TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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.

Residue formula for a pole of order m

Statement

If $a$ is a pole of order $m \geq 1$ of $f$ , then

$Res (f, a) = \frac{1}{(m - 1)!} \lim_{z \to a} \frac{d^{m - 1}}{d z^{m - 1}} ((z - a)^{m} f (z)) .$

Equivalently, if $g$ denotes the holomorphic extension of $(z - a)^{m} f (z)$ across $a$ , then

$Res (f, a) = \frac{g^{(m - 1)} (a)}{(m - 1)!} .$

Facts & Assumptions

Given: A pole of order $m \geq 1$ of $f$ at $a$ .

[L1]

If $a$ is a pole of order $m$ , then $g (z) : = (z - a)^{m} f (z)$ extends holomorphically across $a$ with $g (a) \neq 0$ (Characterizations of poles).

[L2]

The residue is the normalized contour integral on every sufficiently small circle around the pole (The residue is the normalized small-circle integral).

[L3]

For a holomorphic function $g$ , the integral formula $g^{(m - 1)} (a) = \frac{(m - 1)!}{2 π i} \int_{∣ ζ - a ∣ = ρ} \frac{g (ζ)}{(ζ - a)^{m}} d ζ$ holds on every sufficiently small circle around $a$ (The higher-derivative form of the global Cauchy formula).

Proof

technique · direct

1.1

Let $g$ be the holomorphic extension from [L1]. On a sufficiently small punctured circle one has $f (ζ) = g (ζ) (ζ - a)^{- m}$ , so [L2] gives $Res (f, a) = \frac{1}{2 π i} \int_{∣ ζ - a ∣ = ρ} \frac{g (ζ)}{(ζ - a)^{m}} d ζ .$

L1L2algebra

2.1

Applying [L3] to the same circle gives $\frac{1}{2 π i} \int_{∣ ζ - a ∣ = ρ} \frac{g (ζ)}{(ζ - a)^{m}} d ζ = \frac{g^{(m - 1)} (a)}{(m - 1)!},$ so the displayed residue formula follows.

step 1.1L3

3.1

Since $g$ is holomorphic at $a$ , the limit of its $(m - 1)$ st derivative at $a$ is just the value $g^{(m - 1)} (a)$ , so the derivative-limit form is the same statement.

step 2.1∎

Depends on

Used by

The residue of eᶻ/z³ from the pole-derivative formula Example

Dependency tree · two levels

26 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

Jean-Baptiste Campesato, MAT334 course page and notes index (standard reference, not scraped)
Jeremy Orloff, MIT 18.04 Topic 7: Taylor and Laurent Series (standard reference, not scraped)