Modelling exponential growth
Learning Objectives
- Recognise situations whose underlying structure is a geometric progression.
- Translate a verbal description into the parameters \(b_1\) (or \(b_0\)), \(q\), and \(n\).
- Apply the general-term and sum formulas to questions about compound growth, population, and decay.
- Distinguish multiplicative from additive change, and recognise where the geometric model breaks down in real situations.
Motivation
Many real quantities do not change by a fixed amount at each step; they change by a fixed fraction of their current value. A savings account with compound interest earns a percentage of whatever is already in it, so the amount added each year grows. A bacterial population that doubles every hour adds more new bacteria each hour than in the previous hour. A radioactive sample loses the same fraction of its remaining atoms in each half-life, so the absolute number of decays diminishes over time, even though the proportion stays the same.
In every such situation the natural model is a geometric progression. The work of this lesson is to learn how to spot the pattern in a verbal description and how to use the tools of the previous two lessons — the general term and the partial sum — to answer the questions one wants to ask.
A modelling protocol
The translation from words to mathematics, when a geometric progression is the right tool, follows the same outline as for arithmetic ones, with one critical change: the per-step operation is multiplication, not addition.
Identify the changing quantity. What does \(b_n\) represent? — a balance, a population, a remaining mass.
Identify the step. What event takes us from \(b_n\) to \(b_{n+1}\)? — a year of interest accrues, an hour passes, a half-life elapses.
Determine the initial value. Usually \(b_0\) (the value before any step has occurred) or \(b_1\) (the value after the first step), depending on the natural reading of the problem.
Determine \(q\). The fixed multiplicative factor per step. For growth at a rate of \(r\) per step (with \(r\) expressed as a decimal), \(q = 1 + r\). For decay at the same rate, \(q = 1 - r\). For doubling, \(q = 2\); for halving, \(q = 1/2\).
Verify the structure. Is the change really proportional to the current value? If instead it is a fixed amount per step, the model is arithmetic, not geometric.
Translate the question. Use the general term \(b_n = b_0 \cdot q^{\,n}\) for the value at step \(n\), the partial sum for cumulative totals, and the inequality \(b_n \geq T\) (or \(\leq T\)) for threshold questions.
Compound interest
An account is opened with \(\,€1000\) and earns 5% interest annually, compounded once a year. We model the balance at the end of each year as a geometric progression.
Let \(b_n\) denote the balance at the end of year \(n\), with \(b_0 = 1000\) as the opening deposit. Each year the balance is multiplied by \(1 + 0.05 = 1.05\), so the common ratio is \(q = 1.05\) and the general term is
\[b_n = 1000 \cdot (1.05)^{\,n}.\]
Balance after ten years. \(b_{10} = 1000 \cdot (1.05)^{10} \approx 1628.89\) — roughly \(\,€1629\).
When does the balance double? We solve \(1000 \cdot (1.05)^{n} \geq 2000\), that is \((1.05)^{n} \geq 2\). Computing successive powers: \((1.05)^{14} \approx 1.980\), \((1.05)^{15} \approx 2.079\). The threshold is first crossed at \(n = 15\) — the balance doubles in roughly fifteen years.
A useful quick estimate, the rule of 72, says that money doubles in about \(72 / r\) years where \(r\) is the percentage rate. At 5%, this predicts \(72/5 = 14.4\) years — very close to our exact answer of 15. The rule is a small-rate approximation to the exact logarithmic answer, and it works well for the kinds of interest rates that occur in practice.
A doubling population
A bacterial culture starts with 500 cells and doubles every hour. Let \(P_n\) denote the number of cells after \(n\) hours, with \(P_0 = 500\). Each hour multiplies the count by \(2\), so \(q = 2\) and
\[P_n = 500 \cdot 2^{\,n}.\]
After 5 hours: \(P_5 = 500 \cdot 32 = 16\,000\). After 10 hours: \(P_{10} = 500 \cdot 1024 = 512\,000\). After 20 hours: \(P_{20} = 500 \cdot 1\,048\,576 \approx 5.2 \times 10^{8}\) cells. The growth is patient at first and then sudden — by the end of a single day (24 hours), the count is on the order of \(10^{10}\), and the pattern continues to outrun any imaginable container long before the bacteria run out of nutrients.
This kind of behaviour is what Thomas Malthus had in mind in 1798 when he contrasted the geometric growth of populations with the arithmetic growth of food production. The arithmetic against geometric race is the same one we set up at the end of the previous unit: in finite time the arithmetic side falls behind, no matter how favourable its starting position.
Exponential decay
A medical isotope has a half-life of 6 hours: every 6 hours, half of whatever is present has decayed. A patient receives a 200 mg dose. Let \(M_n\) denote the mass remaining after \(n\) six-hour intervals.
The initial mass is \(M_0 = 200\) mg. Each interval multiplies the remaining mass by \(1/2\), so \(q = 1/2\) and
\[M_n = 200 \cdot \left(\tfrac{1}{2}\right)^{\,n}.\]
After one half-life (\(n = 1\)), \(M_1 = 100\) mg. After two (\(n = 2\)), \(M_2 = 50\) mg. After ten (\(n = 10\), i.e. 60 hours), \(M_{10} = 200 / 1024 \approx 0.195\) mg — less than a milligram remains.
Notice how the decay slows: the first half-life removes 100 mg, the second removes 50 mg, the third 25 mg, and so on. The proportion lost is constant, but the absolute amount lost shrinks. This is the multiplicative analogue of the depreciation example in the previous unit, where the absolute amount lost was constant and the value reached zero in finite time. Under exponential decay the value approaches zero but, mathematically, never reaches it.
Connection to Computer Science
Exponential growth and decay appear throughout computing in two distinct roles. The first is as a feared time complexity: a procedure whose work doubles at each level of recursion produces a geometric progression of costs, and the partial sum of those costs is the total running time. An exponential-time algorithm becomes infeasible long before the input gets large; recognising a geometric pattern in a recurrence is often the first clue that a different algorithm is needed.
The second role is more constructive. Divide-and-conquer algorithms — binary search, merge sort, fast exponentiation — work by repeatedly halving the input, generating a geometric progression of problem sizes with \(q = 1/2\). After \(\log_2 n\) halvings the input has shrunk to a base case, which is why these algorithms are so much faster than their linear counterparts. The same multiplicative structure that makes exponential growth unmanageable makes exponential reduction extraordinarily powerful: doubling the input size adds only one extra step to the recursion.
When exponential is the wrong model
Real bacterial populations do not actually double every hour forever. After a few generations they fill the container, consume the available nutrients, and stop growing. The exponential model is accurate for early growth, when resources are abundant, and increasingly wrong as the population approaches its environmental limit.
The same caveat applies to most situations modelled by geometric progressions. Compound interest is exponential only as long as the underlying rate is stable; rates change, accounts get closed, currencies inflate. Radioactive decay is exponential at the population level — given an enormous number of atoms — but eventually one is dealing with a small number of individual atoms, each of which decays at a random time, and the deterministic model gives way to a probabilistic one.
The point is the same as for arithmetic progressions: a model is a tool, useful in the range where its assumptions hold. Pure exponential growth predicts that 500 bacteria multiplied by hourly doubling fill the visible universe in a couple of days. The mathematics is consistent; the physical situation it is meant to describe is not. Knowing when the model applies, and when it fails, is part of using it well — and is itself a topic for further study, in the form of more refined models such as the logistic equation.
Exercises
Exercise 1
An account is opened with €2 000 and earns 4% interest annually, compounded once a year. Let \(b_n\) be the balance at the end of year \(n\), with \(b_0 = 2000\). Use \(b_n = 2000 \cdot (1.04)^n\) and round answers to the nearest euro where needed.
Exercise 2
A colony of cells starts at 200 and triples every day. Let \(P_n\) denote the population after \(n\) days, with \(P_0 = 200\).
Exercise 3
A 320 mg sample of a radioactive isotope has a half-life of 4 hours. Let \(M_n\) be the mass remaining (in mg) after \(n\) half-lives elapse, with \(M_0 = 320\).
Exercise 4
Decide whether each situation is naturally modelled by a geometric progression.
Summary
Geometric progressions arise whenever a quantity changes by a fixed fraction of its current value at each step: compound interest, biological populations, radioactive decay.
To build the model, identify the quantity \(b_n\), the step, the initial value, and the common ratio \(q\). For a growth rate \(r\) per period, \(q = 1 + r\); for decay at the same rate, \(q = 1 - r\); for doubling or halving, \(q = 2\) or \(q = 1/2\).
A geometric model is right precisely when the change per step is proportional to the current value. If instead it is a fixed amount, the right tool is an arithmetic progression.
Exponential growth eventually outruns any linear comparison; exponential decay approaches zero but, mathematically, never reaches it. Real-world situations follow the model only within the range where its assumptions hold — populations meet resource limits, interest rates change, decay laws give way to probabilistic effects at small scales.
The same exponential structure appears in computer science both as a feared time complexity and as the engine behind divide-and-conquer algorithms.