Is it possible to generate a Taylor expansion of $\lvert\sin(x)\rvert$?
I understand that this is not possible around the point $0$, since $\frac{d}{dx}\lvert\sin(x)\rvert$ is undefined at $x = n\pi$. But can we achieve this by generating the series around the points $\frac{n\pi}{2}$ and obtain an expression that is valid for all $0<x<\infty$, except the points $x=n\pi$?
Best Answer
For some neighborhood of any $c \ne k\pi, k \in \Bbb Z$, either $|\sin x| = \sin x$ everywhere in the neighborhood, or $|\sin x| = -\sin x$ in the neighborhood.
All the higher derivatives of $|\sin x|$ at $c$ will therefore be the same as either $\sin x$ or its opposite, and thus the Taylor series at $c$ will either be the Taylor series for $\sin x$ or for $-\sin x$. This Taylor series will indeed converge for all $x$. But it will not converge to $|\sin x|$ everywhere.