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Recurrence Relations

Key Terms

A recurrence relation defines each term using the previous term(s)

Arithmetic: t_n+1 = t_n + d (add a constant)

Geometric: t_n+1 = r × t_n (multiply by a constant)

Combined: t_n+1 = r × t_n + b (geometric + constant, e.g. loan repayments)

The initial value t₁ (or t₀) must be given to use the relation

Iterate by substituting previous values one step at a time

Arithmetic recurrence: t_n+1 = t_n + d, t₁ = a

Geometric recurrence: t_n+1 = r × t_n, t₁ = a

Combined (loan model):
t_n+1 = r × t_n − P
(r = interest factor, P = regular payment)

Worked Example: The recurrence relation t_n+1 = 1.05t_n − 2000, t₁ = 30 000 models a loan. Find t₂, t₃, and t₄.

t₂ = 1.05 × 30 000 − 2000 = 31 500 − 2000 = 29 500
t₃ = 1.05 × 29 500 − 2000 = 30 975 − 2000 = 28 975
t₄ = 1.05 × 28 975 − 2000 = 30 423.75 − 2000 = 28 423.75

Hot Tip: Recurrence relations are perfect for spreadsheet calculations — set up one formula and fill down. For an exam, iterate step by step, showing each calculation clearly.

What is a Recurrence Relation?

A recurrence relation (or difference equation) defines a sequence by specifying how each term is calculated from one or more previous terms, along with an initial value.

The simplest forms in this course are first-order — each term depends only on the immediately preceding term.

Types

Arithmetic: t_n+1 = t_n + d (produces an arithmetic sequence)
Geometric: t_n+1 = r × t_n (produces a geometric sequence)
Combined linear: t_n+1 = r × t_n + b (models loans, repeated processes)

Financial Applications

Loan and annuity problems use the combined form:

t_n+1 = r × t_n − P models a reducing-balance loan (borrow t_n, apply interest rate r, subtract payment P)
t_n+1 = r × t_n + D models an investment with regular deposits D

Strategy: For multi-step iteration, build a table with columns for n, t_n, and intermediate calculations. This reduces errors and is easy for examiners to follow.

Mastery Practice

See Answers ➔

Fluency Write out the first 5 terms of the sequence defined by t_n+1 = t_n + 6, t₁ = 3.
Fluency Write out the first 5 terms of the sequence defined by t_n+1 = 2t_n, t₁ = 5.
Fluency The recurrence relation t_n+1 = 1.1t_n, t₁ = 1000 models a savings account growing at 10% per period. Find t₄.
Fluency Identify whether each relation produces an arithmetic or geometric sequence: (a) t_n+1 = t_n − 5, t₁ = 50 (b) t_n+1 = 0.5t_n, t₁ = 80
Understanding A recurrence relation is t_n+1 = 1.04t_n − 500, t₁ = 10 000. This models a loan with monthly interest and payments. Calculate t₂, t₃, t₄, and t₅.
Understanding Write the recurrence relation for a savings account that starts with $2 000, earns 3% interest per year, and has $300 deposited at the end of each year.
Understanding The relation t_n+1 = 0.8t_n + 40, t₁ = 200 models the amount of a drug in a patient’s bloodstream after each dose. Find t₂, t₃, t₄, and t₅. What value does the drug level appear to approach?
Understanding A sequence is defined by t_n+1 = 3t_n − 2, t₁ = 1. Find the first 4 terms. Does this sequence grow, decay, or remain constant?
Problem Solving A person borrows $20 000 at 6% annual interest (applied each year). They repay $2 500 at the end of each year. Set up the recurrence relation and find the balance at the end of each of the first 4 years. Will the loan ever be paid off at this rate?
Problem Solving A sequence has recurrence relation t_n+1 = rt_n + b, t₁ = 100. After many terms, the sequence appears to approach a constant value (a fixed point).
- (a) If the fixed point is L, explain why L = rL + b.
- (b) Solve for L when r = 0.6 and b = 80.
- (c) Verify using the recurrence relation that starting near L keeps you near L.

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