1.2. Doubling Time and Half-Life 3
tells anything, it says that the rate cannot continue, a statement about the
future of a kind that will be studied in Chapter 8.
Still, the conditional growth rate whose hypothetical continuance is help-
ing us to understand the present can be defined in many ways. In particular,
in the example above, it could be an increase of 20,000 persons per year
just as well as of 0.02 per year. Arithmetical and geometrical projections
are equally easy to make; why should the latter be preferred?
The reason for preferring the model of geometric increase is simple:
constancy of the elements of growth translates into geometric increase. If
successive groups of women coming to maturity have children at the same
ages, and if deaths likewise take place at the same ages, and if in- and
out-migration patterns do not change, then the population will increase
(or decrease) geometrically. Any fixed set of rates that continues over time,
whether defined in terms of individuals, families, or age groups, ultimately
results in increase at a constant ratio. That any fixed pattern of childbear-
ing, along with a fixed age schedule of mortality, implies long-run geometric
increase is what makes this kind of increase central in demography. (The
stationary population and geometric decrease can be seen as special cases,
in which the ratios are unity and less than unity, respectively.)
1.2 Doubling Time and Half-Life
The expression for geometric increase gives as the projection to time t,
N
t
= N
0
(1 + x)
t
, (1.2.1)
where x is now the fraction of increase per unit of time. The unit of time
maybeamonth,ayear,oradecade,solongasx and t are expressed
in the same unit. Any one of N
0
, N
t
, x,ort may be ascertained if the
other three are given. When the quantity x is negative, the population is
decreasing; and when t is negative, the formula projects backward in time.
If x is the increase per year as a decimal fraction, then 100x is the increase
as percentage and 1000x the increase per thousand population.
We are told that the population of a certain country is increasing at
100x percent per year, and would like to determine by mental arithmetic
the population to which this rate of increase would lead if it persisted over
a long interval. Translating the rate into doubling time is a convenience in
grasping it demographically as well as arithmetically.
Since annual compounding looks easier to handle than compounding by
any other period, we will try it first. We will see that it leads to an un-
necessarily involved expression, whose complication we will then seek to
remove.
If at the end of 1 year the population is 1 + x times as great as it was
at the beginning of the year, at the end of 2 years (1 + x)
2
,..., and at the