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Given a functional equation of form $f(f(x))=T(x)$ is there any good ways to solve it numerically? If not then at least approximate in some small region $x\in(-a;a)$.

E.g. with the equation $f(f(x))=x+x^2$ one could prove that the solution is between $l(x)=|x+0.5|+0.5$ and $h(x)=x+x^2$. However I'm unsure how to proceed further.

I've found a solution for equations like these on arxiv. However it seems to converge very badly (or I'm doing something wrong).

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Select the one you like ($f^{[1/2]}(x)$ means half-iteration, iterative square root):

$$f^{[1/2]}(x)=\sum_{m=0}^{\infty} \binom {1/2}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{[k]}(x)$$

$$f^{[1/2]}(x)=\lim_{n\to\infty}\binom {1/2}n\sum_{k=0}^n\frac{1/2-n}{1/2-k}\binom nk(-1)^{n-k}f^{[k]}(x)$$

$$f^{[1/2]}(x)=\lim_{n\to\infty}\frac{\sum_{k=0}^{n} \frac{(-1)^k f^{[k]}(x)}{(1/2-k)k!(n-k)!}}{\sum_{k=0}^{n} \frac{(-1)^k }{(1/2-k) k!(n-k)!}}$$

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  • $\begingroup$ Can you add a source? I'm more interested in the process of arriving to this solution as opposed to the this specific solution. $\endgroup$
    – Warmist
    Commented Apr 13, 2021 at 9:45
  • $\begingroup$ @Warmist This is Newton series. $\endgroup$
    – Anixx
    Commented Apr 13, 2021 at 9:54
  • $\begingroup$ @Warmist see here for more info: mathoverflow.net/questions/17605/how-to-solve-ffx-cosx/… $\endgroup$
    – Anixx
    Commented Apr 13, 2021 at 9:56

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