Births that remember
Tonight's idea
Here is a model of an animal population. Babies arrive once a year, in spring. An animal has babies in the springs when it is aged 1, 2 and 3, on average half a baby each time (picture half the animals being female, with one baby each).
Write \(B(t)\) for the number born in year \(t\). The parents in year \(t\) are the animals born 1, 2 and 3 years earlier.
\(B(t) = 0.5 \times \big(B(t-1) + B(t-2) + B(t-3)\big)\)
The rule looks only backwards. Nothing in it needs the future, so from the start you can work out every later year in turn. The population has a memory three years long.
When time runs smoothly instead of in yearly steps, the sum becomes an integral over the past. Such equations are named after the mathematician Vito Volterra, and they are solved the same way: forward, one small step at a time.
This is an idealisation. It works with averages, so half animals can appear, and real births vary.
Watch one: 1024 founders
In year 0, 1024 newborn animals are released on an empty island. So \(B(0) = 1024\), and \(B\) is 0 for every year before that.
Year 1: only the founders are old enough. \(B(1) = 0.5 \times (1024 + 0 + 0) = 512\).
Year 2: \(B(2) = 0.5 \times (512 + 1024 + 0) = 768\).
Year 3: \(B(3) = 0.5 \times (768 + 512 + 1024) = 1152\).
Year 4: the founders are now 4 and drop out of the sum. \(B(4) = 0.5 \times (1152 + 768 + 512) = 1216\).
| Year \(t\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Births \(B(t)\) | 1024 | 512 | 768 | 1152 | 1216 |
The start is bumpy: down, up, up, then hardly any change. The exercises show what happens next.
Your turn
- Continue the table to year 10. Every answer is a whole number.
- (a) For each year from 5 to 10, divide the births by the births the year before, to 2 decimal places. What number do the ratios settle near? (b) Suppose every animal dies after its third spring of babies. Then the animals alive just after the births of year \(t\) number \(A(t) = B(t) + B(t-1) + B(t-2) + B(t-3)\). Find \(A(9)\) and \(A(10)\), and check that the change \(A(10) - A(9)\) equals \(B(10) - B(6)\). Why should it?
- Change 0.5 to \(\tfrac13\) and start with 729 founders, so \(B(t) = \tfrac13\big(B(t-1) + B(t-2) + B(t-3)\big)\) and \(B(0) = 729\). (a) Find the births in years 1 to 6. (b) Is this population growing? (c) How many babies does one animal have in its whole life, on average, under this rule? Under the first rule?
- Go back to 0.5 and 1024 founders, but now animals have babies only at ages 1 and 2: \(B(t) = 0.5 \times \big(B(t-1) + B(t-2)\big)\). (a) Find the births in years 1 to 10. (b) Write down the change from each year to the next. What is the pattern? (c) What do the births seem to settle towards?
Stretch: where 1.23 comes from
(a) Suppose that, after many years, the births in the first model are multiplied by the same number \(r\) every year. Use the rule to show that \(r^3 = 0.5\,(r^2 + r + 1)\). Test \(r = 1.23\) and \(r = 1.24\) to show that \(r\) lies between them. (b) Do the same for the rule in exercise 4. Solve the equation you get. What do its two solutions tell you?
Answers
- \(B(5) = 0.5 \times (1216 + 1152 + 768) = 1568\). Then 1968, 2376, 2956, 3650 and \(B(10) = 0.5 \times (3650 + 2956 + 2376) = 4491\). Year 11 would be 5548.5: the model deals in averages.
- (a) 1.29, 1.26, 1.21, 1.24, 1.23, 1.23. The ratios swing above and below and settle near 1.23, so in the long run births grow by about 23% a year. (b) \(A(9) = 3650 + 2956 + 2376 + 1968 = 10\,950\) and \(A(10) = 4491 + 3650 + 2956 + 2376 = 13\,473\). The change is 2523, and \(B(10) - B(6) = 4491 - 1968 = 2523\). From one year to the next the population gains the newborns, \(B(10)\), and loses the animals that turned 4, who were born in year 6. The rate of change of \(A\) needs the births of four years ago: a rate with a memory. (With smooth time, equations that mix a rate of change with an integral over the past are called integro-differential.)
- (a) 243, 324, 432, 333, 363, 376. (b) No. After the bumpy start the births hover around 365, about half the founders, and if you carry on they stay there: 357.3, 365.4, 366.3, ... (c) \(3 \times \tfrac13 = 1\) baby, so each animal exactly replaces itself and the population levels off. Under the first rule it is \(3 \times 0.5 = 1.5\) babies, more than a replacement, so the population grows.
- (a) 512, 768, 640, 704, 672, 688, 680, 684, 682, 683. (b) \(-512, +256, -128, +64, -32, +16, -8, +4, -2, +1\). Each change is half the size of the one before, with the opposite sign. (c) The swings shrink towards 0 and the births settle at \(682\tfrac23\), which is two thirds of 1024. Each animal now has \(2 \times 0.5 = 1\) baby in its life, so again the population only replaces itself.
Stretch. (a) If three years ago there were \(b\) births, then the years since had \(br\), \(br^2\) and now \(br^3\). The rule says \(br^3 = 0.5\,(br^2 + br + b)\). Divide by \(b\). At \(r = 1.23\): \(r^3 \approx 1.861\) and \(0.5\,(r^2 + r + 1) \approx 1.871\), so the left side is too small. At \(r = 1.24\): \(1.907\) against \(1.889\), too big. So the two sides are equal somewhere in between, at \(r \approx 1.234\). (b) \(r^2 = 0.5\,(r + 1)\), so \(2r^2 - r - 1 = 0\), which factors as \((2r + 1)(r - 1) = 0\). The solutions are \(r = 1\) and \(r = -\tfrac12\). The first says the births can stay level. The second is the pattern in the changes: multiplied by \(-\tfrac12\) each year, they halve and flip sign.