Iteration of $x_{n+1} = \tan (e^{x_n})$ Consider the following recursive formula:
$$x_{n+1} = \tan (e^{x_n})$$
I have tried iterating this (using a calculator) and it seems to approach the cycle
$$1.5574\ \ -29.3038 \ \ 1.8773 \cdot 10^{-13}$$
Does this iteration actually approach that cycle or is this due to rounding errors?
 A: It's an actual cycle! Here's a high precision computation in Mathematica to verify:
f[x_] = Tan[Exp[x]];
F[x_] = f[f[f[x]]];
{x0} = x /. NSolve[F[x] == x && 0 < x < 1/100, x,
  WorkingPrecision -> 50]

F'[x0]

(* Out: 
  {1.8773044356719480057597965448120393463423922987617*10^-13}
  2.6241404453009944988075920570460257174401823773723234978026*10^-9
*)

Note that the computation of the derivative shows that $x_0$ is attractive for $F$, as we might hope.
If you don't have access to Mathematica, you can perform a similar computation on the Sage Cell Server.
A: This is not a complete answer, but I would like to bring some supplementary light on the issue, in particular, how to obtain the fixed points of the given recurrence, which are unreachable by iteration. Indeed, all fixed point are unstable, as observed by @kingW3 : starting from one of them, recurrence $x_{n+1}$ falls very rapidly always into the upsaid cycle.
Let us set $$\phi:=\tan \circ \exp.$$
The curve of $\phi$ (see below) cannot be considered as very simple, with intervals of continuity $I_k=(a+\ln(2k-1),a+\ln(2k+1))$ (where $a=\ln(\pi/2)$).  In particular, each iteration generally jumps from an interval $I_k$ to another, making any analysis difficult.
This notable instability comes from the fact that, for all $x>0$
$$ \phi'(x)=\dfrac{e^x}{\cos^2(e^x)} > e^x>1,$$
preventing the usual fixed point method to converge.  
Nevertheless, $\phi$ possesses an infinity of fixed points whose values are the abscissas of intersection points of the curve of $\phi$ with the straight line with equation $y=x$, the first one being:
$$\tag{1}f_1=1.409879435228055...$$
How to find them ? By reversing the issue using the reciprocal function: $$\psi:=\phi^{-1}=\ln \circ \ \text{atan},$$ i.e., by considering the sequence :
$$\tag{2}y_{n+1}=\ln(\text{atan}(y_n))$$
which has the same fixed points as those of $\phi$. 
No, not exactly, one has to consider that the reciprocal of $\tan$ is not the single function $\text{atan}$ but the family of functions $\text{atan} + k\pi$ (in the sense of a "multivalued" function, as one considers them in complex function theory):
$$\tag{3}y_{n+1}=\ln(\text{atan}(y_n+K\pi))$$
for a certain fixed value of integer $K$.
Iteration (3), with a suitable initial value $y_0$ gives fixed points like (1). 
Taking for example :


*

*$K=2$, gives fixed point (see figure) $f_2=2.000177743332531...$

*$K=3$, provides fixed point $f_3=2.3603667749663...$ etc.
The convergence of sequence (3) is usually very rapid due to the fact that the derivative of $\psi$ is very small almost everywhere.

Curve of $\phi(x)=tan(exp(x))$
