Recursive Rules vs Functions for Sequences

Recursive Rules vs Functions for Sequences
Function rules (equations) can be used to generate a sequence and/or written to describe a sequence.
Ex: f(x) = 2x – 3 would generate the sequence -1, 1, 3, 5, 7, … when we use a domain of the
natural numbers. (The domain would not include zero because sequence values are based on x
representing their position in the pattern. -1 is the first term, where x=1, not the ‘zeroeth’ term.)
Ex: The sequence 5, 10, 15, 20, 25, … can be represented by the function g(x) = 5x with a domain
of natural numbers.
But there’s another way to write a rule for a sequence – the recursive rule. A recursive rule
describes/defines how to find the nth term of a sequence based on the previous term(s). It must also
define the first term(s) so that the recursion process has a starting point.
Think of the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, …
How is this pattern created? Each term (after the first two) is found by adding the previous two terms
together. This is not so easily described with a function, but we can still represent it recursively.
𝑎1 = 1
𝑎2 = 1
The recursive rule for the Fibonacci sequence is {
𝑎𝑛 = 𝑎𝑛−1 + 𝑎𝑛−2
Let’s look at these parts closely.
𝑎1 refers to the first term of the sequence
𝑎2 refers to the second term
𝑎𝑛 refers to the term in the nth position.
So what does 𝑎𝑛−1 refer to? Remember, the subscript is the position/location of the term, so 𝑎𝑛−1
refers to the term before 𝑎𝑛 .
Look again at the recursive rule. The first two parts tell you that our sequence starts with 1, 1. The last
part tells you that to find any other term (𝑎𝑛 ), you must add the two previous terms together.
Let’s look at some easier recursive rules.
Example: Let 𝑎1 = 5 and 𝑎𝑛 = 𝑎𝑛−1 + 3. Use this recursive rule to generate the sequence.
Start with the first term in the sequence (hint: it’s been given to you!) Then build the rest of the
sequence using the second part of the rule. You get 5, 8, 11, 14, 17, …
Example: Now consider 𝑎1 = -10 and 𝑎𝑛 = 4𝑎𝑛−1 . What sequence does that create? We already know
that the first term is -10. The second term is 𝑎2 = 4𝑎2−1 = 4𝑎1 = 4(−10) = −40. You could repeat
this process, or you could notice that the recursive procedure is to multiply the previous term by 4, so
we get -10, -40, -160, -640, -2560, …
Now it’s your turn. List the first five terms for the sequences described by these recursive rules.
1) 𝑎1 = 0; 𝑎𝑛 = 𝑎𝑛−1 − 7
4) 𝑎1 = 128; 𝑎𝑛 = 𝑎𝑛−1 /4
1
2) 𝑎1 = −8; 𝑎𝑛 = 𝑎𝑛−1 + 2
3) 𝑎1 = 12; 𝑎𝑛 = −2𝑎𝑛−1
5) 𝑎1 = 2; 𝑎𝑛 = (𝑎𝑛−1 )2
Time to reverse the process. Give the recursive rule for each of the following sequences. Don’t forget to
define the first term(s) for each. It can help to first determine the type of sequence (arithmetic,
geometric, etc.) and find the common difference/ratio, then use that info as you write the rule.
6) 19, 13, 7, 1, …
9) 2.4, -7.2, 21.6, -64.8, 194.4, …
7) -4, -6, -8, -10, …
10) 729, 243, 81, 27, 9, …
8) -5, -15, -45, -135, …
Now let’s try to incorporate the function rules into this. We can create a sequence from a function by
using a domain of the natural numbers, then take the same sequence and write a recursive rule for it.
Ex: h(x) = 20 – 4x

The sequence is: 16, 12, 8, 4, 0, …
 𝑎1 = 16; 𝑎𝑛 = 𝑎𝑛−1 − 4
Remember, to create a sequence from a function, your domain is the set of counting numbers!
Now you try. For each of the functions below: (a) List the first five terms of the sequence
(b) Write the recursive rule for that sequence.
11) z(x) = -8x
13) m(x) = 80 ÷ 2x
12) j(x) = x + 17
14) w(x) = -12 – 1.5x
And again, we need to be able to reverse the process. Let’s take a recursive rule, write out some terms
from the beginning of the sequence, then write the function rule for that sequence.
Ex: 𝑎1 = 216; 𝑎𝑛 = 𝑎𝑛−1 /3  The sequence is: 216, 72, 24, 8, …  q(x) = 216/3x-1 or q(x) = 648/3x
Your turn – For each of the recursive rules below: (a) List the first five terms of the sequence
(b) Write the function for that sequence
15) 𝑎1 = −18; 𝑎𝑛 = 𝑎𝑛−1 + 7
17) 𝑎1 = 4.3; 𝑎𝑛 = 𝑎𝑛−1 − 3
16) 𝑎1 = 2; 𝑎𝑛 = 𝑎𝑛−1 ∗ 2
18) 𝑎1 = 10; 𝑎𝑛 = 𝑎𝑛−1 /5