A well presented solution from Richard of The Royal Hospital School reflected those of a number of other solvers including Kevin of Langley Grammar, Jeff from New Zealand and Andrei of Tudor Vianu School. Well done to all of you.

We are given that:

\begin{matrix} F_{0} (x) & = & 1 / (1 - x) \\ F_{n} (x) & = & F_{0} (F_{n - 1} (x)) \end{matrix}

This implies that:

\begin{matrix} F_{n} (x) & = & 1 / (1 - F_{n - 1} (x)) \\ and can be extended to : \\ F_{n} (x) & = & 1 / (1 - (1 / (1 - F_{n - 2} (x)))) \end{matrix}

Creating functions of x when n = 1, 2 and 3 gives:

\begin{matrix} F_{1} (x) & = & 1 / (1 - F_{0} (x)) \\ = & 1 / (1 - (1 / (1 - x))) \\ = & 1 / ((1 - x - 1) / (1 - x)) \\ = & 1 (1 - x) / - x \\ = & (1 - x) / - x \\ F_{2} (x) & = & 1 / (1 - F_{1} (x)) \\ = & 1 / (1 - ((1 - x) / - x)) \\ = & 1 / ((- x - 1 + x) / - x) \\ = & - x / - 1 \\ = & x \\ F_{3} (x) & = & 1 / (1 - F_{2} (x)) \\ = & 1 / (1 - x) \\ = & F_{0} (x) \end{matrix}

Therefore the function of x will repeat itself every three times.

\begin{matrix} F_{0} (x) & = & F_{3} (x) \\ F_{1} (x) & = & F_{4} (x) \end{matrix}

Etc. etc. So, to find

F_{2000} (x)

, we must find the remainder given when 2000 is divided by three.

${Mod}_{3} 2000 = 2$

Thus, $F_{2000} (x) = F_{2} (x)$ , and $F_{2} (x) = x$

Therefore:

$F_{2} 2000 = 2000$