Reciprocal Taylor series e.g. $\sin^{-1}x $ I am completely stuck on how you might find the Taylor series expansions of functions such as $\sin^{-1}x $  and $\tan^{-1}x $ about $x=0$. They are undefined at $x=0$ and their derivatives do not exist at $x=0$ so I definitely can't use the standard Taylor series formula. I have thought about using the known expansions of $\sin(x)$ and $\tan(x)$ and treating them with a power of $-1$ in some way, but I don't know how I would do this as I only know reciprocal expansions for 2 terms (binomial expansion) and not for an infinite series!
 A: Hint
You can define a generalized expansion for $x$ near but $\neq 0$ by
$$\frac{1}{\sin(x)}=\frac{1}{x}\frac{1}{1-\frac{x^2}{6}+\frac{x^4}{5!}-...}$$
$$=\frac{1}{x}\left(1+\frac{x^2}{6}+\frac{7x^4}{360}+...\right).$$
A: If a function is not defined at a point, it does not have a Taylor series expansion at that point.
You can only expand $f$ around a point $a$ if $f$ is defined and infinitely differentiable on some small interval $(a-\epsilon, a+\epsilon)$.
A: It is known that
$$
\frac1{\sin x}=\csc x\quad\text{and}\quad \frac1{\tan x}=\cot x.
$$
On page 42 in the handbook [1] listed below, it is collected that
$$
\cot x=\frac1x-\sum_{k=1}^\infty\frac{2^{2k}|B_{2k}|}{(2k)!}x^{2k-1}
$$
and
$$
\csc x=\frac1x+\sum_{k=1}^\infty\frac{2(2^{2k-1}-1)|B_{2k}|}{(2k)!}x^{2k-1}
$$
for $x^2<\pi^2$, where $B_{2k}$ denotes the Bernoulli number generated by
\begin{equation*}
\frac{z}{\textrm{e}^z-1}=\sum_{n=0}^\infty B_n\frac{z^n}{n!}=1-\frac{z}2+\sum_{n=1}^\infty B_{2n}\frac{z^{2n}}{(2n)!}, \quad \vert z\vert<2\pi.
\end{equation*}
Reference

*

*I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Translated from the Russian, Translation edited and with a preface by Daniel Zwillinger and Victor Moll, Eighth edition, Revised from the seventh edition, Elsevier/Academic Press, Amsterdam, 2015; available online at https://doi.org/10.1016/B978-0-12-384933-5.00013-8.

