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\newcommand{\R}{I\!\!R}
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\renewcommand{\P}{\frac{\pi}{2}}
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\begin{document}

Given $f: \R^+ \rightarrow \R$ and $n \in \N$ with $n \geqslant 2$

Define $g: \R \rightarrow \R$ by

$g(x) = \left\{
\begin{array}{ll@{}c@{}l@{}c@{}l}
\arctan(x) - \pi
        & \mbox{ if } x \in [ & 0 & , & \infty & )      \\
x - \P
        & \mbox{ if } x \in [ & -(n\.-1) \P & , & 0 & ) \\
f(\tan(x + n \P))
        & \mbox{ if } x \in [ & -n \P & , & -(n\.-1) \P & )     \\
\end{array}
\right.$

Then

$\begin{array}{@{}lccl@{\hspace*{10mm}}cl@{}c@{}l@{}c@{}l}
& x &&& \in & [ & 0 & , & \infty & )    \\
%\Rightarrow &&& \arctan(x) & \in & [ & 0 & , & \P & )  \\
\Rightarrow & g(x) & = & \arctan(x) - \pi
        & \in & [ & - \pi & , & - \P & )        \\
\Rightarrow & g(g(x)) & = & \arctan(x) - 3 \P
        & \in & [ & - 3 \P & , & - \pi & )      \\
&& \vdots && \vdots     \\
\Rightarrow & g^{n\.-2}(x) & = & \arctan(x) - (n\.-1) \P
        & \in & [ & - (n\.-1) \P & , & - (n\.-2) \P & ) \\
\Rightarrow & g^{n\.-1}(x) & = & \arctan(x) - n \P
        & \in & [ & - n \P & , & - (n\.-1) \P & )       \\
\end{array}$

Hence

$\begin{array}{@{}lll}
& g^n(x)        \\
= & f(\tan(g^{n\.-1}(x) + n \P))        \\
= & f(\tan(\arctan(x) - n \P + n \P))   \\
= & f(\tan(\arctan(x))  \\
= & f(x)        \\
\end{array}$

That is, $g$ is an $n$th functional root of $f$.

\end{document}