Series of tan inverse x for x > 1

Question

The Maclaurin series of the Tan inverse of $x$ works well for $-1 \leq x \leq 1$ but it breaks down afterwards because the curve takes a different. What function defines the curve after $x = 1$? Thanks Does the Taylor expansion govern the curves' transition?

Answer 1

If you want a series that applies for all real $x$, consider $x=\tan\theta$ and apply the half-angle formula

$\tan(\theta/2)=\dfrac{\sin\theta}{\cos\theta+1}=\dfrac{\tan\theta}{1+\sec\theta}.$

For $|\theta|<\pi/2$ we then have $\sec\theta=\sqrt{1+\tan^2\theta}$ and this leads to

$\theta=\tan^{-1}x=2\tan^{-1}\left(\dfrac{x}{1+\sqrt{1+x^2}}\right).$

Thus, given a preliminary calculation of $x/(1+\sqrt{1+x^2})$, we can use the Taylor series with that argument. Since $|x|<|1+\sqrt{1+x^2}|$ for all real $x$, the series will converge accordingly.