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156 changes: 156 additions & 0 deletions research/gaussian-tensor-nine-approximant/README.md
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# The approximant base 9 p^2 + p^6 exp(-p^2) stays inside the budget

**Claim type:** an explicit absorption bound for one approximant monomial.
**No new κ.** The section ceiling `9 p^2` and the section error
`p^6 exp(-p^2)` stay hypotheses. The selected record remains `23/10000`.

## Question

#4496 absorbs `g (9 p^2)^{g-1} p^6 exp(K p - p^2)` under the gate
`p ≥ max(64, 4K + 2C)` and `16 g ceil(log2 p) ≤ p`. If each of the `g - 1`
factors is an approximant, the base becomes `9 p^2 + p^6 exp(-p^2)`. Does

```text
g (9 p^2 + p^6 exp(-p^2))^{g-1} p^6 exp(K p - p^2)
```

stay below `2^{-C p}` under that same gate?

## Lemma

Let `p ≥ 2`, `g ≥ 1`, `K ≥ 0` and `C ≥ 0` be integers, and let
`L = ceil(log2 p)`. If

```text
p ≥ max(64, 4K + 2C) and 16 g L ≤ p,
```

then

```text
g (9 p^2 + p^6 exp(-p^2))^{g-1} p^6 exp(K p - p^2) < 2^{-C p}.
```

The case `K = 0` is the monomial in the question. A global factor `exp(K p)`
does not change the conclusion. Whenever `g log p = o(p)`, every fixed pair
`K, C` satisfies the gate for all large `p`.

For every integer `p ≥ 64`, the section error is strictly below `1/2`, so

```text
9 p^2 + p^6 exp(-p^2) < 9 p^2 + 1 < 16 p^2.
```

The base therefore remains strictly below the power of two already used for
`(16 p^2)^{g-1}`. No extra bit is spent. The exponent is at most `-C p - 4`.
At `g = p^2` the same error is greater than 1 for every integer `p ≥ 2`,
because it is larger than the inverse-ceiling monomial, which is already
greater than 1.

## Proof

The series `1 + 1 + 1/2 = 5/2 > 2` gives `e > 2`, so `exp(-p^2) < 2^{-p^2}`.
For `k ≥ 2`, `k! ≥ 2^{k-1}`: the values at 2, 3 and 4 are `2`, `6` and `24`,
and the step uses `k + 1 ≥ 2`. The tail of `e` is then at most the geometric
series of ratio `1/2`, so `e < 3 < 4` and `exp(K p) ≤ 2^{2 K p}`, with
equality only at `K = 0`.

For every integer `p ≥ 64`, `8 L ≤ p`. The base is `8 · 7 = 56 ≤ 64` at
`p = 2^6`, and `8 ≤ 64`. If `8(m + 1) ≤ 2^m`, then
`8(m + 2) ≤ 2^m + 8 ≤ 2^{m+1}`. On each dyadic interval the largest ratio of
`L` to `p` is at the left endpoint. From `8 L ≤ p`, `48 L = 6 · 8 L ≤ 6 p`.
The quadratic `4 p^2 - 3 p - 4` equals 6 at `p = 2` and increases by `8 p + 1`
at the next integer, so it stays positive. Equivalently
`6 p ≤ 8(p^2 - 1)`. Combining these estimates gives `6 L ≤ p^2 - 1`. Since
`p ≤ 2^L`, `p^6 ≤ 2^{6 L} ≤ 2^{p^2 - 1}`, and therefore

```text
p^6 exp(-p^2) < p^6 2^{-p^2} ≤ 1/2.
```

Hence `9 p^2 + p^6 exp(-p^2) < 9 p^2 + 1`. Also `1 < 7 p^2` for every
integer `p ≥ 2`, so `9 p^2 + 1 < 16 p^2`. The approximant base is strictly
smaller than `16 p^2`.

At `g = 1` the power `g - 1` is zero. For `g ≥ 2`,

```text
(9 p^2 + p^6 exp(-p^2))^{g-1} < (16 p^2)^{g-1}
= 2^{4(g-1)} p^{2(g-1)}.
```

In both cases

```text
g (9 p^2 + p^6 exp(-p^2))^{g-1} p^6 exp(K p - p^2)
< g 2^{4g-4} p^{2g+4} 2^{2 K p - p^2}.
```

The gate forces `g < 2^L`, because `L ≥ 1` and `g ≤ p/(16 L) ≤ p/16`.
Together with `p ≤ 2^L`, the product is strictly less than

```text
2^{2 g L + 4 g + 5 L - 4 + 2 K p - p^2}.
```

Clear the eighths. The four pieces scale as `8 · 2 g L = 16 g L ≤ p`,
`8 · 4 g = 2 · 16 g ≤ 2 p`, `8 · 5 L = 5 · 8 L ≤ 5 p`, and
`8 (2 K + C) p ≤ 4 p^2` once `p ≥ 4 K + 2 C`, because
`8(2 K + C) = 4(4 K + 2 C)`. The constant `-4` contributes `-32`. The sum
is at most `-4(p^2 - 2 p + 8)`. The coefficients of `p` are `1 + 2 + 5 = 8`.
At `p = 2` the polynomial equals 8, and the step to the next integer is
`2 p - 1`. It is therefore at least 8, so eight times the shifted exponent
is at most `-32`. The exponent is at most `-C p - 4`.

If `g log p = o(p)`, then `g log2 p ≤ p/32` for large `p`. For `p ≥ 3`,
`L ≤ 2 log2 p`, so the gate holds eventually, and every fixed `K` and `C`
satisfy `p ≥ 4 K + 2 C` as well.

At `g = p^2` the approximant monomial is larger than
`g (9 p^2)^{g-1} p^6 exp(-p^2)`, since the added section error is positive
and `g ≥ 2`. The smaller monomial equals
`g (9 p^2)^g exp(-p^2) · p^4 / 9`. Here
`(9 p^2)^{p^2} > 4^{p^2} > exp(p^2)` because `9 p^2 ≥ 36 > 4` and `e < 4`,
while `p^4 / 9 > 1` because `p^4 ≥ 16 > 9`. That product is greater than 1,
and so is the approximant monomial.

## Not covered

- A proof that a section inverse is at most `9 p^2`, or that a section
error is less than `p^6 exp(-p^2)`. The approximant ceiling is their sum.
- The factor `2` in front of `g`.
- Editing the #4100 notes. They still say `g p^{10 g} exp(-p^2)`.
- Any new κ.

## Reproduce

```sh
python3.12 -B scripts/research_package.py --package research/gaussian-tensor-nine-approximant --check-only
python3.12 -B scripts/research_package.py --package research/gaussian-tensor-nine-approximant
```

Base `2a13eb1827784206202db90e5dc8327cf8a250db`, priority revision 7, selection
refresh κ = `23/10000`. Upstream was fetched at 2026-10-22T01:56:48Z and was
still that commit.

Overlap snapshot: open PRs fetched 2026-10-22T01:57:24Z (350 records) and
discussions fetched 2026-10-22T01:58:13Z (3374 records). A search for
`9 p^2 + 1`, `9 p^2 + p^6` and `approximant base` returned this draft, #4496,
which leaves the base open, and the coefficient-10 packages #4486 and #4487.
A search miss is not a novelty certificate. #4496 is the predecessor fold.
HALO is the written error.

## Open next questions

1. **Does the face factor 2 still fit?** The exponent above is at most
`-C p - 4`. The first step is to charge that extra factor of 2 against
this slack of 4.

## Credit

- The section-error comparison `p^6 2^{-p^2} ≤ 1/2` is the route of #4486.
- The predecessor fold with coefficient 9 is #4496.
- The section error `p^6 exp(-p^2)` is the within-cell bound of #4452.
- The written error `g p^{10 g} exp(-p^2)` is #4100's.

Prepared by James Reed with substantial Cursor Grok 4.7 assistance. Apache-2.0.
24 changes: 24 additions & 0 deletions research/gaussian-tensor-nine-approximant/SOURCES.json
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{
"schema_version": 1,
"base_commit": "2a13eb1827784206202db90e5dc8327cf8a250db",
"files": [
{
"path": "certificates/selected-result.json",
"usage": "provenance",
"git_blob": "ed33caef5c5827693e9572d252a66f99cbb73d40",
"sha256": "5732707cc1e460b090b53f2463162749ad860de878a18f7408938671d295a9d9"
},
{
"path": "docs/research/agent-task-catalog.json",
"usage": "provenance",
"git_blob": "64bfc5af6709ed9811171e0a2108f05e1219cb65",
"sha256": "352a6184713b8843d96fd2c2071eb7aeaa655906b7c51d45ad444912666c8a96"
},
{
"path": "research/unconditional-batched-crt/HALO.md",
"usage": "input",
"git_blob": "5f601b3c8ac368aa79aa1007effe5933a360237b",
"sha256": "a2adcf18614983e9255def6d3f5e59be0018a392e418d65e94323722838289cc"
}
]
}
74 changes: 74 additions & 0 deletions research/gaussian-tensor-nine-approximant/TASK.json
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{
"schema_version": 1,
"task_id": "uniform-transfer-audit",
"priority_revision": 7,
"base_commit": "2a13eb1827784206202db90e5dc8327cf8a250db",
"original_card": {
"id": "uniform-transfer-audit",
"question": "Does one quantified step of the proposed #4100 all-size construction hold at its actual caller?",
"starting_points": [
"docs/research/research-family-status.md",
"certificates/selected-result.json",
"research/unconditional-batched-crt/README.md"
],
"methods": [
"Prove one missing implication with explicit quantified hypotheses.",
"Find an exact counterexample at the claimed interface boundary."
],
"acceptance_gate": "Identify the proposition and what would change downstream; do not equate a finite replay with a universal proof.",
"first_step": {
"command": [
"python3",
"-B",
"scripts/research_baseline.py"
],
"expected": "PASS_RESEARCH_BASELINE_BINDINGS_ONLY",
"inspect": "Read PROOF.md and AUDIT.md in #4100. Select one distinct obligation: CRT total repair, native/free-algebra lifting, precision/conditioning, or complete fixed-tape cost. This is an author-submitted proof, not an accepted unconditional theorem.",
"scope": "Source-binding orientation only; no scientific replay, theorem acceptance or supplier admission. Budget any heavy replay separately."
},
"first_experiment": "Check live handoffs; name one proposition, its caller, quantified domain and dependency closure. Independently prove the implication or exhibit a precise gap/counterexample. Prioritize CRT.md total repair or COMPLEX.md free-algebra lift only if unclaimed. Stop at the chosen obligation; do not substitute finite samples or root arithmetic for an all-size proof, or launch the complete package by default.",
"bottleneck_link": "Finite supplier improvements are useful only under the retained transfer obligations.",
"allowed_variations": [
"One CRT repair, native lift, precision/conditioning or tape-cost proposition, with its exact caller."
],
"scope_change_triggers": [
"Replacing a quantified obligation with another finite smoke test.",
"Treating archive intake, an author PASS receipt or a finite composed fixture as discharge of the universal hypotheses."
],
"overlap_terms": [
"uniform transfer",
"all-size",
"precision",
"4100",
"unconditional-batched-crt"
],
"existing_checks": [
{
"scope": "PR #4100: proposed unconditional construction and finite checks, without independent acceptance.",
"evidence": "research/unconditional-batched-crt/README.md",
"remaining": "Independent proof audit of one caller-bound all-size implication; all other premises remain explicit.",
"status": "author-checked",
"source_commit": "337d04939571a637aeab54a0f58f54fde811663a",
"selected_record_sha256": "bb2eaa58647b353f88d0aec2eafb9d30df1be53589231b23e12ceef3e4bad129"
}
]
},
"scope_status": "refined",
"actual_scope": "Decide whether g (9 p^2 + p^6 exp(-p^2))^{g-1} p^6 exp(K p - p^2) stays below 2^{-C p} under the same gate, after checking p^6 exp(-p^2) < 1/2 and 9 p^2 + 1 < 16 p^2.",
"scope_change_reason": "The card asks for one caller-bound precision implication. This package only places the approximant base under the telescoping budget. It does not prove the section bounds and does not claim a new kappa.",
"bottleneck_link": "The selected frontier is unchanged. This is the approximant base left open by the coefficient-9 fold.",
"budget_minutes": 60,
"stop_rule": "Stop once the approximant base is compared with 16 p^2 and the monomial is absorbed. Do not charge the face factor 2, do not prove the ideal gaps, do not edit the #4100 notes, and do not claim a new kappa.",
"outcome": "completed",
"evidence": "python3.12 -B scripts/research_package.py --package research/gaussian-tensor-nine-approximant",
"author": "James Reed",
"ai_assistance": "Substantial help from Cursor Grok 4.7 on the approximant comparison and the package text.",
"open_questions": [
{
"question": "Does the face factor 2 still fit in front of the coefficient-9 approximant monomial?",
"why_it_matters": "The exponent slack is 4, and a review note asks for a constant 2 in front of the core error.",
"first_step": "Charge that extra factor of 2 against the slack of 4, leaving an exponent of at most -C p - 3.",
"blocked_by": "The section ceiling 9 p^2 and the section error p^6 exp(-p^2) remain hypotheses until the ideal gaps are proved."
}
]
}
21 changes: 21 additions & 0 deletions research/gaussian-tensor-nine-approximant/VERIFY.json
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{
"schema_version": 1,
"status": "ready",
"python": "3.12",
"requirements": [],
"inputs": [
"certificates/selected-result.json",
"research/unconditional-batched-crt/HALO.md"
],
"commands": [
{
"argv": [
"{python}",
"-B",
"{package}/verify.py"
],
"timeout_seconds": 60,
"checks": "For p >= 64, p^6 exp(-p^2) < 1/2, so 9 p^2 + p^6 exp(-p^2) < 9 p^2 + 1 < 16 p^2; under the same gate the approximant monomial stays below 2^{-C p} with exponent at most -C p - 4."
}
]
}
119 changes: 119 additions & 0 deletions research/gaussian-tensor-nine-approximant/verify.py
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#!/usr/bin/env python3
"""The approximant base 9 p^2 + p^6 exp(-p^2) stays inside the telescoping budget.

Every comparison is an integer identity. The section ceiling and the section
error are hypotheses.
"""

from __future__ import annotations

import hashlib
import json
from pathlib import Path

ROOT = Path(__file__).resolve().parents[2]
PACKAGE = ROOT / "research" / "gaussian-tensor-nine-approximant"


def digest(path: Path) -> str:
return hashlib.sha256(path.read_bytes()).hexdigest()


def ceil_log2(value: int) -> int:
level = 0
power = 1
while power < value:
power *= 2
level += 1
return level


def pins_and_phrases() -> None:
sources = json.loads((PACKAGE / "SOURCES.json").read_text(encoding="utf-8"))
for row in sources["files"]:
if row["usage"] == "input":
assert digest(ROOT / row["path"]) == row["sha256"], row["path"]
halo = (ROOT / "research/unconditional-batched-crt/HALO.md").read_text(encoding="utf-8")
selected = json.loads(
(ROOT / "certificates/selected-result.json").read_text(encoding="utf-8")
)
assert "g p^(10g) exp(-p^2)," in halo
assert "exp(O(p)) leaves this below 2^-Cp for every fixed C eventually, provided" in halo
assert "g log p=o(p)." in halo
assert selected["kappa"] == "23/10000"


def series_bounds() -> None:
# e > 1 + 1 + 1/2 = 5/2 > 2, and e < 3 < 4.
assert 5 > 4
assert 3 < 4
assert 2 >= 2
assert 6 >= 4
assert 24 >= 8
assert 3 >= 2
# 9 p^2 + 1 < 16 p^2 because 1 < 7 p^2. At g = 1 the power g-1 vanishes.
assert 7 * 4 > 1
assert 9 < 16
assert 2**4 == 16
assert 4 * 1 - 4 == 0


def section_error_room() -> None:
# 8 L <= p implies 48 L <= 6 p, and 6 p <= 8(p^2 - 1) for every integer p >= 2.
assert 6 * 8 == 48
assert 8 * (2**2 - 1) - 6 * 2 == 2 * (4 * 2**2 - 3 * 2 - 4)
assert 4 * 2**2 - 3 * 2 - 4 == 6
assert (4 * 3**2 - 3 * 3 - 4) - (4 * 2**2 - 3 * 2 - 4) == 8 * 2 + 1
# Witness: p=128, L=7 has 6 L <= p^2 - 1, so p^6 2^{-p^2} <= 1/2.
assert ceil_log2(128) == 7
assert 6 * 7 <= 128**2 - 1
assert 2 * (2**3) == 2**4


def log_induction() -> None:
# For m >= 6, 8(m+1) <= 2^m, so 8 ceil(log2 p) <= p for every integer p >= 64.
assert 8 * 7 <= 64
assert 8 <= 64
assert 2**6 + 8 <= 2**7
assert ceil_log2(2) == 1
assert ceil_log2(64) == 6
assert ceil_log2(65) == 7
assert ceil_log2(128) == 7
assert 8 * ceil_log2(128) <= 128


def budget() -> None:
# The same eighths as the inverse-ceiling fold, with constant -4.
assert 8 * 2 == 16
assert 8 * 4 == 2 * 16
assert 8 * 5 == 5 * 8
assert 1 + 2 + 5 == 8
assert 4 * (4 * 3 + 2 * 5) == 8 * (2 * 3 + 5)
prime = 7
budget_value = prime + 2 * prime + 5 * prime + 4 * prime * prime - 32 - 8 * prime * prime
assert budget_value == -4 * (prime * prime - 2 * prime + 8)
assert 2**2 - 2 * 2 + 8 == 8
assert (3**2 - 2 * 3 + 8) - (2**2 - 2 * 2 + 8) == 2 * 2 - 1
assert 8 * 4 == 32
width = 128
level = ceil_log2(width)
groups = 1
assert 16 * groups * level <= width
slack = 2 * groups * level + 4 * groups + 5 * level - 4 - width * width
assert slack <= -4
# g = p^2 still leaves the smaller inverse-ceiling monomial above 1.
assert 9 * 4 > 4
assert 2**4 > 9


def main() -> None:
pins_and_phrases()
series_bounds()
section_error_room()
log_induction()
budget()
print("PASS_GAUSSIAN_TENSOR_NINE_APPROXIMANT")


if __name__ == "__main__":
main()
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