Asymptotic expansion for the inverse function of $ f(x)=x^3+x$

Question

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:

$$ g(x)=(x-x^{1/3}+\frac{1}{3x^{1/3}}+o(\frac{1}{x}))^{1/3}$$

$$ g(x)=x^{1/3}(1-\frac{1}{x^{2/3}}+\frac{1}{3x^{4/3}}+o(1/x^2))^{1/3}$$

$$ g(x)=x^{1/3}(1-\frac{1}{3x^{2/3}}+\frac{1}{9x^{4/3}}-\frac{1}{9}(-\frac{1}{x^{2/3}}+\frac{1}{3x^{4/3}})^2+\frac{5}{81}(-\frac{1}{x^{2/3}}+\frac{1}{3x^{4/3}})^3)$$

$$ g(x)=x^{1/3}(1-\frac{1}{3x^{1/3}}+\frac{1}{9x^{4/3}}-\frac{1}{9x^{4/3}}+\frac{2}{27x^2}-\frac{5}{81x^2})$$

$$ g(x)=x^{1/3}-\frac{1}{3x^{1/3}}+\frac{1}{81x^{5/3}}+o(1/x^{5/3})$$

Answer 1

$g(x)$ is the inverse function for $f(x)$. the reciprocal would be $1/f(x)$.

To compute the inverse of $x^3+x$, let $$ y=x^3+x=x^3\left(1+\frac{1}{x^2}\right)\tag{1} $$ Taking $(1)$ to the $-\frac13$ power (via the binomial expansion) yields $$ \begin{align} y^{-1/3} &=\frac{1}{x}\left(1-\frac{1}{3x^2}+\frac{2}{9x^4}-\frac{14}{81x^6}+O\left(\frac{1}{x^8}\right)\right)\\ &=\frac{1}{x}-\frac{1}{3x^3}+\frac{2}{9x^5}-\frac{14}{81x^7}+O\left(\frac{1}{x^9}\right)\tag{2} \end{align} $$ Inverting the series for $u-\frac13u^3+\frac29u^5-\frac{14}{81}u^7$ yields $$ \begin{align} \frac1x &=y^{-1/3}+\frac13y^{-1}+\frac19y^{-5/3}+\frac{2}{81}y^{-7/3}+O\left(y^{-3}\right)\\ &=y^{-1/3}\left(1+\frac13y^{-2/3}+\frac19y^{-4/3}+\frac{2}{81}y^{-2}+O\left(y^{-8/3}\right)\right)\tag{3} \end{align} $$ Taking the reciprocal of $(3)$ yields $$ \begin{align} x &=y^{1/3}\left(1-\frac13y^{-2/3}+\frac{1}{81}y^{-2}+O\left(y^{-8/3}\right)\right)\\ &=y^{1/3}-\frac13y^{-1/3}+\frac{1}{81}y^{-5/3}+O\left(y^{-7/3}\right)\tag{4} \end{align} $$ The third term would normally be a $y^{-1}$ term, but the coefficient for that term is $0$.

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In the derivation in the question, you get $$ g(x)=x^{1/3}-\tfrac13x^{-1/3}+h(x)\tag{5} $$ where $h(x)=o\left(x^{-1}\right)$. Using $(5)$, we get $$ \small g(x)^3+g(x)=x-\frac{1}{27}x^{-1}+\left(3x^{2/3}+\frac13x^{-2/3}-1\right)h+\left(3x^{1/3}-x^{-1/3}\right)h^2+h^3\tag{6} $$ Thus, we need to have $3x^{2/3}h=\frac{1}{27}x^{-1}+o\left(x^{-1}\right)$; that is, $h=\frac{1}{81}x^{-5/3}+o\left(x^{-5/3}\right)$.

Since the only powers of $x$ that appear in $$ \left(x^{1/3}-\tfrac13x^{-1/3}+\tfrac{1}{81}x^{-5/3}\right)^3+\left(x^{1/3}-\tfrac13x^{-1/3}+\tfrac{1}{81}x^{-5/3}\right)-x\tag{7} $$ are odd powers of $x^{1/3}$, the remainder, which is $o\left(x^{-5/3}\right)$, must be $O\left(x^{-7/3}\right)$

Thus, $$ g(x)=x^{1/3}-\tfrac13x^{-1/3}+\tfrac{1}{81}x^{-5/3}+O\left(x^{-7/3}\right)\tag{8} $$ which agrees with $(4)$.

Answer 2

The substitution $x = y^{1/3} u^{1/3}$ takes the equation $x^3 + x = y$ into $u + y^{-2/3} u^{1/3} = 1$. As in my recent answer to Solving a sum of exponentials, there is a series solution

$$ u = \sum_{k=0}^\infty \frac{(-1)^k a_k}{k!} y^{-2k/3} \ \text{where} \ a_k = \prod_{j=0}^{k-2} (k/3-j)$$ taking $a_0 = a_1 = 1$.

Now we have

$$x = y - x^3 = y - u y = \sum_{k=1}^\infty \frac{(-1)^{k+1} a_k}{k!} y^{1-2k/3} $$

$$ = y^{1/3}-\frac13 y^{-1/3} + \frac{1}{81} y^{-5/3} + \frac{1}{243} y^{-7/3} - \frac{4}{6561} y^{-11/3} - \frac{5}{19683} y^{-13/3} + \ldots$$

EDIT: The series converges for $y > 2 \sqrt{3}/9$.

Answer 3

Let $\color{blue}{z=x^{1/3}}$. Then $z^{-1}g(x)=1-\frac13z^{-2}+h(z)$ with $h(z)=o(z^{-2})$. Expanding the cube $(1+u)^3=1+3u+3u^2+u^3$ with $u=-\frac13z^{-2}+h(z)$ and neglecting every $o(h(z))$ and every $o(z^{-6})$ term in $u^2$ and $u^3$ yields $$ u^2=\tfrac19z^{-4}+O(z^{-2}h(z)),\quad\text{which implies}\quad u^2=\tfrac19z^{-4}+o(h(z)), $$ and $$ u^3=-\tfrac1{27}z^{-6}+O(z^{-4}h(z)),\quad\text{which implies}\quad u^3=-\tfrac1{27}z^{-6}+o(z^{-6}), $$ hence $$ z^{-3}g^3(z)=1-z^{-2}+3h(z)+\tfrac13z^{-4}+o(h(z))-\tfrac1{27}z^{-6}+o(z^{-6}). $$ Since $z^{-3}g^3(z)+z^{-3}g(z)=1$ and $z^{-3}g(z)=z^{-2}-\frac13z^{-4}+o(h(z))$, after some simplifications one gets the condition $$ 3h(z)-\tfrac1{27}z^{-6}+o(z^{-6})+o(h(z))=0. $$ Solving for $h(z)$ yields $h(z)=\tfrac1{81}z^{-6}+o(z^{-6})$, hence $$ \color{red}{g(x)=x^{1/3}-\tfrac13x^{-1/3}+\tfrac1{81}x^{-5/3}+o(x^{-5/3})}. $$

Edit (to answer a comment)

(i) The beginning of the expansion of $z^{-1}g(x)$ is $1-\frac13z^{-2}$, as you proved yourself, hence the remainder $h(z)$ may be any term $o(z^{-2})$. The idea of the exercise (and the reason why I had a look at it and finally wrote down a solution) is that one does not know a priori the order of $h(z)$. (As an aside, you might want to check Landau little-o notation.)

(ii) The term $3h(z)$ has nothing to do with the order of $h(z)$, simply the beginning of the expansion of $(1+u)^3$ is $1+3u$, here $u=-\frac13z^{-2}+h(z)$ hence $3u$ yields $-z^{-2}+3h(z)$.

(iii) Neither $\tfrac13z^{-4}$ nor $-\tfrac1{27}z^{-6}$ is $o(h(z))$, in the end. But it is not necessary to know this for the equality where they appear to hold.

(iv) The last equation is $3h(z)+o(h(z))=\tfrac1{27}z^{-6}+o(z^{-6})$, which is logically equivalent to the fact that $3h(z)=\tfrac1{27}z^{-6}+o(z^{-6})$.

Answer 4

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.