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)#?

Question

1598 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-22T19:27:04

#f_2016(2016)=-1/2015#

#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#