And so on - and on -and on
Can you find the value of this function involving algebraic
fractions for x=2000?
Age
16 to 18
Challenge level
Let $$ f_{0}(x)= \frac{1}{1-x}$$ and $$f_{n}(x)=f_{0}(f_{n-1}(x))$$
where $n = 1,2, 3, 4, ...$
Evaluate $f_{2000}(2000)$
Evaluate $f_{2000}(2000)$
The solution needs you to be systematic.
Start with $ f_{0} $, then work out $ f_{1}$, then work out $ f_{2}$
\begin{eqnarray}f_{0}(2000)&=& \frac{1}{1-2000}\\ &=& \frac{1}{-1999}\end{eqnarray}
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: $$F_{2}2000 = 2000$$
It is worth emphasising the cyclic nature of this recurrence
relation.
What happens with different starting numbers and why?
What happens with different starting numbers and why?