# NS × X Integral × 24/72 Paradigm In Practice
## Round 71 — Pure Continuous Secant Riccati Volterra / Direct $O(\nu)$ Jost Bridge

- Date:  2026-08-18
- Version:  v0.1
- Status:  Proof-Route Experiment / Continuous-Only Secant Jost-Bridge Branch
- Previous round:  Round 70 — Dual Scalar Volterra Kernel
- canonical math delimiters: inline `$...$`; display `$$...$$`

## 0. Why Round 71 Abandons "Proving the Derivative First"

Round 70 reduced the ordinary-viscosity tangent to a scalar Jost kernel, but at extremely small viscosity, the finite-$J$ terminal eigenspace derivative still produces unnecessary cancellation with the local source sum.

However, what Round 68 truly needs is not the derivative itself.

It only requires that the central sign cone does not deviate too far from the endpoint.

Therefore, Round 71 instead compares:

$$
\boxed{
R_n^\nu-R_n^0
}
$$

rather than:

$$
\partial_\nu R_n.
$$

This directly reduces the local source from:

$$
O(\nu)
$$

to:

$$
\boxed{
O(\nu^2).
}
$$

---

# 1. Parity-rescaled transfer

Following Round 70:

$$
r_n=
\begin{cases}
u_n,&n\ {\rm even},\\
u_n/\nu,&n\ {\rm odd}.
\end{cases}
$$

In this chart:

$$
T_n^\nu=T_n^0
$$

for odd $n$, while for even $n$:

$$
\boxed{
T_n^\nu
=
T_n^0
+
\nu^2\mathcal E_n.
}
\tag{1.1}
$$

Only one transfer entry enters $\mathcal E_n$.

---

# 2. Exact Secant Riccati Identity

Write:

$$
T_n^\nu
=
\begin{pmatrix}
A_n^\nu&B_n\\
C_n&D_n
\end{pmatrix}.
$$

Let:

$$
R_n^\nu
=
(X_n^\nu)^{-1}
(
B_n-R_{n+1}^\nu D_n
),
$$

where:

$$
X_n^\nu
=
R_{n+1}^\nu C_n-A_n^\nu.
$$

endpoint graph:

$$
R_n^0
=
(X_n^0)^{-1}
(
B_n-R_{n+1}^0D_n
).
$$

Define:

$$
\Delta R_n
=
R_n^\nu-R_n^0.
$$

Since:

$$
A_n^\nu
=
A_n^0+\nu^2E_n
$$

on even layers, direct subtraction yields:

$$
\boxed{
\Delta R_n
=
(X_n^\nu)^{-1}
\left[
\nu^2E_nR_n^0
-
\Delta R_{n+1}
Y_n^0
\right],
}
\tag{2.1}
$$

where:

$$
\boxed{
Y_n^0
=
C_nR_n^0+D_n.
}
\tag{2.2}
$$

The same equation holds for odd layers, but with zero source.

Named:

$$
\boxed{
\textbf{Secant Riccati Volterra Identity}.
}
$$

This is an exact finite-difference identity, containing no Taylor remainder.

---

# 3. Why Secant is More Suitable than Tangent for the Final Closure

The ordinary tangent local source is:

$$
\partial_\nu(\nu^2)
=
2\nu.
$$

whereas the secant source remains intact:

$$
\nu^2.
$$

Therefore, on the transition scale:

$$
n\sim\nu^{-1/3}
$$

although the local coefficient has:

$$
|b_n/A_4^{(n)}|
\sim
n^4,
$$

the leading scale of the secant kernel becomes:

$$
\nu^2n^4
\times
n^{-2}
=
\nu^2n^2.
$$

Summing over the transition width:

$$
\sum_{n\lesssim\nu^{-1/3}}
\nu^2n^2
=
O(\nu).
$$

Thus, the central graph displacement is naturally:

$$
\boxed{
R_1^\nu-R_1^0
=
O(\nu).
}
$$

This is precisely the linear matching law consistently observed numerically in Round 60.

---

# 4. Corrected affine-Jost central chart

Following the true affine endpoint plane from Round 69, first subtract the even-driven odd particular response.

central base:

$$
\boxed{
(e_0,\widetilde o_2,o_0),
}
$$

central output:

$$
\boxed{
(e_2,e_1,\widetilde o_1).
}
$$

The condition numbers of the endpoint graph are only:

$$
\kappa_-\approx1.447,
\qquad
\kappa_+\approx1.579.
$$

So the final closure does not need to read the central cone in the ill-conditioned companion chart.

---

# 5. Exact central scalar solve

Let:

$$
x
=
\widetilde o_2
=
o_2-o_{2,0}.
$$

In the sheared chart, the $n=1$ central equation can be written as:

$$
\boxed{
c_0(R^\nu)
+
c_1(R^\nu)x
=
0.
}
\tag{5.1}
$$

endpoint:

$$
c_0(R^0)=0.
$$

And the exact endpoint Jost data gives:

### small fibre

$$
\boxed{
c_{1,-}(R^0)
>
0.62,
}
\tag{5.2}
$$

### large fibre

$$
\boxed{
|c_{1,+}(R^0)|
>
4.47.
}
\tag{5.3}
$$

The Lipschitz constants of the two coefficient functions with respect to the graph entries can be roughly taken as:

$$
\boxed{
L_-<5.21,
}
\tag{5.4}
$$

$$
\boxed{
L_+<33.1.
}
\tag{5.5}
$$

---

# 6. Central Secant-Cone Bridge Lemma

Suppose:

$$
\boxed{
\|R_1^\nu-R_1^0\|_{\max}
\le
8\times10^4\,\nu
}
\tag{6.1}
$$

for:

$$
0<\nu\le10^{-6}.
$$

Then:

$$
\delta
:=
\|R_1^\nu-R_1^0\|_{\max}
\le
0.08.
$$

### small fibre

Denominator remains:

$$
c_1>0.20.
$$

The central solve gives:

$$
|x|<2.1.
$$

Using the rigorous endpoint cone:

$$
e_{1,-}^0<-0.917,
$$

$$
4.16<o_{2,-}^0<4.17,
$$

we still obtain:

$$
\boxed{
e_{1,-}<-0.6,
}
$$

and:

$$
\boxed{
2<o_{2,-}<7.
}
$$

This is far inside Round 68's sufficient cone:

$$
e_1\le-0.1,
\qquad
0\le o_2\le100.
$$

### large fibre

Similarly:

$$
|c_1|>1.8,
$$

$$
|x|<1.5,
$$

and:

$$
\boxed{
e_{1,+}<-0.5,
\qquad
o_{2,+}>2.
}
$$

Hence the large sign cone also survives.

Therefore:

$$
\boxed{
\|R_1^\nu-R_1^0\|_{\max}
\le
8\times10^4\nu
\Longrightarrow
a_{3,\pm}(\nu)>0
}
\tag{6.2}
$$

through the entire final strip.

This completely bypasses a derivative theorem.

---

# 7. Corrected secant diagnostics

Using the true affine endpoint plane, the maximum graph-entry secant slopes are:

### small fibre

$$
\boxed{
\frac{
\|R_1^\nu-R_1^0\|_{\max}
}{
\nu
}
\approx
115.8
}
\tag{7.1}
$$

through the tested microscopic strip.

### large fibre

$$
\boxed{
\frac{
\|R_1^\nu-R_1^0\|_{\max}
}{
\nu
}
\approx
12.6
\text{--}
13.0.
}
\tag{7.2}
$$

The verifier extends this diagnostic to:

$$
\nu=10^{-9}.
$$

Compare with the sufficient theorem constant:

$$
8\times10^4.
$$

The proof target is therefore roughly:

- $700$ times looser on the small fibre;
- more than $6000$ times looser on the large fibre.

These remain diagnostics, not the missing uniform theorem.

---

# 8. Exact coefficient tail bound

The secant local source contains:

$$
\nu^2
\frac{
b_n
}{
A_4^{(n)}
}.
$$

Round 71 proves algebraically that for every real:

$$
n\ge100,
$$

the normalized coefficient ratio is strictly decreasing.

### small fibre

$$
\boxed{
\left|
\frac{
b_n
}{
A_4^{(n)}
}
\right|
<
14n^4.
}
\tag{8.1}
$$

At $n=100$ the exact value is:

$$
13.6558370\ldots n^4.
$$

### large fibre

$$
\boxed{
\left|
\frac{
b_n
}{
A_4^{(n)}
}
\right|
<
0.054n^4.
}
\tag{8.2}
$$

At $n=100$:

$$
0.05375034\ldots n^4.
$$

The verifier proves monotonicity by differentiating the exact algebraic ratio, shifting:

$$
n=m+100,
$$

and verifying every coefficient of the derivative numerator is negative while every denominator coefficient is positive in:

$$
\mathbb Q(\sqrt{17})[m].
$$

So this part is rigorous rather than sampled.

---

# 9. Cubic-WKB secant envelope

The exact secant dual formula has a scalar local kernel:

$$
\kappa_n^{\rm sec}.
$$

The remaining directional object is the Jost Green factor multiplying:

$$
\nu^2 b_n/A_4^{(n)}.
$$

Round 70 diagnostics suggest the natural majorant:

$$
\boxed{
|\kappa_n^{\rm sec}|
\lesssim
C
\nu^2n^2
e^{-c\nu n^3}.
}
\tag{9.1}
$$

This is exactly the expected cubic-WKB form.

---

# 10. Analytic sum of the cubic envelope

For:

$$
h(x)
=
\nu x^2
e^{-c\nu(x^3-N^3)},
$$

one has exactly:

$$
\boxed{
\int_N^\infty
h(x)\,dx
=
\frac1{3c}.
}
\tag{10.1}
$$

At:

$$
c=0.01,
$$

the integral is:

$$
33.\overline3.
$$

For:

$$
\nu\le10^{-6},
\qquad
N\ge100,
$$

the unimodal discrete correction is $<0.1$, so:

$$
\boxed{
\sum_{n\ge N}
\nu n^2
e^{-0.01\nu(n^3-N^3)}
<
34.
}
\tag{10.2}
$$

This summation estimate is rigorous and viscosity-uniform.

---

# 11. Candidate final Green-factor cone

Define the directional secant Green factor $\Gamma_n$ so that:

$$
|\kappa_n^{\rm sec}|
=
\nu^2
\left|
\frac{
b_n
}{
A_4^{(n)}
}
\right|
\Gamma_n.
$$

The next theorem only needs a deliberately loose cone such as:

$$
\boxed{
n^2
e^{\,0.02\nu(n^3-N_\ast^3)_+}
\Gamma_n
\le
200,
}
\tag{11.1}
$$

where:

$$
\boxed{
N_\ast
=
2\nu^{-1/3}.
}
\tag{11.2}
$$

Why this is enough:

For the worst small fibre coefficient:

$$
|b_n/A_4|\le14n^4.
$$

Before $N_\ast$:

$$
\sum
|\kappa_n|
\lesssim
14\times200
\nu^2
\sum_{n\le2\nu^{-1/3}}
n^2
<
8.4\times10^3\nu.
$$

After $N_\ast$, using the cubic envelope:

$$
\sum
|\kappa_n|
<
\frac{
14\times200
}{
3(0.02)
}
\nu
<
4.67\times10^4\nu.
$$

Thus:

$$
\boxed{
\text{$n\ge100$ source budget}
<
5.51\times10^4\nu.
}
\tag{11.3}
$$

The Central Secant-Cone Lemma allows:

$$
8\times10^4\nu.
$$

So the first $99$ layers plus the terminal pairing still receive a budget exceeding:

$$
\boxed{
2.4\times10^4\nu.
}
$$

Their actual diagnostics are vastly smaller.

---

# 12. What remains now

Round 70's generic task:

$$
\text{prove scalar kernel summability}
$$

has now been reduced to one concrete inequality:

$$
\boxed{
n^2
e^{0.02\nu(n^3-N_\ast^3)_+}
\Gamma_n
\le200.
}
$$

Everything around it is already accounted for:

- coefficient growth: proved;
- cubic summation: proved;
- central sign tolerance: proved;
- endpoint chart conditioning: proved;
- observed secant slope: $\sim115.8/13$;
- allowed slope: $8\times10^4$.

So the remaining theorem is a **directional Green-factor / Jost roughness estimate**.

---

# 13. STOP-C75 — Directional Green-Factor / Secant Jost-Passage Gap

$$
\boxed{
\begin{aligned}
\text{layer}
&=
\mathrm{parity\text{-}rescaled\ secant\ Riccati\ bridge},
\\
\Delta R_n
&=
(X_n^\nu)^{-1}
[
\nu^2E_nR_n^0
-
\Delta R_{n+1}Y_n^0
],
\\
\text{central sufficient displacement}
&=
8\times10^4\nu,
\\
\text{observed central slope}
&\approx
115.8,\ 13,
\\
|b/A_4|_{K_-}
&<
14n^4,
\\
|b/A_4|_{K_+}
&<
0.054n^4,
\\
\text{cubic envelope sum}
&<
34,
\\
\text{candidate directional cone}
&=
n^2e^{0.02\nu(n^3-N_\ast^3)_+}\Gamma_n\le200,
\\
\text{candidate }n\ge100\text{ budget}
&<
5.51\times10^4\nu,
\\
\text{remaining center+terminal budget}
&>
2.4\times10^4\nu,
\\
\text{remaining theorem}
&=
\mathrm{uniform\ directional\ Green\ factor}
\\
&\quad+
\mathrm{finite\ center/terminal\ closure},
\\
T_{\mathsf C\to\mathsf D}
&=
\mathrm{NOT\ REACHED}.
\end{aligned}
}
$$

Named:

$$
\boxed{
\textbf{STOP-C75:
Directional Green-Factor / Secant Jost-Passage Gap}.
}
$$

---

# 14. 24/72 Ledger — Round 71

| Step | object | $B$ | $U$ | $O$ | $L$ | status |
|---|---|---|---|---|---|---|
| C1114 | secant graph $\Delta R$ | $\mathsf C$ | Jost Grassmannian | relational | $\mathsf F$ | DEFINED |
| C1115 | exact secant Riccati identity | $\mathsf C$ | graph difference | matrix | $\mathsf F$ | PROVED |
| C1116 | $\nu^2$ secant local source | $\mathsf C$ | parameter injection | scalar | $\mathsf F$ | EXACT |
| C1117 | central affine quotient | $\mathsf C$ | endpoint shear | scalar | $\mathsf F$ | EXACT |
| C1118 | central secant-cone lemma | $\mathsf C$ | sign geometry | targeted | $\mathsf F$ | PROVED CONDITIONAL |
| C1119 | corrected secant slopes | $\mathsf C$ | graph diagnostic | profile | $\mathsf F$ | VERIFIED |
| C1120 | small-fibre coefficient bound | $\mathsf C$ | tail coefficient | scalar | $\mathsf F$ | PROVED |
| C1121 | large-fibre coefficient bound | $\mathsf C$ | tail coefficient | scalar | $\mathsf F$ | PROVED |
| C1122 | cubic envelope summation | $\mathsf C$ | WKB majorant | scalar | $\mathsf F$ | PROVED |
| C1123 | directional Green-factor cone | $\mathsf C$ | dual Jost kernel | scalar | $\mathsf F$ | TARGET |
| C1124 | $n\ge100$ budget reduction | $\mathsf C$ | final bridge budget | scalar | $\mathsf F$ | PROVED CONDITIONAL |
| C1125 | finite-center/terminal budget | $\mathsf C$ | closure remainder | targeted | $\mathsf F$ | OPEN / STOP-C75 |

---

# 15. Continuous-versus-discrete status

The secant graph compares two continuous-viscosity admissible subspaces of the same periodic Floquet operator family.

The index $n$ remains a Fourier/Floquet coordinate.

The remaining theorem is a uniform continuous-parameter directional Green estimate.

Therefore:

$$
\boxed{
T_{\mathsf C\to\mathsf D}
=
\text{NOT YET REACHED}.
}
$$

---

# 16. Next round — Directional Green Cone / Final Secant Closure

Concrete targets:

1. move from the ill-conditioned companion graph to the Round 63 fast-Schur/Jost frame;
2. identify the two scalar primal/dual factors inside $\Gamma_n$;
3. prove their product carries the algebraic factor:
   $$
   n^{-2};
   $$
4. use Round 60's stable multiplier to prove:
   $$
   e^{-0.02\nu(n^3-N_\ast^3)}
   $$
   attenuation beyond the transition;
5. certify the first $99$ layers with a fixed endpoint-Jost interval box;
6. kill the terminal secant pairing by the same WKB attenuation;
7. conclude:
   $$
   \|R_1^\nu-R_1^0\|_{\max}
   <
   8\times10^4\nu;
   $$
8. invoke the Central Secant-Cone Lemma;
9. if successful:
   $$
   a_{3,\pm}(\nu)>0
   \qquad
   \forall\nu>0.
   $$

This becomes:

$$
\boxed{
\textbf{Directional Green Cone / Final Secant Closure}.
}
$$

---

# 17. External primary-source anchors

Fresh primary-source check before this round:

1. F. Battelli, M. Franca, K. J. Palmer, *Exponential Dichotomy for Noninvertible Linear Difference Equations*, arXiv:2111.04553.
   - roughness and persistence of difference-equation dichotomies;
   - relevant to lifting the reduced WKB Green bound to the full Jost bundle.

2. Pierre Del Moral, Emma Horton, *A note on Riccati matrix difference equations*, arXiv:2107.12918.
   - time-varying Riccati semigroup representations and uniform Riccati bounds;
   - relevant to the secant graph transform.

3. Fritz Gesztesy, Yuri Latushkin, Kevin Zumbrun, *Derivatives of (Modified) Fredholm Determinants and Stability of Standing and Traveling Waves*, arXiv:0802.1665.
   - develops parameter-derivative/Fredholm-determinant methods for semi-separable kernels;
   - adjacent framework for interpreting the dual Jost sensitivity kernel.

4. Yuri Latushkin, Shibi Vasudevan, *Fredholm determinants, continued fractions, Jost and Evans functions for a Jacobi matrix associated with the 2D-Euler equations*, arXiv:2401.14037.
   - hydrodynamic Jost/Evans/Fredholm/continued-fraction equivalence;
   - directly adjacent to the present Jost-Green formulation.

These are framework anchors only. The NS-specific secant identity, coefficient inequalities, central-cone budget, and numerical slopes above are direct results of this series.