Bummer!

[bo198214]

According to Markus Müller, the superfunction of \( 2x^2 - 1 \) from (-1) is \( \cos(\pi 2^x) \).

Is that right? can I say "from"?

If you mean the fixed point then I would say "at". However -1 is not a fixed point of \( f(x)=2x^2-1 \) but 1 is, so I dont know exactly what you mean.

Yes \( F(x)=\cos(\pi 2^x) \) is a \( 0\mapsto -1 \) superfunction of \( f(x)=2x^2-1 \).

\( F(0)=\cos(\pi)=-1 \)
\( F(x+1)=\cos(2 \pi 2^x)= 2\cos(\pi 2^x)^2-1 = 2F(x)^2-1 = f(F(x)) \)

Is it regular?
The regular iteration is characterized by \( x\mapsto f^t(x) \) being differentiable (but at least asymptotically differentiable) at the fixed point \( \lambda \). (This also implies that \( (f^t)'(\lambda) = f'(\lambda)^t \).)
The iteration is given in terms of the superfunction \( F \) by:
\( f^t(x)=F(t+F^{-1}(x)) \)

\( 2x^2-1 \) has two fixed points 1 and \( -\frac{1}{2} \):
\( \cos(\pi 2^x) \) is not invertible at 1, but it is invertable at \( \lambda=-\frac{1}{2} \).
So the t-th iterate of \( f \) is differentiable at \( \lambda \) and so \( F \) is the regular superfunction at \( -\frac{1}{2} \).