Let f(x) = (x-1)^2 + 1.

If we want the half-iterate of f(x) for x>1 we can use

S(S^[-1](x)+1/2) where S is the superfunction of f(x).

For x>2 this superfunction is F(x) = 2^(2^x) + 1.

Hence I call that the fermat superfunction.

This fermat superfunction F(x) is entire so there are no other branches.

F^[-1] does have branches though but they are easy to understand.

Notice F(x) = 1 has no complex solution.

( 1 is the other fixpoint of f(x) )

But what do we do for the other nonparabolic fixpoint of f(x) ? What superfunction belongs there ?

Is this related to branches of F^[-1] ?

regards

tommy1729

If we want the half-iterate of f(x) for x>1 we can use

S(S^[-1](x)+1/2) where S is the superfunction of f(x).

For x>2 this superfunction is F(x) = 2^(2^x) + 1.

Hence I call that the fermat superfunction.

This fermat superfunction F(x) is entire so there are no other branches.

F^[-1] does have branches though but they are easy to understand.

Notice F(x) = 1 has no complex solution.

( 1 is the other fixpoint of f(x) )

But what do we do for the other nonparabolic fixpoint of f(x) ? What superfunction belongs there ?

Is this related to branches of F^[-1] ?

regards

tommy1729