Lagrange Inversion

Formal Laurent Series

We already know the ring of formal power series \(\mathbb{C}\lbrack\lbrack x\rbrack\rbrack\). Now define the ring of formal Laurent series:

\[ \mathbb{C}\left(\left(x\right)\right):=\left\lbrace \sum_{k\geq N}a_kx^k : N\in\mathbb{Z},a_k\in \mathbb{C}\right\rbrace \]

We can define the multiplicative inverse for elements in \(\mathbb{C}\left(\left(x\right)\right)\) by imitating the definition for formal power series:

If for \(f:=\sum_{k\geq N}f_kx^k\) with \(f_N\neq 0\), there exists \(g=\sum_{k\geq -N}g_kx^k\) such that \(fg=1\), then

\[ g_k:= \begin{cases} f_N^{-1}, &\text{ if }k=-N\text{,} \\ -f_N^{-1}\sum_{i> N}f_ig_{k-i}, &\text{ otherwise} \end{cases} \]

Similar to formal power series, we also define for nonzero \(f(x)=\sum_{k\geq N}f_kx^k\):

\[ \operatorname{ord} f:=\min\lbrace k:f_k\neq 0\rbrace \]

Clearly, for \(g\neq 0\), we have

\[ \operatorname{ord} (fg)=\operatorname{ord}(f)+\operatorname{ord}(g) \]

Formal Residue

The formal residue is the coefficient of the \(x^{-1}\) term in a formal Laurent series. Denote \(\operatorname{res} f:=\lbrack x^{-1}\rbrack f\).

Lemma: For any formal Laurent series \(f\), \(\operatorname{res} f'=0\).

Proof: Consider the definition of the formal derivative \(\left(x^k\right)'=kx^{k-1}\).

Lemma: For any formal Laurent series \(f,g\), \(\operatorname{res}(f'g)=-\operatorname{res}(fg')\).

Proof: Consider the product rule \((fg)'=f'g+fg'\), so \(0=\operatorname{res}((fg)')=\operatorname{res}(f'g)+\operatorname{res}(fg')\).

Lemma: For a formal Laurent series \(f(x)\neq 0\), \(\operatorname{res}(f'/f)=\operatorname{ord}f\).

Proof: Let \(\operatorname{ord}f=k\). Then

\[ \begin{aligned} \operatorname{res}\left(\frac{f'}{f}\right)&=\operatorname{res}\left(\frac{kf_kx^{k-1}+\cdots}{f_kx^k+f_{k+1}x^{k+1}+\cdots}\right) \\ &=\operatorname{res}\left(\frac{kf_kx^{-1}+\cdots}{f_k+f_{k+1}x+\cdots}\right) \\ &=k \end{aligned} \]

Lemma: For a formal Laurent series \(f\) and a formal power series \(g\neq 0\), \(\operatorname{res}(f)\operatorname{ord}(g)=\operatorname{res}(f(g)g')\).

Proof: By linearity, it suffices to prove the case \(f=x^k\) where \(k\in\mathbb{Z}\). If \(k\neq -1\), then

\[ \begin{aligned} \operatorname{res}x^k&=0 \\ \operatorname{res}(g^kg')&=\operatorname{res}\left(\frac{1}{k+1}\left(g^{k+1}\right)'\right) \\ &=\frac{1}{k+1}\operatorname{res}\left(\left(g^{k+1}\right)'\right) \\ &=0 \end{aligned} \]

If \(k=-1\), then

\[ \begin{aligned} \operatorname{res}f&=\operatorname{res}\left(x^{-1}\right)=1 \\ \operatorname{res}(f(g)g')&=\operatorname{res}(g'/g) \\ &=\operatorname{ord}(g) \\ &=\operatorname{res}(f)\operatorname{ord}(g) \end{aligned} \]

Compositional Inverse

Denote \(A(x)\circ B(x):=A(B(x))\).

Proposition: \(f(x):=\sum_{k\geq 1}f_kx^k\) has a compositional inverse \(f^{\langle -1\rangle}(x)\) if and only if \(f(0)=0\neq f'(0)\), and in this case \(f^{\langle -1\rangle}(x)\) is unique. Furthermore: if \(g(x)=\sum_{k\geq 1}g_kx^k\) satisfies \(f(g(x))=x\) or \(g(f(x))=x\), then \(g(x)=f^{\langle -1\rangle}(x)\).

Proof: Consider

\[ \begin{aligned} g(f(x))&=g_1(f_1x+f_2x^2+f_3x^3+\cdots ) \\ &+g_2(f_1x+f_2x^2+\cdots )^2 \\ &+g_3(f_1x+\cdots )^3 \\ &+\cdots \\ &=g_1f_1x+(g_1f_2+g_2f_1^2)x^2+(g_1f_3+2g_2f_1f_2+g_3f_1^3)x^3+\cdots \end{aligned} \]

Since \(g(f(x))=x\), we have the following system of equations

\[ \begin{cases} g_1f_1&=1 \\ g_1f_2+g_2f_1^2&=0 \\ g_1f_3+2g_2f_1f_2+g_3f_1^3&=0 \\ \vdots \end{cases} \]

We can solve the first equation only when \(f_1\neq 0\), then we can solve for \(g_2,\dots\) successively.

In particular, consider \(f(h(x))=x\). Then \(g(f(h(x)))=g(x)\), and thus \(g(x)=g\circ f\circ h(x)=x\circ h(x)=h(x)\).

Lagrange Inversion Formula

Let \(f(x),g(x)\in\mathbb{C}\lbrack\lbrack x\rbrack\rbrack\) satisfy \(f(g(x))=g(f(x))=x\). Take \(\Phi(x)\in\mathbb{C}\lbrack\lbrack x\rbrack\rbrack\) (or \(\Phi(x)\in\mathbb{C}\left(\left(x\right)\right)\)). Then

\[ \begin{aligned} \lbrack x^n\rbrack\Phi(f(x))&=\lbrack x^{n-1}\rbrack\Phi(x)\frac{g'(x)}{g(x)}\left(\frac{x}{g(x)}\right)^n \\ &=\lbrack x^{-1}\rbrack\frac{\Phi(x)g'(x)}{g(x)^{n+1}} \end{aligned} \]

Proof:

\[ \begin{aligned} \lbrack x^n\rbrack\Phi(f(x))&=\operatorname{res}\left(\frac{\Phi(f(x))}{x^{n+1}}\right) \\ &=\operatorname{res}\left(\frac{\Phi(f(g(x)))g'(x)}{g(x)^{n+1}}\right)\cdot \left(\operatorname{ord}(g(x))\right)^{-1} \\ &=\operatorname{res}\left(\frac{\Phi(x)g'(x)}{g(x)^{n+1}}\right) \end{aligned} \]

Some readers might be more familiar with the following version: For \(k\in\mathbb{Z}_{\geq 0},n\in\mathbb{Z}_{>0}\),

\[ \lbrack x^n\rbrack f(x)^k=\frac{k}{n}\lbrack x^{n-k}\rbrack\left(\frac{x}{g(x)}\right)^n \]

or

\[ \begin{aligned} \lbrack x^n\rbrack \Phi(f(x))&=\frac{1}{n}\lbrack x^{n-1}\rbrack \Phi'(x)\left(\frac{x}{g(x)}\right)^n \\ &=\frac{1}{n}\lbrack x^{-1}\rbrack\frac{\Phi'(x)}{g(x)^n} \end{aligned} \]

Observe that

\[ \begin{aligned} \operatorname{res}\left(\frac{\Phi'(x)}{g(x)^n}-n\frac{\Phi(x)g'(x)}{g(x)^{n+1}}\right)&=\operatorname{res}\left(\left(\frac{\Phi(x)}{g(x)^n}\right)'\right) \\ &=0 \end{aligned} \]

This can be derived from the part we have already proven.

References

Richard P. Stanley and Sergey P. Fomin. Enumerative Combinatorics Volume 2 (Edition 1).
Ira M. Gessel. Lagrange Inversion.

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