# Find radius of convergence for a complicated series for $f'(f(x)) = f(f'(x))$

Several months ago, I answered this question asking for solutions to the functional equation $$f'(f(x)) = f(f'(x))$$ by expanding as a formal Taylor series around some arbitrary fixed point of $$f$$. This gives a formal solution, and if the series is convergent it gives an analytic solution. I'm wondering if there's a way to prove that this series has nonzero radius of convergence? I tried some crude tricks (like inductively proving $$|f^{(n)}(c)| < n! a^n$$ for some $$a$$) but didn't get anywhere. Proving existence of an analytic solution would also suffice, but I haven't found a simpler way to prove that either, except in some special cases. Is there a way to prove non-zero radius of convergence for this series given arbitrary $$c$$ and $$\lambda$$?

• Note that you can see several analytical solutions to the equation in the original question. – Botond May 2 '20 at 16:58
• I'm asking about in general, for arbitrary $c$ and $\lambda$ in my notation from the original question. – Dark Malthorp May 2 '20 at 17:01
• I felt like that is what you wanted, but the sentence "Proving existence of an analytic solution would also suffice" sounded like it's enough for you if someone proves that there exists an analytical solution. – Botond May 2 '20 at 17:02
• Yes. I mean, an analytic solution for general $c$ and $\lambda$, not a particular example of an analytic solution, i.e. given $c,\lambda\in\mathbb{C}$, is there an analytic function $f$ in a neighborhood of $c$ such that $f(c) = c$ and $f''(c) = \lambda$. – Dark Malthorp May 2 '20 at 17:04