Math Core

Many processes are easiest to describe one step at a time: "this month's balance is last month's balance plus interest, minus a payment." A recursive rule captures exactly that idea. In this lesson you'll read and write recursive rules, convert between recursive and explicit forms, and use recursion to model situations that settle toward a long-run value.

What makes a rule recursive

An explicit formula gives $a_n$ from $n$ alone. A recursive formula gives $a_n$ from earlier terms.

Definition

Recursive formula

A recursive formula for a sequence has two parts:

one or more initial conditions that give the first term (or terms), and
a recurrence relation that tells how to find each later term from the terms before it.

For example,

a_1 = 4, \qquad a_n = 3a_{n-1} - 5 \ \text{ for } n \ge 2

says: start at $4$ ; to get any term, triple the previous term and subtract $5$ . Both parts matter. The same recurrence with $a_1 = 1$ produces a completely different sequence.

You'll also see function notation, such as $f(1) = 4$ and $f(n) = 3f(n-1) - 5$ . It means the same thing.

Worked example: Generating terms from a recursive rule

Find the first five terms of $a_1 = 4$ , $a_n = 3a_{n-1} - 5$ .

Apply the rule over and over:

\begin{aligned} a_1 &= 4 \\ a_2 &= 3(4) - 5 = 7 \\ a_3 &= 3(7) - 5 = 16 \\ a_4 &= 3(16) - 5 = 43 \\ a_5 &= 3(43) - 5 = 124 \end{aligned}

Rules that look back two steps

Some recurrences use the two previous terms. Then you need two initial conditions to get started. The most famous is the Fibonacci sequence: $F_1 = 1$ , $F_2 = 1$ , $F_n = F_{n-1} + F_{n-2}$ , which gives $1, 1, 2, 3, 5, 8, 13, \dots$

Worked example: A Fibonacci-style sequence

Find $a_6$ if $a_1 = 2$ , $a_2 = 5$ and $a_n = a_{n-1} + a_{n-2}$ .

Each term is the sum of the two before it:

a_3 = 5 + 2 = 7, \quad a_4 = 7 + 5 = 12, \quad a_5 = 12 + 7 = 19, \quad a_6 = 19 + 12 = 31.

Common mistake

Read the subscripts carefully. In $a_n = a_{n-1} + 2a_{n-2}$ , the term two back gets doubled, not the term just before. Writing the terms in a row and labeling each with its subscript prevents this mix-up.

Converting between recursive and explicit

Arithmetic and geometric sequences have simple recursive rules, and you can switch forms easily:

Recursive and explicit forms

sequence	recursive	explicit
arithmetic	$a_1$ given, $a_n = a_{n-1} + d$	$a_n = a_1 + (n - 1)d$
geometric	$a_1$ given, $a_n = r \cdot a_{n-1}$	$a_n = a_1 \cdot r^{\,n-1}$

Adding the same number each step means arithmetic; multiplying by the same number each step means geometric. A rule like $a_n = 3a_{n-1} - 5$ does both, so it is neither arithmetic nor geometric.

Worked example: Switching forms

(a) Write an explicit formula for $a_1 = -3$ , $a_n = a_{n-1} + 6$ .

Each step adds $6$ , so it's arithmetic with $d = 6$ :

a_n = -3 + (n - 1)(6) = 6n - 9.

(b) Write a recursive formula and an explicit formula for $80, 20, 5, \tfrac54, \dots$

Each term is $\tfrac14$ of the previous one, so it's geometric with $r = \tfrac14$ :

\text{recursive: } a_1 = 80,\ \ a_n = \tfrac14 a_{n-1} \qquad \text{explicit: } a_n = 80\left(\tfrac14\right)^{n-1}.

Which form is better? The recursive form shows how the sequence changes; the explicit form lets you jump to any term. To find $a_{50}$ , the explicit form wins by a mile.

Modeling with recursion

Recursion shines when each step combines multiplying and adding. Think of a medicine: each day the body removes a fixed percent of the drug, and then the patient takes a new dose.

Worked example: Medication level

A patient takes a $100$ mg dose every morning. By the next morning, the body has removed $40\%$ of the medicine in the bloodstream. Let $a_n$ be the amount right after the $n$ th dose. Write a recursive rule, find $a_5$ , and find the long-run level.

After the first dose, $a_1 = 100$ . Each day $60\%$ remains, and then $100$ mg more is added:

a_1 = 100, \qquad a_n = 0.6\,a_{n-1} + 100.

The terms are

a_2 = 160, \quad a_3 = 196, \quad a_4 = 217.6, \quad a_5 = 230.56.

The amounts are rising, but by less each day. If the level settles at some value $L$ , then applying the rule to $L$ must give $L$ back:

L = 0.6L + 100 \quad\Rightarrow\quad 0.4L = 100 \quad\Rightarrow\quad L = 250.

In the long run, the amount right after each dose approaches $250$ mg.

The medication levels aₙ = 0.6aₙ₋₁ + 100 climb toward the dashed long-run level of 250 mg.Open in grapher →

Tip

For a recurrence $a_n = r\,a_{n-1} + c$ with $|r| < 1$ , the terms approach the value $L$ that solves $L = rL + c$ . You can check that this is a fixed point: if $a_{n-1} = L$ , the rule gives $a_n = L$ again.

Practice

Practice 1

Find $a_5$ if $a_1 = 5$ and $a_n = a_{n-1} - 4$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Find $a_4$ if $a_1 = 2$ and $a_n = 3a_{n-1}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Find $a_5$ if $a_1 = 1$ and $a_n = 2a_{n-1} + 3$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Find $a_6$ if $a_1 = 1$ , $a_2 = 3$ and $a_n = a_{n-1} + 2a_{n-2}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

A sequence is defined by $a_1 = 7$ and $a_n = a_{n-1} + 4$ . Write an explicit formula for $a_n$ in terms of $n$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 6

Which recursive formula describes $3, 12, 48, 192, \dots$ ?

Practice 7

A sequence is defined by $a_1 = 6$ and $a_n = \tfrac12 a_{n-1}$ . Write an explicit formula for $a_n$ in terms of $n$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 8

A lake is stocked with fish. Each year $25\%$ of the fish die or are caught, and then $30$ (hundred) new fish are added, so the population in hundreds follows $a_n = 0.75\,a_{n-1} + 30$ . In the long run, what value (in hundreds of fish) does the population approach?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.