Download 11.1 Notes

Obj: The student will demonstrate the ability
to evaluate the first five terms of explicit
and recursive sequences.
Drill
What is the next shape/number for each?
1.
2. 5, 3, 1, -1, -3, ____
3. 1, 4, 9, 16, 25, ____
4. 2, 4, 8, 16, 32, ____
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-1
11.1 Sequences
Sequences are ordered lists generated by a
function, for example f(n) = 100n
f (1), f (2), f (3),...
100, 200,300,...
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-2
11.1 Sequences
A sequence is a function that has a set of natural
numbers (positive integers) as its domain.
•
•
•
f (x) notation is not used for sequences.
Write an  f (n)
Sequences are written as ordered lists
a1 , a2 , a3 , ...
•
a1 is the first element, a2 the second element,
and so on
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-3
11.1
Graphing Sequences
The graph of a sequence, an, is the graph of the
discrete points (n, an) for n = 1, 2, 3, …
Example Graph the sequence an = 2n.
Solution
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-4
11.1 Sequences
A sequence is often specified by giving a formula for
the general term or nth term, an.
Example Find the first four terms for the sequence
n 1
an 
n2
Solution
a1  (1  1) /(1  2)  2 / 3, a2  (2  1) /(2  2)  3/ 4
a3  (3  1) /(3  2)  4 / 5, a4  (4  1) /(4  2)  5 / 6
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-5
11.1
Sequences
• A finite sequence has domain the finite set
{1, 2, 3, …, n} for some natural number n.
Example 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
• An infinite sequence has domain
{1, 2, 3, …}, the set of all natural numbers.
Example 1, 2, 4, 8, 16, 32, …
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-6
11.1 Convergent and Divergent Sequences
• A convergent sequence is one whose terms get
closer and closer to a some real number. The
sequence is said to converge to that number.
• A sequence that is not convergent is said to be
divergent.
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-7
11.1 Convergent and Divergent Sequences
1
Example The sequence an 
converges to 0.
n
The terms of the sequence 1, 0.5, 0.33.., 0.25, …
grow smaller and smaller approaching 0. This can be
seen graphically.
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-8
11.1 Convergent and Divergent Sequences
Example The sequence an  n is divergent.
The terms grow large without bound
2
1, 4, 9, 16, 25, 36, 49, 64, …
and do not approach any one number.
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-9

1
a
n
n
n
2
Replacing n with n = 1, 2, 3, 4, and 5 will give you the first
five terms.
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-10
11.1 Sequences and Recursion Formulas
• A recursion formula or recursive definition
defines a sequence by
– Specifying the first few terms of the sequence
– Using a formula to specify subsequent terms in
terms of preceding terms.
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-11
11.1 Using a Recursion Formula
Example Find the first four terms of the
sequence a1 = 4; for n >1, an = 2an-1 + 1
Solution We know a1 = 4.
Since an = 2an-1 + 1
a2  2  a1  1  2  4  1  9
a3  2  a2  1  2  9  1  19
a4  2  a3  1  2 19  1  39
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-12
11.1
Applications of Sequences
Example The winter moth population in thousands
per acre in year n, is modeled by
a1  1, an  2.85an1  0.19an21 for n > 2
(a) Give a table of values for n = 1, 2, 3, …, 10
(b) Graph the sequence.
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-13
11.1
Solution
(a)
n
an
n
an
(b)
Applications of Sequences
1
1
7
9.31
2
2.66
8
10.1
3
6.24
9
9.43
4
5
6
10.4 9.11 10.2
10
9.98
Note the population
stabilizes near a value
of 9.7 thousand insects
per acre.
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-14
Class work
Name:____________
Write the first five terms of each sequence.(explicit)
1. an = 2n + 3
2. an = n3 + 1
3. an = 3(2n )
4. an = (-1)n (n)
Find the third, fourth and fifth terms of each.(recursive)
5. a1 = 6; an = an-1 + 4 6. a1 = 1; an = an-1 + 2n – 1
7. a1 = 9; an =1 an-1
3
Copyright © 2011 Pearson Education, Inc.
8. a1 = 4; an = (an-1 )2 - 10
Slide 11.1-15
11.1 Series and Summation Notation
• Sn is the sum a1 + a2 + …+ an of the first n terms
of the sequence a1, a2, a3, … .
•  is the Greek letter sigma and indicates a sum.
n
• The sigma notation
a
i 1
i
means add the terms ai
beginning with the 1st term and ending with the
nth term.
• i is called the index of summation.
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-16
11.1 Series and Summation Notation
A finite series is an expression of the form
n
Sn  a1  a2  a3  ...  an   ai
i 1
and an infinite series is an expression of the form

S  a1  a2  a3  ...  an  ...   ai .
i 1
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-17
11.1 Series and Summation Notation
Example Evaluate
6
6
k
(2
  1)
(a)
(b)
a
j 3
k 1
j
Solution
(a)
6
k
1
2
3
4
(2

1)

(2

1)

(2

1)

(2

1)

(2
 1)

k 1
 (25  1)  (26  1)
 3  5  9  17  33  65  132
(b)
6
a
j 3
Copyright © 2011 Pearson Education, Inc.
j
 a3  a4  a5  a6
Slide 11.1-18
6 Examples
1.
k
of Finite Series
k2
7
4n7

2.
n3
5
3.
4
k
k 1
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-19
11.1 Series and Summation Notation
Summation Properties
If a1, a2, a3, …, an and b1, b2, b3, …, bn are two
sequences, and c is a constant, then, for every positive
integer n,
n
(a)
n
 c  nc
 ca
(b)
i 1
i 1
n
(c)
 c  ai
i 1
n
 (a  b )   a   b
i 1
Copyright © 2011 Pearson Education, Inc.
n
i
n
i
i
i 1
i
i 1
i
Slide 11.1-20
11.1 Series and Summation Notation
Summation Rules
n(n  1)
i  1  2  ...  n 

2
i 1
n
n(n  1)(2n  1)
2
2
2
2
i  1  2  ...  n 

6
i 1
n
2
2
n
(
n

1)
3
3
3
3
i

1

2

...

n


4
i 1
n
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-21
11.1 Series and Summation Notation
Example Use40 the summation
properties
to
22
14
2
2
i
5
(2
i
 3)
evaluate (a) 
(b) 
(c) 
i1
i 1
i 1
Solution
40
(a)
 5  40(5)  200
i1
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-22
11.1 Series and Summation Notation
Solution
22(22  1)
 506
(b)  2i  2 i  2
2
i 1
i 1
22
22
14
14
14
14
14
i 1
i 1
i 1
i 1
i 1
2
2
2
(2
i

3)

2
i

3

2
i
(c) 
    3
14(14  1)(2 14  1)
2
 14(3)  1988
6
Copyright © 2011 Pearson Education, Inc.
Slide 11.1-23