Download Lesson 8.1 - Coweta County Schools

8.1
Sequences
Quick Review
x
Let f ( x) 
. Find the values of f .
x4 5
1. f (5)
1
9

2. f (-1)
3
Evaluate the expression a   n  1 d for the given values of
a, n, and d .
3. a  -2, n  2, d  3
4. a  1, n  2, d  2
1
5
Quick Review
n -1
Evaluate the expression ar for the given values of a, r , and n.
1
5. a  , r  2, n  3
2
6. a  2, r  1.5, n  4
2
 6.75
Find the value of the limit.
2x  2
7. lim
4x  x 1
sin  4 x 
8. lim
x
2
x 
x 0
2
1
2
4
What you’ll learn about
Defining a Sequence
 Arithmetic and Geometric Sequences
 Graphing a Sequence
 Limit of a Sequence

Essential Question
How can we use calculus to define and
evaluate sequences?
Defining a Sequence
a  is a list of numbers written in an explicit order.
For example: a   a , a , a ,..., a ,... , where a is the first term
A sequence
n
n
1
2
3
n
1
and a is the nth term of the sequence.
n
Let a , a , a ,..., a ,... be a function with domain the set of positive
1
2
3
n
integers and range a , a , a ,..., a ,.... If the domain is finite, then
1
2
3
n
the sequence is a finite sequence. If the domain is infinite, then
the sequence is an infinite sequence.
Example Defining a Sequence Explicitly
1. Find the first four terms and the 100th term of the sequence {an} where
 1
n
an 
n 2
2
.
Set n equal to 1, 2, 3, 4, and 100.

 1

1
a1
1

2
1 2
3
 1
3
1
a3  2

3 2
11
 1
100
1
a100 

2
100  2 10,002
 1
2
1
a2  2

2 2 6
 1
4
1
a4  2

4  2 18
Example Defining a Sequence Recursively
2. Find the first three terms and the 7th term of the sequence defined
recursively by the conditions: b1 = 4 and bn = bn – 1 – 2 for all n > 2.
b1  4
b2  b21  2  b1  2  4  2  2
b3  b31  2  b2  2  2  2  0
b7  b71  2  b6  2  6  2  8
Arithmetic Sequence
A sequence {a} is an arithmetic sequence if it can be written in the
form {a, a + d, a + 2d, . . . , a + (n – 1)d, . . .} for some constant d.
The number d is the common difference.
Each term in an arithmetic sequence can be obtained recursively from
its preceding term by adding d:
an  an1  d for all n  2.
Example Defining Arithmetic Sequences
3. Given the arithmetic sequence: – 3, 1, 5, 9, . . . find
a. the common difference,
b. the ninth term,
c. a recursive rule for the nth term,
d. an explicit rule for the nth term.
a. The common difference is a2  a1.
b. an  a1  n  1d
1   3  4
a9   3  9  14   29
c. The recursive rule is : a1  3, an  an 1  4
d. The explicit rule is : an  a1  n  1d
an  3 n  14   4n  7
Geometric Sequence
A sequence {a} is an geometric sequence if it can be written in the
form {a, a . r, a . r2, . . . , a . r n – 1 , . . .} for some nonzero constant r.
The number r is the common ratio.
Each term in an geometric sequence can be obtained recursively from
its preceding term by multiplying by r:
an  an1  r for all n  2.
Example Defining Geometric Sequences
4. Given the geometric sequence: 1, – 3, 9, – 27, . . . find
a. the common ratio,
b. the tenth term,
c. a recursive rule for the nth term,
d. an explicit rule for the nth term.
a2  3
a. The common ratio is
.
a1
1
n 1
b. an  a1  r
a9  1 3
101
 3
 19,683
c. The recursive rule is : a1  1, an
d. The explicit rule is : an  a1  r n1
an 1 3   3
n 1
n 1
  3an1
Example Constructing a Sequence
5. The second and fifth term of a geometric sequence are –6 and 48,
respectively. Find the first term, common ratio and an explicit rule for
the nth term.
5 1
4
a1r
a1r
48


2 1
1
a1r
a1r
6
r  8
r  2
3
a1r  6
 2a1  6
a1  3
The explicit rule is : an  a1  r
an  3 2
n 1
n 1
  1 32
n 1
n 1
Example Graphing a Sequence Using
Parametric Mode
n n 1
6. Draw a graph of the sequence {an} with an   1
, n  1, 2, 3, . . .
n
Change the mode on your calculator to parametric and dot.
Let X1T
 T 1 
 T, Y1T   1 

 T 
T
Set your window for the following:
Tmin  1, Tmax  20, Tstep  1
X min  0, X max  20, Xscl  2
Ymin  2, Ymax  2, Yscl  1
Example Graphing a Sequence Using
Sequence Graphing Mode
7. Graph the sequence defined recursively by b1 = 4 and bn = bn – 1 + 2
for all n > 2.
Change the mode on your calculator to sequence and dot.
Replace bn by u(n).
Select nMin = 1, u(n) =u(n – 1) + 2, and u(nMin) = {4}.
Example Graphing a Sequence Using
Sequence Graphing Mode
7. Graph the sequence defined recursively by b1 = 4 and bn = bn – 1 + 2
for all n > 2.
Set nMin = 1, uMax = 10, PlotStart = 1, PlotStep = 1, and
graph in the [0, 10] by [–5, 25] viewing window.
Limit
Let L be a real number. The sequence a has limit L as n approaches ∞
if, given any positive number e, there is a positive number M such that
for all n > M we have an  L  e .
We write lim an  L and say that the sequence converges to L.
n 
Sequences that do not have limits diverge.
Properties of Limits
If L and M are real numbers and lim an  L and lim bn  M , then
n 
n 
1. Sum Rule: lim an  bn   L  M
n 
2. Difference Rule: lim an  bn   L  M
n 
3. Product Rule: lim anbn   L  M
n 
4. Constant Multiple Rule: lim can   c  L
n 
 an  L
5. Quotient Rule: lim 


,M  0


n  b
 n M
Example Finding the Limit of a Sequence
8. Determine whether the sequence converges or diverges. If it
converges, find its limit.
2n  1
an 
n
Graph it, changing the mode to parametric and dot.
Find the limit analytically, using the Properties of Limits:
2n  1
1
1


lim
 lim  2    lim 2   lim  
n 
n 
n  n
n
n  n 

 
 2 0  2
The Sandwich Theorem for Sequences
If lim an  lim cn  L and if there is an integer N for which
n 
n 
an  bn  cn for all n  N , then lim bn  L
n 
Absolute Value Theorem
Consider t he sequence an . If lim an  0, then lim an  0.
n
n
Example Using the Sandwich Theorem to
find the Limit of a Sequence
9. Show that the following sequence converges and find its limit.
cos n
an 
n
cos x  1
cos n 1
cos n


n
n
n
1 cos n 1
 

n
n
n
1
 1
lim    0
lim     0
cos
n


n  n
n 
 
lim 
 n

0

n 
 n 
Pg. 441, 8.1 #1-43 odd