# Applying stochastic witness calculus to the Riemann hypothesis

Research attempt dated **6 September 2026**. Status: **RH remains unresolved**.

The [second attempt](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/CONTINUATION.md) adds exact lower bounds on the finite-basis optima, proves seven finite instances of the proposed contraction, and develops further infinite-domain proof attempts and obstructions.

I applied Kenju Tomita's SWC v6 framework to an established approximation criterion for RH. The result is a finite arithmetic certificate system, a proved bound for its infinite tail, and eight checked instances. The strongest instance establishes a squared approximation error below **0.007525031**, together with positive accepted mass over a distribution with **2^511 seeds**. These are finite mathematical statements; they establish neither RH nor a new record concerning zeta zeros.

The decisive missing result is a proof that such certificates exist at **every** accuracy level. The work below isolates that obligation and explains why several plausible ways of bypassing it fail.

## SWC source and literature checked

The SWC source was read from the v6 source archive in `/Users/kt/Documents/webeph/assets/research/`, together with the v5 refinements, v6 assurance section and induction audit. The archive fingerprint is recorded in [source-provenance.json](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/source-provenance.json). The relevant SWC rules are exact positive-mass soundness, expectation certificates, finite scope, and pointwise compilation. The existing website and publication package were not edited.

The literature search was performed on 6 September 2026. This is a targeted review, not an exhaustive survey. Dates below distinguish publication from preprint revision.

| Source | Verified status and result | Consequence for this attempt |
|---|---|---|
| [Clay Mathematics Institute](https://www.claymath.org/millennium-problems/) | RH remains listed among the unsolved Millennium problems. | No accepted solution was identified. |
| Guth and Maynard, [*New large value estimates for Dirichlet polynomials*](https://annals.math.princeton.edu/2026/203-2/p06) | Published in *Annals of Mathematics* 203 (2026), 623–675; online 1 March 2026. Gives the zero-density exponent 30/13 and prime interval exponent 17/30. [arXiv v2](https://arxiv.org/abs/2405.20552v2) was posted 7 April 2026. | Stronger control of the frequency of possible exceptions does not eliminate each exception. |
| Turnage-Butterbaugh, [July 2026 exposition](https://arxiv.org/abs/2607.04632) | Explains the Guth–Maynard result and its historical context; announced for the *Bulletin of the AMS*. | Useful account of what the density estimate actually proves. |
| Baluyot, Goldston, Suriajaya and Turnage-Butterbaugh, [*Pair Correlation … I*, v3](https://arxiv.org/abs/2501.14545v3) | Revision posted **1 September 2026**, with manuscript dated June 2026. The stated two-thirds result for simple critical-line zeros assumes a narrow box condition, with parameter b = 0.3185. The authors describe the manuscript as under review. | The box condition must remain an assumption when using this theorem. |
| Goldston and Suriajaya, [*Zeta Zeros on the Critical Line*, v2](https://arxiv.org/abs/2511.20059v2) | Revised 5 February 2026. Explains how removing RH from Montgomery's simple-zero argument would also locate two thirds of zeros on the line. | Provides the background for the next item. |
| Claude / Anthropic, [August 2026 paper](https://www-cdn.anthropic.com/95c246936988e43127bc6b2ceb7077c1dad2d68e.pdf), linked by the [10 August announcement](https://www.anthropic.com/research/riemann-zeta) | Theorem A states an unconditional asymptotic proportion of at least 0.67250… for simple critical-line zeros. The announced mechanism uses rank, trace, inertia and a finite compression of Weil's form. The announcement was updated 13 August. | This is a recent publicly released result, not a full RH proof. I inspected its statement and limitations; I did not independently validate its complete analytic argument or Lean dependency chain. |
| Báez-Duarte, [*A strengthening of the Nyman–Beurling criterion*](https://arxiv.org/abs/math/0202141) | Preprint 2002, published 2003. Restricts the fractional-part approximation criterion to integer dilations. | The established equivalence used in the construction below. |
| Darses and Hillion, [*On probabilistic generalizations …*](https://arxiv.org/abs/1805.06733v4) | Final arXiv revision 3 June 2021. Develops random-dilation criteria; Theorem 4 explicitly uses positive measure to extract an approximation. | A probabilistic existence approach to this criterion predates this SWC application. |
| Alouges, Darses and Hillion, [*Polynomial approximations …*](https://arxiv.org/abs/2006.02953) | Develops generalized approximation criteria and identifies control of approximation coefficients as a remaining obstacle. | Smoothing or approximation alone does not remove the coefficient-control problem. |
| Platt and Trudgian, [*The Riemann hypothesis is true up to 3·10^12*](https://arxiv.org/abs/2004.09765) | Rigorous computation verifies zeros through that height. | A benchmark for the difference between finite zero verification and a universal theorem. This report does not claim it is the latest computational record. |

I also screened the October 2025 [Carvill preprint on smoothed Gram matrices](https://arxiv.org/abs/2510.18132). Its abstract concerns decay and compression of a smoothed system. No theorem from that preprint is assumed here. Online assertions of complete RH proofs were not adopted as established results.

## 1. Turning RH into the right SWC witness problem

For integer k ≥ 2 define, on 0 < x < 1,

\[
 g_k(x)=\left\{\frac1{kx}\right\}-\frac1k\left\{\frac1x\right\}.
\]

For finitely many real coefficients c = (c₂,…,c_N), put

\[
 F_c=\sum_{k=2}^N c_k g_k,
 \qquad E(c)=\|1-F_c\|_{L^2(0,1)}^2.
\]

The integer-dilation Nyman–Beurling–Báez-Duarte criterion gives

\[
 \mathrm{RH}\quad\Longleftrightarrow\quad
 \forall j\ge1\;\exists N\ge2\;\exists c\in\mathbb Q^{N-1}:
 E(c)<2^{-j}. \tag{1}
\]

Here is the normalization argument, to avoid silently changing the established criterion. Write ρ_k(x) = {1/(kx)} on (0,∞), and χ = 1_(0,1). Báez-Duarte's criterion is approximation of χ by finite sums f = Σ a_kρ_k in L²(0,∞). On x > 1, f(x) = S/x, where S = Σ a_k/k. Thus |S| ≤ ||χ−f||₂. Replace f by f−Sρ₁. This new sum vanishes on x > 1 and equals Σ_(k≥2) a_k g_k. Its error is at most (1+||ρ₁||₂)||χ−f||₂, which tends to zero with the original error. Conversely every such balanced sum is an allowed sum of the ρ_k. Finally rational coefficients suffice by continuity in each finite-dimensional coefficient space. This proves the passage to (1).

The SWC instance index is therefore **an accuracy level j**, not a zero height or a count of tested zeros. A witness is a finite rational coefficient vector plus finite arithmetic data certifying E(c) < 2^−j.

## 2. An exact arithmetic verifier for the infinite integral

The derivations in this and the next two sections were worked out for this application. No claim of literature priority is made for the elementary identities or bounds.

On the interval 1/(m+1) < x < 1/m, write 1/x = m+u with 0 < u < 1. The fractional linear terms cancel, giving

\[
 g_k(x)=\frac{m\bmod k}{k}.
\]

The interval length is 1/[m(m+1)]. Consequently

\[
 E(c)=\sum_{m=1}^{\infty}
 \frac{\bigl(1-\sum_{k=2}^N c_k(m\bmod k)/k\bigr)^2}{m(m+1)}. \tag{2}
\]

This identity includes the entire interval near x = 0; a finite sum alone would omit its tail.

Choose an integer D > 0 and integers b_k with c_k = k b_k/D. Every finite rational c admits this representation after choosing D appropriately. Define

\[
 r_m=D-\sum_{k=2}^N b_k(m\bmod k).
\]

For a cutoff M and precision p, compute

\[
 S^+=2^{-p}\sum_{m=1}^M
 \left\lceil\frac{2^p r_m^2}{D^2m(m+1)}\right\rceil.
\]

Replacing ceilings by floors gives S^−. These are exact rational bounds for the head of (2), with width at most M2^−p. The checker performs integer division and exact rational comparisons; decimal output is only a display format.

A simple tail bound follows from

\[
 A_b=\frac1D\max\left(
 \left|D-\sum_{b_k>0}b_k(k-1)\right|,
 \left|D-\sum_{b_k<0}b_k(k-1)\right|\right).
\]

Since |r_m|/D ≤ A_b and Σ_(m>M) 1/[m(m+1)] = 1/(M+1),

\[
 E(c)\le S^++\frac{A_b^2}{M+1}. \tag{3}
\]

This is sound, but it is too loose to achieve the strongest certificate at the chosen cutoff.

## 3. A sharper tail from periods and gcds

Write u_k(m) = (m mod k)/k. Over a complete common period,

\[
 \langle u_i\rangle=\frac{i-1}{2i},\qquad
 \langle u_i u_k\rangle=
 \frac{(i-1)(k-1)}{4ik}
 +\frac{\gcd(i,k)^2-1}{12ik}. \tag{4}
\]

To prove the covariance term, condition on the common residue modulo d = gcd(i,k). The remaining quotient residues are independent and uniform by the Chinese remainder theorem. The common residue contributes variance (d²−1)/12. This proves (4) without enumerating the least common multiple of all basis indices.

Let μ_b be the period average of the squared residual in (2). Equation (4) yields the exact rational expression

\[
 \mu_b=
 \left(1-\frac{\sum_k b_k(k-1)}{2D}\right)^2
 +\frac{\sum_{i,k}b_i b_k(\gcd(i,k)^2-1)}{12D^2}. \tag{5}
\]

Although individual terms in the second sum can be negative, the whole expression is nonnegative because it is an average of squares.

Define

\[
 K_b=2\sum_k |c_k|(k-1)
 +\sum_{i,k}|c_i c_k|
   \frac{(i-1)(k-1)}{\gcd(i,k)}. \tag{6}
\]

**Tail lemma.** With w_m = 1/[m(m+1)],

\[
 \left|
 \sum_{m>M}\frac{(r_m/D)^2}{m(m+1)}-
 \frac{\mu_b}{M+1}
 \right|
 \le\frac{K_b}{(M+1)(M+2)}. \tag{7}
\]

**Proof.** A mean-zero sequence of period q bounded in absolute value by B has every partial sum, starting at any phase, bounded by qB: full periods cancel and fewer than q terms remain. Expand the centered square as linear combinations of u_i−⟨u_i⟩ and u_i u_k−⟨u_i u_k⟩. Their periods are i and lcm(i,k). Their absolute bounds are (i−1)/i and (i−1)(k−1)/(ik), respectively. The triangle inequality gives exactly K_b as a bound for every partial sum of the centered square. Summation by parts against the positive decreasing weights w_m gives a tail bound K_b w_(M+1), since the differences of successive weights telescope. This is (7). ∎

Combining the head calculation and the two tail estimates gives

\[
 E(c)\le U(c):=S^++\min\left\{
 \frac{A_b^2}{M+1},
 \frac{\mu_b}{M+1}+\frac{K_b}{(M+1)(M+2)}\right\}. \tag{8}
\]

A lower enclosure is S^− + max{0, μ_b/(M+1)−K_b/[(M+1)(M+2)]}. All entries in these bounds are rational.

For N = 512 and M = 2,000,000, the coarse tail bound is about 0.00171265. The periodic bound is about 0.000131247, roughly thirteen times smaller. At this cutoff the coarse bound does not establish E < 1/128; the periodic bound does. This is a concrete improvement to the finite verification procedure, not an asymptotic RH result.

## 4. A nontrivial SWC mass certificate without enumerating the seeds

Fix a positive integer a. Draw independent uniform signs ξ_k ∈ {−1,+1} and set

\[
 B_k=b_k+a\xi_k,\qquad C_k=k B_k/D.
\]

There are 2^(N−1) sign seeds, each with the exact same probability. Independence here is part of the specified product measure, not inferred from observed samples.

For a sign-generated vector B, use the acceptance guard

\[
 V(B)=\mathbf1\{Q(B)<\varepsilon\},\qquad
 Q(B)=S^+(B)+\frac{\mu_B}{M+1}
       +\frac{K_*}{(M+1)(M+2)}, \tag{9}
\]

where K_* is (6) evaluated using coefficient absolute bounds k(|b_k|+a)/D. Thus K_* ≥ K_B for every seed. The tail lemma proves E(C) ≤ Q(B), so every accepted output is sound.

Set

\[
 V_a=\frac{a^2}{D^2}\sum_{k=2}^N(k-1)^2.
\]

**Moment certificate.** The exact rational number

\[
 A=S^+(b)+M2^{-p}+V_a+
       \frac{\mu_b}{M+1}+\frac{K_*}{(M+1)(M+2)} \tag{10}
\]

satisfies E_ξ[Q(B)] ≤ A. Therefore, whenever A < ε,

\[
 \Pr\{V(B)=1\}\ge\delta:=1-\frac{A}{\varepsilon}>0. \tag{11}
\]

**Proof.** Expanding the squared residual, all mixed sign terms and all linear sign terms have expectation zero. The added variance at residue m is a²D^−2 Σ_k(m mod k)². Bound each square by (k−1)². The head weights have total M/(M+1), and the tail-average weight is 1/(M+1); together their variance contribution is at most V_a. Rounding the random head upward adds at most M2^−p. Its unperturbed head is at most S^+(b). The remaining terms in (10) now bound the expected acceptance score. The score is nonnegative. Markov's inequality gives Pr(Q ≥ ε) ≤ A/ε, proving (11). ∎

This instantiates the SWC tuple: the product sign measure is μ; (9) is the deterministic verifier; the tail lemma supplies soundness; (10)–(11) supply the mass proof; exact rational comparison supplies δ > 0. No Monte Carlo estimate occurs in this certificate.

## 5. What the computation actually established

The producer used floating-point weighted least squares on m = 1,…,65,536 only to propose centers. It then rounded the coefficients to c_k = k b_k/2^30. The verifier recomputed the finite head on a dyadic grid of spacing 2^−64, the infinite-tail bounds and the mass obligations using arbitrary-precision integers and rational arithmetic.

The table rounds error upper bounds upward and mass lower bounds downward. Full rational values and all coefficients are in the individual JSON certificates.

| N | Head cutoff M | Certified upper bound for E(center) | Target ε | Certified accepted mass at least | Sign seeds |
|---:|---:|---:|---:|---:|---:|
| 4 | 100,000 | 0.065951698 | 1/8 | 0.413338 | 2^3 |
| 8 | 100,000 | 0.024244539 | 1/32 | 0.196152 | 2^7 |
| 16 | 100,000 | 0.017936394 | 1/32 | 0.372780 | 2^15 |
| 32 | 100,000 | 0.014078440 | 1/64 | 0.086605 | 2^31 |
| 64 | 100,000 | 0.011408759 | 1/64 | 0.236078 | 2^63 |
| 128 | 250,000 | 0.009703284 | 1/64 | 0.331557 | 2^127 |
| 256 | 1,000,000 | 0.008274945 | 1/64 | 0.411558 | 2^255 |
| 512 | 2,000,000 | 0.007525031 | 1/128 | 0.032152 | 2^511 |

![Finite approximation errors and comparison of two rigorous tail bounds](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/certificate-results.png)

The final center satisfies, conservatively,

\[
 0.0073937837405 < E(c) < 0.0075250303436 < 2^{-7}.
\]

The last mass certificate proves at least 3.2152% accepted sign mass at ε = 1/128. This is a statement about a finite specified distribution. It is not a probability that RH is true.

Because the final witness meets 2^−7, it also meets the larger targets 2^−j for j = 1,…,6. Thus the first seven accuracy instances of (1) have witnesses. The integers 7 and 512 do not refer to checked zeta zeros or zero heights. The displayed values are bounds for chosen approximants, not certified optima of their subspaces.

All eight saved certificates pass the standalone checker. Twelve tests pass, including direct residue enumeration, complete small-period averages, all phases of a discrepancy bound, a rational enclosure of the known single-basis error 1−log 2, exhaustive checking of a small sign distribution, and rejection of forged mass and universal-scope claims. These checks support the implementation. They are not a Lean formalization or external peer review.

## 6. Why this has not proved RH

### The precise unresolved obligation

The construction would prove RH if one established

\[
 \boxed{\forall j\ge1\;\exists(N,D,b,M,p,a):
             A(N,D,b,M,p,a)<2^{-j}.} \tag{12}
\]

For every such instance, (11) and SWC soundness produce the witness required by (1). No lower bound for δ uniform in j is necessary: any proved positive δ(j) suffices.

Conversely, RH implies the existence of certificates of this type. Given a strict approximation with error well below a target, rationalize its coefficients. Make the sign perturbation arbitrarily small by increasing D and scaling b while keeping a = 1. For these fixed coefficients and perturbation, increase M so that the periodic discrepancy term tends to zero, and then increase p so that M2^−p tends to zero. Formula (10) will eventually be below the target. This explains why the scheme has the right logical scope; it does not discharge (12) unconditionally.

### An off-line zero creates a fixed positive error floor

The obstruction can be derived directly. For 0 < Re(s) < 1, the Mellin identity gives

\[
 \int_0^1 g_k(x)x^{s-1}\,dx
 =-\frac{\zeta(s)}s\left(k^{-s}-k^{-1}\right). \tag{13}
\]

If ζ(ρ) = 0 at ρ = β+iγ with β > 1/2, every finite approximation satisfies

\[
 \int_0^1(1-F_c(x))x^{\rho-1}\,dx=\frac1\rho.
\]

Cauchy–Schwarz, with ∫₀¹|x^(ρ−1)|² dx = 1/(2β−1), therefore yields

\[
 \boxed{E(c)\ge\frac{2\beta-1}{|\rho|^2}\quad\text{for every }c.} \tag{14}
\]

This is the classical Mellin obstruction, explicitly derived in our normalization. It shows why errors tending to zero would exclude every right-of-line zero; the functional equation then handles the left side. It also shows why a finite small error cannot do so. A hypothetical zero with β = 3/4 and |γ| near 3·10^12 would impose a floor only about 5.6·10^−26. Our finite bounds are vastly larger. No improvement on the published zero-verification range follows from these certificates.

### Random coefficient variance cannot supply the missing convergence

For a fixed basis with Gram matrix G and random coefficients C with finite second moments,

\[
 \mathbb E E(C)=E(\mathbb E C)+\operatorname{tr}(G\operatorname{Cov}(C))
 \ge E(\mathbb E C). \tag{15}
\]

This follows by expanding the Hilbert-space square around the mean. Thus adding random coefficient noise cannot lower the best expected error. In this experiment the center already has a sound deterministic certificate; randomization demonstrates a compositional mass certificate for many additional possible witnesses. It does not create the missing asymptotic approximation.

### Trace and density information leave room for exceptions

The August paper's discussion of limitations identifies restrictions on the trace/second-moment approach and on its frequency support. This is consistent with a simple exact obstruction to promoting aggregate information into positivity of every direction.

Take H_d = diag(1,…,1,−1), of size d ≥ 3. Then

\[
 \operatorname{tr}H_d=d-2,\qquad
 \operatorname{tr}(H_d^2)=d,\qquad
 \frac{(\operatorname{tr}H_d)^2}{d\operatorname{tr}(H_d^2)}
 =\left(1-\frac2d\right)^2\longrightarrow1.
\]

Yet H_d has a negative direction for every d. This example does not model all arithmetic constraints of Weil's form or refute that research program. It proves that these aggregate moments alone cannot certify positive semidefiniteness. Likewise, even a density-one statement about critical-line zeros could leave an exceptional set.

There is a useful SWC nuance: on a countable domain with **positive mass at each point**, an exact zero failure probability does exclude every failure. The difficulty is proving exact zero mass. Finite sampling does not do that. Nor does the limiting uniform distribution on zeros provide a fixed positive atomic mass for every particular zero.

## 7. The next mathematical target

A useful continuation should seek a theorem controlling progress of the approximation as the basis expands. Let d_N² be the true minimum error in span(g₂,…,g_N), and write e_j = d_(2^j)². A sufficient estimate would be

\[
 e_{j+1}\le\left(1-\frac\kappa{j+1}\right)e_j
 \quad(j\ge j_0),\qquad \kappa>0. \tag{16}
\]

For j₀ large enough that the factors are positive, their product tends to zero, so (16) would imply RH. The exact projection identity identifies the required gain: when a new basis direction has nonzero orthogonal component h, the squared error drops by |⟨r,h⟩|²/||h||², where r is the current projection residual. The unproved part is a lower bound on the aggregate gain from new directions, uniform at all later scales.

This estimate has **not** been proved or inferred from the eight computations. Extending the table alone would still leave that issue open. The present concrete contribution is a way to test proposed coefficient constructions with a sound infinite-tail bound and to express their success in SWC's exact mass language.

## Reproduce and inspect

Run these commands from `/Users/kt/Documents/ChatGPT/mathstuff`:

```bash
python3 research/swc-rh/verify.py
python3 -m unittest discover -s research/swc-rh -p 'test_*.py' -v
```

Verification uses only the Python standard library. To regenerate proposals, NumPy is required:

```bash
python3 research/swc-rh/search.py
```

The producer may generate slightly different centers on another floating-point platform. The stored integer certificates can be checked directly without rerunning it.

- [Exact checker](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/verify.py)
- [Proposal generator](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/search.py)
- [Arithmetic and certificate tests](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/test_verify.py)
- [Strongest certificate, including all 511 coefficients](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/certificates/nb-N0512.json)
- [Display results](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/results.json)

No proof of RH, new critical-line proportion, or novelty priority is claimed by this report.
