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 = a 1, a 2, a 3, …, a n, where a 1 = f(1), a 2 = f(2), a 3 = f(3), …, a n = f(n) : 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 : 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 : 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 : Sequences As Functions

Key Concepts, continued If a common difference exists, a n – a n – 1 = constant. So, a 4 – a 3 = a 3 – a 2 = a 2 – a 1. 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. If a common ratio exists, then. So, : Sequences As Functions

Key Concepts, continued Explicit Sequences Explicitly defined sequences provide the function that will generate each term. For example: a n = 2n + 3 or b n = 5(3) n. In each case, we simply plug in the n representing the nth term and we get a n or b n : 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: a n = a n – or b n = b n – 1 3, where n is the number of the term : 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, a n = n – 1, while the second graph is of the line f(x) = x – : Sequences As Functions

Key Concepts, continued 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 a n, while the line is in terms of x and y : Sequences As Functions Sequence graphLine graph

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 : Sequences As Functions

Guided Practice Example 2 Find the missing terms in the sequence using recursion. A = {8, 13, 18, 23, a 5, a 6, a 7 } : Sequences As Functions

Guided Practice: Example 2, continued 1.First look for the pattern. Is there a common difference or common ratio? a 2 – a 1 = 5 a 3 – a 2 = 5 a 4 – a 3 = 5 The terms are separated by a common difference of 5. From this, we can deduce a n = a n – : Sequences As Functions

Guided Practice: Example 2, continued 2.Use the formula to find the missing terms. a n = a n – a 5 = a = = 28 a 6 = a = = 33 a 7 = a = = 38 The missing terms are 28, 33, and : Sequences As Functions ✔

: Sequences As Functions Guided Practice: Example 2, continued 14

Guided Practice Example 5 Find the seventh term in the sequence given by a n = 3 2 n – 1. Then, graph the first 5 terms in the sequence : Sequences As Functions

Guided Practice: Example 5, continued 1.Substitute 7 for n. a n = 3 2 n – 1 a 7 = 3 2 (7) – 1 = = 3 64 = : Sequences As Functions

Guided Practice: Example 5, continued 2.Generate the first 5 terms of the sequence. a n = 3 2 n – 1 a 1 = 3 2 (1) – 1 = = 3 1 = 3 a 2 = 3 2 (2) – 1 = = 3 2 = 6 a 3 = 3 2 (3) – 1 = = 3 4 = 12 a 4 = 3 2 (4) – 1 = = 3 8 = 24 a 5 = 3 2 (5) – 1 = = 3 16 = : Sequences As Functions

Guided Practice: Example 5, continued 3.Create the ordered pairs from the sequence. n corresponds to x, and a n corresponds to y. (n, a n ) (1, 3) (2, 6) (3, 12) (4, 24) (5, 48) : Sequences As Functions

Guided Practice: Example 4, continued 4.Plot the ordered pairs. Do not connect the points : Sequences As Functions ✔

: Sequences As Functions Guided Practice: Example 5, continued