First-order linear recurrence relations
Master first-order linear recurrence relations in Year 11 VCE Specialist Mathematics. A first-order linear recurrence builds a sequence with the rule \(t_{n+1}=a\,t_n+b\) (with \(a\neq 0\)) from a given first term, generating each term from the one before. It sits in the Discrete mathematics area of study of the VCE Mathematics Study Design (VCAA), within the Sequences and series topic of Unit 1.
You will learn to generate terms by recursion, recognise the arithmetic case \(a=1\) and the geometric case \(b=0\), and find the steady-state (fixed-point) value \(t^*=\dfrac{b}{1-a}\) and whether it is stable — the tools for modelling savings, loans and populations.
Theory
A first-order linear recurrence relation builds a sequence by the rule \(t_{n+1}=a\,t_n+b\) (with \(a\neq 0\)) from a given first term. In Year 11 Specialist Mathematics you generate terms by applying the rule step by step, recognise the arithmetic case \(a=1\) and the geometric case \(b=0\), and find the steady-state (fixed-point) value \(t^*=\dfrac{b}{1-a}\) and whether it is stable. These ideas model savings, loans and populations.
A recurrence relation defines each term of a sequence from the term (or terms) before it. A first-order recurrence uses just the previous term, and a linear one combines it with a multiply and an add: \(t_{n+1}=a\,t_n+b\), where \(a\neq 0\) and \(b\) are constants. To start the sequence you also need a first term, given as \(t_1\) (or sometimes \(t_0\)).
To generate the terms you apply the rule repeatedly — this is called recursion: substitute the current term to get the next one, then repeat. For example \(t_{n+1}=2t_n+3\) with \(t_1=4\) gives \(t_2=2(4)+3=11\), then \(t_3=2(11)+3=25\), and so on.
Two familiar sequences are special cases. When \(a=1\) the rule is \(t_{n+1}=t_n+b\): a fixed amount \(b\) is added each step, so the sequence is arithmetic with common difference \(b\). When \(b=0\) the rule is \(t_{n+1}=a\,t_n\): each term is a fixed multiple of the one before, so the sequence is geometric with common ratio \(a\).
When \(a\neq 1\) the sequence may settle toward a steady-state (or fixed-point) value \(t^*\), the value that repeats unchanged: if a term equals \(t^*\), the next term does too. Whether the terms actually move toward \(t^*\) is the question of stability, and it depends on the size of \(a\).
A first-order linear recurrence relation is defined by a rule and a first term:
The special cases are the arithmetic and geometric sequences:
When \(a\neq 1\), the steady-state (fixed-point) value is found by setting \(t_{n+1}=t_n=t^*\) and solving:
Working with a recurrence relation
- Read off \(a\), \(b\) and the first term. In \(t_{n+1}=a\,t_n+b\), \(a\) is the multiplier and \(b\) is the amount added; note the given \(t_1\) (or \(t_0\)).
- Generate terms by recursion: substitute the current term into the rule to get the next term, then repeat — write each step on its own line.
- Spot a special case: \(a=1\) means arithmetic (add \(b\) each step); \(b=0\) means geometric (multiply by \(a\) each step).
- Find the steady state (when \(a\neq 1\)): set \(t_{n+1}=t_n=t^*\) and solve \(t^*=a\,t^*+b\), giving \(t^*=\dfrac{b}{1-a}\).
- Decide stability: if \(0
For a worded problem, build the rule from the context first. A balance earning \(5\%\) interest then a \(\$200\) deposit gives \(t_{n+1}=1.05\,t_n+200\); a population growing \(20\%\) a year with \(300\) removed gives \(t_{n+1}=1.2\,t_n-300\). Then generate terms or find the steady state as needed.
Apply the rule \(t_{n+1}=3t_n-4\) one step at a time, starting from \(t_1=3\):
| \(t_2\) | \(=\) | \(3t_1-4\) |
| \(=\) | \(3(3)-4\) | |
| \(=\) | \(9-4\) | |
| \(=\) | \(5\) |
Now use \(t_2=5\) to get the next term:
| \(t_3\) | \(=\) | \(3t_2-4\) |
| \(=\) | \(3(5)-4\) | |
| \(=\) | \(15-4\) | |
| \(=\) | \(11\) |
The next two terms are \(t_2 = 5\) and \(t_3 = 11\).
(a) Here \(b=0\), so the sequence is geometric — multiply by \(a=2\) each step:
| \(t_1\) | \(=\) | \(3\) |
| \(t_2\) | \(=\) | \(2(3)=6\) |
| \(t_3\) | \(=\) | \(2(6)=12\) |
| \(t_4\) | \(=\) | \(2(12)=24\) |
(b) Here \(a=1\), so the sequence is arithmetic — add \(b=5\) each step:
| \(t_1\) | \(=\) | \(2\) |
| \(t_2\) | \(=\) | \(2+5=7\) |
| \(t_3\) | \(=\) | \(7+5=12\) |
| \(t_4\) | \(=\) | \(12+5=17\) |
(a) The geometric terms are \(3,\ 6,\ 12,\ 24\). (b) The arithmetic terms are \(2,\ 7,\ 12,\ 17\).
At the steady state the terms stop changing, so set \(t_{n+1}=t_n=t^*\) and solve:
| \(t^*\) | \(=\) | \(0.5\,t^*+6\) |
| \(t^*-0.5\,t^*\) | \(=\) | \(6\) |
| \(0.5\,t^*\) | \(=\) | \(6\) |
| \(t^*\) | \(=\) | \(12\) |
Check stability from the size of \(a\):
| \(a\) | \(=\) | \(0.5\) |
| \(0\) | \(<\) | \(0.5<1\) |
| \(\Rightarrow\) | \(\text{terms approach } t^*=12 \text{ (stable)}\) |
The steady-state value is \(t^*=12\); since \(0
Model the balance: \(5\%\) interest multiplies by \(1.05\), then add the \(\$200\) deposit:
| \(t_{n+1}\) | \(=\) | \(1.05\,t_n+200\) |
| \(t_0\) | \(=\) | \(1000\) |
Balance after \(1\) year, using \(t_1=1.05\,t_0+200\):
| \(t_1\) | \(=\) | \(1.05(1000)+200\) |
| \(=\) | \(1050+200\) | |
| \(=\) | \(1250\) |
Balance after \(2\) years, using \(t_2=1.05\,t_1+200\):
| \(t_2\) | \(=\) | \(1.05(1250)+200\) |
| \(=\) | \(1312.50+200\) | |
| \(=\) | \(1512.50\) |
The model is \(t_{n+1}=1.05\,t_n+200\); the balance is \(\$1250\) after \(1\) year and \(\$1512.50\) after \(2\) years. Because \(a=1.05>1\), the balance keeps growing (no stable steady state).
Common pitfalls
Frequently asked questions
What is a first-order linear recurrence relation?
It is a rule of the form \(t_{n+1}=a\,t_n+b\) (with \(a\neq 0\)) that gives each term from the one before it, together with a first term to start the sequence. "First-order" means it uses only the previous term; "linear" means the term is multiplied by \(a\) and then \(b\) is added.
How do I generate the terms of a recurrence?
Start from the given first term and apply the rule repeatedly (recursion): substitute the current term to get the next one, then repeat. For \(t_{n+1}=2t_n+3\) with \(t_1=4\), \(t_2=2(4)+3=11\), then \(t_3=2(11)+3=25\), and so on.
When is the sequence arithmetic or geometric?
When \(a=1\) the rule is \(t_{n+1}=t_n+b\), which adds a constant each step, so the sequence is arithmetic with common difference \(b\). When \(b=0\) the rule is \(t_{n+1}=a\,t_n\), which multiplies by a constant each step, so the sequence is geometric with common ratio \(a\).
What is the steady-state (fixed-point) value?
It is the value \(t^*\) that repeats unchanged: if a term equals \(t^*\), so does the next. Setting \(t_{n+1}=t_n=t^*\) in \(t_{n+1}=a\,t_n+b\) gives \(t^*=\dfrac{b}{1-a}\), valid when \(a\neq 1\).
How do I tell whether the steady state is stable?
Look at the multiplier \(a\). The gap between a term and \(t^*\) is multiplied by \(a\) each step: if \(0
How are recurrence relations used to model real situations?
They model any quantity that is scaled then adjusted each period. A balance earning \(5\%\) interest with a \(\$200\) deposit is \(t_{n+1}=1.05\,t_n+200\); a population growing \(20\%\) a year with \(300\) harvested is \(t_{n+1}=1.2\,t_n-300\). Generating terms predicts the balance or population year by year.