Linear Recurrence Relations

Video lesson · from 1:222 worked examples

One lesson video covers all of Recurrence Relations, so it opens at 1:22 for this topic — not from the beginning.

Linear recurrence relations are of the form:

U_{n+1} = aU_n + b

where $U_0$ is the initial value

U_n = aU_{n-1} + b

where $U_1$ is the initial value

$a \neq 0 \text{ and } b \in \mathbb{R}$

Identify $a$ and $b$ from the words: $a$ is the multiplier and $b$ the amount added each time. Percentage changes belong in $a$ , fixed quantities in $b$ .
A decrease of 20% means $a = 0.8$ : not $-0.2$ . The multiplier is what remains.
Apply the operations in the stated order: whether the fixed amount is added before or after the percentage change changes the answer.
Define your terms: say what $u_n$ represents and in what units — a bare recurrence relation does not answer a context question.

Example 1

A sequence is defined by the recurrence relation $U_{n+1} = 0.6U_n + 4$ with $U_0 = 7$ .

a) Calculate the value of $U_3$ .

b) Find the smallest value of $n$ for which $U_n > 9.7$ .

U_1 = 0.6(7) + 4 = 8.2

U_2 = 0.6(8.2) + 4 = 8.92

U_3 = 0.6(8.92) + 4 = 9.352

b) Continue calculating terms:

U_4 = 0.6(9.352) + 4 = 9.6112

U_5 = 0.6(9.6112) + 4 = 9.76672

We see that $U_5 > 9.7$ . The smallest value of $n$ is $5$ .

Example 2

A patient is injected with 156 ml of a drug. Every 8 hours, 22% of the drug passes out of his bloodstream.

To compensate, a further 25 ml dose is given every 8 hours.

a) Find a recurrence relation for the amount of drug in his bloodstream.

b) Calculate the amount of drug remaining after 24 hours.

a) 22% is lost, meaning 78% remains ( $0.78$ ).

An extra 25 ml is added.

U_{n+1} = 0.78U_n + 25, \quad U_0 = 156

b) 24 hours implies 3 periods of 8 hours (so we need $U_3$ ).

U_1 = 0.78(156) + 25 = 146.68

U_2 = 0.78(146.68) + 25 = 139.4104

U_3 = 0.78(139.4104) + 25 = 133.740112

Amount of drug remaining after 24 hours is approximately 133.74 ml.