TITLE: Asymptotic approximation of the arctangent?
QUESTION [5 upvotes]: That is, I am looking for an algebraic function $f(x)$ that approximates $\arctan x$ for large values of $x$.
The approximation could be reasonably modest -- perhaps something like 
$$\tan (f(x)) = \frac{\pi}{4} + O\left(\frac{1}{x^2}\right).$$

REPLY [4 votes]: This approximation converges much faster:
$$
\arctan(x)=2\sum_{n=1}^{\infty}{\frac{1}{2n-1}\frac{{{a}_{n}}\left(x\right)}{a_{n}^{2}\left(x\right)+b_{n}^{2}\left(x\right)}},
$$
where
$$
\begin{align}
& a_1(x)=2/x,\\ 
& b_1(x)=1,\\ 
& a_n(x)=a_{n-1}(x)\,\left(1-4/x^2\right)+4b_{n-1}(x)/x,\\ 
& b_n(x)=b_{n-1}(x)\,\left(1-4/x^2\right)-4a_{n-1}(x)/x.
\end{align}
$$
Therefore, by taking the first term only, the following asymptotic approximation can be used:
$$
\arctan\left(x\right)
={\pi \over 2}\,{\rm sgn}\left(x\right) - \arctan\left(1 \over x\right)
\approx \frac{\pi}{2}{\rm sgn}(x)-\frac{4 x}{4 x^2+1}\,,\qquad
\left\vert\,x\,\right\vert \gg 1.
$$