Published online by Cambridge University Press: 20 November 2018
We prove a result concerning power series   $f\left( Z \right)\,\in \,\mathbb{C}\left[\!\left[ Z \right]\!\right]$  satisfying a functional equation ofthe form?
 $f\left( Z \right)\,\in \,\mathbb{C}\left[\!\left[ Z \right]\!\right]$  satisfying a functional equation ofthe form?
 1 $$f\left( {{Z}^{d}} \right)\,=\,\sum\limits_{k=1}^{n}{\frac{{{A}_{k}}\left( Z \right)}{{{B}_{k}}\left( Z \right)}\,f{{\left( Z \right)}^{k}}},$$
 $$f\left( {{Z}^{d}} \right)\,=\,\sum\limits_{k=1}^{n}{\frac{{{A}_{k}}\left( Z \right)}{{{B}_{k}}\left( Z \right)}\,f{{\left( Z \right)}^{k}}},$$
where   ${{A}_{k}}\left( Z \right),\,{{B}_{k}}\left( Z \right)\,\in \,\mathbb{C}\left[ Z \right]$ . In particular, we show that if
 ${{A}_{k}}\left( Z \right),\,{{B}_{k}}\left( Z \right)\,\in \,\mathbb{C}\left[ Z \right]$ . In particular, we show that if   $f\left( Z \right)$  satisfies a minimal functional equation of the above form with
 $f\left( Z \right)$  satisfies a minimal functional equation of the above form with   $n\,\ge \,2$ , then
 $n\,\ge \,2$ , then   $f\left( Z \right)$  is necessarily transcendental. Towards a more complete classification, the case
 $f\left( Z \right)$  is necessarily transcendental. Towards a more complete classification, the case   $n=\,1$  is also considered.
 $n=\,1$  is also considered.