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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!

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  • $\begingroup$ The Taylor theorem assumes at least differentiability at the point, and infinite differentiability if you are to obtain a series. Without this premise, the function cannot be expanded in a neighborhood of that point. Case in point is the function you have. $\endgroup$ Nov 17, 2016 at 12:33
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    $\begingroup$ You're almost certainly misinterpreting the symbols. I bet by $\sin^{-1}$ and $\tan^{-1}$ they mean $\arcsin$ and $\arctan$. Those have Taylor expansions about $0$. $\endgroup$ Nov 17, 2016 at 12:33
  • $\begingroup$ @DanielFischer You were right! Thank you! The answers given were still really insightful. $\endgroup$
    – Meep
    Nov 17, 2016 at 16:05
  • $\begingroup$ At math.stackexchange.com/a/4331451/945479, it was written that "These formulas can be applied to establish general formulas of the $n$th derivatives for functions of the types $f(\sin x)$ and $f(\cos x)$, such as $\sin^\alpha x$, $\cos^\alpha x$, $\sec^\alpha x$, $\csc^\alpha x$, $e^{\pm\sin x}$, $e^{\pm\cos x}$, $\ln\cos x$, $\ln\sin x$, $\ln\sec x$, $\ln\csc x$, $\sin\sin x$, $\cos\sin x$, $\sin\cos x$, $\cos\cos x$, $\tan x$, and $\cot x$." When taking $\alpha=-1$, we know that $\sin^\alpha x=\frac{1}{\sin x}$, which is the function discussed in this question. $\endgroup$
    – qifeng618
    Dec 14, 2021 at 11:16

3 Answers 3

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Hint

You can define a generalized expansion for $x$ near but $\neq 0$ by

$$\frac{1}{\sin(x)}=\frac{1}{x}\frac{1}{1-\frac{x^2}{6}+\frac{x^4}{5!}-...}$$

$$=\frac{1}{x}\left(1+\frac{x^2}{6}+\frac{7x^4}{360}+...\right).$$

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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 $(a-\epsilon, a+\epsilon)$.

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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

  1. 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.
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