# 0.3 Exponential growth

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Since it is the dynamic nature of a system that we want to model and understand, the simplest form will be considered. This will involve onestate variable and will give rise to so-called "exponential" growth.

## Two examples

First consider a mathematical model of a bank savings account. Assume that there is an initial deposit but after that, no deposits orwithdrawals. The bank has an interest rate ${r}_{i}$ and service charge rate ${r}_{s}$ that are used to calculate the interest and service charge once each time period. If the net income (interest less service charge)is re-invested each time, and each time period is denoted by the integer $n\phantom{\rule{0.166667em}{0ex}}$ , the future amount of money could be calculated from

$M\left(n\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}1\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}M\left(n\right)\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}{r}_{i}\phantom{\rule{4pt}{0ex}}M\left(n\right)\phantom{\rule{4pt}{0ex}}-\phantom{\rule{4pt}{0ex}}{r}_{s}\phantom{\rule{4pt}{0ex}}M\left(n\right)\phantom{\rule{2.em}{0ex}}$

A net growth rate $r$ is defined as the difference

$r\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{r}_{i}\phantom{\rule{4pt}{0ex}}-\phantom{\rule{4pt}{0ex}}{r}_{s}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{2.em}{0ex}}$

and this is further combined to define $R$ by

$R\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\left(r\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}1\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$

The basic model in [link] simplifies to give

$M\left(n\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}1\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\left(1\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}{r}_{i}\phantom{\rule{4pt}{0ex}}-\phantom{\rule{4pt}{0ex}}{r}_{s}\right)\phantom{\rule{4pt}{0ex}}M\left(n\right)\phantom{\rule{2.em}{0ex}}$
$=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\left(1\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}r\right)\phantom{\rule{4pt}{0ex}}M\left(n\right)$
$M\left(n\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}1\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}R\phantom{\rule{4pt}{0ex}}M\left(n\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}.$

This equation is called a first-order difference equation , and the solution $M\left(n\right)$ is found in a fairly straightforward way. Consider the equation for the first fewvalues of $n\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}0,\phantom{\rule{0.166667em}{0ex}}1,\phantom{\rule{0.166667em}{0ex}}2,\phantom{\rule{0.166667em}{0ex}}...$

$M\left(1\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}R\phantom{\rule{4pt}{0ex}}M\left(0\right)$
$M\left(2\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}R\phantom{\rule{4pt}{0ex}}M\left(1\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{R}^{2}\phantom{\rule{4pt}{0ex}}M\left(0\right)$
$M\left(3\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}R\phantom{\rule{4pt}{0ex}}M\left(2\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{R}^{3}\phantom{\rule{4pt}{0ex}}M\left(0\right)$
$\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}·\phantom{\rule{4pt}{0ex}}·\phantom{\rule{4pt}{0ex}}·$
$M\left(n\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}M\left(0\right)\phantom{\rule{4pt}{0ex}}{R}^{n}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$

The solution to [link] is a geometric sequence that has an initial value of $M\left(0\right)$ and increases as a function of $n$ if $R$ is greater than 1 $\left(r\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}0\right)$ , and decreases toward zero as a function of $n$ if $R$ is less than 1 $\left(r\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}0\right)\phantom{\rule{0.166667em}{0ex}}$ . This makes intuitive sense. One's account grows rapidly with a high interest rate and low service charge rate, andwould decrease toward zero if the service charges exceeded the interest.

A second example involves the growth of a population that has no constraints. If we assume that the population is a continuous function of time $p\left(t\right)\phantom{\rule{0.166667em}{0ex}}$ , and that the birth rate ${r}_{b}$ and death rate ${r}_{d}$ are constants ( $\phantom{\rule{4pt}{0ex}}$ not functions of the population $p\left(t\right)$ or time $t\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$ ), then the rate of increase in population can be written

$\frac{dp}{dt}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\left({r}_{b}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}-\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{r}_{d}\right)\phantom{\rule{4pt}{0ex}}p\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$

There are a number of assumptions behind this simple model, but we delay those considerations until later and examine the nature of the solution ofthis simple model. First, we define a net rate of growth

$r\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{r}_{b}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}-\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{r}_{d}\phantom{\rule{2.em}{0ex}}$

which gives

$\frac{dp}{dt}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}rp\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$

which is a first-order linear differential equation. If the value of the population at time equals zero is ${p}_{o}\phantom{\rule{4pt}{0ex}}$ , then the solution of [link] is given by

$p\left(t\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{p}_{o}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{e}^{rt}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{p}_{o}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}p\left(0\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$

The population grows exponentially if $r$ is positive (if ${r}_{b}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{r}_{d\phantom{\rule{0.166667em}{0ex}}}$ ) and decays exponentially if $r$ is negative ( ${r}_{b}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{r}_{d\phantom{\rule{0.166667em}{0ex}}}$ ). The fact that [link] is a solution of [link] is easily verified by substitution. Note that in order to calculate future values ofpopulation, the result of the past as given by $p\left(0\right)$ must be known. ( $\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}p\left(t\right)$ is a state variable and only one is necessary.)

## Exponential and geometric growth

It is worth spending a bit of time considering the nature of the solution of the difference [link] and the differential [link] . First, note that the solutions of both increase at the same "rate". If we samplethe population function $p\left(t\right)$ at intervals of $T$ time units, a geometric number sequence results. Let ${p}_{n}$ be the samples of $p\left(t\right)$ given by

${p}_{n}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}p\left(nT\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}n\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}0,\phantom{\rule{4pt}{0ex}}1,\phantom{\rule{4pt}{0ex}}2,\phantom{\rule{4pt}{0ex}}...\phantom{\rule{2.em}{0ex}}$

${p}_{n}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}p\left(nT\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{p}_{o\phantom{\rule{0.166667em}{0ex}}e}^{rnT}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{p}_{o}{\left({e}^{rT}\right)}^{n}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$

which is the same as [link] if

$R\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{e}^{rT}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$

This means that one can calculate samples of the exponential solution of differential equations exactly by solving the difference [link] if $R$ is chosen by [link] . Since difference equations are easily implemented on a digital computer, this is an important result; unfortunately,however, it is exact only if the equations are linear. Note that if the time interval $T$ is small, then the first two terms of the Taylor's series give

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