Let the sequence $u\_1, u\_2, \ldots$ satisfy $u\_{n+1} = u\_n - u\_n^2 + O(u\_n^3)$. Then it can be shown that if $u\_n \to 0$ as $n \to \infty$, then $u\_n = n^{-1} + O(n^{-2} \log n)$. (See N. G. de Bruijn, *Asymptotic methods in analysis*, Section 8.5.)


This can be used to obtain asymptotics for $v\_{n+1} = Av\_n - Bv\_n^2 + O(v\_n^3)$, where $A$ and $B$ are constants. Let $w\_n = A^{-n} v\_n$; this gives
$$ A^{n+1} w\_{n+1} = A^{n+1} w\_n - B A^n w\_n^2 + O(A^n w\_n^3)$$
and so
$$ w\_{n+1} = w\_n - BA^{-1} w\_n^2 + O(w\_n^3). $$
Then let $w\_n = Ax\_n/B$ to get
$$ Ax\_{n+1}/B = Ax\_n/B - B/A \cdot (Ax\_n/B)^2 + O(x\_n^3) $$
and after simplifying
$ x\_{n+1} = x\_n - x\_n^2 + O(x\_n^3)$. This satisfies the initial requirements for $u\_n$ (with some checking of the side condition); then substitute back.


But say I actually know that $u\_{n+1} = P(u\_n)$ for some polynomial $P$, with $P(z) = z - z^2 + a\_3 z^3 + \cdots + a\_d z^d.$ In this case it seems like it should be possible to get more explicit information about $u\_n$. Is there a known algorithm for computing an asymptotic series for $u\_n$ as $n \to \infty$?



Asymptotics of iterated polynomials
Have you tried the method of section 8.7, i.e., solving Abel's equation $\phi(P(x))-\phi(x)=1$? Here we expect $\phi(t)=t^{-1}+\sum\_{n=1}^{\infty}c\_n t^n$, and you can find the coefficients of $\phi$ one by one. For example, I took $P(x)=x-x^2+x^3+x^4$ and immediately found $c\_1=-2$, $c\_2=-5/2$, $c\_3=-7/2$, $c\_4=-17/4$... Not a general formula, but you can get as many terms as you want for a given $P$.


