Proof there is no fixpoint

Question

I want to proof that a certain function (in this case tan(x) on (0, 1/4)) doesn't have a fixpoint. Is there a general approach to try that? For example an "iff" fixpoint theorem?

Answer 1

For $f$ a real function, proving that $f$ has no fixed point, means that $f(x) \neq x$ for all the definition set $D$ of $f$. For a continuous function, this means that $g(x)=f(x)-x$ keeps the same sign on $D$. This is not true if $f$ is not continuous.

So you can study the sign of $g$ to look at fixed point(s) of $f$.

Coming back to your example, you can prove that $\tan x - x >0$ on $(0,\frac{1}{4})$ to prove that $\tan$ has no fixed point on this interval.