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Discrete and continuous growth

We have seen many examples of a quantity \(x\) which continuously grows (or decays) over time by an amount proportional to \(x\), with continuous growth rate \(k\). Such an \(x\) obeys a differential equation with an exponential solution:

\[ \dfrac{dx}{dt} = kx \quad \iff \quad x(t) = C e^{kt}, \text{ where \(C\) is a constant.} \]

In discussing interest rates, however, we saw an example of a quantity \(x\) which grows in discrete steps by an amount proportional to \(x\). Such an \(x\) obeys a difference equation with an exponential solution:

\[ x(t+1) - x(t) = r \, x(t) \quad \iff \quad x(t) = (1+r)^t \, x(0). \]

Thinking of this difference equation as \(\Delta x = rx\), by analogy with the continuous case we call \(r\) the discrete growth rate. At each step, \(x\) is multiplied by \(1+r\), and \(x(t)\) is obtained from \(x(0)\) by \(t\) such multiplications. As we have derived it, this equation only holds for integers \(t\). However, if we imagine it is valid for all real \(t\), then \(x(t)\) is a continuous function. Rewriting using log laws,

\[ x(t) = e^{t \, \log_e (1+r)} \, x(0). \]

We may then differentiate, noting that \(x(0)\) is a constant:

\begin{align*} \dfrac{dx(t)}{dt} &= \log_e (1+r) \, e^{t \, \log_e (1+r)} \, x(0) \\ &= \log_e (1+r) \, x(t). \end{align*}

We conclude that the discrete growth rate \(r\) corresponds to a continuous growth rate of

\[ k = \log_e (1+r). \]

We saw this fact in exercise 8.

Discrete and continuous interest

In the module Exponential and logarithmic functions, we discussed a related concept. We imagined that, instead of paying interest once a year at a rate of \(r\), a bank might add interest twice a year at the rate \(\dfrac{r}{2}\), or three times a year at the rate \(\dfrac{r}{3}\). In general, we considered interest added \(n\) times a year at the rate of \(\dfrac{r}{n}\). In this case, each year your money is multiplied by

\[ \Bigl( 1 + \dfrac{r}{n} \Bigr)^n. \]

We saw that, as \(n \to \infty\), this quantity approaches \(e^r\):

\[ \lim_{n \to \infty} \Bigl( 1 + \dfrac{r}{n} \Bigr)^n = e^r. \]

So the quantity \(x(t)\) of money is multiplied by \(e^r\) each year, and

\[ x(t) = e^{rt} \, x(0). \]

Therefore

\begin{align*} \dfrac{dx(t)}{dt} &= r \, e^{rt} \, x(0) \\ &= r \, x(t), \end{align*}

and the continuous growth rate is \(r\).

Hence, if we take a discrete growth rate (or interest rate) of \(r\), it corresponds to a continuous growth rate of \(\log_e (1+r)\). But if that discrete growth is divided into an interest rate of \(\dfrac{r}{n}\), applied \(n\) times as often, in the limit we obtain continuously compounded interest at a continuous growth rate of \(r\).

Discrete and continuous population growth

The distinction between discrete and continuous growth also applies to populations. We have seen models of exponential population growth with constant continuous growth rates — usually unrealistic models.

However, in a non-technical context, the 'growth rate' of a population is sometimes taken to mean something like the percentage increase in population from the previous year. We can translate this discrete growth rate into a continuous growth rate, as in the following example.

Note that in this example the discrete and continuous growth rates are very close. This is because \(\log_e (1+x) \approx x\) for \(x\) close to 0. The two growth rates are close when they are small, but not when they are large.

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