(06/10/2022, 11:35 PM)Catullus Wrote: On the topic of function composition. Is the successor function the only function f(x) such that f(f(x)) = f(x)+1?
This depends on how you want to phrase this question. I believe you could probably find solutions to this equation in tropical algebra, and crazy areas like that.
In Complex Function theory, the answer is yes, but with a caveat.
If you assume that \(x \in \widehat{\mathbb{C}}\), then \(f = \infty\) is also a solution.
But, there's much much much more from this equation.
\[
f(f(x)) = f(x) + 1\\
\]
Since \(\infty\) is a fixed point (assuming \(f(\infty) = \infty\)), then... let's do leibniz variable change: \(F(x) = 1/f(1/x)\). This function satisfies \(F(0) = 0\), assuming that \(f\) is meromorphic at infinity. Now take \(F'(0) = C\). We are asking that:
\[
F(F(x)) = F(x) + 1\\
\]
In a neighborhood of \(x \approx 0\). This is kind of like a golden ratio/fibonacci equation, which I've never seen before. I think It's unique though.
\[
\begin{align*}
F'(0)\cdot F'(0) = F'(0) = 1\\
F'(F(x))\cdot F'(x) = F'(x)\\
\end{align*}
\]
Therefore \(F' =1\). Therefore \(F(x) = x+1\). Therefore there are two solutions in complex analysis.
There's the constant solution \(\infty\), and the successor function \(x \mapsto x+1\).
So you are right, but there's an exception in complex analysis. The "successorship" hyperoperator could be an essential singularity... it looks like how \(e^{1/z}\) looks like \(0\) and \(\infty\) near \(z\approx 0\). We can have a hyperoperator successorship that looks like \(\infty\) and \(x \mapsto x +1\). So, SINGULAR solutions to the above question can exist.