Unity, Refinement, and Signed Capture

Seeing more detail does not automatically improve an explanation. Two readings can see the same thing. A new reading can cancel against something the existing explanation already contains. If we count every visible feature as new information, we can claim an improvement that never takes place.

Arithmetic gives this difficulty a concrete form. A collection of remainder clocks tries to reproduce one at every positive integer. We admit more clocks, retain the earlier ones, and ask what the finer description can account for. The target stays fixed. The relationships between the readings decide what we gain.

This is what I mean by signed capture. A reading earns its place by supplying a correction that reduces what remains unresolved. Its sign and its overlap with the other readings are part of that correction.

The construction below proves that a prescribed family can make these corrections at a bounded joint cost, even as the number of eligible refinement levels grows. It also identifies a gap that conservation alone does not close. Locating unresolved information and capturing it with the available clocks are different mathematical acts.

One, divided into addresses

Give address one half the observation weight. Give address two a sixth, address three a twelfth, and continue with

$$\omega_m=\frac{1}{m(m+1)}=\frac1m-\frac1{m+1}.$$

These shares fill one whole. On the unit interval, the share belonging to address $m$ is simply the interval between $1/(m+1)$ and $1/m$. Larger addresses occupy smaller intervals. The entire tail from address $L$ onward has weight $1/L$.

This measure has an arithmetic origin in The Clocks Beneath Collision Energy and The Weight of a Carry. The primitive collision kernel assigns mass $1/[2ab(a+b)]$ to each coprime positive pair $(a,b)$. Its total mass is one. The slice with $a=1$ has mass one half. Conditioning on that slice gives exactly the weights above. The observation measure is already present inside the collision structure.

Now let the period-two clock run. At successive addresses its fractional readings are one half, zero, one half, zero. Double them. We obtain one at every odd address and zero at every even address.

That is the best fit available from this clock alone. Nothing remains unresolved at the odd addresses. Everything at the even addresses remains. Their combined weight is $1-\log 2$, about $0.30685$.

Here unity is both the fixed target and the total observation mass. The mathematical question is how increasingly rich arithmetic descriptions fit that same whole. The weights do not change when a new clock arrives, and the early clocks do not disappear.

The parent remains in the child

A clock of period $j$ reads the fractional part $\phi_j(m)=\{m/j\}$. Admitting every period from two through $N$ gives a finite collection of patterns. Their best weighted combination is $p_N$. The residual $r_N=1-p_N$ records what that combination leaves unresolved. Its squared size is $D_N$.

Calling this a best fit has a precise consequence. No adjustment using only the clocks already admitted can improve it. A useful correction must bring a direction those clocks do not yet represent.

Binary copying makes the relation between resolutions visible. Copy the value at parent address $m$ to the two child addresses $2m$ and $2m+1$. At address one, where there is no positive parent, the copy is zero. The child weights obey

$$\omega_{2m}+\omega_{2m+1}=\frac12\omega_m.$$

The copied value is unchanged, while its contribution to squared norm is halved. This holds for the whole sequence, including its infinite tail.

The copy is itself arithmetic. A copied period-$j$ clock can be written using two clocks at the finer scale,

$$T\phi_j=\phi_{2j}-\frac1j\phi_2.$$

For example, the period-three value $1/3$ at parent address one becomes $1/3$ at both child addresses two and three. The period-six clock alone gives different values there. Subtracting the appropriate period-two reading restores the inherited value at both children.

This is the sense in which refinement retains ancestry. The finer description has more parts, but those parts remain linked by the carry identities.

Copying is not the whole refinement step. A new fit also introduces a source beyond the copy, and that source interacts with it. If $G=TF+\eta$, then

$$\|G\|^2=\tfrac12\|F\|^2+\|\eta\|^2+2\langle TF,\eta\rangle.$$

The first term is controlled by copying. The last two still need their own accounting.

Many readings without an ever-growing charge

We can organize readings of the residual along the binary ancestry. The paper proves that a definite amount of unresolved energy appears within an initial range of these levels. In symbols, at least $D_N^2/4$ appears before depth $\lceil\log_2(4/D_N)\rceil$ on the distinguished binary chain.

But a visible pattern need not be a combination of the clocks we are allowed to use. The construction therefore makes a separate family of permitted probes. At each eligible depth it takes the part of the root address available in a fourfold band of clock periods, copies it to the current scale, and removes everything already represented by the old clock space. Each resulting probe, written $w_{N,h}$, belongs to the clocks admitted by cutoff $4N$ and is perpendicular to the old space.

The cost of using several probes is the squared size of their combined correction. Overlapping probes can increase that cost. We must keep their interactions.

For this particular family, the probes’ squared sizes add to at most two. Consequently, any linear combination costs at most twice the sum of its squared coefficients. The figure shows why. Before copying, alternate depths use disjoint projection bands. One set of alternate depths spends at most one unit of root energy. The other set spends at most one more. Normalized copying preserves the norms, and removing the old space cannot increase them.

Let $b_{N,h}=\langle r_N,w_{N,h}\rangle$ be the complete signed reading. A simple correction works at every cutoff,

$$\widetilde p_N=p_N+\frac12\sum_h b_{N,h}w_{N,h}.$$

The bound on joint cost gives

$$\boxed{D_N-D_{4N}\ \ge\ \frac12\sum_h |b_{N,h}|^2.}$$

This is a proved improvement rule. Any nonzero complete reading gives a positive improvement. A negative reading tells the correction to move in the opposite direction; its square still contributes positively to the bound.

The constant does not grow with the number of eligible levels. Here cost means the size of a correction in the weighted space. It does not mean that calculating the projections takes constant computer time.

A reading includes the fit already made

The signs become consequential when a reading is assembled from simpler expressions. Its clock contribution, conditioning term and transported old fit must all be present.

At cutoff $25$, one of the paper’s nfield calculations gives a clock contribution of about $0.082296$. Conditioning removes about $0.073788$. The result so far is positive, about $0.008507$.

The old fit then removes about $0.012247$. The actual reading is negative, about $-0.003740$.

Omitting the transported fit sends the correction in the wrong direction in this example. It would amount to asking a descendant clock to explain the original target while forgetting what its ancestors already explain.

The example illustrates an exact response identity. It is a finite calculation, not the proof of a uniform lower bound. The symbolic cost theorem uses the complete reading, whatever its sign. What remains difficult is proving that sufficiently strong readings must continue to survive.

Palindromic, torsional, radial

The digit geometry supplies another family of patterns to read. In decimal, fix the final two-digit sum at nine. The resulting block is

$$09,\ 18,\ 27,\ 36,\ 45,\ 54,\ 63,\ 72,\ 81,\ 90.$$

Reversing the two digits reverses this entire list. That is its palindromic reflection. An address such as $36$ need not itself be a palindrome; it is paired with $63$ inside a symmetric block.

Consecutive addresses move between residue classes modulo nine. I call these classes spokes. Nine address steps return to the same spoke at a later address.

The paper assigns a polarity using the digit sum and the full earlier prefix. Below $100$, it is positive for digit sums below nine and negative for sums from nine through seventeen. Half the change in polarity between neighboring addresses is a torsional reading. Adding the nine readings measures the radial increment between the two endpoints on the same spoke.

From $36$ through $45$, the polarity changes twice. It rises at $40$ and falls at $45$. The two nonzero torsional readings are $+1$ and $-1$. Their individual squares add to two. The square of their sum is zero.

The full radial energy retains that interaction. It is not a third independent reservoir beside the torsional readings. Reflection also does not make the address weights equal. The pair $36,63$ has a symmetric digit relation and unequal observation weights.

These names describe operations we can calculate. The identities behind their accounting are weighted mean–difference decomposition and telescoping. The arithmetic determines where the changes occur and how their signs meet the actual residual.

Copied signed patterns have a further uniform cost bound. With copies separated by a scale factor $q\ge2$, the optimal universal constant is

$$\frac{\sqrt q+1}{\sqrt q-1}.$$

At fourfold spacing it is three, regardless of how many copies are used. This is a classical geometric Gram bound applied to the native copies. It controls their cost; it does not guarantee a stronger reading. Indeed, the paper’s finite comparisons favor simpler address probes over the more elaborate polarity family in the tested bands.

Where Bagchi enters

The weighted approximation problem is established mathematics. Bagchi’s 2006 exposition of the Nyman–Beurling–Báez-Duarte criterion uses precisely the weights $1/[m(m+1)]$, the fractional-part clocks and the constant target one. His Theorem 1 says that approximating this target arbitrarily closely is equivalent to the Riemann hypothesis.

So the connection here is literal. It is the same approximation problem.

The collision program reaches those weights by conditioning its primitive unit mass. The present paper then selects particular finite readings, controls the cost of their corrections, and derives their complete responses under arithmetic refinement. Those constructions and their response estimates are the work offered here. The approximation criterion, orthogonal projection and the geometric Gram estimate retain their classical attribution.

The unity perspective directs the questions. It asks how the same whole persists through finer arithmetic descriptions, which relationships refinement preserves, and how much the descriptions actually account for together. Its mathematical content has to be judged in those constructions and estimates.

Two bounds with different jobs

There are already bounds on both sides of the research. Burnol’s lower-bound theorem transfers to these clocks and gives $D_N\ge c/\log N$ for sufficiently large $N$. It limits how fast the best-fit error can disappear. It does not establish that the error disappears.

A separate arithmetic construction prescribes a signed refinement packet. Its complete energy, called $E_t$, includes the infinite observation tail. The paper develops this construction after the capture criterion and proves

$$E_t\le C_B\frac{t}{(\log t)^B} \qquad\text{for every fixed }B>0.$$

The same argument also gives a stronger stretched-exponential saving. These estimates control the entire packet, but still permit growth. For example, $t^{0.2}$ eventually lies below $t/(\log t)^B$ for every fixed $B$. Increasing $B$ does not remove the gap: its constant and the point where the comparison takes hold can change with $B$.

The optimized error, the packet’s energy and the gain from our selected readings are different quantities. We have not sandwiched one quantity between two bounds tending to zero. The packet route needs a slower growth rate; the detector route needs enough accumulated signed capture. The paper gives the precise sufficient conditions and keeps both questions open.

What must keep surviving

Let $J_N$ be the best improvement obtainable from the prescribed family when all its overlaps are included. We know that $J_N\le D_N-D_{4N}$. Along a fixed chain $N_k=N_*4^k$, the paper proves the implication

$$\sum_k\frac{J_{N_k}}{D_{N_k}^2}=\infty \quad\Longrightarrow\quad D_N\longrightarrow0.$$

The implication is a short telescoping argument. Establishing its premise is the open arithmetic problem. Weak steps are allowed. The question concerns how the gains accumulate through continued refinement.

We have a location theorem, a family of permitted corrections and a cost that stays bounded as more levels are admitted. We also have the exact cancellation that prevents us from substituting visible energy for captured gain.

The whole remains fixed. The partitions become finer. What must still be proved is that the signed readings cannot keep revealing structure without capturing enough of it.