# Using the Langrange Inversion Formula

I am currently going through the proof of the Theorem $$5.4.2$$ on page 38 of Enumerative Combinatorics Vol $$2$$ by R. Stanley and I think I understand the proof. The proof of Corollary $$5.4.3$$ as shown in the attached proof is just a sketch and I am unable to prove it using the hint given. All my reasoning does not lead me anywhere. Could some please show me the proof? or perhaps a hint? I saw some proofs that strictly used the Cauchy-Residue Theorem but I want to avoid using this.

• I have to say that this looks pretty clear to me. As Stanley says, for $H(x)=x^k$ this reduces to the previous theorem. In general if $H(x)=\sum h_k x^k$, then for a given $n$, the $x^n$ coefficient in $H(f(x))$ will depend only on finitely many of the $h_k$, so one can assume $H(x)$ is a polynomial etc. – Lord Shark the Unknown Oct 18 '18 at 3:47

A little elaboration on Lord Shark the Unknown's comment. Fix $$n$$ and $$f$$ and define an operator $$T: K[[x]] \rightarrow K$$ by
$$T(H) = n[x^n]H(f(x)) - [x^{n-1}]H'(x) G(x)^n.$$
We want to show $$T(H) = 0$$ for any $$H$$. Note that $$T(H) = 0$$ when $$H(x) = x^k$$ for any $$k$$. This is exactly the statement of Theorem 5.4.2. Note also that this operator is linear. You can show directly that $$T(\alpha H_1 + \beta H_2) = \alpha T(H_1) + \beta T(H_2)$$.
Thus $$\ker T$$ contains every polynomial. Now for any $$H$$, let $$H_n$$ be the truncation of $$H$$ to an $$n$$-th degree polynomial. Then $$T(H) = T(H_n)$$, since the $$x^{n+1}$$ or higher terms do not affect the coefficient of $$x^n$$ in $$H(f(x))$$ or the coefficient of $$x^{n-1}$$ in $$H'(x) G(x)^n$$. Since $$H_n$$ is a polynomial, it follows that $$T(H_n) = 0$$, and so $$T(H) = 0$$.