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How do recurrence relations model step-by-step change, and when does such a sequence settle towards a limit?

Recurrence relations of the form u sub n plus 1 equals a u sub n plus b, generating terms, the condition for a limit to exist, finding the limit, and interpreting a recurrence relation in a real context.

A focused answer to the SQA Higher Mathematics sequences and recurrence relations content, covering linear recurrence relations, generating terms, the condition for a limit, finding the limit, and interpreting a recurrence relation in a real-world context.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Generating terms
The condition for a limit
Finding the limit
Examples in context
Try this

What this dot point is asking

The SQA wants you to work with linear recurrence relations of the form $u_{n+1} = au_n + b$ , generate terms, state the condition for a limit, find the limit when it exists, and interpret a recurrence relation in a real context such as drug dosage or pollution.

Generating terms

A recurrence relation defines a sequence implicitly: it tells you how to get the next term from the current one, rather than giving a formula for the $n$ th term directly. Given a starting value $u_0$ , you apply the rule repeatedly. The order of operations matters: multiply by $a$ first, then add $b$ .

The condition for a limit

The reason is that the gap between successive terms is multiplied by $a$ each step. When $|a| < 1$ , repeated multiplication drives that gap towards zero, so the terms settle. When $|a| \ge 1$ the gap holds steady or grows, so there is no limit. You must state this condition in an exam before quoting a limit.

Finding the limit

The idea behind the limit equation is that, in the long run, $u_{n+1}$ and $u_n$ are both essentially equal to $L$ , so the recurrence becomes $L = aL + b$ . You then solve this linear equation for $L$ .

Find the limit of

u_{n+1} = 0.8u_n + 5

Find the long-term limit of the recurrence relation $u_{n+1} = 0.8u_n + 5$ .

Step 1: Check the condition for a limit to exist

A limit only exists when $-1 < a < 1$ . Here $a = 0.8$ , which lies in that range, so repeated multiplication by $0.8$ shrinks the gap between successive terms to zero and the sequence settles. You must state this condition explicitly before claiming a limit, because examiners award a mark for it.

Step 2: Set up and solve the limit equation

At the limit, consecutive terms are both equal to $L$ , so the recurrence becomes $L = 0.8L + 5$ . Rearranging:

L - 0.8L = 5 \implies 0.2L = 5 \implies L = \dfrac{5}{0.2} = 25

Final answer: The limit is $L = 25$ . The sequence we generated earlier, rising past $17.32$ , continues to close in on $25$ from below.

A recurrence relation with a negative multiplier

A sequence is defined by $u_{n+1} = -0.5u_n + 12$ with $u_0 = 0$ . Determine whether it has a limit and, if so, find it.

Step 1: Test the limit condition

Here $a = -0.5$ . Since $-1 < -0.5 < 1$ , the condition is satisfied and a limit exists. A negative multiplier does not prevent a limit: the terms still converge, but they oscillate above and below the limiting value rather than approaching it from one side.

Step 2: Find the limit

At the limit, $u_{n+1}$ and $u_n$ are both $L$ , so the recurrence becomes $L = -0.5L + 12$ . Adding $0.5L$ to both sides:

1.5L = 12 \implies L = \dfrac{12}{1.5} = 8

Step 3: Verify with the first few terms

Starting from $u_0 = 0$ : $u_1 = -0.5(0) + 12 = 12$ , $u_2 = -0.5(12) + 12 = 6$ , $u_3 = -0.5(6) + 12 = 9$ , $u_4 = -0.5(9) + 12 = 7.5$ . The terms $12, 6, 9, 7.5, \ldots$ oscillate around $8$ with the gap halving each step, confirming convergence to $8$ .

Final answer: A limit exists because $|a| = 0.5 < 1$ , and the limit is $L = 8$ .

Examples in context

Recurrence relations model any process where a fixed proportion of a quantity carries over and a fixed amount is added each step. In a lake, suppose $40\%$ of a pollutant is broken down each year, but $24$ tonnes are discharged annually. With $u_n$ the amount at the start of year $n$ , the model is $u_{n+1} = 0.6 u_n + 24$ . Since $a = 0.6$ lies in $(-1, 1)$ a long-term level exists: $L = 0.6L + 24$ gives $0.4L = 24$ , so $L = 60$ tonnes. This steady value tells an environmental regulator the pollution the lake settles to if discharge continues unchanged.

Try this

Q1. Write down the first three terms of $u_{n+1} = 3u_n - 2$ with $u_0 = 1$ . [2 marks]

Cue. $u_1 = 3(1) - 2 = 1$ , $u_2 = 1$ , $u_3 = 1$ ; this sequence stays at $1$ , which is its fixed point.

Q2. Explain why $u_{n+1} = 0.5u_n + 6$ has a limit, and find it. [3 marks]

Cue. $a = 0.5$ is between $-1$ and $1$ ; $L = \dfrac{6}{1 - 0.5} = 12$ .

Q3. A recurrence relation $u_{n+1} = a u_n + 9$ has limit $30$ . Find $a$ . [3 marks]

Cue. $30 = 30a + 9$ , so $30a = 21$ and $a = \dfrac{21}{30} = 0.7$ , which lies in $(-1, 1)$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20205 marksA patient is given a drug. Each day

30\%

of the drug present in the body is eliminated, and a fresh dose of

20

mg is then taken. The amount of drug in the body just after the dose on day

n

u_n

mg. Write down a recurrence relation for

u_{n+1}

in terms of

u_n

, explain why a long-term amount exists, and calculate it.

Show worked answer →

If $30\%$ is eliminated then $70\%$ remains, so before the new dose there is $0.7 u_n$ , and adding $20$ mg gives $u_{n+1} = 0.7 u_n + 20$ (2 marks).

A limit exists because the multiplier $a = 0.7$ satisfies $-1 < 0.7 < 1$ , so the change between terms shrinks towards zero (1 mark).

At the limit $L = 0.7L + 20$ , so $0.3L = 20$ and $L = \dfrac{20}{0.3} = 66.\dot{6}$ mg, that is $66.7$ mg to one decimal place (2 marks). Markers reward the correct multiplier of $0.7$ , stating the $-1 < a < 1$ condition explicitly, and solving the limit equation by dividing by $1 - a$ .

SQA Higher 20235 marksTwo sequences are defined by

u_{n+1} = 0.5u_n + 16

with

u_0 = 4

and

v_{n+1} = a v_n + 6

. Determine the limit of the first sequence, and find the value of

a

for which the second sequence has the same limit.

Show worked answer →

For the first sequence $a = 0.5$ lies in $(-1, 1)$ , so the limit $L$ satisfies $L = 0.5L + 16$ , giving $0.5L = 16$ and $L = 32$ (2 marks).

For the second sequence to share this limit, $32 = 32a + 6$ (1 mark), so $32a = 26$ and $a = \dfrac{26}{32} = \dfrac{13}{16} = 0.8125$ (1 mark).

Check the condition: $0.8125$ lies in $(-1, 1)$ , so the limit is valid (1 mark). Markers reward the first limit, setting the second limit equal to $32$ , and confirming $a$ lies in the valid range.

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Sources & how we know this

SQA Higher Mathematics Course Specification — SQA (2018)