# Finding a three-term asymptotic expansion of the inverse function of $f(x)=x^3+x$

I would like to find a three-term asymptotic expansion of $g$ the reciprocal 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:

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

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

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

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

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

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You mean inverse, not reciprocal. – lhf Apr 9 '12 at 23:27
Lagrange's inversion theorem may help. – lhf Apr 9 '12 at 23:28
If I may ask, what is going on under "Moreover" from the fourth to the tenth lines? Compare the fourth and the tenth. – Antonio Vargas Apr 11 '12 at 16:44

$g(x)$ 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 $0$.

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 $(4)$.

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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 $y > 2 \sqrt{3}/9$.

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You are the master of computation. – Pedro Tamaroff Apr 9 '12 at 20:00

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 $$\color{red}{g(x)=x^{1/3}-\tfrac13x^{-1/3}+\tfrac1{81}x^{-5/3}+o(x^{-5/3})}.$$

(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 $3h(z)=\tfrac1{27}z^{-6}+o(z^{-6})$.

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In your first line why do you take $h(z)=o(z^{-2})$ and not $h(z)=o(x^{-1})=o(z^{-3})$? As $h(z)=o(z^{-2})$, shouldn't $3h(z)$ be $h(z)$? And as the precision of the asymptotic expansion of $z^{-3}g(z)^3$ is $o(h(z))$ why do you keep the terms $\frac{1}{3}z^{-1/4}$ and $-\frac{1}{27}z^{-1/6}+o(z^{-1/6})$ which are $o(h(z))$ terms? Same question in the last equation: shouldn't we have $h(z)=o(z^{-1/6})$ (which is wrong?) to get $h(z)=\frac{1}{81}z^{-1/6}+o(z^{-1/6})$ ? – Chon Apr 8 '12 at 7:44
See Edit.   – Did Apr 8 '12 at 8:07
In the first equation, $$u^2=\tfrac19z^{-4}+O(z^{-2}h(z)),\quad\text{which implies}\quad u=\tfrac19z^{-4}+o(h(z))$$ did you intend $u=\tfrac13z^{-2}+o(h(z))$? I also have a similar question regarding the next equation. – robjohn Apr 8 '12 at 13:34
@robjohn: I intended what is written. Note that $u^2=\frac19z^{-4}+v$ with $v=-\frac23z^{-2}h(z)+h(z)^2$. First step: $h(z)=o(z^{-2})$ hence $h(z)^2=o(z^{-2}h(z))$ and $v=O(z^{-2}h(z))$. Second step: $z^{-2}h(z)=o(h(z))$ hence $v=o(h(z))$. – Did Apr 8 '12 at 21:29
Perhaps I am missing something, but how can both $u^2$ and $u$ be close to $\tfrac19z^{-4}$? In the next equation, it seems that both $u^3$ and $u^2$ are close to $-\frac{1}{27}z^{-6}$ (which is ridiculous since $u^2>0$). Perhaps the exponent of $u$ is off in the conclusions. It also seems that $o(h(z))$ is a much bigger error than $O(z^{-2}h(z))$. In keeping with the other estimates, I would have expected concluding $o(z^{-4})$ from $O(z^{-2}h(z))$. – robjohn Apr 9 '12 at 19:04

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.

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