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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

  1. Erdélyi, arXiv:1702.05823, Theorem 1.7. For integers λ1 < ⋯ < λm and complex a_j,

    ∫_0^{2π} |∑ a_j e^{i λ_j t}| dt  ≥  (1/30) ∑ |a_j|/j.
    

    On [0,1] with e(t) = exp(2π i t) this is 1/(60π) times the same sum. Paying 1/2 from the cosine-to-exponential conversion (α_k = c_j/2) we use

    c_L = 1/(120 π)
    

    as the constant in ||g̃||_{L¹[0,1]} ≥ c_L P^{-2} log N.

  2. Erdélyi Lemma 3.7. m = ⌊32 d log log(2d+3)⌋, d_m := lcm[1..m] < 3^m (Nair, cited as Erdélyi [65]).

  3. Erdélyi Lemma 3.8 on the algebraic lift. A {0,1}-cosine polynomial lifts to a self-reciprocal G with coefficients in {0,1,2} (Bedert §4, footnote: the middle coefficient lies in 2S). We take S = {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}.
    
  4. Kernel. φ(t) = (2/P ∑_{n=0}^{P-1} cos(2π n t))² ≥ 0, φ(0) = 4. Fourier coefficients of φ lie in P^{-2} ℤ ∩ [-4, 4] (compute/check_kernel.py). Hence so do the coefficients c_j of g̃ = g φ after 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})

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.