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 $f(x)=\tan x$, there is a fixed point of $f$ in each interval $((n-\frac12)\pi,(n+\frac12)\pi)$, hence the series in question is essentially the harmonic series, diverging.
For $f(x)=\tan^2 x$ and $f(x)=\tan 2x$, the same argument applies.
$f(x)=\sqrt{\tan x}$ makes no difference compared to $\tan x$ (except that the irrelevant negative parts become undefined).
For $f(x)=\tan\sqrt x$, however, we find one fixed points in $((n-\frac12)^2\pi^2,(n+\frac12)^2\pi^2)$ instead, so the growth is quadratic and the series converges, just as $\sum \frac1{n^2}$ does.
