Question

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)`?

Answer 1

`f_2016(2016)=-1/2015`

Explanation:

`f_2(x) = 1/(1-f_1(x))`

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

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

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

`=1-1/f_0(x)`

`=1-1/(1/(1-x))`

`=1-(1-x)`

`=x`

Note, then, that
`f_3(x) = f_0(x)`
`f_4(x) = f_1(x)`
`f_5(x) = f_2(x) = x`
`...`

In general:
`f_k(x) = {(f_0(x) if k -= 0" (mod 3)"), (f_1(x) if k -= 1" (mod 3)"), (x if k -= 2" (mod 3)"):}`

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

`f_2016(2016) = f_0(2016)`

`=1/(1-2016)`

`=-1/2015`