Extract coefficients for a formal power series using Lagrange Inversion Formula

Question

Given $f(x)$ is a formal power series that satisfies $f(0) = 0$

$(f(x))^{3} + 2(f(x))^{2} + f(x) - x = 0$

I know that the Lagrange inversion formula states given f(u) & $\varphi(u)$ are formal power series with respect to u, and $\varphi(0) = 1$ then the following is true.

$[x^{n}](f(u(x))) = \frac{1}{n}[u^{n-1}](f'(u)\varphi(u)^{n})$

How do I find the coefficient of $x^{n}$ in $f(x)$ using Lagrange inversion formula?

Answer 1

Suppose we have

$$f(z)^3 + 2f(z)^2 + f(z) = z$$

and we seek $[z^n] f(z).$

While we wait for a contribution from LIF experts in the meantime we can use Poor Man's Lagrange Inversion which is the Cauchy Residue Theorem. We have

$$[z^n] f(z) = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} f(z) \; dz.$$

Now put $w=f(z)$ so that $w^3+2w^2+w = z$ and $3w^2 + 4w + 1 \; dw = dz$ to get

$$\frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1} (1+w)^{2n+2}} \times w \times (3w^2 + 4w + 1) \; dw.$$

Now $$w \times (3w^2 + 4w + 1) = 3 (1+w)^3 - 5 (1+w)^2 + 2 (w+1).$$

Extracting coefficients we thus have

$$3 (-1)^n {n+2n-2\choose n} - 5 (-1)^n {n+2n-1\choose n} + 2 (-1)^n {n+2n\choose n} \\ = (-1)^n \times \left(3{3n-2\choose n} - 5 {3n-1\choose n} + 2{3n\choose n}\right).$$

This is

$$(-1)^n {3n-2\choose n} \left(3 - 5\frac{3n-1}{2n-1} + 2 \frac{(3n-1)3n}{(2n-1)2n}\right) \\ = \frac{(-1)^n}{1-2n} {3n-2\choose n}.$$

Remark. The fact that we are given $f(0) = 0$ determines the choice of branch so that $z=0$ corresponds to $w=0$ (which is needed when we make the substitution). Also, if we choose $\epsilon$ sufficiently small we also get a small circle for $w$ (with a different $\epsilon.$)

Addendum, five years later. A better choice of integral is

$$[z^n] f(z) = \frac{1}{n} [z^{n-1}] f'(z) = \frac{1}{n} \frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{1}{z^n} f'(z) \; dz.$$

We put $w = f(z)$ and obtain

$$\frac{1}{n} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^n (1+w)^{2n}} \; dw = \frac{(-1)^{n-1}}{n} {n-1+2n-1\choose n-1} \\ = \frac{(-1)^{n-1}}{n} {3n-2\choose n-1}.$$

Answer 2

Here we use a somewhat simpler but equivalent variant of the Lagrange Inversion Formula. (See Theorem A.2 in Analytic Combinatorics by P. Flajolet and R. Sedgewick for the equivalence of the variants).

Lagrange Inversion Formula:

Let $g(x), f(x)\in x\mathbb{C}[x]$ be inverses: $g(f(x))=x$. If $g(x)=\frac{x}{\phi(x)}$ and $f(x)=x\phi\left(f(x)\right)$, then \begin{align*} [x^n]f(x)=\frac{1}{n}\left[x^{n-1}\right]\left(\phi(x)\right)^n\tag{1} \end{align*}

The functional relation \begin{align*} (f(x))^{3} + 2(f(x))^{2} + f(x) - x = 0 \end{align*} can be written as $g\left(f(x)\right)=x$ with \begin{align*} g(x)&=x^3+2x^2+x\\ &=x(1+x)^2 \end{align*}

We can write $$\phi(x)=\frac{x}{g(x)}=\frac{1}{(1+x)^2}$$

and obtain

\begin{align*} [x^n]f(x)&=\frac{1}{n}\left[x^{n-1}\right]\left(\phi(x)\right)^n\tag{2}\\ &=\frac{1}{n}\left[x^{n-1}\right]\frac{1}{(1+x)^{2n}}\\ &=\frac{1}{n}\binom{-2n}{n-1}\tag{3}\\ &=\frac{(-1)^{n-1}}{n}\binom{3n-2}{n-1} \end{align*}

Comment:

• In (2) we apply the Lagrange Inversion Formula (1)

• In (3) we use the binomial identity \begin{align*} \binom{-p}{q}=\binom{p+q-1}{q}(-1)^q \end{align*}