\ \end{array} \right)$$