Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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  • 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 normalized large-radius loop of a monic degree-n polynomial has degree n

Statement

Let p(z)=zn+∑j<najzj be a monic complex polynomial of degree n≥1, and put S=∑j<n∣aj∣. If R>max⁡{1,S}, then the based normalized circle loop

αR(u)=h−1(p(Rh(u))/p(R)∣p(Rh(u))/p(R)∣),u∈R/Z,

is well defined and has degree n.

Facts & Assumptions

Given: A monic polynomial p(z)=zn+∑j<najzj of degree n≥1, the number S=∑j<n∣aj∣, and a real R>max⁡{1,S}.

[F1]

For a nonzero complex polynomial, degree is the final coefficient index and monic means that its leading coefficient is 1 (Formal complex polynomials, evaluation, degree, leading coefficient, and monic polynomials).

[L2]

The homeomorphism h:R/Z→S1 sends [0] to 1∈S1 ([t]↦(cos⁡2πt,sin⁡2πt) is a homeomorphism from R/Z to the unit circle).

[L4]

Path-homotopic based circle loops have the same degree (Path-homotopic based circle loops have the same degree).

[L5]

The standard loop ωm has degree m for every integer m (deg⁡(ωn)=n for every integer n).

Proof

technique · constructive
1.1givenF1L1algebra

If ∣z∣=R, then ∣∑j<najzj∣≤∑j<n∣aj∣Rj≤SRn−1<Rn=∣zn∣. The estimate includes n=1 and S=0, since R>1 and S<R.

2.1step 1.1L1L2L3construct

For s∈[0,1] put ps(z)=zn+s∑j<najzj. Step 1.1 remains strict with sS≤S, so ps never vanishes on ∣z∣=R, in particular ps(R)≠0. The formula H(s,u)=h−1(ps(Rh(u))/ps(R)∣ps(Rh(u))/ps(R)∣) is therefore a continuous based homotopy by [L2] and [L3]. At s=1 it is αR, while at s=0 it is u↦h−1(h(u)n)=ωn(u).

3.1step 2.1L4L5discharge-construct∎

Homotopy invariance and the standard-loop calculation give deg⁡(αR)=deg⁡(ωn)=n.

Depends on

Used by

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Sources