TITLE: Asymptotic expansion for the inverse function of $ f(x)=x^3+x$
QUESTION [7 upvotes]: I would like to find a three-term asymptotic expansion of $g$, the inverse function of:
$$ f(x)=x^3+x$$
We have:
$$ f(g(x))=x=g(x)^3+g(x)$$
As $$\ g(x) \rightarrow_{x\rightarrow \infty} \infty$$
$$ x=g(x)^3+g(x) \sim g(x)^3$$
$$ g(x) \sim x^{1/3}$$
$$ g(x)=x^{1/3}+o(x^{1/3})$$
Moreover
$$ g(x)=(x-g(x))^{1/3}=(x-x^{1/3}+o(x^{1/3}))^{1/3}$$
$$ g(x)=x^{1/3}(1-\frac{1}{x^{2/3}}+o(1/x^{2/3}))^{1/3}$$
$$ g(x)=x^{1/3}(1-\frac{1}{3x^{2/3}}+o(1/x^{2/3}))$$
$$ g(x)=x^{1/3}-\frac{1}{3x^{1/3}}+o(1/x^{2/3}) $$
$$ g(x)=(x-x^{1/3}+\frac{1}{3x^{1/3}}+o(\frac{1}{x^{1/3}}))^{1/3}$$
$$ g(x)=x^{1/3}(1-\frac{1}{x^{2/3}}+\frac{1}{3x^{4/3}}+o(\frac{1}{x^{4/3}}))^{1/3}$$
$$ g(x)=x^{1/3}(1-\frac{1}{3x^{2/3}}+\frac{1}{9x^{4/3}}-\frac{1}{9}(-\frac{1}{x^{2/3}}+\frac{1}{3x^{4/3}})^2+o(\frac{1}{x^{4/3}}))$$
$$ g(x)=x^{1/3}(1-\frac{1}{3x^{2/3}}+\frac{1}{9x^{4/3}}-\frac{1}{9x^{4/3}}+o(\frac{1}{x^{4/3}}))$$
$$ g(x)=x^{1/3}(1-\frac{1}{3x^{2/3}}+o(\frac{1}{x^{4/3}}))$$
$$ g(x)=x^{1/3}-\frac{1}{3x^{1/3}}+o(\frac{1}{x}) $$
A solution is: 
$$ g(x)=(x-x^{1/3}+\frac{1}{3x^{1/3}}+o(\frac{1}{x}))^{1/3}$$
$$ g(x)=x^{1/3}(1-\frac{1}{x^{2/3}}+\frac{1}{3x^{4/3}}+o(1/x^2))^{1/3}$$
$$ g(x)=x^{1/3}(1-\frac{1}{3x^{2/3}}+\frac{1}{9x^{4/3}}-\frac{1}{9}(-\frac{1}{x^{2/3}}+\frac{1}{3x^{4/3}})^2+\frac{5}{81}(-\frac{1}{x^{2/3}}+\frac{1}{3x^{4/3}})^3)$$
$$ g(x)=x^{1/3}(1-\frac{1}{3x^{1/3}}+\frac{1}{9x^{4/3}}-\frac{1}{9x^{4/3}}+\frac{2}{27x^2}-\frac{5}{81x^2})$$
$$ g(x)=x^{1/3}-\frac{1}{3x^{1/3}}+\frac{1}{81x^{5/3}}+o(1/x^{5/3})$$

REPLY [2 votes]: I really like this question! I’ll say absolutely nothing here that hasn't been said by others, but I'd like to treat this in a more formal-series-theoretical way. The situation is analytic in the neighborhood of $(\infty,\infty)$, by which I mean that when you substitute $x=1/\xi$, $y=1/\eta$, you get $\eta=1/(\xi^{-1}+\xi^{-3})=\xi^3-\xi^5+\xi^7-\cdots=G(\xi)$. Now $G$ is a cube in ${\mathbb{R}}[[x]]$, in fact when you extract its cube root, you see that the coefficients of the cube root are in $A={\mathbb{Z}}[1/3]$, the ring of rational numbers with only powers of $3$ in the denominator. Use Binomial Theorem if you like, but you can get the same result out of Newton’s method, because you’re dividing by $3f(\xi)^2$ for a different $f$ at each stage of the approximation, and you start with $f(\xi)=\xi\in A[[\xi]]$. So let's put $Y=\eta^{1/3}$ formally, and get $Y=G(\xi)^{1/3}=H(\xi)\in A[[\xi]]$. In fact the first term of $H(\xi)$ is just $\xi$, so that you can revert, i.e. take the inverse function, to get $\xi=H^{-1}(Y)=Y+\cdots\in A[[Y]]$. (Again, you may find $H^{-1}$ by the use of Newton’s method.) Now what you’re interested in is $x$, in terms of $Y=y^{-1/3}$, so you take the reciprocal of $H$ as Laurent series over $A$, and it comes out $x=1/Y - (1/3)Y + (1/81)Y^5-\cdots$. It's a series with coefficients still in $A$, so you get an explanation of why the only denominators you see in the explicit computations are powers of $3$. I like this treatment ’cause it shows, for instance, that you can ask the same question $p$-adically and get $p$-adic convergence outside the closed unit disk for all $p$ different from $3$. And you can use the properties of the derivative of $x+x^3$ to see what the domain of convergence is $3$-adically as well.