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And So on - and on -and On

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

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{eqnarray} F_0(x) &=& 1/(1 - x) \\ F_n(x) &=& F_0 (F_{n-1}(x)) \end{eqnarray}
This implies that: $$F_n(x) = 1 / (1 - F_{n-1}(x))$$ and can be extended to: $$F_n(x) = 1 / (1 - (1 / (1 - F_{n-2}(x))))$$
Creating functions of $x$ when $n = 1, 2$ and $3$ gives:
\begin{eqnarray} 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{eqnarray} Therefore the function of x will repeat itself every three times. \begin{eqnarray} F_0(x) &=& F_3(x)\\ F_1(x) &=& F_4(x) \end{eqnarray} 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: $$F_{2}2000 = 2000$$