Originally Posted by

**ymar** Let $\displaystyle f_n\colon[0,\pi/2]\longrightarrow[0,1]$ for $\displaystyle n=0,1,...$ and

$\displaystyle \begin{cases}f_0=\sin\\f_{n+1}=\sin\circ f_n\end{cases}$

I would like to prove that for $\displaystyle n=1,2,...$ the third derivative $\displaystyle f_n^{(3)}$ has a unique zero in $\displaystyle (0,\pi/2)$. I thought calculating the third derivative and solving an equation, even if doable, doesn't seem very useful. I already "know" it's true after plotting many graphs, so all I would like to know is an elegant proof, not a proof that no sane person would like to read. (I have almost finished calculating the third derivative and if someone thinks it's a good idea, I can post the calculations. I can't guarantee they're correct though.)