# If f_0(x)=1/(1-x) and f_k(x)=f_0(f_(k-1)(x)) what is the value of f_(2016)(2016)?

Oct 22, 2016

${f}_{2016} \left(2016\right) = - \frac{1}{2015}$

#### Explanation:

${f}_{2} \left(x\right) = \frac{1}{1 - {f}_{1} \left(x\right)}$

=1/(1-1/(1-f_0(x))

$= \frac{1 - {f}_{0} \left(x\right)}{1 - {f}_{0} \left(x\right) - 1}$

$= \frac{1 - {f}_{0} \left(x\right)}{f} _ 0 \left(x\right)$

$= 1 - \frac{1}{f} _ 0 \left(x\right)$

$= 1 - \frac{1}{\frac{1}{1 - x}}$

$= 1 - \left(1 - x\right)$

$= x$

Note, then, that
${f}_{3} \left(x\right) = {f}_{0} \left(x\right)$
${f}_{4} \left(x\right) = {f}_{1} \left(x\right)$
${f}_{5} \left(x\right) = {f}_{2} \left(x\right) = x$
$\ldots$

In general:
${f}_{k} \left(x\right) = \left\{\begin{matrix}{f}_{0} \left(x\right) \mathmr{if} k \equiv 0 \text{ (mod 3)" \\ f_1(x) if k -= 1" (mod 3)" \\ x if k -= 2" (mod 3)}\end{matrix}\right.$

As $2016$ is divisible by $3$, we have

${f}_{2016} \left(2016\right) = {f}_{0} \left(2016\right)$

$= \frac{1}{1 - 2016}$

$= - \frac{1}{2015}$