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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

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    Notice if the lim f(z,-oo) exists it is f(z) = z + 1 since this is the fixpoint of f(z,x-1) = f(f^-1(z,x)+1). Notice -oo is a solution of x-1 = x in a way. I wonder if it is possible to solve for x such that f(z,x) = z ? Dont know if it helps ...2012-09-09

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