A harmonic is a step taken twice

The previous essay reads a pair out of a divergence sequence by sorting the significant lags into residue classes. It is worth being clear about why that is the only thing that works, because three more obvious methods were tried first and all three failed, and what they failed on says something about the object.

The problem, stated exactly. At a rise of 0.005 the correlations that clear the band are at lags 8, 13, 16, 21, 24, 26 and 29, with values 0.82, 0.56, 0.62, 0.48, 0.46, 0.31 and 0.39. The parastichy pair on that stem is 8 and 13. So the task is to look at those fourteen numbers and pick out 8 and 13 — not 16, which is larger than 13 in correlation; not 21, which is the sum and is the most family-looking number in the list.

Why the peak heights do not do it #

Rank the lags by correlation and the order at this rise is 8, 16, 13, 21, 24, 29, 26. The second-ranked lag is a harmonic. Taking the top two would return 8 and 16 and be wrong, and taking the top two that are not multiples of each other would return 8 and 13 here and 5 and 3 one rung coarser, where the pair is 5 and 8.

The reason is visible once the lattice is drawn beside it. The hop lengths on a stem at a rise of 0.013 are 0.111 at offset eight, 0.120 at offset five, 0.145 at offset three, 0.180 at offset thirteen. The correlation is not simply a decreasing function of that: 13 is a nearer neighbour than 3 and correlates less than it, because correlation travels along chains of steps and 3 is reached in two steps from the pair while 13 is one step but a weaker one.

So peak height is a function of two things at once — how short the hop is, and how many steps the correlation had to travel — and no threshold on one quantity separates a set defined by the other.

The two obvious repairs, and what they cost #

Partial autocorrelation. The textbook instrument for exactly this problem: an autoregressive process has an autocorrelation with echoes in it and a partial autocorrelation that cuts off at the true lags. Since the placement rule really is an autoregression on the two neighbours, the partial autocorrelation should be large at m and at n and nothing else.

It is not. Computed on the same sequences it puts its weight at lags 1, 2, 4 and 7 — the first few — and nothing at 8 or 13 at all. The reason is that the object being autocorrelated is a sequence of differences: a divergence is the gap between two azimuths, so a deviation at one node enters two consecutive divergences with opposite signs, and the differencing manufactures exactly the low-order structure the partial autocorrelation then reports. An AR(2) with complex roots produces a periodic autocorrelation, and the fitter finds one.

Fitting two lags directly. If node i’s deviation is a weighted sum of the deviations at i − m and i − n, then fitting that regression over every candidate pair and taking the best-fitting pair should return the parastichy numbers. Run over all pairs up to twenty-six, the best fit at a rise of 0.005 is (1, 8) and at 0.013 is (1, 2). The lag-one term wins every time, for the same reason: it is the differencing, and it explains more variance than the physics does.

Both failures have the same cause and it is worth stating as a caution rather than as a defeat. The quantity a botanist writes down is not the quantity the rule acts on. The rule places azimuths; the notebook records gaps between azimuths. Every method that models the recorded series directly is modelling the gaps, and the gaps have a strong artefact at lag one that has nothing to do with the lattice.

The residue-class reading works because it never models the series at all. It asks only which lags carry correlation, and that question is unaffected by the differencing — a lag either has structure at it or it does not.

The arithmetic, and the closure that makes it a claim #

Take the significant lags, take the spacing of the strongest comb through the origin, and sort the rest by remainder. The claim is that they fall into exactly two classes and that the second class’s residue is the difference of the pair.

That is a stronger statement than “the pair can be extracted”, because it can be wrong in a visible way: a third populated residue class, or a second class whose residue is not n − m, would both show up as a set that does not close. The readout checks the residue against the position counter’s pair on every run at every rise, which is a comparison between two instruments rather than a consistency check inside one.

There is a second way of putting the same fact, and it is the one that connects this essay to the tissue thread. The set of lags with correlation is closed under addition, up to the length of the window: 8 and 13 are there, so 21 is there, so 29 is there; 8 is there twice, so 16 and 24 are there. A set closed under addition and generated by two elements is a numerical semigroup, and its two generators are exactly the two smallest members. The pair is the two generators of the lag set.

What a harmonic is, physically #

The word is borrowed and the borrowing is exact. A loop with a delay of m rings at m, and a ringing has overtones: a disturbance felt at i + m is felt again at i + 2m, weakened by whatever the coupling weakened it by the first time. So the correlation at 2m is roughly the square of the correlation at m — 0.82 squared is 0.67, and the measured value at lag sixteen is 0.62 — and the correlation at 3m is roughly the cube.

That prediction is not exact and is not asserted as a law, because the coupling is not a single number and there are several chains reaching each lag. What it does explain is the ordering that defeats the height ranking. A one-step hop to the larger family is weaker than a two-step hop along the smaller one whenever the coupling to the larger family is weaker than the square of the coupling to the smaller — which at this rise it is, 0.56 against 0.67, and one rung coarser it is not.

So the ordering flips between rungs while the residue classes do not. A method built on the ordering would work at one rise and fail at the next, which is what the ranking method does, and it would be indistinguishable from a method that works — five runs at one rise, all right, and no indication that the next rise is where it breaks.

The third class that is not there #

Two residue classes is a claim, not a description, and the way to see that is to ask what a third would mean.

A lag congruent to 2n modulo m would be reached by two steps of the larger family with no step of the smaller — at a rise of 0.005 that is 26, which is 2 × 13 and is congruent to 2 mod 8. It is in the significant set: the correlation there is 0.31, above the band of 0.11. So there is a third class, at the level of individual lags, and the readout does not use it.

The reason it does not is worth stating because it is the sort of thing that would otherwise look like a gap. A comb needs three members inside thirty lags to be called one, and the class at 2 mod 8 has members at 2, 10, 18 and 26, of which only 26 has anything in it — the others are reached by no short chain. Its mean over the class is 0.07 against a threshold of 0.16, so it does not clear. The classes that clear are the ones a one-step chain populates, and there are exactly two of those because a chain of one step goes forwards or backwards along one family or the other.

That is the sense in which two classes is a prediction. A model in which each primordium responded to three families rather than two would populate a third class densely enough to clear, and the readout would report a spacing whose second comb was ambiguous between two residues — which is a distinguishable outcome and not a silent one.

What the window length does to the classes #

Thirty lags is the previous phase’s number and it was kept rather than re-chosen, which turned out to matter. Forty was tried, on the reasonable ground that a longer window holds more members of every comb and the tests are means over members.

It is worse. At forty lags the search over spacings reaches thirteen, and a comb of three members at spacing thirteen — lags 13, 26, 39 — can be assembled out of the same handful of significant lags that the true reading uses. At a rise of 0.008 the readout goes from 5 and 8 on two runs in five to 13 and 18 on four. A longer window buys members of the true comb and candidates for false ones, and past a point the second wins.

The condition is arithmetic and it is why thirty is the right side of the boundary: a comb needs three members, so the largest spacing the search reaches is the window divided by three, and the reading is safe while that stays below the larger parastichy number. At thirty lags the search stops at ten and cannot mistake thirteen for a spacing. At forty it stops at thirteen and can.

Which means the window is not a free parameter. It has to be long enough to hold three multiples of the smaller number and short enough not to reach the larger, and for the rungs this instrument works on — 5/8 and 8/13 — that is a narrow range with thirty comfortably inside it. One rung finer, at 13/21, there is no window that satisfies both, and the instrument as written cannot go there. That is a real limit and it is arithmetic rather than statistical: no amount of stem fixes it.

The lag the readout refuses to look at #

Lag one is excluded from the search, by both this readout and the one before it, and the reason is the artefact described above. A reading error on a single organ enters two consecutive divergences with opposite signs, so any error at all puts a large negative correlation at lag one and none anywhere else. On a noiseless run at a fixed rise it is −0.63 and there is no protractor involved: the rule’s own self-correction does the same thing, since a node placed to one side leaves a gap the next falls into.

That number is the previous phase’s statistic and it is a good one — it separates noise that arrives before the rule’s choice from noise that arrives after it. It is simply not about the lattice, and including it in a comb search would put a tooth at lag one on every stem ever measured.

What a wrong answer looks like, and why one is left in #

The margin test — the winning spacing has to beat its best rival by a sampling band — was added after the readout returned a pair the positions contradicted. It is worth showing what that looked like, because a refusal is only worth something if there was a wrong answer available to refuse.

At a rise of 0.008 the stem sits where the third-shortest hop is only a ninth longer than the second, which is the tightest of the four rises measured. The combs at spacing five and spacing eight come out at 0.22 and 0.17, then 0.19 and 0.15, then 0.27 and 0.10 — the first two in one order and the third in the other. The argmax takes whichever is larger and the pair follows from that choice, so three runs said 8 and 13 and two said 5 and 8, on stems that differ only in a noise seed.

With the margin required, the three whose margin was under a band refuse, and the two whose margin was 0.13 and 0.17 report, and both of those agree with the counter. Five wrong-or-right became two right and three silent, which is the same trade the previous phase’s clearance test made and the same reason for making it.

One wrong answer does survive all three tests, at a disturbance near the quiet end of the window, and it is left in the record rather than tuned out. A sweep in which nothing ever fails is a sweep whose passing means nothing, and the alternative — moving the cap until that run refuses — would also refuse a run that is right.

What this leaves #

The distinction between a family and a harmonic is arithmetic and not statistical, and that is the useful form of it: it needs no threshold, no fitted model and no assumption about the noise. It needs only that the lags be sorted.

What it does assume is that the sequence came from a rule with a delay. A set of angles that merely describes a lattice — the right divergence, the right rise, independent errors — has no correlation at any lag, so it has no residue classes and nothing to sort. That is a control rather than a caveat, and it turns out to be the strongest claim in this thread: the comb is not evidence about the shape of the arrangement at all.