Lagrange Inversion

Wildon's Weblog 2018-03-12

In this post I’ll give two short proofs of the Lagrange Inversion formula, both based on the ideas in these notes. The first uses nothing more than Cauchy’s Integral Formula, and Cauchy’s Theorem to justify a change of variable. Rather than use Rouché’s Theorem, or the Open Mapping Theorem, as in the linked notes, I’ve added the necessary assumption on bijectivity into the hypotheses. In the standard combinatorial applications this is often obvious. The inverse is assumed to be analytic; this is easily proved, by trivial changes to the usual real variable argument, once it is known to exist. The second is even more elementary, but a little harder to motivate: some version of it will probably be the one I use for my planned textbook (expected, at the current rate, some time in 2022).

Theorem. Let $f : \mathbb{C} \rightarrow \mathbb{C}$ be an analytic function such that $f(0) = 0$ and $f$ restricts to a bijective function on a domain containing $0$ . Then

$\displaystyle [w^n] f^{-1}(w) = \frac{1}{n} [z^{n-1}] \bigl( \frac{z}{f(z)}\bigr)^n.$

Proof 1. We use Cauchy’s Integral Formula to find the coefficient of $w^n$ in the Taylor Series of $f^{-1}(w)$ , change variables by $s = f(z)$ (so $\mathrm{d}s = f'(z)\mathrm{d}z$ ), and then integrate by parts. The change of variables is justified by Cauchy’s Theorem since the approximation $f(z) \approx z$ for small $z$ shows that the contour $t \mapsto f(r\mathrm{e}^{it})$ for $t \in [0,2\pi]$ winds once around the origin provided $r$ is sufficiently small. Thus

$\displaystyle \begin{aligned} \ [w^n] f^{-1}(w) &= \frac{1}{2\pi \mathrm{i}} \oint \frac{f^{-1}(s)}{s^{n+1}} \mathrm{d} s \\ &= \frac{1}{2\pi \mathrm{i}} \oint \frac{z}{f(z)^{n+1}} f'(z) \mathrm{d} z \\ &= \frac{1}{2\pi \mathrm{i}} \oint - \frac{z}{n} \frac{\mathrm{d}}{\mathrm{d}z} \frac{1}{(f(z))^{n}} \\ &= \frac{1}{n} \frac{1}{2\pi \mathrm{i}} \oint \frac{1}{f(z)^n} \mathrm{d} z \\ &= \frac{1}{n} [z^{-1}] \frac{1}{f(z)^n} \\ &= \frac{1}{n} [z^{n-1}] \bigl( \frac{z}{f(z)}\Bigr)^n \end{aligned}$

as required. $\Box$

Since $[z^n]f(z) = \frac{1}{n!}\frac{\mathrm{d}^n}{\mathrm{d}z^n} f(z)$ , evaluated at $0$ , an equivalent statement of the result is

$\displaystyle f^{-1}(w) = \sum_{n=0}^\infty \frac{w^n}{n!} \frac{\mathrm{d}^{n-1}}{\mathrm{d}z^{n-1}} \bigl( \frac{z}{f(z)} \bigr)^n \bigr|_{z=0}$ .

Proof 2. Let $c_n = [w^n] f^{-1}(w)$ . Since $z = f^{-1}\bigl( f(z) \bigr)$ , the Taylor series for $f^{-1}$ gives

$\displaystyle z = \sum_{m=1}^\infty c_m f(z)^m.$

Differentiate and then divide through by $f(z)^n$ to get

$\displaystyle \frac{1}{f(z)^n} = n c_n \frac{f'(z)}{f(z)} + \sum_{m \not= n} m c_m f(z)^{m-1-n} f'(z).$

The formal residue of $f'(z)/f(z)$ is $1$ (this is the count of zeros and poles). The summand for $m$ where $m\not= n$ is, up to scalars, the derivative of $f(z)^{m-n}$ , so has zero residue. Therefore, taking residues, we obtain $[z^{-1}] \frac{1}{f(z)^n} = n c_n$ , which is equivalent to $c_n = [z^{n-1}] (z/f(z))^n$ , as required. $\Box$

Not a proof. It seems impossible to replace the use of residues in Proof 2 with the more elementary derivative. From

$\displaystyle \frac{z}{f(z)^n} = c_n + \sum_{m\not= n} c_m f(z)^{m-n}$

we can multiply through by $z^{n-1}$ to get

$\displaystyle c_n z^{n-1} + \sum_{m=1}^{n-1} c_m \bigl(\frac{z}{f(z)} \bigr)^n \frac{f(z)^m}{z} + \sum_{r=1}^\infty c_{n+r} z^{n-1} f(z)^r.$

It is now tempting to differentiate $n-1$ times to extract $c_n$ : the third summand is no obstacle, since it is $\mathrm{O}(z^n)$ . The second summand is now holomorphic (this is clear from the rewriting above, since the singularity at $0$ is removable). But of course it contributes erroneous terms to the derivative. Indeed, this procedure gives $(n-1)!c_n$ , whereas the theorem predicts that $n! c_n = \frac{\mathrm{d}^{n-1}}{\mathrm{d}z^{n-1}} \bigl( \frac{z}{f(z)} \bigr)^n \bigr|_{z=0}$ , so something is surely wrong.