Requesting basic explanation of the Lagrange inversion theorem

Solution 1:

The Lagrange inversion theorem in a nutshell.

Assume to have a holomorphic function which is $z+o(z)$ in a neighbourhood of the origin, like $$\sin(z) = \sum_{n\geq 0}\frac{(-1)^n z^{2n+1}}{(2n+1)!} \tag{1}$$ and to want to compute the coefficients of the Maclaurin series of its inverse function $\arcsin(z)$.
Say the coefficient of $z^7$. Well, by Cauchy's integral formula

$$[z^7]\arcsin(z) = \frac{1}{2\pi i}\oint_{|z|=\varepsilon}\frac{\arcsin(z)}{z^8}\,dz \tag{2}$$ and something nice happens$^{(*)}$ if we enforce the substitution $z=\sin u$ in the RHS of $(2)$. The simple contour around the origin $|z|=\varepsilon$ is mapped into a similar (homeomorphic) simple contour around the origin by a conformal map, hence $$ [z^7]\arcsin(z) = \frac{1}{2\pi i}\oint_{|u|=\varepsilon}\frac{u\cos(u)}{\sin(u)^8}\,du \tag{3}$$ and the problem boils down to evaluating the residue of $\frac{u\cos u}{\sin(u)^8}$ at the origin, which is a pole of order $7$ for such a function. In particular $$\operatorname*{Res}_{u=0}\frac{u\cos u}{\sin(u)^8} = \lim_{u\to 0}\frac{1}{6!}\frac{d^6}{du^6}\left(u^7\cdot \frac{u\cos u}{\sin(u)^8}\right)=\lim_{u\to 0}\frac{1}{7!}\frac{d^6}{du^6}\left(\frac{u}{\sin u}\right)^7=\frac{5}{112}\tag{4}$$ and the whole tour proves a connection between the Maclaurin coefficients of $\arcsin$ and the derivatives of $\left(\frac{u}{\sin u}\right)^k$ at the origin.

Now you might wonder if to compute the derivatives at the origin of $\left(\frac{u}{f(u)}\right)^k$ for some holomorphic $f(u)=u+o(u)$ is a simple task. Well, in general it is not. For instance the Maclaurin series of $\arcsin$ can be computed with considerably fewer efforts by applying the extended binomial theorem to $\frac{d}{du}\arcsin(u)=\frac{1}{\sqrt{1-u^2}}$. On the other hand something really nice is produced by this approach by considering $f(u)=u e^u$, i.e. the Maclaurin series of the Lambert $W$ function: $$ W(x) = \sum_{n\geq 1}\frac{(-1)^{n+1} n^{n-1}}{n!} x^n\quad \Longrightarrow\quad \sum_{n\geq 1}\frac{n^{n-1}}{n!e^n}=\color{red}{1} \tag{!}$$ and the crucial part of the argument $(*)$ can be used for finding the Maclaurin series of $\arcsin^2$, $\arcsin^3$, $\arcsin^4$ etcetera, leading to some non-trivial hypergeometric identities.

A concise form of the statement and another example.

Lagrange's inversion formula. If $f(z)$ is a holomorphic function in a neighbourhood of the origin, such that $f(z)=z+o(z)$ as $z\to 0$, we have $$ f^{-1}(z) = \sum_{n\geq 1}\frac{z^n}{n}\cdot [z^{n-1}]\left(\frac{z}{f(z)}\right)^n$$ where $[z^m]g(z)$ stands for the coefficient of $z^m$ in the Maclaurin series of $g(z)$.

More generally, if $f,h$ are holomorphic functions in a neighbourhood of the origin and $f(z)=z+o(z)$, $$ h(f^{-1}(z))=h(0)+\sum_{n\geq 1}\frac{z^n}{n}\cdot [z^{n-1}]\left(h'(z)\cdot\left(\frac{z}{f(z)}\right)^n\right).$$

Another celebrated application is given by Catalan numbers. It is straightforward to prove in a combinatorial fashion that they fulfill $ C_{n+1}=\sum_{k=0}^{n} C_k C_{n-k}$, hence their ordinary generating function multiplied by $z$ is given by the inverse function of $f(z)=z-z^2$. By Lagrange's inversion formula

$$ f^{-1}(z)=\sum_{n\geq 1}\frac{z^n}{n}[z^{n-1}]\left(\frac{1}{1-z}\right)^n $$ and by stars and bars $\frac{1}{(1-z)^{n}}=\sum_{m\geq 0}\binom{m+n-1}{m}z^n$, hence $$ f^{-1}(z) = \sum_{n\geq 1}\frac{z^n}{n}\binom{2n-2}{n-1} $$ and $$ C_n = \frac{1}{n+1}\binom{2n}{n}.$$

Ref. A ridiculously simple and explicit implicit function theorem, A.D. Sokal, 2009.