Estimates and Taylor expansion for $|\tan(x)| $ What is the Taylor expansion for $|\tan(x)|$? In particular, does one of the following estimates hold in $[-\pi,\pi]$?
$$|\tan(x)| \le |x| \text{ or } |\tan(x)| \le \frac{1}{|x|}.$$
 A: The function $x\mapsto\left|\tan x\right|$ does not have a Taylor expansion centered at $x=0$ because the function is not differentiable at that point.
You could find a Taylor expantion centered at, for example, $x=0.1$, but in a neighborhood of that point you have $\tan x>0,$ so you can drop the absolute value sign and the expansion is just that of $x\mapsto\tan x.$ That expansion would converge to $\tan x$ in an interval that has $0$ in its interior, and where $x<0$ and so $\tan x<0$ it would still converge to $\tan x$ and not to $\left|\tan x\right|.$
Since $x\mapsto\left|\tan x\right|$ has a pole (or vertical asymptote if you want to call it that) at each of $x=\pi/2,$ the radius of convergence of the series for $x\mapsto\tan x$ cannot exceed $\pi/2.$ (In fact, it is exactly $\pi/2.)$ So there can be no such series that converges in the interval $(-\pi,\pi).$
In fact, you have $\left|\tan x\right| > |x|$ if $0<x<\pi/2.$ One way to see that is that the slope at $x=0$ is $1$ and the tangent function is concave upward everywhere in the interval from $0$ to $\pi/2.$ Similarly, $\left|\tan x\right| > |x|$ if $-\pi/2<x<0,$ by the symmetry of the tangent function (it's an odd function: $\tan(-x) = -\tan x).$
A: The radius of convergence of the power series expansion of $\tan x$ around $x=0$ is $\pi/2$, and $\tan x$ is an odd function on $(-\pi/2,\pi/2)$.
So the Taylor series for $\tan x$ converges on $[0,\pi/2)$ and gives the same values as $|\tan x|$ there.  The coefficients of this Taylor series are not as easy to express as those for $\sin x$ and $\cos x$, so maybe a link to the Wikipedia pages on series definitions of trigonometric functions is in order.  The coefficients of the tangent power series are related to the Bernoulli numbers.
In any case $|\tan x|$ tends to infinity at $\pm \pi/2$, so bounds of the kind you hoped for are impossible.
