The Common Reset in the Cubic Law

The digit tables in this notebook are finite. Every entry counts something, how many carries one multiplication makes, and every count is a whole number that anyone patient could check by hand. Counts like that can be trusted completely. The same finiteness makes them hard to think about all at once, because a table of a million counts is just a million entries, with no formula to hold them together.

So the tables get traded for something smoother to work with. Behind each one sits a curve that takes the table’s values at evenly spaced points and fills in every point between them. The curve is the more abstract of the two, since no one can ever list its values at every real number. It is also the version calculus can handle, and areas under it come with formulas. It is the same trade Gauss made when he replaced the count of primes with the area under a smooth curve.

Reading a curve at evenly spaced points is called sampling, and the table is a sample of its curve in exactly that sense. A digital photograph does this too, each pixel recording the scene at one spot. The bargain is that the finer the sampling, the more alike the two become, so a photograph with small enough pixels looks like the scene itself. One bad pixel hardly shows, because its share of the photograph shrinks as the pixels multiply.

Digit-collision energy breaks that bargain in one exact place. Its carry table samples a curve built from sawtooth clocks, readings that climb steadily and then reset, with one more clock for every step up in base. As the base grows the samples crowd closer together. Zero is their common reset, the one place where every clock starts over at once. Just beside zero the clocks agree and their sum is as large as it can ever be, while the table, reading exactly at zero, records a zero. It is a dark pixel at the brightest spot in the scene, and the brightness there grows at the same rate the pixels shrink.

At base 1,009 the table has more than a million entries, and beside zero the curve’s 1,008 clocks add up to about a thousand. The energy is built from the squares of the entries, so the square that belongs at zero is about a million. Spread over a million entries, that one missing square pulls the table’s average down by 0.998, almost a full unit.

The Secondary Term of the Cubic Law leaves the gap between the two averages open. Every table it computes points to one, but its proof can only hold the gap below a bound that keeps growing with the base. The paper behind this article proves that the gap tends to exactly one, and that the dark pixel accounts for the whole unit. Everywhere else, the finite table and the continuous curve agree in the limit. So the abstract curve does tell the truth about the whole numbers once one unit is set aside, and that unit sits at a single point that the curve’s average, on its own, cannot even see.

Two marks cross zero

This is the same table that Digit Collisions and the Cubic Law builds, and the picture below repeats its smallest case. In base three the words with matching digits, 00, 11 and 22, are the numbers zero, four and eight. Space nine places evenly around a circle and mark those three. Doubling sends them to zero, eight and seven. Then move every mark two more places. The zero goes to two. The eight passes zero on its way to one. The seven lands exactly on zero, and landing there is counted as a carry as well. That gives two carries.

A step of two crosses zero from two of the nine starting places, so three marks placed with no pattern would expect two thirds of a crossing between them. The excess, four thirds, is the gold line under the circle. Doing this for every multiplier from zero to eight fills the carry table. The excesses are its centered readings, and their squares add up to the collision energy.

The picture’s lower half reads the same table a second way. Picture a quantity that climbs at a steady rate and falls back each time it reaches a whole number. Center it so it averages zero, and give it the value zero at the moment of the drop. That repeating ramp is a sawtooth, and I call it a clock. In base $p$ I run $p-1$ of them, the first going around once per unit, the second twice, the last $p-1$ times. Doubling each clock and adding them gives a single curve. Read that curve at the $p^2$ evenly spaced points of the grid and the readings are exactly the centered carry table, shuffled into a different order. The stems in the picture are the nine readings at base three, from $-4/3$ at the second point to $4/3$ at the last.

The energy now has two averages to compare. Square the nine stems and average them and you get $16/27$. Average the square of the whole curve and you get exactly $1$. The gold $11/27$ at the bottom is their difference. It is the gap from the opening, and the paper calls it the sampling defect, since it measures how far the average over the sample points falls from the average over the whole curve.

The two averages in the paper’s notation

Let $\psi(x)=\{x\}-1/2$ away from integers and $\psi(x)=0$ at integers. Here $\{x\}$ is the fractional part. Put $N=p-1$ and

$$B_N(x)=2\sum_{j=1}^{N}\psi(jx).$$

The continuous average and the grid average are

$$\mathcal J(N)=\int_0^1 B_N(x)^2 dx, \qquad \frac{E_p}{p^2}=\frac1{p^2}\sum_{a=0}^{p^2-1}B_N(a/p^2)^2.$$

Their difference is $\mathcal S_p=\mathcal J(p-1)-E_p/p^2$. The theorem proves $\mathcal S_p\to1$ through odd primes.

At base five the continuous average is $23/9$, the grid average is $48/25$, and the defect is exactly $143/225$.

Four clocks, one reset

Base five has four clocks and a grid of twenty-five points. In the upper half of the next picture each row is one doubled clock. The first rises once across the window and the fourth rises four times. Every one of them resets at zero, the shaded band on the left, because each goes around a whole number of times per unit.

Just to the right of zero each doubled clock sits near minus one, the open circle at the foot of its first ramp. Just to the left, coming around from the far end, each sits near plus one. The four add to nearly minus four on one side and nearly plus four on the other, so their square is close to sixteen on both sides. That is the open circle at the top of the peak in the lower graph. At zero itself each clock takes its assigned value zero, the gold dots in the shaded band, and the squared sum is zero. The grid samples exactly there. Its reading is the gold dot where the dashed line meets the axis, sixteen units below the curve on either side.

The curve’s average is the same whatever value sits at zero, since an integral cannot see one point. The grid gives that point a twenty-fifth of its average, and putting sixteen there in place of zero would raise the average by $16/25$, the number at the right. At base $p$ the peak is $(p-1)^2$ and the sample’s share is $1/p^2$, so the missing amount is

$$\frac{(p-1)^2}{p^2}\longrightarrow1.$$

This one point keeps its weight because the clocks are added before they are squared. Near zero they all agree, so the height of the peak grows with the square of the number of clocks, while the grid’s weight on one point shrinks as $1/p^2$. The two run at the same pace. With a fixed set of clocks and an ever finer grid, the missing sample would count for less and less, like any single point under an integral.

Twenty-five rectangles

The common reset at zero is far from the only reset. A clock restarts wherever its frequency carries it to a whole number, and at each such point some clocks restart while the others carry on. At base five these are the smaller peaks and breaks at $1/4$, $1/3$, $1/2$, $2/3$ and $3/4$. Each one bends the curve in a way the grid may or may not catch, depending on where the reset falls between two sample points.

To see what they amount to together, put sixteen in place of the zero at the origin and leave every other sample alone. Draw each sample as a rectangle $1/25$ wide, standing on its own point and reaching to the right. The rectangles’ total area is the corrected grid average, and comparing them with the curve cell by cell shows every part of the difference that remains.

On the falling stretch at the left, each rectangle takes its height from the high end of its cell and overshoots the curve. Those cells are violet. On the rising stretch at the right, each rectangle takes the low end and falls short, and those cells are teal. The squared curve is symmetric about one half, so the tall violet bars opening the lower chart nearly mirror the tall teal bars that close it. The short bars between them belong to the interior resets.

Add all twenty-five areas with their signs and the result is

$$\frac{23}{9}-\frac{48+16}{25}=-\frac1{225}.$$

Once the origin is corrected, the rectangles hold slightly more than the curve does. At other bases the leftover can fall on either side of zero, and the theorem proves that it tends to zero.

Watch the clocks multiply, from 25 samples to more than a million

Four clocks make a pattern we can follow by eye. At 1,008 clocks, the same ramps form a much finer weave. Each horizontal band below is one clock. Violet is a negative reading and teal is a positive one. Read straight down a column, add those readings, and square the total. That gives the curve beneath the weave.

The large returns near one half and one third have a simple cause. Every second clock resets at one half; every third resets at one third. At zero they all do. The six pictures use the same horizontal interval, so those meeting places stay put while more clocks crowd around them. Their heights are scaled to each panel’s common peak, not to one shared energy scale.

Select the image to open the full-resolution view. At base 1,009, every one of the 1,008 clock rows is retained.

The figures under each panel separate the reset at zero from everything else. Its contribution rises from 0.64000 to 0.99802. The signed remainder can fall on either side of zero; these six bases are not a decreasing sequence of errors. The proof has to control that remainder through all odd primes, not just the ones pictured here.

Two orientations in one wedge

Proving that the leftover vanishes is the hard part, and the proof spends nearly all its length there. The combined contribution of the interior resets can be written exactly as a signed sum of Dedekind sums, small finite sums that Richard Dedekind introduced in 1892, in his commentary on two of Riemann’s fragments about modular functions. They obey a reciprocity law. The Dedekind sum for a pair $(h, k)$ plus the one for $(k, h)$ equals a simple fraction.

Each use of that law is a division with remainder, much like a step of Euclid’s algorithm, and it trades one sum for an explicit correction and a new sum with a smaller modulus. Following every term down this way produces a large family of division paths. They have different lengths, and each comes in one of two orientations with opposite signs. Bounding the paths one at a time, by size alone, would throw away the cancellation between the orientations, and the proof needs that cancellation.

Each path can be rebuilt by a two-by-two matrix, and the top row of that matrix gives two whole numbers, $A$ and $B$. Their ratio always lands strictly inside one interval,

$$-\frac1{\varphi^2}<\frac BA<\frac1\varphi, \qquad \varphi=\frac{1+\sqrt5}{2},$$

from about −0.382 to 0.618. In the other direction, every reduced fraction $B/A$ inside that interval, with $|B|$ at least two, comes from exactly one path. The endpoints come from the extreme division steps repeated forever, which turn $B/A$ into ratios of Fibonacci numbers closing in on the golden limits.

Each point in the picture is one path, with $B$ across and $A$ up. The two gold edges are the golden limits, and the lowest points sit at $A = 5$, the smallest denominator for a path of two or more steps. Circles and triangles are the two orientations. They are mixed together everywhere in the wedge, and neither claims a region of its own. An exact identity moves each path’s orientation sign into the reading it carries, so the two kinds can be summed as one population.

Once they are, hold $B$ fixed and walk up a single column of the wedge. Along that column the readings swing like a phase that depends on the reciprocal of $A$ plus a small shift, and classical estimates for sums of that kind show that they cancel by a small power of the base. The saving survives the sum over every starting cutoff and every depth of descent, which is what the limit requires.

The interval itself is classical. It is the shape of a family of continued fractions introduced by Hitoshi Nakada, and the paper proves directly the finite list of paths it needs. The golden ratio decides which paths exist. It has no part in the size of the unit, which comes from the reset at zero alone.

Most of the tools along the way are classical too. Dedekind sums and their reciprocity law, Weil’s bound for Kloosterman sums and the derivative estimates for exponential sums all have long histories. The new parts are the exact identity that sets the reset at zero apart from everything else, and the joining of the two orientations that holds the signs together long enough for those tools to apply.

The rate the proof gives

The proved estimate is

$$\mathcal S_p=1+O\left(p^{-\eta}\log^4(2p)\right), \qquad \eta>0,$$

with an absolute implied constant. The exponent comes from two choices in the proof. Derivative estimates through order fourteen give a saving of $1/(2^{14}-2)$, and splitting the denominators at $p^{1/28}$ converts it to

$$\eta=\frac1{28(2^{14}-2)}=\frac1{458696}.$$

The exponent is a conservative choice, enough to make the error tend to zero. The golden interval is $((\sqrt5-3)/2,(\sqrt5-1)/2)$, and the unit itself comes from the separate ratio $(p-1)^2/p^2$.

Closing in on one

The last picture sets the theorem beside the tables. In the top panel the teal points are the full defect for every odd prime through 1,009 and for seven larger bases up to 10,007. They start at 0.41 for base three, climb, and then scatter above and below the dashed line marking one, from about 0.87 at base 73 to about 1.12 at base 373. The gray curve beneath them is the reset alone, $(p-1)^2/p^2$, rising toward one. The lower panel takes that curve away. The violet points that remain are everything the interior resets do. They straddle zero, swinging by as much as a tenth among the bases below 1,000 and lying close to the line at the larger checkpoints.

In energy units, writing $\mathcal J(p-1)$ for the curve’s average, the theorem says

$$E_p=p^2\bigl[\mathcal J(p-1)-1\bigr]+o(p^2).$$

Digit Collisions and the Cubic Law gives the energy’s leading size, $p^3$, and The Secondary Term of the Cubic Law gives the next correction, of size $p^2(\log p)^2/\pi^2$. The unit found here sits beneath both, a correction of size $p^2$ measured against the exact continuous curve. It does not by itself give a third term of the energy’s expansion, which would also need a finer expansion of the curve’s own average.

The complete proof is in The Unit Sampling Defect in Digit-Collision Energy (PDF, preprint).

The proof’s rate is very slow. Its exponent, $1/458696$, is enough to force the limit and says almost nothing about speed. The tables suggest something much faster, a square-root rate,

$$\left|\mathcal S_p-1\right| =O_\varepsilon\left(p^{-1/2+\varepsilon}\right) \qquad\text{for every }\varepsilon>0.$$

The conjecture lets the defect wander above and below one. If it holds, the envelope around that wandering narrows about tenfold for every hundredfold increase in the base, apart from the small allowance $\varepsilon$. Proving it would take stronger cancellation within the same signed family of interior resets.

A million samples at base 1,009

Back at base 1,009, the grid has 1,018,081 samples and the curve has 308,925 interior resets inside the unit interval. The defect there is 1.03746, a little above one. The single sample at zero accounts for 0.99802 of it, and the interior resets together add the other 0.03944. At other bases their leftover comes out negative and the defect dips below one, which is the scatter in the last picture. Either way the leftover shrinks toward zero as the base grows, while the share of the sample at zero rises toward exactly one. In the limit, the whole difference between the table and its curve comes from the one sample taken where every clock starts over.