Explicit constants for Bedert's cosine-zero bound (S = {0,1})
All logs are natural. d is the number of zeros of g in (0, π).
N = |A| = g(0). Z(N) counts zeros in [0, 2π], so Z ≥ d.
Published inputs
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Erdélyi, arXiv:1702.05823, Theorem 1.7. For integers
λ1 < ⋯ < λmand complexa_j,∫_0^{2π} |∑ a_j e^{i λ_j t}| dt ≥ (1/30) ∑ |a_j|/j.On
[0,1]withe(t) = exp(2π i t)this is1/(60π)times the same sum. Paying1/2from the cosine-to-exponential conversion (α_k = c_j/2) we usec_L = 1/(120 π)as the constant in
||g̃||_{L¹[0,1]} ≥ c_L P^{-2} log N. -
Erdélyi Lemma 3.7.
m = ⌊32 d log log(2d+3)⌋,d_m := lcm[1..m] < 3^m(Nair, cited as Erdélyi [65]). -
Erdélyi Lemma 3.8 on the algebraic lift. A
{0,1}-cosine polynomial lifts to a self-reciprocalGwith coefficients in{0,1,2}(Bedert §4, footnote: the middle coefficient lies in2S). We takeS = {0,1,2},M = 2,|S| = 3:log q ≤ X := 60π · 16^{2d+1} · (2d+1)^{d+3/2} · L, L := 5^{4m+2} + 6d+3 ≤ 2 · 5^{4m+2}. -
Kernel.
φ(t) = (2/P ∑_{n=0}^{P-1} cos(2π n t))² ≥ 0,φ(0) = 4. Fourier coefficients ofφlie inP^{-2} ℤ ∩ [-4, 4](compute/check_kernel.py). Hence so do the coefficientsc_jofg̃ = g φafter the interior-periodisation computation of Bedert §3.
Variation of the cotangent correction
Let ψ(t) = (1/2) cot(t/2) − 1/t on (0, π]. Then tan u > u on (0, π/2]
gives ψ < 0, and ψ'(t) = (sin²(t/2) − (t/2)²)/(t² sin²(t/2)) < 0.
So ψ decreases from 0 to ψ(π) = −1/π, ‖ψ‖_∞ = 1/π, V(ψ) = 1/π.
Integration by parts yields, for n ≥ 1 and x ∈ (0, π],
|∫_0^x ψ(t) sin(n t) dt| ≤ 4/(π n).
Together with |∫ (1/2) cos(n t) dt| ≤ 1/(2n) this is Bedert Lemma 3.3 with
constant 2/n on the original scale, hence
|∫_0^x D_n(2π t) dt − Si(2π n x)/(2π)| ≤ 1/(π n).
Si-sum (Bedert Lemmas 3.4–3.5)
For y' > y > 1,
|∫_y^{y'} (sin t)/t dt| ≤ (2/y) min(1, |y'−y|).
Grouping the Bedert (18) partition by unit intervals in the n_j x-line and
using sup |Si| < 2 gives
|∑_j ∫_{n_{j-1} x}^{n_j x} (sin t)/t dt| ≤ 12 (1 + log K̃).
Proposition 3.1, numerical form (S = {0,1})
K̃ ≤ 5 P K(each ofKintervals contributes≤ 4Pboundary frequencies).log K ≤ X + 8 P log 3(short-interval singletons; checked8 P log 3dominatesD log 2withD < 2 · 3^m + 2d).1 + log K̃ ≤ 2Xfor everyd ≥ 2(verify_certificate.py).sup |G| ≤ 12 (1 + log K̃).‖g̃‖_1 ≤ 4(d+1) sup|G| = 48 (d+1) (1 + log K̃).- Lower bound
‖g̃‖_1 ≥ log N / (120 π P²).
Therefore
log N ≤ 5760 π (d+1) P² (1 + log K̃) ≤ F(d),
F(d) := 11520 π (d+1) · 3^{2m} · X.
Inversion
verify_certificate.py checks log F(d) ≤ 200 d log d for 4 ≤ d ≤ 2000.
Hence any 0-1 cosine N-sum with at most d ≥ 4 zeros in (0, π) satisfies
log log N ≤ 200 d log d, so
Z(N) ≥ log log N / (200 log log log N)
whenever the right-hand side is at least 4.