Basis Lab

Answers · The mathematics underneath

What does it mean if a system ends up on an attractor?

In shortIt means the long run is pulled onto a set rather than a single value: a resting point, a repeating loop, or a spiral closing onto one of those.

Ending up on an attractor means the state has been drawn into a set that neighbouring states are drawn into as well. That set can be a single point the system comes to rest at, a closed loop it goes round forever without ever repeating a value at a given instant, or a spiral that closes onto either of those. The claim is about the shape of the long run and about the rate of approach, not about a final number. A system on a loop attractor is exactly as attached to its attractor as one sitting still is, and quoting a single value for it would be wrong.

attractor A:dist⁡(xk, A)→0  as k→∞,  for every x0 started near A.\text{attractor } A: \quad \operatorname{dist}(x_k,\, A) \to 0 \ \text{ as } k \to \infty, \ \text{ for every } x_0 \text{ started near } A.

One term decides which

Take a valley where rabbits breed in proportion to their number and foxes eat them, and where the pair circles a closed loop forever. Now put a ceiling on the rabbits only, so their breeding term becomes aR(1−R/M)−bRFaR(1 - R/M) - bRF with M=150M = 150, against a cycle that was running between about sixty and a hundred. Nothing else is touched. The foxes' equation is unchanged, so wherever the foxes stop changing, cRF−dF=0cRF - dF = 0, which gives R=d/c=80R = d/c = 80: the resting rabbit count is the one the old rule had. The fox count moves, because it comes from the rabbit equation, and it moves a long way.

cRF−dF=0  ⇒  R=dc=80,aR(1−RM)−bRF=0  ⇒  F=ab(1−80150).cRF - dF = 0 \;\Rightarrow\; R = \frac{d}{c} = 80, \qquad aR\left(1 - \frac{R}{M}\right) - bRF = 0 \;\Rightarrow\; F = \frac{a}{b}\left(1 - \frac{80}{150}\right).

The uncorrected resting fox count was 25, which fixes a/b=25a/b = 25. At eighty rabbits the share of the ceiling still unused is 1−80/150=1−0.5333=0.46671 - 80/150 = 1 - 0.5333 = 0.4667, so F=25×0.4667=11.67F = 25 \times 0.4667 = 11.67. A correction that changes the rates by a few per cent has changed the answer by more than half, and has changed something else besides.

FIG 1
steprabbitsfoxesswing
095.0022.00not yet
7553.7815.9634.01
26287.7910.389.40
45478.3911.931.98
64280.3711.610.35
102080.0211.660.03
150080.0011.67at rest

The guess that fails is that the long run is one final value. On the uncorrected rule the pair visits a whole closed curve and never settles anywhere, so the honest answer is the curve, and the band it occupies, sixty to a hundred rabbits with the matching foxes. Even where the attractor is a single point, the point is not the whole answer, because which point and how fast the approach is both matter. A second failure hides inside the same guess: the attractor is being read off a finite run. Stopped at step three hundred the corrected valley still swings from 87.79 down to 53.78 and back, and anybody watching would call it a cycle. Whether it cycles forever or spirals in is a fact about the infinite future, inferred from a few hundred steps, which is the inference most likely to be wrong.

Attractors made by arithmetic

There is another way to end up on an attractor that belongs to nobody's rule. Advance a spring with the obvious step and the quantity that must not change obeys Ek+1=(1+h2)EkE_{k+1} = (1 + h^2)E_k, so after kk steps it is E0(1+h2)kE_0(1+h^2)^k. At h=0.3h = 0.3 the factor is 1.091.09 and 1.0920≈5.61.09^{20} \approx 5.6: an energy that started at one is 5.6 after twenty steps. Turning the step down looks like a cure. At h=0.05h = 0.05 the growth per step is 0.00250.0025, and doubling takes k=ln⁡2/ln⁡1.0025≈278k = \ln 2 / \ln 1.0025 \approx 278 steps, which is not many. Nor does the drift per unit of simulated time vanish: over one unit of time at h=0.02h = 0.02 the drift is 1.000450−1=0.02021.0004^{50} - 1 = 0.0202, and at h=0.01h = 0.01 it is 1.0001100−1=0.01011.0001^{100} - 1 = 0.0101. Halving the step halves the drift, so the error falls in proportion to hh rather than disappearing, and a long enough run is still wrong.

The plainest case is a ball bounced by reflecting it. The book loses about one part in seven over 240 steps, some thirty seven times the tolerance, and the bounces shrink until the ball sits on the floor. That is an attractor, and it is in the arithmetic rather than in the rule. It is also convincing, because balls dropped on real floors do lose height, so the picture confirms a belief instead of testing one. Made with a floor that gives, where the sinking is real and gives the energy back, the same scene keeps the book flat and keeps bouncing.

What survives both audits is where the system rests, whether it comes back after a nudge, what it averages over a cycle, and the band it cannot leave. What does not survive is whether the long run cycles forever or spirals in, how long until the first peak, and where the system sits at a named step. The resting rabbit count survived the ceiling because it came from the equation nobody touched; the resting fox count did not. And settling is not evidence of anything until the same conclusion has been drawn again at half the step and refused to move.

Worth keeping

  • An attractor is a set, so a system can be settled on it while its state is changing at every instant.
  • A resting value that comes from an equation the correction never touched survives that correction, and a value that comes from a touched equation does not.
  • A run can settle onto an attractor that exists only in the step size, and halving the step is what tells the two apart.

Where this goes next

This question is one step of Systems That Evolve, a course that works through its subject from the beginning. The page that teaches this part of it is What the Long Run Looks Like, and it reads the whole way through whether or not you have an account.

What an account changes is the practice: problems standing under every idea, marked with a reason rather than a tick, in a copy of the course that is yours.

Read that pageThe whole course

Other questions people ask about this

  1. What stays constant when a system just goes around in a loop forever?A fixed combination of the changing counts stays the same round the loop, and that is what keeps the system from settling or escaping.
  2. What is the difference between variance and standard deviation?They measure the same spread: variance is the average squared distance from the mean, standard deviation is its square root, so you report the standard deviation and calculate with the variance.
  3. How do you make a vague claim precise enough to tell whether it's true?Fix what is being talked about, what is asserted, which cases are covered and the cut-off, then name an observation that would prove it false.
  4. How do you find the parabola that best fits a function near a point?Match the function's height, slope and bending at the chosen point; those three conditions single out exactly one parabola.
  5. Why does a test for a rare disease give so many false positives if the test is accurate?A positive result means the test fired, not that you are likely to be ill; when the disease is rare, healthy people vastly outnumber ill ones, so a small false-alarm rate on them swamps the few true cases.
  6. When should I give up on finding an exact formula and simulate instead?Give up on an exact formula when the quantity can be written as an average of something drawable, since simulation then gives an estimate with a measurable error.

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