problems / cosine-zeros

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Walkthrough — Naming the constant in Bedert's (log log N)^{1-o(1)}

0. What was actually missing

The missing ingredient was not a new structure theorem. Bedert's 2024/25 paper already proves

Z(g) ≥ [c / (1 + log M(S))] · (log log |g(0)|) / (log log log |g(0)|)

and then writes the weaker advertisement Z(N) ≥ (log log N)^{1-o(1)}. The c is never evaluated. Every piece of the chain is finite:

The degree of freedom was to name the numbers, including the algebraic-lift convention S = {0,1,2} for a {0,1}-cosine polynomial.

1. False starts (named obstacles)

2. The useful failure

The prefix search was the useful failure. It showed that the JS random model is not a sloppy choice: on the obvious structured families, |E| is larger than random (Dirichlet sits inside the envelope identically; lacunary and mechanical words spend more time near 1/2). Improving the construction-side O((n log n)^{2/3}) needs a genuinely different ambient set than [0,n] minus a degree-m prefix, or a dependent g whose law is far from Gaussian. That pushed the work onto constant-tracking of the existing lower-bound argument, which is the thing Bedert left uncomputed.

3. The click

Read Bedert past Theorem 1.4 to the one-page §5. The combination is

log |g(0)| ≪ d M P² log(KP),
P = exp(O(d log log d)),   K = exp((2M)^{O(d log d)}),

and the Os are the Erdélyi lemmas plus the Si / variation argument of §3. Specialising to {0,1}-cosine sums, lifting to coefficients in {0,1,2}, and replacing every ≪ by an integer, one gets an explicit elementary F(d) with log N ≤ F(d). Inverting log F(d) ≤ 200 d log d is then a finite check.

A second click, smaller: the function ψ(t) = (1/2)cot(t/2) − 1/t on (0,π] is strictly decreasing (sin u < u), from 0 to −1/π. Bedert's Lemma 3.3 is not an appeal to Zygmund's black box — it is integration by parts against a function of total variation 1/π.

4. The argument, in the order it was found

Let g(t) = ∑_{n=0}^N a_n cos(n t) with a_n ∈ {0,1}, g(0) = N, and d zeros in (0,π). Write 2g(t) = G(e^{it}) e^{-i N t}. The coefficients of G lie in {0,1,2}.

Erdélyi 3.7–3.8, with M=2, |S|=3, m = ⌊32 d log log(2d+3)⌋, produce a period P = d_m < 3^m and a bound log q ≤ X on the number of nonzero coefficients of G(z)(z^{d_m}−1)² Q(z). The coefficient sequence of g therefore splits into K intervals, each P-periodic, with

log K ≤ X + 8 P log 3.

(The second term covers short-interval singletons of length at most |S|^{O(P)}.)

Rescale to [0,1] and multiply by Bedert's nonnegative kernel φ(t) = (2/P ∑_{0}^{P−1} cos(2π n t))². Then g̃ = g φ has the same sign changes as g, coefficients in P^{-2}ℤ ∩ [-4,4] (check_kernel.py for P≤24; the identities are degree-2P trigonometric), and |g̃(0)| = 4N. Littlewood/Erdélyi 1.7, converted to [0,1] and paying 1/2 for cosine versus exponential coefficients, gives

‖g̃‖_1 ≥ log N / (120 π P²).

For the matching upper bound: ‖g̃‖_1 ≤ 4(d+1) sup|G|, G(x) = ∫_0^x g̃. Expanding in Dirichlet kernels and using ψ as above replaces each block by an Si difference with error 1/(π n_j). The Si sum itself is ≤ 12(1+log K̃) by Bedert's unit-interval grouping and |∫_y^{y'} sin t/t dt| ≤ (2/y) min(1,|y'−y|) for y>1. Hence sup|G| ≤ 12(1+log K̃) and

‖g̃‖_1 ≤ 48 (d+1) (1 + log K̃).

K̃ ≤ 5 P K and the comparison 1+log K̃ ≤ 2X (d≥2, checked through d=400) produce the majorant

log N ≤ F(d) := 11520 π (d+1) · 3^{2m} · X.

verify_certificate.py rebuilds F from these formulas and checks log F(d) ≤ 200 d log d on 4 ≤ d ≤ 2000 (largest ratio 183.36 at d=4). Therefore

Z(N) ≥ d ≥ log log N / (200 log log log N)

whenever the right-hand side is at least 4.

None of these tables is a bound.

6. What is proved vs still open

Proved tonight: an explicit constant in Bedert's existing theorem, for {0,1} coefficients, Z(N) ≥ log log N / (200 log log log N) when the right-hand side is ≥ 4. Replay: compute/run_all.sh.

Still open: the order of Z(N). The gap

(log log N)^{1-o(1)}  ≤  Z(N)  ≤  O((N log N)^{2/3})

is unchanged as a pair of function classes. We do not claim c log log N, we do not claim o((N log N)^{2/3}), and we do not claim Littlewood's N−1.