Example If the function for a sequence is f (n) = 2n 1 then the values are found by substituting the domain values into the function in order

Transcription

1 Section 12 1A: Sequences A sequence is a function whose domain is the positive integers Z +. Z + represents the counting numbers 1, 2, 3, 4, 5, 6, 7,... We use the letter n to represent the domain of the function. To find the range for the function you substitute the values for n into the function f (n) and then list the outcomes in order with a comma between the values. The first term is found by substituting the first domain element n = 1 into the function that defines the sequence. The second term is found by substituting the second domain element n = 2 into the function and so on. The domain does not end so the list uses ellipsis... to show this Example If the function for a sequence is f (n) = 2n 1 then the values are found by substituting the domain values into the function in order Domain: n =1 n = 2 n = 3 n = 4 n = 5... Range: f (1) = 1, f (2) = 3, f (3) = 5, f (4) = 7, f (5) = 9... The value of each f (n) is called a term of the sequence The sequence is written in the following order first second third fourth fifth term term term term term f (n) = 1, 3, 5, 7, 9,... Sequences use special notation for the function and the terms of the function The a n notation allows us to denote each term as a specific a n the function for a sequence is a n = 2n 1 a 1 is the first term, n = 1 and a 1 = 2(1) 1= 1 a 2 is the second term, n = 2 and a 2 = 2(2) 1= 3 a 3 is the third term, n = 3 and a 3 = 2(3) 1= 5 a 4 is the fourth term, n = 4 and a 4 = 2(4) 1= 7 a 5 is the fifth term, n = 5 and a 5 = 2(5) 1= 9 1, 3, 5, 7, 9,... Section 12 1A" Page 1" 2016 Eitel

2 Finding the first 5 terms of a sequence Example 1!! Example 2 a n = 3n 2 a n = 4n +1 a 1 = 3(1) 2 = 1 a 2 = 3(2) 2 = 4 a 3 = 3(3) 2 = 7 a 4 = 3(4) 2 = 10 a 5 = 3(5) 2 = 13 a n = 1, 4, 7, 10, 13,..., ( 3n 2)! a 1 = 4(1) +1= 3 a 2 = 4(2) +1= 7 a 3 = 4(3) +1= 11 a 4 = 4(4) +1= 15 a 4 = 4(5) +1= 19 a n = 3, 7, 11, 15, 19,..., ( 4n +1) Example 3!! Example 4 a n = 2n 2 n a n = ( 2) n + 3 a 1 = 2(1) 2 1= 1 a 2 = 2(2) 2 1= 7 a 3 = 2(3) 2 1= 17 a 4 = 2(4) 2 1= 31 a 4 = 2(5) 2 1= 49 a 1 = ( 2) = 1 a 2 = ( 2) = 7 a 3 = ( 2) = 5 a 4 = ( 2) = 19 a 4 = ( 2) = 29 a n = 1, 7, 17, 31, 49,..., ( 2n 2 n)! a n = 1, 7, 5, 19, 29,..., ( 2) n + 3 a n = Example 5 n if n is odd n 2 if n is even a 1 :n =1 is odd = 1 a 2 :n = 2 is even = 4 a 3 :n = 3 is odd = 3 a 4 :n = 4 is even = 16 a 5 :n = 5 is odd = 5 a n = 1, 4, 3, 16, 5,... Section 12 1A" Page 2" 2016 Eitel

3 Sequences with alternating signs Example 6!! Example 7 a n = ( 5n 4) ( 1) n a n = ( 5n 4) ( 1) n +1 a 1 = ( 5(1) 4) ( 1) 1 = 1 a 2 = ( 5(2) 4) ( 1) 2 = 6 a 3 = ( 5(3) 4) ( 1) 3 = 11 a 4 = ( 5(4) 4) ( 1) 4 = 16 a 5 = ( 5(5) 4) ( 1) 5 = 21 a 1 = ( 5(1) 4) ( 1) 2 = 1 a 2 = ( 5(2) 4) ( 1) 3 = 6 a 3 = ( 5(3) 4) ( 1) 4 = 11 a 4 = ( 5(4) 4) ( 1) 5 = 16 a 5 = ( 5(5) 4) ( 1) 6 = 21 a n = 1, 6, 11, 16, 21,..., ( 5n 4) ( 1) n! a n = 1, 6, 11, 16, 21,..., ( 5n 4) ( 1) n +1 Finding what term a number in the sequence is. Example 1!! Example 2 a n = 7n 6 What term has a value of 50 a n = 50 for some value of n Set a n = 7n 6 equal to 50 and solve for n 7n 6 = 50 7n = 56 n = 8 50 is the 8 th term a n = n 2 2 What term has a value of 14 a n = 4 for some value of n Set a n = n 2 2 equal to 14 and solve for n n 2 2 = 14 n 2 = 16 n = 4 or 4 n Z + so n = 4! 14 is the 4 th term Section 12 1A" Page 3" 2016 Eitel

4 Recursive Formulas The normal way to define a sequence is with the a n notation. This form is called the closed form. There is a second method but it has some limitations. We can state the actual value of the first ( or first 2 or 3 terms). Each term thereafter is defined by a formula that includes the values of the terms just before that term. A sequence defined this way is said to be defined recursively and the formula is called a recursive formula. Specific subscripts of a n define where a term is in the sequence based on any given term a n. If a n is any term in the sequence then... a n 2, a n 1, a n, a n +1,... the SECOND term the FIRST term any term the FIRST term BEFORE a n BEFORE a n of the AFTER a n in the sequence in the sequence sequence in the sequence Example: a 1 = 2 and a n = a n for n > 1 The first term has n =1 and is written as a 1. The value of a 1 given as a 1 = 2 The rule for finding the terms for n > 1 is given by a n = a n for n > 1 The formula says that every term past the first term is found by adding 6 to the term just before it The first term is given as a 1 = 2 The second term a 2 has n = 2. It is found by adding 6 to the first term. a 2 = a = a The third term a 3 has n = 3. It is found by adding 6 to the second term. a 3 = a = a The fourth term a 4 has, n = 4, is found by adding 6 to the third term. a 4 = a = a The fifth term a 5 has, n = 5, is found by adding 6 to the fourth term. a 5 = a = a The tenth term is found by adding 6 to the ninth term, but to do that you would need to find the 8th term which requires finding the 7th term which requires finding the 6th and so on. You can see that you need to find every term before the 10th tern to find the 10th term. The limitation to a recursive formula is that to find terms later in the sequence you must start at the first term and work your way out to the latter terms in the sequence. This works well with a computer program or an excel spreadsheet but not so well by hand. The closed form for a n is the best form. It allows you to find any term by substituting the number of that term directly into the closed form formula. The problem is that not every sequence can be defined using the closed form formula. Historically many sequences were first defined in the recursive form. At a later date, in some cases much later, a closed form was discovered for the sequence and the use of that sequence was much improved. Section 12 1A" Page 4" 2016 Eitel

5 Many famous sequences are only defined recursively The Fibonacci Sequence is 1, 1, 2, 3, 5, 8, 13, 21, It is defied recursively by stating the first two terms are both 1. a 1 = 1 and a 2 = 1 The recursive formula is a n = a n 2 + a n 1 The formula says that every term past the first 2 terms is found by adding the two terms just before it. The 3rd term is found by adding the the 1st and 2nd terms. The 4th term is found by adding the 2nd and 3rd terms. The 5th term is found by adding the 4th and 3rd terms. a 1 = 1 the 1st term is given as 1 a 2 = 1 the 2nd term is given as 1 The 3rd term is the sum of the 1st and 2nd terms a 3 = a 1 + a 2 = 1 +1= 2 a n = 1,1, 2,... The 4th term is the sum of the 2nd and 3rd terms a 4 = a 2 + a 3 = = 3 a n = 1,1, 2, 3,... The 5th term is the sum of the 2nd and 3rd terms a 5 = a 3 + a 4 = = 5 a n = 1,1, 2, 3, 5,... The 6th term is the sum of the 4th and 5th terms a 6 = a 4 + a 5 = = 8 a n = 1,1, 2, 3, 5, 8,... The 7th term is the sum of the 5th and 6th terms a 7 = a 5 + a 6 = = 13 a n = 1,1, 2, 3, 5, 8,13,... The 8th term is the sum of the 6th and 7th terms a 8 = a 6 + a 7 = = 21 a n = 1,1, 2, 3, 5, 8,13, a n = 1,1, 2, 3, 5, 8,13,21, 34, 55, 89, Section 12 1A" Page 5" 2016 Eitel

6 When you make squares with those widths, you get a nice spiral: Fibonacci Spiral The Fibonacci numbers are Nature's numbering system. They appear everywhere in Nature, from the leaf arrangement in plants, to the pattern of the florets of a flower, the bracts of a pinecone, or the scales of a pineapple. The Fibonacci numbers are therefore applicable to the growth of every living thing, including a single cell, a grain of wheat, a hive of bees, and even all of mankind. Plants do not know about this sequence - they just grow in the most efficient ways. Many plants show the Fibonacci numbers in the arrangement of the leaves around the stem. Some pine cones and fir cones also show the numbers, as do daisies and sunflowers. Sunflowers can contain the number 89, or even 144. Many other plants, such as succulents, also show the numbers. Some coniferous trees show these numbers in the bumps on their trunks. And palm trees show the numbers in the rings on their trunks. Why do these arrangements occur? In the case of leaf arrangements, or phyllotaxis, some of the cases may be related to maximizing the space for each leaf, or the average amount of light falling on each one. Even a tiny advantage in light would cause that outcome to dominate, over many generations. In the case of close-packed leaves in cabbages and succulents the correct arrangement may be crucial for availability of space. In the seeming randomness of the natural world, we can find many instances of mathematical order involving the Fibonacci numbers themselves and the closely related "Golden" elements. Section 12 1A" Page 6" 2016 Eitel

7 Find the first 5 terms of the sequence written in recursive form Example 1 Find the first 5 terms for a n = a n for n 2 if a 1 = 4 Using the formula!! In english a 1 = 4 a 2 = a = a = = 10 a 3 = a = a = = 16 a 1 = 4 the first term is 4 a n = a n find any term by adding 6 to the term before it a 4 = a = a = = 22 a 1 = 4 a 5 = a = a = = 28 a n = 4, 10, 16, 22, 28...! a 2 = add 6 to a 1 a 2 = = 10 a 3 = add 6 to a 2 a 3 = = 16 a 4 = add 6 to a 3 a 4 = = 22 a 5 = add 6 to a 4 a 5 = = 28 a n = 4, 10, 16, 22, Section 12 1A" Page 7" 2016 Eitel

8 Example 2 Find the first 5 terms for a n = a n 2 + a n 1 for n > 2 if a 1 = 7 and a 1 = 9 Using the formula!! In english a 1 = 7, a 2 = 9 a n = a n 2 + a n 1 for n > 2 a 1 = 7 a 2 = 9 a 3 = a a 3 1 = a 1 + a 2 = = 16 a 1 = 7 the first term is 7 a 2 = 9 the second term is 9 a n = a n 2 + a n 2 find any term by adding the two terms before it a 4 = a a 4 1 = a 2 + a 3 = = 25 a 5 = a a 5 1 = a 3 + a 4 = = 41 a n = 7, 9, 16, 25, 41, 66...! a 1 = 7 a 2 = 9 a 3 = add a 1 and a 2 a 3 = = 16 a 4 = add a 2 and a 3 a4 = = 25 a 5 = add a 3 and a 4 a 5 = = 41 a n = 7, 9, 16, 25, 41, Section 12 1A" Page 8" 2016 Eitel

9 Integers a n = 1, 2, 3, 4, 5,..., ( n) Famous Sequence Examples Integers starting with 4 a n = 4, 5, 6, 7, 8,..., ( n + 3) Integers starting with 3 a n = 3, 2, 1, 0, 1, 2, 3,..., ( n 4) n = n Even Integers a n = 2, 4, 6, 8,..., ( 2n) Even Integers starting with 6 a n = 6, 8, 10, 12,..., ( 2n + 4) Even Integers starting with 4 a n = 4, 2, 0, 2, 4,..., ( 2n 6) Odd Integers a n = 1, 3, 5, 9,..., ( 2n 1) Odd Integers starting with 9 a n = 9, 11, 13, 15,..., 2n + 7 Odd Integers starting with 5 a n = 5, 3, 1, 1, 3,..., ( 2n 7) Section 12 1A" Page 9" 2016 Eitel

10 Powers of 2 a n = 2, 4, 8, 16, 32,..., 2 n a n = 2 1, 2 2, 2 3, 2 4,..., ( 2 n ) 1 more than a Power of 2 a n = 3, 5, 9, 17, 33,..., 2 n +1, ( ), ( ), ( ),..., ( 2 n +1) a n = less than a Power of 2 a n = 1, 3, 7, 15, 31,..., 2 n 1, ( 2 2 1), ( 2 3 1), ( 2 4 1),..., ( 2 n 1) a n = Powers of 3 a n = 3, 9, 27, 81, 243,..., 3 n a n = 3 1, 3 2, 3 3, 3 4,..., ( 3 n ) 2 more than a Power of 3 a n = 5, 11, 29, 83, 245,..., 3 n + 2, ( ), ( ), ( ),..., ( 3 n + 2) a n = less than a Power of 3 a n = 2, 8, 26, 80, 242,..., 3 n 1, ( 3 2 1), ( 3 3 1), ( 3 4 1),..., ( 3 n 1) a n = Section 12 1A" Page 10" 2016 Eitel

11 a n = 2, 6, 12, 20, 30,..., ( n(n +1) ) Triangular numbers (factors) a n = ( 1 2), ( 2 3), ( 3 4), ( 4 5), ( 5 6),..., ( n(n +1) ) a n = 2, 12, 30, 56, 90,..., ( 2n(2n 1) ) A pattern based on factors a n = ( 2 1), ( 4 3), ( 6 5), ( 8 7), ( 10 9),..., ( 2n(2n 1) ) a n = 1, +1, 1, +1, 1, +1,..., ( 1) n Alternating signs of 1 starting with 1 a n = ( 1) 1, ( 1) 2, ( 1) 3, ( 1) 4, ( 1) 5, ( 1) 6,..., ( 1) n n = n a n = +1, 1, +1, 1, +1, 1,..., ( 1) n +1 Alternating signs of 1 starting with + 1 a n = ( 1) 2, ( 1) 3, ( 1) 4, ( 1) 5, ( 1) 6, ( 1) 7..., ( 1) n +1 n = n Section 12 1A" Page 11" 2016 Eitel