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Statement

Notation (pp. 1, 2 and 15). p0(z)=zn−1p_0(z)=z^n-1, ∂E1(p0)={z:∣p0(z)∣=1}\partial E_1(p_0)=\{z:\lvert p_0(z)\rvert=1\} its lemniscate, ℓ\ell arclength, D(0,r)D(0,r) the open disk of radius rr about the origin, and BB the beta function.

Lemma 3.2 (The lemniscate for p0p_0, p. 20). One has

ℓ(∂E1(p0))=∫−ππ∣1+eiα∣−n−1n dα=21/nB(12,12n)(3.5)\ell(\partial E_1(p_0))=\int_{-\pi}^{\pi}\lvert1+e^{i\alpha}\rvert^{-\frac{n-1}{n}}\,d\alpha =2^{1/n}B\Bigl(\frac12,\frac1{2n}\Bigr) \qquad (3.5)

and, for every 0<r0≤10<r_0\le1,

ℓ(∂E1(p0)∩D(0,r0))=2nr0+O(r02n+1).(3.6)\ell(\partial E_1(p_0)\cap D(0,r_0))=2nr_0+O(r_0^{2n+1}). \qquad (3.6)

Consequently, with Ir0I_{r_0} the set of α∈(−π,π)\alpha\in(-\pi,\pi) with ∣1+eiα∣1/n≥r0\lvert1+e^{i\alpha}\rvert^{1/n}\ge r_0,

ℓ(∂E1(p0)∖D(0,r0))=∫Ir0∣1+eiα∣−n−1n dα=ℓ(∂E1(p0))−2nr0+O(r02n+1).(3.7)\ell(\partial E_1(p_0)\setminus D(0,r_0))=\int_{I_{r_0}}\lvert1+e^{i\alpha}\rvert^{-\frac{n-1}{n}}\,d\alpha =\ell(\partial E_1(p_0))-2nr_0+O(r_0^{2n+1}). \qquad (3.7)

The OO constant is absolute (Section 2.1, p. 15).

On the factor 21/n2^{1/n}: the print's (3.5) ends with B(12,12n)B(\frac12,\frac1{2n}) without the factor 21/n2^{1/n}. The factor is restored here because (1.1) (p. 2) states ℓ(∂E1(p0))=21/nB(12,12n)\ell(\partial E_1(p_0))=2^{1/n}B(\frac12,\frac1{2n}) and the proof (p. 21) rewrites the integral as 2n+1n∫0π/2(cos⁡t)−n−1n dt2^{\frac{n+1}n}\int_0^{\pi/2}(\cos t)^{-\frac{n-1}n}\,dt, which the beta identity it cites turns into 21/nB(12,12n)2^{1/n}B(\frac12,\frac1{2n}). By Stirling's formula the paper records (1.2) (p. 2): ℓ(∂E1(p0))=2n+4log⁡2+O(1/n)\ell(\partial E_1(p_0))=2n+4\log2+O(1/n) as n→∞n\to\infty.

Proof pointer

Pp. 20--21. Apply the arclength formula (3.1) to p0p_0 outside a small disk and let its radius shrink: each α\alpha contributes the nn roots of zn=1+eiαz^n=1+e^{i\alpha}, each with ∣z∣n−1=∣1+eiα∣(n−1)/n\lvert z\rvert^{n-1}=\lvert1+e^{i\alpha}\rvert^{(n-1)/n}, which gives the integral; the substitution t=α/2t=\alpha/2 and the beta integral give its value. For (3.6), formula (3.3) on an annulus, with zp0′(z)=n(1+p0(z))zp_0'(z)=n(1+p_0(z)), shows that inside D(0,r0)D(0,r_0) the lemniscate is 2n2n curves from the origin, each meeting every circle ∣z∣=r<r0\lvert z\rvert=r<r_0 once transversally, with radial density 1+O(r2n)1+O(r^{2n}). Subtracting gives (3.7).

Read depth

Claims checked: Lemma 3.2, (1.1), (1.2) and the proof on pp. 20--21 were read on the page images of arXiv:2512.12455v2, and the value of the beta integral was recomputed. The general formulae (3.1) and (3.3) of Lemma 3.1 were not checked. Nothing here is independently reviewed.

Dependencies

Lemma 3.1 of the same paper (p. 18), the arclength formulae (3.1) and (3.3); no page of this corpus.

Source. Terence Tao, The maximal length of the Erdős–Herzog–Piranian lemniscate in high degree, arXiv:2512.12455 (2025), version v2 of 22 December 2025; the edition read is named on the source card.

Bears on

  • Problem 114: computes the length of the conjectured extremal lemniscate, the value that Theorem 1.1 compares every monic polynomial against; it proves no bound for other polynomials.