ARM B: window asymptotics for the depth-1 damage kernel

Support run 5418c63e, 2026-08-24, for run 872d7dce (hunts/r_c7f779). Instrument: probe_arm_b.py (runs end to end in ~7 s; --quick in <1 s). Data: results.json. Nothing here bears on RH (docs/08).

Answer, in one paragraph

Option (ii), and stronger than asked. The window centres minus 2 pi d are not linear in d; they decay like K/(2 pi d) with K = 0.893732172363013, so the limiting offset is exactly 0 and the windows track 2 pi with an O(1/s) correction whose sign is positive (the window sits slightly to the right of the lattice point). No multiple of 2 pi with d >= 1 falls outside a depth-1 window, and this is not a horizon statement: on the lattice D(1, .) collapses to an explicit rational function whose numerator is a cubic in s^2 with one nonnegative root at s = 1.3307, so D(1, 2 pi d) > 0 for every integer d >= 1, with no d_max. The minimum over d >= 1 of the distance from 2 pi d to the nearest window edge is 0.817252917249998, attained at d = 2, and it is bounded away from 0: the distance increases monotonically for d >= 2 toward w0 = arccos(sech 1) = 0.8657694832396585.

For the caller's purpose: the ceiling rho* <= 0.153216295 is not threatened by drift. The measured input it rested on (P = 0 on the critical lattice out to d = 4000) is now derived for all d, not measured to a horizon.

One caveat worth stating because it is the only exception: d = 0 is outside, D(1, 0) = -0.911564713 < 0. That is harmless here, because P(T,y) sums over p != q, so tau = 0 never occurs, but a reader who takes "every multiple of 2 pi" literally will trip over it.

What is measured and what is derived

1. The kernel has an exact one-term form

sinh(zp/2) + sinh(zm/2) = 2 cos(1/sqrt2) sinh(z/2) and sinh(zm/2) - sinh(zp/2) = -2i cosh(z/2) sin(1/sqrt2), so putting the two terms of ghat over the common denominator z^2 + 2 gives, exactly,

ghat(z) = [ alpha z sinh(z/2) + beta cosh(z/2) ] / (z^2 + 2), alpha = 2 cos(1/sqrt2) = 1.5204891941512604, beta = 2 sqrt2 sin(1/sqrt2) = 1.8374507397311368 = 2 * A_CONST.

Checked against hunts/frontier_math/gram_form.damage (read-only import) at 24 (a, s) points: worst relative difference 2.92e-12, which is the two-term form's own cancellation, not a discrepancy. Against an independent mpmath evaluation of the two-term form at dps = 50: worst relative 6.52e-16. Squaring and using sinh^2 = (cosh z - 1)/2:

ghat(z)^2 = [ (alpha^2/2) z^2 (cosh z - 1) + alpha beta z sinh z

• (beta^2/2)(cosh z + 1) ] / (z^2+2)^2.

2. The asymptotic, with the next-order term the caller asked for

Expanding (z^2+2)^{-2} = z^{-4}(1 - 4z^{-2} + ...) and z^{-n} for z = 1 + is, and taking D = -Re ghat^2:

`D(1,s) = (alpha^2/2)(cosh1 cos s - 1)/s^2 + alpha C sin s / s^3

• (E cos s + F)/s^4 + O(s^-5)`

C = beta cosh1 - alpha sinh1 = 1.0484539380169988 E = cosh1 (alpha^2 - beta^2)/2 + 3 alpha beta sinh1 = 9.028736331391707 F = -(alpha^2 + beta^2)/2 = -2.8440563052346253

The caller's hint is right at leading order. (alpha^2/2)(cosh1 cos s - 1) is exactly (4 cos^2(sqrt2/2)/s^2)[sinh^2(1/2) cos^2(s/2) - cosh^2(1/2) sin^2(s/2)], 2 pi-periodic with maxima exactly at s in 2 pi Z.

The next-order term is alpha C sin s / s^3, and it is odd. That is the whole answer to the drift question: an odd perturbation of an even profile translates it, it does not widen it. Relative error of the truncations against the exact D:

3. The edge law

Put s = 2 pi d + x, multiply by S^2 with S = 2 pi d + x. Zeroth order the edges satisfy cos x = sech 1, so x = +/- w0. First order the shift is alpha C sin x0 / (c2 cosh1 sin x0), and the sin x0 cancels, so both edges move the same way and the window translates without changing width. Second order the two shifts are equal and opposite, so they cancel in the centre and add in the width:

centre(d) - 2 pi d = K/s + O(s^-3), no 1/s^2 term half_width(d) = w0 + W2/s^2 + O(s^-3) margin(d) = w0 - K/s + W2/s^2 + O(s^-3), s = 2 pi d

w0 = arccos(sech 1) = 0.8657694832396585 K = alpha C/(c2 cosh1) = 0.893732172363013 W2 = Geff/(c2 sinh1) = 1.779638303047668

Geff keeps the term -alpha C x sin x that comes from expanding 1/S about 1/s. Dropping it is the one mistake this run made and caught: it gives W2 = 2.5534, 43% too large, and the numerics rejected it before the number was written down. The corrected W2 matches high-precision measurement:

The residual falls by two decades per decade of s, i.e. it is O(s^-2), so the half-width's next correction after W2/s^2 is O(s^-4) and not O(s^-3): the odd order drops out of the width, as it did at first order. W2 is confirmed to ten digits. (This check has to be done at dps = 60: half_width - w0 is 4.5e-12 at d = 1e5, below what the double scan can resolve.)

Measured against the double-precision scan, worst |half_width - (w0 + W2/s^2)| over 10 <= d <= 15915 is 2.72e-06.

4. The scan, out to s = 1e5

D(1, .) scanned on [1e-6, 1e5] at step 0.002 (5.0e7 samples), every sign change bracketed and bisected to double precision:

• 31,830 sign changes, i.e. 1.999938 per 2 pi period over 15915.49 periods. No anomalous window anywhere: D is positive at every window midpoint and negative at every midpoint between consecutive windows (both checks in results.json).
• 15,915 windows containing the 15,915 lattice points 2 pi d, 1 <= d <= 15915. Zero lattice points outside a window.
• The first seven sign changes are 5.398373, 7.284924, 11.749118, 13.506670, 18.023030, 19.765262, 24.298419. s = 0 sits in a negative stretch, and the first window opens at 5.398373.

Edges confirmed independently at dps = 50 with mpmath, through a different code path (the un-recombined two-term ghat): agreement 8.7e-16 at d = 1, 8.5e-12 at d = 15915 (the latter is the double result's own ulp scale).

(centre - 2 pi d) * s goes from 0.367336 at d = 1 to 0.893732 at d = 15915, against the derived K = 0.893732172363; the tail mean over d > 1591 is 0.8937322997 with standard deviation 4.1e-07. A linear fit of the offset against d returns slope -2.55e-08 per d, consistent with zero and with the wrong functional form, which is the point: the offset is not linear in d.

5. The lattice is exactly solvable, so there is no horizon

At s = 2 pi d we have cos s = 1, sin s = 0, so cosh z and sinh z are real at z = 1 + is. Then |(z^2+2)^2|^2 = (s^4 - 2s^2 + 9)^2 and taking -Re gives the exact identity

D(1, 2 pi d) = P(u) / (u^2 - 2u + 9)^2, u = (2 pi d)^2, P(u) = p u^3 - (3p - 3q + r) u^2 - (5p + 2q - 10r) u - 9(p + q + r),

p = (1 + cos sqrt2)(cosh 1 - 1) = 0.6277706355638578 q = 2 sqrt2 sin(sqrt2) sinh 1 = 3.2833052932216624 r = 2 (1 - cos sqrt2)(cosh 1 + 1) = 4.293006489071762

checked against direct evaluation at d = 1, 2, 3, 10, 100, 4000, 15915, 1e6 to 8.4e-16 relative or better (worst case d = 1). Numerically

P(u) = 0.6277706 u^3 + 3.6735975 u^2 + 33.2246011 u - 73.8367418.

u^2 - 2u + 9 = (u-1)^2 + 8 > 0, so sign D = sign P. Every coefficient of P but the constant is positive, so P is strictly increasing on u >= 0 and has exactly one nonnegative root, u* = 1.7707490180 (s* = 1.3306949380; the cubic discriminant is -2.224e+05 < 0, confirming the other two roots are complex). For every integer d >= 1, u = (2 pi d)^2 >= 4 pi^2 = 39.478 > u*, hence

D(1, 2 pi d) > 0 for every integer d >= 1.

That is a derived statement about all d, not a scan to a horizon, and it is exactly what P(T,y) = 0 on the critical 2 pi lattice requires (P sums [-D(2y, tau_pq)]^+ over p != q, and on that lattice tau_pq = 2 pi d with d != 0; D is even in s, so d >= 1 covers it).

6. What this does to rho* <= 0.153216295

RESULTS-37fb06a9.md §2 derives the ceiling from the measured fact that P = 0 on the critical lattice out to d = 4000. Section 5 replaces that measurement with a derivation valid for all d. So, on this ingredient, the ceiling is exact and not "slightly below": there is no drift that eventually carries a lattice point out of a window, because the drift is +K/s -> 0 and the margin converges upward to 0.86577.

Two limits on that, stated plainly:

1. This settles only the P = 0 ingredient. The ceiling also rests on the Poisson-summation limit B/k -> c2(0) - A^2 and on Shq(1/2), neither of which this run touched.
2. The derivation in section 5 is a real-coefficient cubic evaluated in double precision. Its sign structure (three positive coefficients, one negative constant) is robust to that by a wide margin (the smallest coefficient is 0.628 and the constant is -73.8), but the reserved word is not used and this is not enclosure-carrying arithmetic.

7. What was not settled

• No remainder bound on the O(s^-5) term. The edge law in section 3 is an asymptotic series checked numerically, not a theorem with explicit constants. The margin >= 0.8172 claim is therefore measured for d <= 15915 and derived-asymptotically beyond it. The sign claim (section 5) has no such gap; only the margin size does.
• Only a = 1 (y = 1/2) was studied, because that is what the caller asked. The expansion in section 2 is written for general a and the leading window half-width is arccos(sech a), which degenerates as a -> 0; nothing here says where that becomes a problem.

The doors

1. Active constraint at the optimum. For the P = 0 question the binding object is the single cubic root u* = 1.7707, and it is not close: the nearest lattice point sits at u = 39.478, a factor of 22.3 above it. This constraint has enormous slack and is not what limits rho*. The ceiling's binding objects are elsewhere: B/k -> c2(0) - A^2 and Shq(1/2), and this run says so rather than pretending its own result is the wall.
2. Frozen-constant inventory. (i) The scan horizon s_max = 1e5, now redundant for the sign question, still live for the margin question. (ii) The scan step 0.002, which bounds the width of a feature that could have been missed at 1.0e-3; the 1.999938 sign changes per period says nothing was. (iii) a = 1, i.e. y = 1/2; the whole calculation is parametric in a and nothing here explores the trade. (iv) The truncation of the expansion at s^-4, which fixes the accuracy of W2 and nothing else.
3. Information class. The section-5 result reads less data than the scan did, not more: it needs only the value of D on 2 pi Z, where the kernel is rational. Extending it from the sign to the margin, a lower bound on the distance to the nearest edge for all d, requires reading D off the lattice, where cos x and sin x re-enter and the object stops being rational. That is the next door, and it is a genuinely different information class.

Grade: the asymptotic series and the lattice identity are derived; every edge location, window count and margin is measured in double precision with mpmath confirmation at dps = 50. The reserved word is not used. Nothing here is evidence for or against RH.