Show interval that maps $x \rightarrow \mathbb{R}$ by $ \tan(x)$ converges to zero. I have difficulty to word my problem nicely but here is my best effort:
Let $f(x) = \tan{x}$. Now I am interested in iterations of $f(x)$. Specifically, I would like to show that the interval around zero which is mapped to the entire real line converges to zero. In other words: the piece of $\tan{x}$ that maps to the entire real line around 0 is, after one iteration, the interval $( \pi/2, \pi/2)$. Now if we take $f^{2}(x)$ this interval shrinks. The piece becomes smaller.
Now is there a way to prove we can make this interval arbitrarily small? Intuitively I would argue there exists some integer $N$ such that $f^{N}(x)(U) \rightarrow \mathbb{R}$ with $U$ an interval arbitrarily small around zero.
 A: Here is an outline of a proof.
Because the tan function is increasing and continuous, it suffices to show that for a given interval $(-\delta, \delta)$, that there exists $N$ such that $\tan^N(-\delta)\leq-\pi/2$ and $\tan^N(\delta)\geq \pi/2$. (Also because it is increasing/continuous we can assume symmetric intervals, since the result for $(-\delta_1,\delta_2)$ would be implied by the result for $(-\delta,\delta)$ with $\delta=\min(\delta_1,\delta_2)$.)
To see this, observe that $|\tan(x)| > x + \frac{x^3}{3}$ for all $x\in(-\pi/2, \pi/2)$, which can be seen by taking a Taylor series for $\tan$ about 0 and noticing that all even terms are 0 and all odd terms are positive. I've just truncated the series after the 3rd term. So it will suffice to show the result for $f(x) = x + x^3/3$.
Then let $N = \lceil\frac{\pi}{2}\frac{3}{\delta^3}\rceil$. Starting from $\delta$, each iteration will increase the value by at least $\delta/3 \geq (\pi/2)/N$, so after $N$ iterations you'll be past $\pi/2$. Similarly starting from $-\delta$ within $N$ iterations you'll be past $-\pi/2$.
Completing the proof is mostly a matter of adding fiddly algebra to pretty-much every step.
A: Let's define $g(x)=\arctan x$. We'll make repeated use of the fact that $g(x)\leq x$ if $x\geq  0$.
Let $U=[-a, a]$ with $0 < a$, be an interval centered around $0$.
What you're asking to prove is equivalent to proving that $g^{(n)}(U)$ decreases to $\{0\}$ as $n\rightarrow +\infty$.
Clearly, $g(U)=[-\arctan a, \arctan a]$.
In fact $$g^{(n+1)}(U)=[-g^{(n+1)}(a), g^{(n+1)}(a)]\subset [-g^{(n)}(a), g^{(n)}(a)]= g^{(n)}(U)$$
To prove what you want, it suffices to show that the sequence $u_n=g^{(n)}(a)$ converges to $0$.
Clearly, $0\leq  u_{n+1}\leq u_n$, so this proves that the intervals shrink at all iterations. And since the sequence $\{u_n\}$ decreases while staying non-negative, it converges to a limit $l$. That limit verifies $l=g(l)=\arctan l$, and thus it must be that $l=0$. This proves that the intervals shrink to the $\{0\}$ set.
You can go further and estimate the speed at which they shrink by writing $$u_{n+1}=\arctan u_n = u_n-\frac {u_n^3}3+\mathcal O(u_n^5)$$
thus
$$\frac 1 {u_{n+1}^2}=\frac 1 {u_n^2}+\frac 2 3 +\mathcal O(u_n^2)$$
and $$u_n\simeq\sqrt{\frac 3 {2n}}$$
