3
$\begingroup$

Cross-posted from Math Stackexchange.

In an older question to which I provided an answer it was asked how to compute a particular limit involving the roots of a transcedental function around its extremum. This limit required the evaluation of several terms of a power series for each one of the two roots the function possesses around the minimum, which is reminiscent of a procedure akin to the Lagrange inversion theorem. However, as explicitly stated in the assumptions of the theorem, the derivative of the function has to be non-zero for the theorem to apply (the function has to be locally invertible). That requirement however, did not stop me from deriving a term by term expansion for both functional inverses around the maximum.

The reason why I was surprised is because I couldn't find any references on the subject, and this calculation seems to be a low-hanging fruit of a simple generalization to a well-known theorem.

My question is twofold, but an answer to either components will suffice:

  1. Assuming that around a point $x_0$, where $f'(x_0)=0, f''(x_0)\neq 0$ and given that $f(x)=\sum_n a_n(x-x_0)^n$, what are the coefficients $b^{\pm}_{n}$ of the series expansions of the two functional inverses $r^\pm(x)=\sum_{n=0}^{\infty}b_n^{\pm}(x-f(x_0))^{n/2}$, that satisfy $f(r^\pm (x))=x$? They also can be seen as the two solutions to the equation $f(x)=c$ in a neighborhood around $c=f(x_0)$. Is there a general formula for them (however formal?) Here I define $r^+(x)$ to be the root satisfying $r^+(x)>x_0$ and the other one such that $r^-(x)<x_0$.
  1. Has this structure been studied in the literature before? Does it come under a certain name? If not, what is it that makes this theorem difficult to establish or uninteresting?

My work: Experimenting with a couple of simple functions and their behavior around double zeroes indicates that these coefficients are related by $b_n^-=(-1)^n b_n^+$. Also, these series expansions are obviously one-sided: they are defined for $x\in [f(x_0),R)$ if $f''(x_0)>0$ and for $x\in (R, f(x_0)]$ if $f''(x_0)<0$, for some value of $R$ representing a radius of convergence. I also computed the first few coefficients for arbitrary $f$ with a minimum at $x_0$

$$b_0=x_0,~~ b_1=\sqrt{\frac{2}{f''(x_0)}}, ~~ b_2=-\frac{f'''(x_0)}{3(f''(x_0))^2}$$

I also noticed that in a further generalization of the problem, when I demand that $f'(x_0)=f''(x_0)= \dots =f^{(n-1)}(x_0)=0, f^{(n)}(x_0)\neq 0$, there are now $n$ functional inverses, but most of them represent complex roots around the extremum ($n-1$ if $n$ is odd and $n-2$ if $n$ is even). This hint may be useful if one actually tries to formalize the theorem, since in the complex plane all branches will be included.

$\endgroup$
2
  • 1
    $\begingroup$ When crossposting from m.se, it is generally considered polite to wait about a week, in order to prevent unnecessary duplication of effort. $\endgroup$
    – user44191
    Commented Sep 13, 2021 at 17:21
  • 3
    $\begingroup$ These things are well known. Look up "singular inversion" and "singular expansion", see e.g. Lemma IV.3 and Thm VI.6 in the book "Analytic Combinatorics" by Flajolet and Sedgewick. $\endgroup$
    – esg
    Commented Sep 13, 2021 at 18:44

1 Answer 1

3
$\begingroup$

This is easily reduced to the ordinary Lagrange inversion.

Indeed, without loss of generality, $x_0=0=f(x_0)=f'(x_0)<f''(x_0)$, so that for all real $z$ close enough to $0$ we have $f(z)=z^2 h(z)$, where $h$ is a function analytic near $0$ such that $h(0)>0$. So, the equation $f(z)=w$ (for real $z$ near $0$ and small enough $w\ge0$) can be rewritten as $$g(z):=z\sqrt{h(z)}=\sqrt w\,\text{sgn}\,z,$$ so that $g$ is analytic near $0$: $g(z)=\sum_{n=0}^\infty a_n z^n$ for real $z$ near $0$. The $a_n$'s can be obtained by Faà di Bruno's formula.

Using now the ordinary Lagrange inversion, we get $$z=\sum_{n=0}^\infty b_n w^{n/2}\text{sgn}^n z$$ for some explicitly written $b_n$'s and all real $w$ close enough to $0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .