From these generating functions is it easy to see that parabolic iteration does not work for \( f_2 = 0 \), which I believe has already been proven by someone, somewhere. What is interesting is that there are actually 2 reasons for this. The first reason is that there are many \( f_2 \) in the denominator, which cannot be zero, and the second reason is that if \( f_2 = 0 \), then \( z=1 \) which means t plays no part in the equations at all.
Also, I wonder if studying the special case \( f^{\circ 1/x}(x) \) would yield more insights, as this would imply that \( z = 1 - f_2 \), so there wouldn't be an x in the denominator. This would make finding more diagonals easier.
Andrew Robbins
PS. I think it was either Bennet or Jabotinsky that showed \( f_2 = 0 \) doesn't work.
Also, I wonder if studying the special case \( f^{\circ 1/x}(x) \) would yield more insights, as this would imply that \( z = 1 - f_2 \), so there wouldn't be an x in the denominator. This would make finding more diagonals easier.
Andrew Robbins
PS. I think it was either Bennet or Jabotinsky that showed \( f_2 = 0 \) doesn't work.