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Dec 16, 2023 at 20:37 comment added Pietro Majer As to just “find $\phi$ solving $\phi(x)=f(x+\phi(x))$ given $f$, if $f$ is analytic you can use the analytic implicit function theorem to solve locally $\phi(x)=f(x+\phi(x))$, in a nbd of a point $x_0$, with $\phi(x_0)=y_0$, provided $x_0=f(x_0+y_0)$ and $f’(x_0+y_0)\neq0$; so that the series solution has indeed positive radius of convergence. Similarly for $\psi$. But if the problem is finding $\phi,\psi,f ,F$ also with $\phi’=\psi$ and $F’=f$, then I’m not sure how one can prove that the formal series solutions do converge…
Dec 16, 2023 at 20:05 comment added Sam Hopkins Possibly Lagrange inversion can be used to connect your answer to the answer of Fred Hucht
Dec 16, 2023 at 17:28 comment added Pietro Majer Yes, exactly $\phantom{} $, and $(I+\varphi)^{-1}$ denotes the compositional inverse of $I+\varphi$
Dec 16, 2023 at 17:26 history edited Pietro Majer CC BY-SA 4.0
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Dec 16, 2023 at 17:26 comment added Sam Hopkins Are you using $I$ for the identity map $I(x) = x$ here?
Dec 16, 2023 at 17:23 history answered Pietro Majer CC BY-SA 4.0