For which of the following functions the $\sum_{x\in S(f)}\frac{1}{x}$ converges? For real valued function $f$ define 
$$S(f)=\{x:x>0,f(x)=x\}$$
For which of the following functions the $\sum_{x\in S(f)}\frac{1}{x}$ converges?
$\tan x,\tan^2x,\tan{\sqrt{x}},\sqrt{\tan x},\tan 2x$
I do not have any idea to solve this problem,I think all these have infinitely many fixed points.I need help.
Thanks.

For each of the given functions, other than $\tan \sqrt{x}$, there is at least one fixed point between $(k-1)\pi$ and $k\pi$ for all positive integer $k$.  So the partial sum for any range up to $n\pi$ is greater than $\sum_{k=1}^n \frac{1}{k\pi}$, which we know diverges, so the sum cannot converge. 
For the case of $f(x) = \tan \sqrt{x}$, there is precisely one 
fixed point between $k^2\pi^2$ and $(k+1)^2\pi^2$ for all non-negative integer $k$.
So the partial sum $S_n$ for any range up to $(n+1)^2\pi^2$ is less than 
$$
\frac{1}{x_0}  + \sum_{k=1}^n \frac{1}{k^2\pi^2}
$$
where $x_0$ is the fizxed point lying in $(0,\pi/2)$.  Those sums approach $x_0 + \frac{1}{6}$ as $n$ goes to infinity. 
Since the sequence of $S_n$ is monotonic increasing and bounded  (by  $x_0 + \frac{1}{6}$, in must converge.  So only the third of the five given functions gives a convergent sum.
