Transcript Sequences as Functions - Social Circle City Schools

Introduction
A sequence is an ordered list of numbers. The numbers,
or terms, in the ordered list are determined by a formula
that is a function of the position of the term in the list. So
if we have a sequence A determined by a function f, the
terms of the sequence will be:
A = a1, a2, a3, …, an, where a1 = f(1), a2 = f(2), a3 = f(3), …, an = f(n)
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3.2.1: Sequences As Functions
Introduction, continued
Unlike a typical function on a variable, there are no
fractional terms between the first and second term,
between the second and third term, etc. Each term is a
whole number. Just like no one can place 1.23rd in a
race, there is no 1.23rd term in a sequence. Therefore,
the domain of f is at most {1, 2, 3, …, n}. This sub-set of
the integers is called the natural number system.
Natural numbers are the numbers we use for counting.
Because every element of the domain of a sequence is
individually separate and distinct, we say a sequence is
a discrete function.
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3.2.1: Sequences As Functions
Key Concepts
• Sequences are ordered lists determined by functions.
• The domain of the function that generates a sequence
is all natural numbers.
• A sequence is itself a function.
• There are two ways sequences are generally
defined—recursively and explicitly.
• An explicit formula is a formula used to find the nth
term of a sequence. If a sequence is defined explicitly
(that is, with an explicit formula), the function is given.
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3.2.1: Sequences As Functions
Key Concepts, continued
• A recursive formula is a formula used to find the next
term of a sequence when the previous term or terms
are known. If the sequence is defined recursively (that
is, with a recursive formula), the next term is based on
the term before it and the commonality between terms.
• For this lesson, sequences have a common difference
or a common ratio.
• To determine the common difference, subtract the
second term from the first term. Then subtract the third
term from the second term and so on.
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3.2.1: Sequences As Functions
Key Concepts, continued
• If a common difference exists, an – an – 1 = constant.
So, a4 – a3 = a3 – a2 = a2 – a1.
• If the difference is not constant, then the commonality
might be a ratio between terms. To find a common
ratio, divide the second term by the first term. Then,
divide the third term by the second term, and so on.
an
• If a common ratio exists, then
= constant .
an-1
a4 a3 a2
So, = = .
a3 a2 a1
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3.2.1: Sequences As Functions
Key Concepts, continued
Explicit Sequences
• Explicitly defined sequences provide the function that
will generate each term. For example:
an = 2n + 3 or bn = 5(3)n .
• In each case, we simply plug in the n representing the
nth term and we get an or bn.
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3.2.1: Sequences As Functions
Key Concepts, continued
Recursive Sequences
• The second way a sequence may be defined is
recursively. In a recursive sequence, each term is a
function of the term, or terms, that came before it. For
example: an = an – 1 + 2 or b n = bn – 1 • 3, where n is the
number of the term.
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3.2.1: Sequences As Functions
Key Concepts, continued
Graphing Sequences
• Sequences can be graphed with a domain of natural
numbers. Compare the graphs on the following slide.
The first graph is of a sequence, an = n – 1, while the
second graph is of the line f(x) = x – 1.
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3.2.1: Sequences As Functions
Key Concepts, continued
Sequence graph
•
Line graph
Notice the sequence only has values where n = 1, 2,
3, 4, etc. Also notice the labels on the axes of each
graph. The sequence is in terms of n and an, while
the line is in terms of x and y.
3.2.1: Sequences As Functions
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Common Errors/Misconceptions
• not realizing that a sequence is not a function
• not understanding that a sequence is graphed without
a line connecting the points
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3.2.1: Sequences As Functions
Guided Practice
Example 2
Find the missing terms in the sequence using recursion.
A = {8, 13, 18, 23, a5, a6, a7}
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3.2.1: Sequences As Functions
Guided Practice: Example 2, continued
1. First look for the pattern.
Is there a common difference or common ratio?
a2 – a1 = 5
a3 – a2 = 5
a4 – a3 = 5
The terms are separated by a common difference of 5.
From this, we can deduce an = an – 1 + 5.
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3.2.1: Sequences As Functions
Guided Practice: Example 2, continued
2. Use the formula to find the missing terms.
an = an – 1 + 5
a5 = a4 + 5 = 23 + 5 = 28
a6 = a5 + 5 = 28 + 5 = 33
a7 = a6 + 5 = 33 + 5 = 38
The missing terms are 28, 33, and 38.
✔
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3.2.1: Sequences As Functions
Guided Practice: Example 2, continued
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3.2.1: Sequences As Functions
Guided Practice
Example 5
Find the seventh term in the sequence given by
an = 3 • 2n – 1. Then, graph the first 5 terms in the
sequence.
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3.2.1: Sequences As Functions
Guided Practice: Example 5, continued
1. Substitute 7 for n.
an = 3 • 2n – 1
a7 = 3 • 2(7) – 1 = 3 • 26 = 3 • 64 = 192
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3.2.1: Sequences As Functions
Guided Practice: Example 5, continued
2. Generate the first 5 terms of the
sequence.
an = 3 • 2n – 1
a1 = 3 • 2(1) – 1 = 3 • 20 = 3 • 1 = 3
a2 = 3 • 2(2) – 1 = 3 • 21 = 3 • 2 = 6
a3 = 3 • 2(3) – 1 = 3 • 22 = 3 • 4 = 12
a4 = 3 • 2(4) – 1 = 3 • 23 = 3 • 8 = 24
a5 = 3 • 2(5) – 1 = 3 • 24 = 3 • 16 = 48
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3.2.1: Sequences As Functions
Guided Practice: Example 5, continued
3. Create the ordered pairs from the
sequence.
n corresponds to x, and an corresponds to y.
(n, an )
(1, 3)
(2, 6)
(3, 12)
(4, 24)
(5, 48)
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3.2.1: Sequences As Functions
Guided Practice: Example 4, continued
4. Plot the ordered
pairs. Do not
connect the points.
✔
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3.2.1: Sequences As Functions
Guided Practice: Example 5, continued
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3.2.1: Sequences As Functions