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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) 1 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. 2 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. 3 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. 4 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 5 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. 6 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. 7 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. 8 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 9 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 10 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} 11 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. 12 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. ✔ 13 3.2.1: Sequences As Functions Guided Practice: Example 2, continued 14 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. 15 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 16 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 17 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) 18 3.2.1: Sequences As Functions Guided Practice: Example 4, continued 4. Plot the ordered pairs. Do not connect the points. ✔ 19 3.2.1: Sequences As Functions Guided Practice: Example 5, continued 20 3.2.1: Sequences As Functions