There's a Limit

Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?

Not Continued Fractions

Which rational numbers cannot be written in the form x + 1/(y + 1/z) where x, y and z are integers?

Comparing Continued Fractions

Which of these continued fractions is bigger and why?

And So on - and on -and On

Stage: 5 Challenge Level:

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: $$F_{2}2000 = 2000$$