Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-30
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  • 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.
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The annulus shows Runge approximation needs a pole in each bounded complementary component

Statement refuted

A single pole placed only in the unbounded complementary component always suffices for Runge approximation on a compact annulus.

Facts & Assumptions

Given: The compact annulus K={z:1≤∣z∣≤2} and the function 1/z on a neighbourhood of K.

[L1]

A rational function with all poles outside the unit disc is holomorphic on ∣z∣<1, so its integral around ∣z∣=1 is 0 (The line integral of a continuous function admitting a primitive is that primitive's endpoint increment along every rectifiable path).

Counterexample

technique · direct
1.1

Let r be any rational function whose poles all lie in the unbounded complementary component of K. Then r is holomorphic on the closed unit disc, so [L1] gives ∫∣z∣=1r(z) dz=0.

givenL1
2.1

But ∫∣z∣=1dz/z=2πi. Therefore no such rational function can approximate 1/z uniformly on K, because the contour integral on the inner circle would preserve the limit. So the inner complementary component also needs a pole representative.

step 1.1algebra∎

Depends on

Used by

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Dependency tree · two levels

16 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