I'm interested in the Lagrange inversion theorem (https://en.wikipedia.org/wiki/Lagrange_inversion_theorem#Theorem_Statement), despite several attempts on different pdfs, on this site and on wikipedia, (checked already its bibliography, i could reach just Abramowitz's Handbook on mathematical functions, that, unfortunately, is not exhaustive on the argument). So I'm asking you guys to explain to me this theorem in as simple words as you can. I'm expecially interested in the "reversion of serie" aspect (quoting from wikipedia). Thank you a lot

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    $\begingroup$ It's not clear what you are asking (at least not to me). Do you want an explanation of the statement of the theorem? What part of it is not clear to you? Do you know that an analytic function is locally one-to-one at a point where it's derivative doesn't vanish? $\endgroup$ – saulspatz Apr 17 '18 at 16:56

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.

  • $\begingroup$ This is such a complete answer that I'm not fully understanding it yet, so, for the great effort, your answer as been flagged as the correct one. $\endgroup$ – Lyn Cassidy Apr 18 '18 at 7:52
  • $\begingroup$ Do you mind writing me down a second example please? $\endgroup$ – Lyn Cassidy Apr 18 '18 at 7:52
  • $\begingroup$ @LynCassidy: which part(s) is(/are) difficult to grasp? Anyway, I have added a concise statement and a further example. $\endgroup$ – Jack D'Aurizio Apr 18 '18 at 11:51
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    $\begingroup$ You do not need Lagrange's inversion formula for computing $$ \sum_{n\geq 1}\frac{\cos(nx)}{n^2} = \text{Re}\,\text{Li}_2(e^{ix}).$$ Since $\sum_{n\geq 1}\frac{\sin(nx)}{n}$ is the Fourier series of the sawtooth-wave, $$ \sum_{n\geq 1}\frac{\cos(nx)}{n^2} $$ is a continuous and piecewise-quadratic function. See Bernoulli polynomials. $\endgroup$ – Jack D'Aurizio Apr 18 '18 at 14:02
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    $\begingroup$ @JackD'Aurizio: Extraordinary nice presentation. (+1) $\endgroup$ – Markus Scheuer Jun 20 '18 at 18:49

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