CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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Treating a double pole as simple gives the wrong answer

Statement refuted

Refuted claim: the simple-pole residue rule can be used unchanged at a pole of order two.

Facts & Assumptions

Given: The function $f (z) = e^{z} / (z - 1)^{2}$ .

[L1]

A simple pole has residue $\lim_{z \to a} (z - a) f (z)$ (At a simple pole the residue is the limit of (z-a)f(z)).

[L2]

A pole of order $2$ has residue $\frac{d}{d z} ((z - a)^{2} f (z)) ∣_{z = a}$ (Residue formula for a pole of order m).

Counterexample

technique · computation

1.1

The point $a = 1$ is a pole of order $2$ of $f$ , and [L2] gives the correct residue $Res (f, 1) = \frac{d}{d z} (e^{z}) ∣_{z = 1} = e .$

L2algebra

2.1

If one incorrectly applies the simple-pole rule from [L1], one obtains $\lim_{z \to 1} (z - 1) \frac{e^{z}}{(z - 1)^{2}} = \lim_{z \to 1} \frac{e^{z}}{z - 1},$ which does not exist as a finite complex number. So the simple-pole rule does not recover the residue at a double pole.

L1step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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

Jeremy Orloff, MIT 18.04 Topic 7: Taylor and Laurent Series (standard reference, not scraped)