12.5 Use Recursive Rules with Sequences and Functions

 

Explicit rule

A rule for a sequence that gives an as a function of the term's position number n

 

Recursive rule

A rule for a sequence that gives the beginning term or terms of a sequence and then a recursive equation that tells how an is related to one or more preceding terms

 

Iteration

The repeated composition of a function f with itself

 

Example 1

Evaluate recursive rules

 

Write the first six terms of the sequence.

a0 = 2, an = an - 1 -3

 

Solution

a0 = 2

a1 = a0 - 3 = _2-3_ = _-1_

a2 = a1 - 3 = _-1-3_ = _-4_

a3 = a2 - 3 = _-4-3_ = _-7_

a4 = a3 - 3 = _-7-3_ = _-10_

a5 = a4 - 3 = _-10 - 3_ = _-13_

 

 

RECURSIVE EQUATIONS FOR ARITHMETIC AND GEOMETRIC SEQUENCES

Arithmetic Sequence an = an-1+ d where d is the common difference

Geometric Sequence an = r · an-1 where r is the common ratio

 

Example 2

Write recursive rules

 

Write a recursive rule for the sequence.

a. 1, 7, 13, 19, 25, . . .                   b. 4, 12, 36, 108, 324, . . .

 

Solution

a.   The sequence is _arithmetic_ with first term

a1 =_1_ and common difference d = _7 - 1_ = _6_.

an = an-1 + d            General recursive equation for an

= an-1 + 6             Substitute for d.

So, a recursive rule for the sequence is a1 = _1_,

an = an-1+6

b.   The sequence is _geometric_ with first term a1 = _4_

and common ratio r =_____      =_3_

an = r · an-1                 General recursive equation for an.

= 3an-1                          Substitute for r.

So, a recursive rule for the sequence is a1 = _4_,

an = 3an-1

 

Example 3

Write recursive rules for special sequences

 

Write a recursive rule for the sequence 3, 5, 2, -3, -5,. . ..

 

Solution

Beginning with the _third_ term in the sequence, each term is the _difference_ between the two previous terms.

So, a recursive rule is: a1 = _3_, a2 = _5_,

an = an-1 - an-2

 

Example 4

Iterate a function

 

Find the first three iterates x1, x2, and x3 of the function f{x) = 2x - 5 for an initial value of x0 = 3.

 

Solution

x1 = f(x0)

x2 = f(x1)

x2 = f(x2)

= f(_3_)

= f(_3_)

= f(_3_)

= 2(_3_) -5

= 2(_1_) -5

= 2(_-3_) -5

= _1_

= _-3_

= _-11_

The first three iterates are _1, -3, -11_.