> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification.

# ACTM Round 28
## Extremal Interpolation Approximate Jets, Uniform Height, and Algebraic Degree Escape
### Extremal Interpolation Approximate Jets, Uniform Height, and Algebraic Degree Escape

**Research program:** Algebraic Combinatorial Transcendence Methodology (ACTM)  
**Authors:** Neo.K; Aletheia (GPT-5.6 Sol)  
**Institution:** EveMissLab / 一言諾科技有限公司  
**Date:** 2026-10-03  
**Version:** Round 28 / v0.1  
**Document type:** Formal research notes / approximate-one-jet round  
**Status statement:** This round does not claim completion of the One-Jet Bridge or impossibility of real BADT$_2$. It proves that the Round 25 extremal interpolation polynomial automatically produces a high-order approximate one-jet under rational flattening, with an exact nonzero error asymptotic; the jet approximant has uniformly bounded Weil height for fixed grid size, but its algebraic degree must escape to infinity. Waldschmidt's 1978 quotient-of-logarithms transcendence measure further gives an effective lower bound on degree escape. This precisely locates AOJBP's new bottleneck in arithmetic degree control rather than analytic approximation order.

---

## Abstract

Fix the first layer of real BADT / Type P:

$$
A=e^\lambda,
\qquad
B=e^{s\lambda},
\qquad
C=e^{s^2\lambda},
$$

where:

$$
A,B,C\in\overline{\mathbb Q}_{>0},
$$

$$
\lambda=\log A\neq0,
$$

$$
s=\frac{\log B}{\log A}
$$

is transcendental.

Fix:

$$
M\ge2,
$$

and let:

$$
L=M^2.
$$

The extremal vanishing polynomial for the first-layer $L$-point grid in Round 25 uses triangular support:

$$
\{(0,0),(1,0),\ldots,(L-1,0),(0,1)\}.
$$

Now take rational flattening:

$$
\rho_Q=\frac1Q,
\qquad
Q\in\mathbb Z_{\ge1},
$$

and:

$$
\mu_Q=\frac{\lambda}{Q}.
$$

Choose the positive real algebraic roots:

$$
A_Q=A^{1/Q},
$$

$$
B_Q=B^{1/Q},
$$

$$
C_Q=C^{1/Q}.
$$

For:

$$
0\le a,b<M,
$$

Define:

$$
r_{a,b}=a+bs,
$$

$$
x_{a,b}^{(Q)}
=
A_Q^aB_Q^b
=
e^{\mu_Qr_{a,b}},
$$

$$
y_{a,b}^{(Q)}
=
B_Q^aC_Q^b
=
e^{\mu_Qs r_{a,b}}.
$$

Since:

$$
y_{a,b}^{(Q)}
=
\left(
x_{a,b}^{(Q)}
\right)^s
$$

holds on the positive real branch, the Round 25 extremal polynomial can be normalized uniquely as:

$$
\boxed{
P_Q(T_1,T_2)
=
T_2-H_Q(T_1),
}
$$

where:

$$
\deg H_Q\le L-1
$$

and:

$$
\boxed{
H_Q
\left(
x_{a,b}^{(Q)}
\right)
=
\left(
x_{a,b}^{(Q)}
\right)^s.
}
$$

Thus:

$$
H_Q
$$

is the interpolation polynomial for the function:

$$
f(x)=x^s
$$

at $L$ algebraic nodes.

Since:

$$
(a,b)=(0,0)
$$

gives:

$$
x_{0,0}=1,
$$

$$
H_Q(1)=1.
$$

The first-layer logarithmic derivative:

$$
\mathcal D_s
=
\Theta_1+s\Theta_2
$$

at:

$$
(1,1)
$$

gives:

$$
\boxed{
J_Q
:=
(\mathcal D_sP_Q)(1,1)
=
s-H_Q'(1).
}
$$

The approximate One-Jet Bridge is therefore reduced to:

$$
\boxed{
\text{polynomial interpolation derivative approximation to }f'(1)=s.
}
$$

---

The first main theorem of this round is:

## Exact Approximate-Jet Order Theorem

Let:

$$
(s)_{\underline L}
=
s(s-1)\cdots(s-L+1).
$$

then:

$$
\boxed{
J_Q
=
c_{M,s,\lambda}
Q^{-(L-1)}
+
O(Q^{-L}),
}
$$

where:

$$
\boxed{
c_{M,s,\lambda}
=
(-1)^{L-1}
\frac{
(s)_{\underline L}
}{
L!
}
\lambda^{L-1}
\prod_{
\substack{
0\le a,b<M\\
(a,b)\neq(0,0)
}
}
(a+bs).
}
$$

And:

$$
\boxed{
c_{M,s,\lambda}\neq0.
}
$$

The reasons are:

- $s$ is transcendental, so:
  $$
  (s)_{\underline L}\neq0;
  $$
- $a+bs\neq0$ for any nonzero integer pair:
  $$
  (a,b).
  $$

Therefore:

$$
\boxed{
|J_Q|
\asymp
Q^{-(M^2-1)}.
}
$$

If:

$$
M=2N,
$$

then:

$$
\boxed{
|J_Q|
\asymp
Q^{-(4N^2-1)}.
}
$$

Thus, rational flattening does create an extremely high-order approximate jet; its order is exact, not merely an upper bound.

---

The second main result studies arithmetic complexity.

Let:

$$
K
=
\mathbb Q(A,B,C),
$$

$$
d_0=[K:\mathbb Q],
$$

and:

$$
K_Q
=
K(A_Q,B_Q,C_Q).
$$

Since each:

$$
A_Q,B_Q,C_Q
$$

respectively satisfy degree-$Q$ polynomials:

$$
X^Q-A,
\quad
X^Q-B,
\quad
X^Q-C,
$$

Thus:

$$
\boxed{
[K_Q:\mathbb Q]
\le
d_0Q^3.
}
$$

By algebraic interpolation:

$$
H_Q\in K_Q[T].
$$

Therefore:

$$
\boxed{
\alpha_Q
:=
H_Q'(1)
\in K_Q.
}
$$

That is:

$$
\boxed{
J_Q=s-\alpha_Q,
\qquad
\alpha_Q\in\overline{\mathbb Q}.
}
$$

---

## Uniform Height Theorem

For fixed:

$$
M,A,B,C,
$$

there exists a constant:

$$
C_{\mathrm{ht}}
$$

such that:

$$
\boxed{
h(\alpha_Q)
\le
C_{\mathrm{ht}}
\qquad
\forall Q\ge1.
}
$$

The proof uses only standard inequalities for absolute logarithmic Weil height:

$$
h(xy)\le h(x)+h(y),
$$

$$
h(x^{-1})=h(x),
$$

$$
h(x-y)\le h(x)+h(y)+\log2.
$$

Since:

$$
h(A_Q)=\frac{h(A)}Q,
$$

and analogously for $B_Q,C_Q$,

the heights of all interpolation nodes / values are uniformly bounded.

The Lagrange derivative coefficients consist only of finitely many:

- Nodes;
- Node differences;
- Inverses of differences;
- Finite sums / products;

as listed above.

And $L=M^2$ is fixed.

Their heights therefore also have a uniform bound.

Therefore:

$$
\boxed{
\alpha_Q
\text{ approaches }s
\text{ at high analytic order while its Weil height stays bounded}.
}
$$

---

This produces the third core result.

## Algebraic Degree Escape Theorem

By the Exact Approximate-Jet Order:

$$
\alpha_Q\to s.
$$

and:

$$
s
$$

is transcendental.

If:

$$
[\mathbb Q(\alpha_Q):\mathbb Q]
$$

is bounded along an infinite subsequence,

then:

- The degree is bounded;
- The Weil height is bounded by the Uniform Height Theorem.

Northcott's theorem states that there are only finitely many algebraic numbers of bounded degree and bounded height.

Thus:

$$
\alpha_Q
$$

can therefore take only finitely many values along that subsequence.

However:

$$
\alpha_Q\to s
$$

and:

$$
s
$$

is not algebraic.

a contradiction.

Thus:

$$
\boxed{
[\mathbb Q(\alpha_Q):\mathbb Q]
\longrightarrow
\infty.
}
$$

In other words:

$$
\boxed{
\text{Rational flattening transfers approximate-jet complexity from height to algebraic degree.}
}
$$

---

This round then quantifies degree escape using Waldschmidt's classical transcendence measure from 1978.

Waldschmidt considers:

$$
\omega
=
\frac{
\log\alpha_1
}{
\log\alpha_2
}
$$

where:

$$
\alpha_1,\alpha_2
$$

are nonzero algebraic numbers and the logarithm quotient is irrational.

His Corollary 4.5 gives:

$$
\boxed{
2C_{21}
n^3
(\Log H+n\Log n)
(1+\Log n)^{-2}
}
$$

as a transcendence measure for:

$$
\omega
$$

the quotient above.

Here, we take precisely:

$$
\omega=s,
$$

$$
\alpha_1=B,
$$

$$
\alpha_2=A.
$$

---

Let:

$$
d_Q
=
[\mathbb Q(\alpha_Q):\mathbb Q].
$$

Take:

$$
F_Q(T)\in\mathbb Z[T]
$$

to be the primitive minimal polynomial of:

$$
\alpha_Q
$$

this approximant.

The Uniform Height Theorem gives:

$$
h(\alpha_Q)\le C_{\mathrm{ht}}.
$$

The standard Mahler-measure / coefficient-height inequality therefore gives the effective:

$$
\boxed{
\log H(F_Q)
\le
C_0d_Q
}
$$

for some fixed:

$$
C_0=C_0(M,A,B,C).
$$

Waldschmidt's measure then gives:

$$
\boxed{
\log|F_Q(s)|
>
-
C_1
d_Q^4
}
$$

in a coarsened form sufficient for this round, where:

$$
C_1
$$

is effective and depends only on:

$$
M,A,B,C.
$$

More precisely, one can retain a gain of the form:

$$
(1+\Log d_Q)^{-1}
$$

as written above; this round does not need it.

---

On the other hand:

$$
F_Q(T)
=
a
\prod_{i=1}^{d_Q}
(T-\alpha_{Q,i}),
$$

where one conjugate is:

$$
\alpha_{Q,1}=\alpha_Q.
$$

By Mahler measure and:

$$
|s-\alpha_{Q,i}|
\le
2(1+|s|)
\max(1,|\alpha_{Q,i}|),
$$

we obtain:

$$
\boxed{
\log|F_Q(s)|
\le
\log|s-\alpha_Q|
+
C_2d_Q
}
$$

for fixed：

$$
C_2.
$$

The Exact Approximate-Jet Order gives:

$$
\log|s-\alpha_Q|
=
-(L-1)\log Q
+
O(1).
$$

Consistency therefore forces:

$$
(L-1)\log Q
\le
C_1d_Q^4
+
C_2d_Q
+
O(1).
$$

Therefore:

## Effective Degree-Escape Corollary

There exist:

$$
c>0,
$$

and:

$$
Q_0,
$$

such that:

$$
\boxed{
d_Q
\ge
c
(\log Q)^{1/4}
\qquad
(Q\ge Q_0).
}
$$

This exponent is not claimed to be optimal; it is merely a convenient effective form obtained directly from Waldschmidt's 1978 measure.

---

On the other hand, a simple field upper bound already gives:

$$
\boxed{
d_Q
\le
d_0Q^3.
}
$$

Thus, the current quantitative information is:

$$
\boxed{
c(\log Q)^{1/4}
\lesssim
d_Q
\lesssim
d_0Q^3.
}
$$

There is no contradiction here.

---

This round therefore gives AOJBP a precise diagnosis:

## Arithmetic-Degree Compensation Barrier

Analytic side:

$$
\boxed{
|J_Q|
\asymp
Q^{-(L-1)}.
}
$$

Height side:

$$
\boxed{
h(\alpha_Q)=O(1).
}
$$

Degree side:

$$
\boxed{
d_Q\to\infty.
}
$$

That is:

> The high-order approximate jet obtained by flattening is not free: its cost is algebraic degree escape.

An off-the-shelf quotient-of-logarithms transcendence measure can only quantify this cost; it cannot yet force:

$$
J_Q\neq0
$$

into:

$$
J_Q=0.
$$

---

This also explains why direct AOJBP exactification has not closed as quickly as ASRP in Round 22.

The barycentric rational functions of Round 22:

$$
f(T)
$$

arise directly from integer support, with degree / height controlled by support geometry.

The jet approximant of Round 28:

$$
\alpha_Q
$$

lies instead in a radical field varying with:

$$
Q
$$

the parameter above:

$$
K_Q.
$$

The smaller the analytic error,

the greater the arithmetic degree.

The genuine new bottleneck is therefore:

$$
\boxed{
\text{Can second-layer BADT data produce an approximate jet in a field of controlled degree?}
}
$$

rather than:

$$
\boxed{
\text{Can first-layer interpolation produce a small jet?}
}
$$

This round has fully proved that the latter is possible.

---

# 1. Extremal interpolation form

The extremal support of Round 25 is:

$$
\{(j,0):0\le j\le L-1\}
\cup
\{(0,1)\}.
$$

Normalize the coefficient of:

$$
T_2
$$

to:

$$
1.
$$

Thus:

$$
P_Q(T_1,T_2)
=
T_2-H_Q(T_1).
$$

The evaluation conditions are:

$$
H_Q(x_{a,b})=y_{a,b}.
$$

---

# 2. Interpolated function is exactly $x^s$

On the positive real branch:

$$
x_{a,b}
=
e^{\mu_Qr_{a,b}},
$$

Thus:

$$
x_{a,b}^s
=
e^{\mu_Qs r_{a,b}}
=
y_{a,b}.
$$

Therefore:

$$
H_Q
$$

is the ordinary Lagrange interpolation polynomial for:

$$
f(x)=x^s
$$

the function above.

---

# 3. One-jet formula

$$
P_Q(1,1)=0.
$$

And:

$$
\Theta_1P_Q(1,1)
=
-H_Q'(1),
$$

$$
\Theta_2P_Q(1,1)
=
1.
$$

Thus:

$$
\boxed{
(\mathcal D_sP_Q)(1,1)
=
s-H_Q'(1).
}
$$

---

# 4. Exact derivative remainder

Let:

$$
x_0=1
$$

and the remaining interpolation nodes:

$$
x_1,\ldots,x_{L-1}.
$$

The standard divided-difference remainder is:

$$
f(x)-H_Q(x)
=
f[x_0,\ldots,x_{L-1},x]
\prod_{i=0}^{L-1}
(x-x_i).
$$

At:

$$
x=x_0
$$

differentiate:

$$
\boxed{
f'(1)-H_Q'(1)
=
f[1,1,x_1,\ldots,x_{L-1}]
\prod_{i=1}^{L-1}
(1-x_i).
}
$$

---

# 5. Flat divided-difference limit

When:

$$
Q\to\infty,
$$

All:

$$
x_i\to1.
$$

Thus, the repeated-node divided difference:

$$
f[1,1,x_1,\ldots,x_{L-1}]
\longrightarrow
\frac{
f^{(L)}(1)
}{
L!
}.
$$

And:

$$
f^{(L)}(1)
=
(s)_{\underline L}.
$$

---

# 6. Node product asymptotic

For:

$$
x_i=e^{\lambda r_i/Q},
$$

we have:

$$
1-x_i
=
-\frac{\lambda r_i}{Q}
+
O(Q^{-2}).
$$

Thus:

$$
\prod_{i=1}^{L-1}
(1-x_i)
=
(-1)^{L-1}
\lambda^{L-1}
Q^{-(L-1)}
\prod_{i=1}^{L-1}r_i
+
O(Q^{-L}).
$$

---

# 7. Exact approximate-jet asymptotic

Combining:

$$
\boxed{
s-H_Q'(1)
=
(-1)^{L-1}
\frac{
(s)_{\underline L}
}{
L!
}
\lambda^{L-1}
\left(
\prod_{i=1}^{L-1}r_i
\right)
Q^{-(L-1)}
+
O(Q^{-L}).
}
$$

---

# 8. Nonzero leading constant

$$
s
$$

is transcendental, so:

$$
s\notin
\{0,1,\ldots,L-1\}.
$$

Therefore:

$$
(s)_{\underline L}\neq0.
$$

Each:

$$
r_i=a+bs
$$

with:

$$
(a,b)\neq(0,0)
$$

is also nonzero.

Thus, the leading constant is nonzero.

---

# 9. Uniform two-sided order

There therefore exist:

$$
c_-,c_+>0,
$$

$$
Q_1
$$

such that:

$$
\boxed{
c_-Q^{-(L-1)}
\le
|s-\alpha_Q|
\le
c_+Q^{-(L-1)}
}
$$

for:

$$
Q\ge Q_1.
$$

---

# 10. Algebraic field of the approximant

$$
x_{a,b}
=
A_Q^aB_Q^b,
$$

$$
y_{a,b}
=
B_Q^aC_Q^b.
$$

Lagrange interpolation uses field operations.

Thus:

$$
H_Q\in K_Q[T],
$$

$$
\alpha_Q=H_Q'(1)\in K_Q.
$$

---

# 11. Degree upper bound

$$
[K(A_Q):K]\le Q,
$$

$$
[K(A_Q,B_Q):K(A_Q)]\le Q,
$$

$$
[K_Q:K(A_Q,B_Q)]\le Q.
$$

Thus:

$$
\boxed{
[K_Q:\mathbb Q]
\le
d_0Q^3.
}
$$

---

# 12. Node height bounds

Let:

$$
H_X
=
(M-1)(h(A)+h(B)),
$$

$$
H_Y
=
(M-1)(h(B)+h(C)).
$$

then:

$$
h(x_{a,b})
\le
\frac{H_X}{Q},
$$

$$
h(y_{a,b})
\le
\frac{H_Y}{Q}.
$$

---

# 13. Difference height bounds

The standard inequality:

$$
h(x-y)
\le
h(x)+h(y)+\log2
$$

gives:

$$
h(x_i-x_j)
\le
\frac{2H_X}{Q}
+
\log2.
$$

Likewise:

$$
h(1-x_j)
\le
\frac{H_X}{Q}
+
\log2.
$$

Inversion does not change height.

---

# 14. Lagrange derivative heights

For:

$$
i\neq0,
$$

$$
\ell_i'(1)
=
\frac{
\prod_{j\neq i,0}
(1-x_j)
}{
(x_i-1)
\prod_{j\neq i,0}
(x_i-x_j)
}.
$$

Therefore:

$$
h(\ell_i'(1))
$$

is built from:

$$
O_M(1)
$$

uniformly bounded-height differences / inverses.

For:

$$
i=0,
$$

$$
\ell_0'(1)
=
\sum_{j\neq0}
\frac1{1-x_j},
$$

is likewise a fixed-length sum.

Thus:

$$
\boxed{
h(\ell_i'(1))
\le
C_{\ell}
}
$$

uniformly in:

$$
Q.
$$

---

# 15. Uniform height of $\alpha_Q$

$$
\alpha_Q
=
\sum_i
y_i\ell_i'(1).
$$

Fix:

$$
L=M^2.
$$

The finite-sum height inequality therefore gives:

$$
\boxed{
h(\alpha_Q)
\le
C_{\mathrm{ht}}
}
$$

uniformly in:

$$
Q.
$$

---

# 16. Northcott degree escape

If there exist:

$$
D
$$

and an infinite sequence:

$$
Q_\nu
$$

such that:

$$
[\mathbb Q(\alpha_{Q_\nu}):\mathbb Q]
\le D,
$$

then:

- The degree is bounded;
- The height is bounded.

Northcott's theorem states that this set is finite.

Thus:

$$
\alpha_{Q_\nu}
$$

has a constant subsequence.

However:

$$
\alpha_Q\to s
$$

and:

$$
s
$$

is transcendental.

a contradiction.

Hence:

$$
\boxed{
\deg_{\mathbb Q}\alpha_Q\to\infty.
}
$$

---

# 17. Waldschmidt quotient-of-logs measure

Waldschmidt 1978, Theorem 4.4 / Corollary 4.5 considers:

$$
\omega
=
\frac{
\log\alpha_1
}{
\log\alpha_2
}.
$$

For:

$$
\omega=s=\log B/\log A,
$$

he gives an explicit approximation / transcendence measure of order:

$$
\boxed{
C_s
n^3
(\Log H+n\Log n)
(1+\Log n)^{-2}.
}
$$

---

# 18. Minimal-polynomial height

Let:

$$
F_Q
$$

to be the primitive minimal polynomial of:

$$
\alpha_Q
$$

this approximant, with degree:

$$
d_Q.
$$

The uniform Weil height:

$$
h(\alpha_Q)\le C_{\mathrm{ht}}
$$

gives:

$$
M(F_Q)
=
e^{d_Qh(\alpha_Q)}
\le
e^{C_{\mathrm{ht}}d_Q}.
$$

The standard coefficient / Mahler bound gives:

$$
H(F_Q)
\le
2^{d_Q}M(F_Q).
$$

Thus:

$$
\boxed{
\log H(F_Q)
\le
(C_{\mathrm{ht}}+\log2)d_Q.
}
$$

---

# 19. Waldschmidt lower bound

Apply Corollary 4.5:

$$
\boxed{
\log|F_Q(s)|
>
-
C_1d_Q^4
}
$$

using a slightly weaker but concise effective form.

---

# 20. Upper bound through the close conjugate

Since:

$$
F_Q(s)
=
a
(s-\alpha_Q)
\prod_{i\neq1}
(s-\alpha_{Q,i}),
$$

and:

$$
|s-\alpha_{Q,i}|
\le
2(1+|s|)
\max(1,|\alpha_{Q,i}|),
$$

Thus:

$$
|F_Q(s)|
\le
|s-\alpha_Q|
[2(1+|s|)]^{d_Q}
M(F_Q).
$$

Therefore:

$$
\boxed{
\log|F_Q(s)|
\le
\log|s-\alpha_Q|
+
C_2d_Q.
}
$$

---

# 21. Effective degree escape

From:

$$
|s-\alpha_Q|
\le
c_+Q^{-(L-1)},
$$

we obtain:

$$
-(L-1)\log Q
+
O(1)
+
C_2d_Q
>
-
C_1d_Q^4.
$$

Thus, for large:

$$
Q,
$$

$$
\boxed{
d_Q
\ge
c(\log Q)^{1/4}.
}
$$

---

# 22. Why this does not close AOJBP

The field construction gives only:

$$
d_Q\le d_0Q^3.
$$

Thus:

$$
c(\log Q)^{1/4}
\le
d_Q
\le
d_0Q^3
$$

leaves enormous room.

Waldschmidt's measure therefore fully allows:

$$
|s-\alpha_Q|
\asymp
Q^{-(L-1)}.
$$

without forcing the exact equality:

$$
\alpha_Q=s.
$$

---

# 23. Arithmetic-Degree Compensation Barrier

Round 28 therefore obtains:

$$
\boxed{
\text{analytic approximation order}
\leftrightarrow
\text{algebraic degree growth}.
}
$$

Flattening makes:

$$
J_Q
$$

small to high order,

but the approximant leaves every bounded-degree algebraic set.

---

# 24. Second-layer status

This round's construction actually uses only:

$$
A,B,C.
$$

It proves that the first layer alone can already produce an approximate jet.

To actually complete OJBP, BADT's second layer must therefore supply not:

$$
\boxed{
\text{better analytic Taylor order}
}
$$

but rather:

$$
\boxed{
\text{better arithmetic degree control}
}
$$

or:

$$
\boxed{
\text{a fixed-field / bounded-degree approximation}.
}
$$

---

# 25. Fixed-Field Approximate Jet Problem

This round proposes:

**Fixed-Field Approximate Jet Problem (FFAJ)**

Can:

$$
D=e^{s^4\lambda}
$$

and the nested BADT structure be used to construct an algebraic:

$$
\widetilde\alpha_Q
$$

such that:

$$
|\widetilde\alpha_Q-s|
$$

with comparable high-order decay,

while:

$$
[\mathbb Q(\widetilde\alpha_Q):\mathbb Q]
$$

stays bounded, or at least far smaller than the rational-root field scale:

$$
Q^3?
$$

If bounded degree and bounded height hold, Northcott alone contradicts convergence to transcendental $s$.

FFAJ is therefore narrower than direct Baker exactification.

---

# 26. A Northcott-level target

Ideally, one needs only:

$$
\boxed{
\sup_Q
h(\widetilde\alpha_Q)<\infty,
}
$$

and:

$$
\boxed{
\sup_Q
[\mathbb Q(\widetilde\alpha_Q):\mathbb Q]
<\infty.
}
$$

together with:

$$
\widetilde\alpha_Q\to s.
$$

to obtain a contradiction directly.

An explicit linear-form lower bound is not even required.

---

# 27. Relation to second-layer elimination barriers

Round 25 proved:

- Coefficient enrichment alone does not help;
- Finite torus elimination does not help.

Thus, if FFAJ exists, it does not come from an ordinary algebraic trace / resultant.

It must produce degree compression while logarithmic information is still available.

The new bridge can therefore be described as:

$$
\boxed{
\text{logarithmic relation}
\Longrightarrow
\text{arithmetic degree compression of approximate tangent}.
}
$$

---

# 28. Round 28 Verdict

**PASS — Rational flattening produces an exact high-order approximate one-jet; its height stays uniformly bounded, while its algebraic degree necessarily escapes**

This round completes:

1. Normalize the Round 25 extremal polynomial as:
   $$
   P_Q=T_2-H_Q(T_1).
   $$

2. Identify:
   $$
   H_Q
   $$
   as the algebraic-node interpolation polynomial for $x^s$.

3. Obtain:
   $$
   J_Q=s-H_Q'(1).
   $$

4. Establish the exact divided-difference derivative remainder.

5. Derive the nonzero leading asymptotic:
   $$
   J_Q
   =
   cQ^{-(L-1)}
   +
   O(Q^{-L}).
   $$

6. Obtain:
   $$
   |J_Q|
   \asymp
   Q^{-(M^2-1)}.
   $$

7. Prove:
   $$
   \alpha_Q=H_Q'(1)
   $$
   is algebraic.

8. Establish the radical-field degree upper bound:
   $$
   \le d_0Q^3.
   $$

9. Establish uniform node / difference height bounds.

10. Prove:
    $$
    h(\alpha_Q)\le C_{\mathrm{ht}}
    $$
    independently of $Q$.

11. By Northcott, prove:
    $$
    \deg\alpha_Q\to\infty.
    $$

12. Apply Waldschmidt's 1978 quotient-of-logarithms transcendence measure.

13. Obtain the crude effective:
    $$
    \deg\alpha_Q
    \gg
    (\log Q)^{1/4}.
    $$

14. Identify the Arithmetic-Degree Compensation Barrier.

15. Explain that the first layer solves analytic approximation, while the second must actually address arithmetic degree control.

16. Propose the Fixed-Field Approximate Jet Problem (FFAJ).

---

# 29. The Formal Entry Point to Round 29

Round 29 no longer attempts to increase the approximation order of:

$$
Q^{-(4N^2-1)}
$$

the decay above.

What actually needs studying is:

$$
\boxed{
\text{Can BADT second layer compress the algebraic degree of the approximate tangent?}
}
$$

Three priority routes follow.

## R29-A: Trace / Norm Descent

Study:

$$
\alpha_Q=H_Q'(1)
$$

through its traces / symmetric functions in:

$$
K_Q/K
$$

this extension.

Check:

- Whether trace normalization still approximates:
  $$
  s;
  $$
- Whether Galois conjugates cancel through root-of-unity twists;
- Whether a bounded-degree symmetric approximation exists.

---

## R29-B: Two-Layer Tangent Approximants

Establish simultaneously:

$$
\alpha_Q\approx s,
$$

and, for the second layer:

$$
\beta_Q\approx s^2.
$$

Study:

$$
\beta_Q-\alpha_Q^2
$$

for its algebraic degree / height.

If the nested structure forces it to be exactly zero, then check whether this can in turn reduce the degree of $\alpha_Q$.

---

## R29-C: Kummer Degree Accounting

Using:

$$
A,B,C
$$

being multiplicatively independent,

Study:

$$
[K(A^{1/Q},B^{1/Q},C^{1/Q}):K]
$$

for its actual growth.

If the degree is nearly:

$$
Q^3,
$$

this proves that first-layer interpolation intrinsically carries a large arithmetic-degree cost, rather than merely being loosely estimated.

---

# 30. The One-Sentence Conclusion

> **Round 28 proves that the first layer is already very effective at producing approximate jets: the tangent slope of the Round 25 extremal interpolation polynomial converges at exact order $Q^{-(4N^2-1)}$ to the transcendental slope $s$. More counterintuitively, these algebraic tangent approximants can retain uniformly bounded Weil height. Only algebraic degree must diverge. Northcott and Waldschmidt's measure both detect this degree escape, but the available lower bounds do not yet force an approximate jet into an exact jet.**

The genuine new bottleneck is therefore:

$$
\boxed{
\text{Approximate Jet}
\text{ is available;}
\quad
\text{Degree Compression}
\text{ is not available.}
}
$$

Round 29 will directly study whether the second layer can supply degree compression.

---

# References

1. Michel Waldschmidt, **Transcendence measures for exponentials and logarithms**, *Journal of the Australian Mathematical Society, Series A* 25 (1978), 445–465.  
   Theorem 4.4 and Corollary 4.5 give approximation and transcendence measures for:
   $$
   \frac{\log\alpha_1}{\log\alpha_2}.
   $$
   PDF: https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/JAustralMS25-1978-445-465.pdf

2. D. G. Northcott, classical finiteness theorem: algebraic numbers of bounded degree and bounded absolute Weil height form a finite set.

3. Standard Lagrange / Hermite divided-difference interpolation remainder formulas.

4. Standard absolute logarithmic Weil height inequalities:
   $$
   h(\alpha\beta)\le h(\alpha)+h(\beta),
   $$
   $$
   h(\alpha\pm\beta)
   \le
   h(\alpha)+h(\beta)+\log2.
   $$

5. Michel Waldschmidt, **Diophantine Approximation on Linear Algebraic Groups**, Springer, 2000.  
   Contains linear-independence measures for logarithms, zero estimates, multiplicity estimates, and the Linear Subgroup Theorem.

---

**End of ACTM Round 28**