<s>[INST] Find $f(f(f(f(f(f(\cdots f(x)))))))$ $2018$ times 
I was given a problem to calculate $f(f(f(\dots f(x))))$ $2018$ times if $f(x)=\frac{x}{3} + 3$ and $x=4$. 

Is it even possible to be done without a computer?
 [/INST] Suppose there exists an $x$ such that $f(x) = x$
$f(x) = \frac {x}{3} + 3 = x\\
x = \frac 92$
It would be a reasonable guess that we get something very close to $\frac {9}{2}$
$f(4) = 4 +\frac 13\\
f(4+ \frac 13) = 4 + \frac 13 + \frac 19\\
f(4+ \frac 13 + \frac 19) = 4 + \frac 13 + \frac 19 + \frac 1{27}$
Now we are on to something.
$f(4+ \sum_\limits{i=1}^{n} (\frac {1}{3})^i) = \frac {4}{3} + \frac{1}3\sum_\limits{i=1}^{n} (\frac {1}{3})^i + 3 =
4 + \sum_\limits{i=1}^{n+1}(\frac{1}3)^i$
$4 +\sum_\limits{i=1}^{n} \frac {1}{3}^n = 4 + \frac {(1-(\frac 13)^n)}{2}\\
4 +\sum_\limits{i=1}^{2018} \frac {1}{3}^n = 4 + \frac {(1-(\frac 13)^{2018})}{2}\\
4\frac 12 - \frac 12(\frac 13)^{2018}$
</s>