Superfunctions with complex iteration indices: Interpretation Superfunctions are a fascinating concept, allowing us to generalize functional iteration to arbitrary real and complex iteration indices. We have
$$
\begin{equation}
\begin{split}
S_f(0) & =\text{identity} \\
S_f(1) & =f \\
S_f(2) & =f\circ f \\
S_f(-1) & =f^{-1}
\end{split}
\end{equation}$$
For integer values of $z$, the definition of $S_f(z)$ matches simple iterations of $f$ or its inverse. Likewise, it is easy enough to form an intuition for fractional real values: $S_f(\frac{1}{2})$ indicates a "half" application of $f$, such that $S_f(\frac{1}{2})\circ S_f(\frac{1}{2})=f$.
What, then, does a complex value of $z$ mean intuitively? Even for the simplest cases it seems incomprehensible what is actually happening except by referencing the definition. What does it mean to iterate a function $i$ times?
 A: Let me respond as someone who does not at all have the total picture. I would say this:
In most cases, it wouldn’t mean anything to speak of a “complex-fold” iteration, because in most cases it wouldn’t be possible to define it. But if your starting function is extremely simple, you may be able to see what “$\,f^{\circ i}\,$” can mean. Here, I’m using the (fairly) common notation $f^{\circ2}$ for $f\circ f$, and $f^{\circ(n+1)}$ for $f\circ f^{\circ n}$.
For a “simple” function of the kind I’m thinking of, take $f(x)=x+1$. Then $f^{\circ n}(x)=x+n$, so certainly $f^{\circ i}(x)=x+i$. Perhaps you might conclude that if $f^{\circ n}$ can be described in nice formulas involving $n$, then you might extend these formulas by replacing your $n$ by $i$ and seeing what happens.
Let me point out in closing that it seems well known that though you can fairly easily write down what the power-series expansion of $\sin^{\circ n}(x)$ turns out to be, what you get for $\sin^{\circ1/2}$ is not convergent anywhere except at zero.
And maybe one truly final remark: If you can find $g=f^{\circ i}$, then it should be that $g^{\circ i} = f^{\circ(-1)}=f^{-1}$.
