LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Formal residues satisfy integration by parts, logarithmic differentiation, and change of variables

Statement

Let $K$ be a field. For $f, g \in K ((x))$ ,

${res}_{x} (D f) = 0, {res}_{x} ((D f) g) = - {res}_{x} (f D g) .$

For nonzero $f$ ,

${res}_{x} ⁣ (\frac{D f}{f}) = v_{x} (f),$

where the integer on the right acts by repeated addition and additive inverses in $K$ . If, in addition, $K$ contains $Q$ and $g \in x K ⟦ x ⟧$ has nonzero linear coefficient, then, for every $F \in K ((x))$ for which $F \circ g$ is formed by Laurent substitution,

${res}_{x} ((F \circ g) D g) = {res}_{x} (F) .$

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[F1]

Formal Laurent series have support bounded below, coefficientwise addition, finite convolution in each degree, least exponent $v_{x}$ , termwise derivative, and residue ${res}_{x} (f) = [x^{- 1}] f$ (Formal Laurent series $K ((x))$ , their order, derivative, and residue).

[F2]

The formal derivative is additive, obeys the product rule, and satisfies $D (f^{m}) = m f^{m - 1} D f$ for $m \geq 1$ while $D (1) = 0$ (Formal differentiation is linear and satisfies product, power, quotient, chain, and coefficient-recovery laws).

[F3]

A formal power series is a unit exactly when its constant coefficient is a unit (A formal power series is a unit exactly when its constant coefficient is a unit).

Proof

technique · inspect Laurent monomials and leading terms

1.1

The same finite coefficient calculation as for power series gives $D (f g) = (D f) g + f D g$ in $K ((x))$ . The coefficient of $x^{- 1}$ in $D f$ could only come from differentiating the constant term, and that contribution is $0$ . Applying this to the product rule gives integration by parts.

givenF1F2

1.2

Write $f = x^{m} u$ , where $m = v_{x} (f)$ and $u \in K ⟦ x ⟧$ has nonzero constant coefficient. Then $u$ is a unit by the unit criterion, and $D f / f = m x^{- 1} + D u / u$ . Since $D u / u \in K ⟦ x ⟧$ , it has no $x^{- 1}$ coefficient. The residue is therefore $m$ .

givenF1F3

2.1

For the change of variables, linearity reduces the claim coefficientwise to $F = x^{m}$ . If $m \neq - 1$ , then $g^{m} D g = D (g^{m + 1}) / (m + 1)$ and its residue is $0$ by step 1.1; division is legitimate because $K$ contains $Q$ . If $m = - 1$ , write $g = x u$ with $u (0) \neq 0$ ; then $g^{- 1} D g = x^{- 1} + u^{- 1} D u$ has residue $1$ by step 1.2. Thus the residue equals that of $x^{m}$ in every case, and locally finite summation extends the identity to $F$ .

step 1.1step 1.2givenF1F2F3

3.1

Steps 1.1-2.1 prove the derivative, integration-by-parts, logarithmic-derivative, and substitution identities.

step 1.1step 1.2step 2.1∎

Depends on

Used by

Lagrange–Bürmann inversion extracts coefficients of a compositional inverse and of functions of it Theorem

Dependency tree · two levels

20 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Benjamin Sambale, An Invitation to Formal Power Series (standard reference, not scraped)