Lagmental Vicfred

Lagrange Inversion Extracts Coefficients from Recursive Series by Vicfred

When T equals x times phi of T, coefficients of functions of T can be read directly from powers of phi. I want the notation, the mechanism, and the failure mode visible at the same time.

Start locally

A sequence \((a_n)_{n\ge0}\) becomes a formal series \(A(x)=\sum_{n\ge0}a_nx^n\). Algebra on \(A(x)\) translates recurrences, convolution, and recursive constructions into coefficient identities.

$$ T(x)=x\phi(T(x)),\qquad\phi(0)\ne0 $$

The first display fixes the mathematical data. I label it \(\mathsf{data}\) mentally, while the next is the \(\mathsf{claim}\); the bridge between them is the displayed \(\Longrightarrow\), not an automatic implication.

$$ [x^n]F(T(x))=\frac1n[u^{n-1}]F'(u)\phi(u)^n $$

Compute before generalising

The computation below is not a second theorem. It is a checksum for the definitions and a place to inspect the difficult LaTeX at full size.

$$ T=x(1+T)^2\Longrightarrow[x^n]T=\frac1n[u^{n-1}](1+u)^{2n}=\frac1n\binom{2n}{n-1} $$

The global view

The compact alignment is a local map of the argument: assumptions on the first row, consequence on the second. Any generalisation must preserve that dependency.

$$ \begin{aligned} \mathsf{D}\;&:\quad T(x)=x\phi(T(x)),\qquad\phi(0)\ne0,\\[5pt] \mathsf{C}\;&:\quad [x^n]F(T(x))=\frac1n[u^{n-1}]F'(u)\phi(u)^n. \end{aligned} $$

Edge conditions

Formal power series permit algebra without analytic convergence, but substitution and inversion still require the correct constant terms.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] [x^n]F(T(x))=\frac1n[u^{n-1}]F'(u)\phi(u)^n \end{gathered}} $$

The result is compact enough to reuse without pretending that the caveat has disappeared. The worked line remains the quickest consistency check.

This article was posted on Wed 10 July 2024. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.