Let $z$ be complex and $x$ real. Define $f(z,0) = f(z)$ where $f(z)$ is an entire function. Define $f(z,x)$ as the $x$ th superfunction of $f(z)$. We know that $f(z,x-1) = f(f^{-1}(z,x)+1)$ where $^{-1}$ means inverse with respect to $z$. How do we find $f(z,x)$ for non-integer $x$ when given $f(z) , f(z,-1)$ and $f(z,1)$ and continu with $x$ ?
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Usefull link perhaps : http://en.wikipedia.org/wiki/Superfunction