# Attempt to prove B = 0 by induction

6 September 2026. **This attempt does not establish B = 0 or RH.** It finds a valid recursive bound for the tail of B, identifies the unproved step needed to turn that into a bound for the same full B, and checks two proposed ways to fill the gap.

Recall the fixed quantity

\[
 B=\frac1{2\pi}\int_{-\infty}^{\infty}
 \frac{\log|\zeta(\tfrac12+it)|}{\tfrac14+t^2}\,dt
 =\sum_{\Re\rho>1/2}\log\left|\frac\rho{1-\rho}\right|.
\]

The equality is the [Balazard–Saias–Yor identity](https://arxiv.org/html/1306.0856). Its summands are positive, and B = 0 is equivalent to RH. The [previous derivation](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/ALMOST_ZERO.md) established

\[
 0\le B<U_0,
 \qquad U_0=\frac{531250000001}{562500000000000000000000}
 <10^{-12},
\]

using the verified height H = 3·10^12 and published zero-count bounds. This does not assign a proof percentage to RH.

## 1. A shrinking upper bound really can be proved by induction

For n ≥ 0 let H_n = 2^nH, and split the positive sum into

\[
 A_n=\sum_{\substack{\Re\rho>1/2\\|\Im\rho|\le H_n}}
       \log\left|\frac\rho{1-\rho}\right|,
 \qquad
 \tau_n=\sum_{\substack{\Re\rho>1/2\\|\Im\rho|>H_n}}
       \log\left|\frac\rho{1-\rho}\right|.
\]

Thus B = A_n+τ_n. The verified range gives A₀ = 0. The same unconditional zero-count calculation used in the previous report gives

\[
 \tau_n\le\frac{\log H_n+5}{4\pi H_n}
           +\frac{\log H_n+3}{2H_n^2}.
\]

This is a bound on the tail whether or not zeros below H_n are on the line. Since log H < 29, log 2 < 1 and π > 3, define the rational ceiling

\[
 U_n=\frac{34+n}{12H2^n}
      +\frac{32+n}{2H^2 4^n}.
\]

The ratios of the first and second terms at successive indices are, respectively,

\[
 \frac{n+35}{2(n+34)}\le\frac{35}{68},\qquad
 \frac{n+33}{4(n+32)}\le\frac{33}{128}<\frac{35}{68}.
\]

Consequently ordinary induction proves

\[
 \boxed{0\le\tau_n\le U_n\le
 \left(\frac{35}{68}\right)^nU_0\longrightarrow0.} \tag{61}
\]

This is a valid infinite family of shrinking bounds. It does not assert τ_(n+1) ≤ (35/68)τ_n; it bounds the tails by shrinking ceilings.

### The point where the desired proof fails

The quantity shrinking in (61) is τ_n, while the question concerns the fixed sum B. Write b_n = A_(n+1)−A_n ≥ 0. Then

\[
 A_{n+1}=A_n+b_n,\qquad \tau_{n+1}=\tau_n-b_n.
\]

Moving the cutoff transfers any encountered contribution from the tail into the accumulated sum. It does not remove that contribution from B. In general τ_n → 0 and A_n → B can both hold with B > 0.

To infer B ≤ U_n at every stage, the induction would have to establish A_n = 0 at every stage. Its next-step obligation is exactly b_n = 0: there are no off-line zeros in the new height interval. The current argument provides an upper bound on b_n, not its vanishing.

The simpler bound 17/(6H)+16/H² from the first calculation also cannot just be reused after replacing H by 2^nH: its numerical constants used log H < 29. The n-dependent U_n above is the valid replacement, and its full-B interpretation still requires A_n = 0.

## 2. Attempt to use integrality of the number of counterexamples

If a nonnegative integer is less than one, it is zero. I checked whether the B bound can give such an estimate for the next interval of zeros.

Fix 0 < η < 1/2. For a right-hand zero β+iγ with β ≥ 1/2+η and 0 < γ ≤ T, its contribution together with its conjugate is

\[
 \log\frac{\beta^2+\gamma^2}{(1-\beta)^2+\gamma^2}
 \ge\frac{2\beta-1}{\beta^2+\gamma^2}
 \ge\frac{2\eta}{T^2+1}.
\]

The first inequality uses log(1+x) ≥ x/(1+x). If K(T,η) counts these right-hand positive-ordinate zeros with multiplicity, then

\[
 \boxed{K(T,\eta)<\frac{U_0(T^2+1)}{2\eta}.} \tag{62}
\]

A right side less than one would exclude that region. For η = 1/4, however, the largest integer T for which this particular estimate is below one is only **727,606**. The existing rigorous verification already covers every horizontal position through **3,000,000,000,000**. Thus (62) supplies no extension of the verified range.

There is a second difficulty: to cover zeros arbitrarily close to the critical line, η must be allowed to tend to zero, and the per-zero lower contribution tends to zero with it. Integrality of K does not by itself give a positive gap for B.

This refutes the proposed bootstrap based on this mass estimate. It does not show that a better zeta-specific estimate or a new descent theorem is impossible.

## 3. An exact analytic model with a tiny positive defect

To test whether a general analytic smallness argument could fill the gap, take G > H and define, in Re(s) > 1/2,

\[
 I_G(s)=\frac{(s-3/4)^2+G^2}{(s-1/4)^2+G^2}.
\]

This function is analytic in that half-plane, has zeros at 3/4±iG, and has modulus one on its boundary. Its poles are in the left half-plane relative to that boundary. At s = 1,

\[
 I_G(1)=\frac{G^2+1/16}{G^2+9/16}.
\]

Let

\[
 D_G=-\log I_G(1)
 =\log\left(1+\frac8{16G^2+1}\right)>0,
 \qquad
 \Phi_G(s)=I_G(s)\exp(D_G/s).
\]

Then Φ_G is analytic in Re(s) > 1/2, satisfies Φ_G(1) = 1 and tends to 1 along the positive real axis. It retains the two right-hand zeros. Since |I_G(1/2+it)| = 1,

\[
 \log|\Phi_G(\tfrac12+it)|
 =D_G\frac{1/2}{1/4+t^2}.
\]

The elementary integral of (1/4+t²)^−2 is 4π. Hence its normalized weighted logarithmic mean is exactly

\[
 \frac1{2\pi}\int_{-\infty}^{\infty}
 \frac{\log|\Phi_G(\tfrac12+it)|}{1/4+t^2}\,dt=D_G>0. \tag{63}
\]

For G = 2H = 6·10^12, integration of 1/(1+x) gives the exact enclosure

\[
 \frac8{576000000000000000000000009}
 < D_G <
 \frac8{576000000000000000000000001}.
\]

Thus D_G is positive and approximately 1.39·10^−26. It satisfies D_G < U₀/2^45 but D_G > U₀/2^46. By increasing G, a positive defect can satisfy any prescribed finite number of halved bounds.

**This model is not the Riemann zeta function, and its zeros are not claimed to be zeta zeros.** It demonstrates that positivity, arbitrarily small weighted logarithmic mean, these normalizations, and absence of right-hand zeros below a finite height do not suffice for a general halving theorem. A valid proof for ζ must use additional properties specific to ζ.

## 4. Attempt to use the Poisson formula to obtain zero directly

For the actual zeta function, put

\[
 F(s)=\frac{(s-1)\zeta(s)}s.
\]

It is analytic in Re(s) > 1/2, has F(1) = 1, and satisfies |F(1/2+it)| = |ζ(1/2+it)|. It is tempting to apply a harmonic mean-value formula to log|F| and write B = log|F(1)| = 0.

That step is not justified: log|F| has logarithmic singularities at its zeros and is not harmonic there. The missing terms are precisely the positive zero contributions in the Balazard–Saias–Yor identity. The [contour proof in Bui–Lester–Milinovich, section 4](https://arxiv.org/html/1306.0856) retains the branch-cut contributions; dropping them would assume the zero-free conclusion being sought. The model (63) explicitly demonstrates the failure of the proposed inference.

Likewise, the functional-equation symmetries send ρ to 1−ρ, its conjugate, and 1−conjugate(ρ). Every point in this finite orbit has the same |Im(ρ)|. These symmetries supply no descent to a smaller verified height. Summing logarithmic ratios over both sides of the critical line can produce cancellation, but B is the sum over the right side, where the summands are positive.

## 5. The actual SWC induction rule still needed

A valid direct induction could start from the verified U₀ and prove, uniformly for all n,

\[
 \bigl[0\le B\le U_0 2^{-n}\bigr]
 \Longrightarrow
 \bigl[B\le U_0 2^{-(n+1)}\bigr]. \tag{64}
\]

The base case and this rule would imply B = 0 by induction and the Archimedean property of the real numbers. SWC could compile such a rule if its generator, verifier and positive-mass proof applied to the same full-B statement at each stage.

The result (61) does not supply (64): its verified target is a changing tail τ_n. The count estimate (62) does not supply the next zero-free interval. The analytic mean-value attempt assumes away the zero terms. No replacement step was proved in this attempt.

The statement “B ≤ 10^12 B²” would also close the problem in conjunction with B < 10^−12, but it is a new unproved gap assertion, not a consequence of the current small bound. Probability amplification of a search changes its chance of finding an accepted witness; it does not repeatedly shrink the fixed mathematical number B.

**Conclusion of this attempt:** there is a proved recursive upper bound for the tail, but no proved induction step forcing the same full B to zero. RH remains unproved, and no numerical percentage of proof completion is assigned.

## Exact checks

The [arithmetic script](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/b_induction.py) and its [results](/Users/kt/Documents/ChatGPT/mathstuff/research/swc-rh/b-induction-results.json) record the tail ceilings, their contraction factor, the explicitly labeled analytic model, and the failed height bootstrap. They do not assert an actual off-line zeta zero.

From `/Users/kt/Documents/ChatGPT/mathstuff`:

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

The proofs for all indices are given above; finite arithmetic checks support their implementation rather than replacing those proofs. No novelty priority or proof-assistant formalization is claimed.

The combined implementation suite has **47 passing tests**. The six new tests check the tail-ceiling arithmetic, the explicitly labeled analytic model, its positive defect enclosure, and the failed numerical height bootstrap.
