Ah now I see what you meant by "from -1", namely that 0 is mapped to -1 by the superfunction. As indicated above we just write \( 0\mapsto -1 \) where unfortunately the \( \mapsto \) should be a \mapsto having a vertical bar at the left, which is however not shown in this TeX-derivate.
Note that regular superfunctions are determined up to translations along the x-Axis in this case \( F(x)=\cos(\pi 2^x) = \cos(2^{\log_2(\pi)+x}) \).
Instead of \( \log_2(\pi) \) we can put an arbitrary different constant, via which we can choose initial conditions different from \( 0\mapsto -1 \).
For example \( 0\mapsto 1 \) would be reached by \( \cos(2\pi 2^x) \) and by \( \cos(4\pi 2^x) \), etc.
Regarding the iteration of quadratic polynomials there is also a very interesting article about the impossibility to do so in the whole complex plane.
[1] R. E. Rice, B. Schweizer, and A. Sklar. When is f(f(z))=az^2 + bz + c? Am.Math.Mon., 87 : 252 −−263,1980.
Not only the impossibility to have analytic, or continous halfiterates; no, there are no halfiterates (functions on \( \mathbb{C} \)) at all!
Note that regular superfunctions are determined up to translations along the x-Axis in this case \( F(x)=\cos(\pi 2^x) = \cos(2^{\log_2(\pi)+x}) \).
Instead of \( \log_2(\pi) \) we can put an arbitrary different constant, via which we can choose initial conditions different from \( 0\mapsto -1 \).
For example \( 0\mapsto 1 \) would be reached by \( \cos(2\pi 2^x) \) and by \( \cos(4\pi 2^x) \), etc.
Regarding the iteration of quadratic polynomials there is also a very interesting article about the impossibility to do so in the whole complex plane.
[1] R. E. Rice, B. Schweizer, and A. Sklar. When is f(f(z))=az^2 + bz + c? Am.Math.Mon., 87 : 252 −−263,1980.
Not only the impossibility to have analytic, or continous halfiterates; no, there are no halfiterates (functions on \( \mathbb{C} \)) at all!