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Sequences and Series From Simple Patterns to Elegant and Profound Mathematics

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David W. Stephens. The Bryn Mawr School. Baltimore, Maryland. PCTM 28 October 2005 ... of Henry Briggs, John Napier, Jobst Burgi, John Wallis, and Johann Bernoulli ... – PowerPoint PPT presentation

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Title: Sequences and Series From Simple Patterns to Elegant and Profound Mathematics


1
Sequences and SeriesFrom Simple Patterns to
Elegant and Profound Mathematics
  • David W. Stephens
  • The Bryn Mawr School
  • Baltimore, Maryland
  • PCTM 28 October 2005

2
Contact Information
  • Email
  • stephensd_at_brynmawrschool.org
  • The post office mailing address is
  • David W. Stephens
  • 109 W. Melrose Avenue
  • Baltimore, MD 21210
  • 410-323-8800
  • The PowerPoint slides will be available on my
    school website
  • http//207.239.98.140/UpperSchool/math/stephensd
    /StephensFirstPage.htm , listed under PCTM
    October 2005

3
Teaching Sequences and Series
  • We will look at some ideas for teaching sequences
    and series as well as some applications in
    mathematics classes at THREE different levels
  • I. Early (Algebra 1, Algebra 2, and
    Geometry)
  • II. Intermediate (Advanced Algebra,
    Precalculus)
  • III. Advanced (AP Calculus, esp. BC
    Calculus)

4
Teaching Sequences and Series
Many of the topics and examples used today will
not be new to you, but I want you to consider
thinking of them and talking about them with
students as sequences and series. It can be a
good way for them to think about these diverse
topics as bring linked mathematically. Just as
functions link a lot of what we teach, the
patterns of sequences and series can tie these
ideas together for better comprehension.
5
Early Sequences and Series (Algebra 1, Algebra
2, and Geometry)
  • 1. Looking for patterns
  • 2. Identifying kinds of sequences
  • 3. Describing patterns in sequences
  • 4. Using variables
  • 5. Summation notation
  • 6. Strategies for summing
  • 7. Applications with geometry ideas
  • 8. Graphing patterns
  • 9. Data analysis functions as sequences

6
Intermediate Sequences and Series (Advanced
Algebra, Precalculus)
  • 1. More geometric sequences and exponential
    functions
  • 2. Infinite series
  • 3. Convergence and divergence
  • 4. Informal limits
  • 5. More advanced data analysis (straightening
    data)
  • 6. Applications (compound interest, astronomy,
    chemistry, biology, economics, periodic motion or
    repeating phenomena)

7
Advanced Sequences and Series (AP Calculus, esp.
BC Calculus)
  • 1. Newtons method for locating roots
  • 2. Riemann sums
  • 3. Trapezoid rule and Simpsons rule
  • 4. Eulers method for differential equations
  • 5. Power series (Maclaurin and Taylor series
    polynomials)
  • 6. Convergence tests for series

8
Early Topics (Algebra 1 , Algebra 2, and
Geometry)
  • I sometimes have begun my Algebra 2 classes in
    September with this topic because
  • a) New students to the school (and the
    class) do not feel
  • new.
  • b) I can use algebraic language.
  • c) I can review linear functions in a
    new context.
  • d) I can sneak in some review which
    does not feel like
  • review!

9
Activity 1
  • Find the next three numbers in these sequences
  • A) 6, 9, 13, 18, 24,
  • B) 12, 17, 13, 16, 14, 15, 15,
  • C) 5, 10, 20, 40,
  • D) 7, -21, 63, -189,
  • E) 2, 3, 4, 5, 4, 3, 2, 1, 0, -1, 0, 1, 2, 3,

10
Activity 2
  • Students build their own sequences, and they
    challenge their classmates to guess the next few
    entries. This can be a neat homework assignment.
    (It can be extended to later activities where
    they have to code their sequence patterns with
    variables, too.)

11
Activity 3
  • Describe the pattern in words
  • A) 7, 5, 3, 1, -1, -3,
  • B) 70, 68, 64, 56, 40, 32, 28, 26, 25,
  • C)
  • D) 12, 13, 14, 15, 16, 15, 14, 13, 12, 13, 14,
    15, ...
  • E) 1, 2, 3, 4, 4,3, 2, 1, 3, 5, 7, 7 , 5, 3, 1,
    12, 23, 34, 23,
  • F) 4, 9, 32, 50, 53, 54, 54, 54, 54, 54.1,
    54.13, 54.135,
  • 54.1356 , ...

12
Activity 4
  • Learn to code the pattern with variables
  • A) 9, 13, 17, 21, 25, 29 ,
  • Let a0 9 an an-1 4 or an a0
    (n-1)d
  • (Some texts use tn, where t term, instead of
    an)
  • It could also be coded that a6 9 and then a7
    13,
  • if you decided to start the count at item 6

13
Activity 4 (continued)
  • B) 3, 6, 12, 24, 48,
  • Let a0 3 an a0r n-1
  • (first term)(ratioterm - 1)
  • ( or let t0 3 tn a0r n-1)

14
Activity 5
  • Introduction to Fibonacci sequences
  • A) 1, 1, 2, 3, 5, 8, 13, 21, 34
  • B) 2, 5, 7, 12, 19, 31, 50
  • an an-1 an-2 (recursively defined functions)

15
Activity 6
  • Series and Summation Notation
  • 1. a) sequence 1 , 3 , 5 , 7 , 9 , 11
  • b) series 1 3 5 7 9
    11
  • c) series notation
    or or

16
Activity 6 (continued)
  • 2. a) sequence 2, 6, 18, 54
  • b) series 2 6 18 54
  • c) series notation
  • 3. a) sequence 4, -2, 1, ,
  • b) series 4 2 1 - -
  • c) series notation

17
Activity 7
  • Interleaved and other creative sequences
  • Find the next three terms, and describe the two
    sequences that are interleaved.
  • A) 1, 3, 4, 9, 7, 27, 10, 81, 13, 243 ,
  • 1, 3, 4, 9, 7, 27, 10, 81, 13, 243 ,
  • B) 5, 1, 7, 4, 9, 7, 11, 10, 13, 13,
  • 5, 1, 7, 4, 9, 7, 11, 10, 13, 13,

18
Introduction to Series
  • How to add up an arithmetic series efficiently
  • Example sn 6 9 12 15 18 219
  • Add the first and last terms, the second
    and the
  • second to the last, etc. What do you
    notice?
  • How many pairs are there?
  • What if there are an odd number of
    terms to add?
  • Sn

19
Introduction to Series
  • How to add up a geometric series efficiently
  • Example sn 5 10 20 40 80 160 320
  • a0 a0r a0r2 a0r3 a0rn-1
  • rsn a0r a0r2 a0r3 a0r4
    a0rn
  • Then sn rsn a0 a0rn
  • This provides us with the usual formula for a
    geometric series
  • sn

20
Activity 8
  • For the series,
  • sn 5 10 20 40 320
  • calculate s28

21
Activity 9
22
Activity 10
23
Data Analysis Building Functions from Data
24
Data Analysis Building Functions from Data
25
Data Analysis Building Functions from Data
26
GeometryAngles of Polygons
  • What is the general formula for the sum of the
    interior angles of a polygon with n sides?
  • (n, measures of interior angles)
  • (3, 180) , (4, 360), (5 , 540) , (6 , 720) ,
  • (n , 180(n-2))

27
GeometryA Modeling Application
  • Handshake Problem
  • If n people shakes hands with everyone else at a
    meeting, how many handshakes occur?
  • 1. Visualize this as a geometry problem.
  • 2. Consider a simpler version with just a few
    number of people.
  • 3. Generalize the data, and consider the data
    as sequence.

28
GeometryA Modeling Application
Handshake Problem
29
GeometryA Modeling Application
  • n number of people
  • h(n) number of handshakes
  • n 1 2 3 4 5 6 7
  • h(n) 0 1 3 6 10 15 21

30
GeometryA Derivation of
Find the perimeter of a sequence of regular
polygons which are inscribed in a unit circle,
and emphasize that the sequence of results is
important to watch. s length of one side of
the polygon p perimeter of the polygon
31
GeometryA Derivation of
s length of one side p perimeter of inscribed
polygon
s p 4 5.657
s p 3 5.196
32
GeometryA Derivation of
s 1 p 6
s 2 sin(36) 1.176 p 5(1.176) 5.878
33
GeometryA Derivation of
In general, the length of one-half of a side of
an inscribed regular polygon is So a side
measures and the perimeter of the polygon
measures Since p ? 2 , then can be
calculated.
34
GeometryA Derivation of
The central angle for each side is
Each half-side has length equal to the sine of
one-half the central angle.
35
GeometryA Derivation of
Here are the perimeters of the polygons from the
TI-83 as a list (L2) Note Ignore L3.
36
Intermediate Sequences and Series (Advanced
Algebra, Precalculus)
  • 1. More geometric sequences and exponential
    functions
  • 2. Infinite series
  • 3. Convergence and divergence
  • 4. Informal limits
  • 5. More advanced data analysis (straightening
    data)
  • 6. Applications (compound interest, astronomy,
    chemistry, biology, economics, periodic motion or
    repeating phenomena)

37
Data AnalysisBuilding Functions from Data
  • Example 4 x 1 2 3 4 5 6
    7 8
  • y 3 9 27 81 243
    729 2187 6561
  • x is an arithmetic sequence
  • y is a geometric sequence
  • This is sometimes called an
    add-multiply property
  • So y f(x) is EXPONENTIAL
  • What is the actual function?
  • Ans f(x) 3x
  • ( where r 3 in
    the geometric sequence)

38
Data AnalysisBuilding Functions from Data
  • Example 5 x 1 2 3 4 5
    6 7 8
  • y 5 11 29 83 245 731 2189
    6563
  • y 3 9 27 81 243
    729 2187 6561
  • x is an arithmetic sequence
  • y is not exactly a geometric
    sequence
  • But if the sequence of y-values is compared
    with the last set of ys, then we see that this
    sequence is 2 more than a geometric sequence.
  • So y 3x 2

39
Data AnalysisBuilding Functions from Data
  • Example 6 x 1 4 7 10 13
  • y 6 48 384 3072 24,576
  • x is an arithmetic sequence
  • y is not exactly a geometric
    sequence
  • Since the two sequences have the add-multiply
    property, then y is a geometric sequence, and it
    is exponential. Notice that the xs do not have
    to be consecutive.
  • We have to find the r value as if we are
    calculating geometric means

40
Data AnalysisBuilding Functions from Data
  • Example 7 x 1 4 7 10 13
  • y 6 48 384 3072 24,576
  • a1 6 and a4 48, and we need to fill in the
    sequence so that we know the y-values for terms 2
    and 3. Since the desired sequence is geometric,
    we need to know what to multiply a1 by repeatedly
    three times to get 48. This suggests that rrr
    48/6.
  • So r 2 , and y 3 2x

41
Data AnalysisBuilding Functions from Data
  • Example 7
  • x 1 2 3 4 5 6 7
    8 9 10 11
  • y 6 12 24 48 96 192 384 768
    1536 3072 6144

r 2 , and y 3 2x
42
An Historical Diversion
Lets take a look at the pairing of an arithmetic
and a geometric sequence. n 1 2 3 4 5 6 an 2 4 8
16 32 64 Lets suppose that we wanted
intermediate terms n 1 3/2 2 5/2 3
7/2 4 9/2 5 6 an 2 4 8 16
32 64
43
An Historical Diversion
  • n 1 3/2 2 5/2 3 7/2 4 9/2 5
  • an 2 4 8 16 32
  • Thinking about an as a geometric sequence, we
    need a geometric mean to fill in the missing
    terms. Our desired multiplier, r, is .
  • an 2 4 8 16 32
  • an 2 2 3/2 4 2 5/2 8 2 7/2 16 2 9/2
    32

44
An Historical Diversion
  • So when we write an 2n , then the sequence, n,
    becomes the exponents, or the logarithms, for the
    geometric sequence.
  • This is part of the history of Henry Briggs, John
    Napier, Jobst Burgi, John Wallis, and Johann
    Bernoulli from 1620 to 1749 in the development of
    logarithms.

45
Function Transformations using Sequences
  • If functions are considered as lists of data, and
    one function is a transformation of another one,
    then the alterations to the sequence of function
    values is the key to decoding the transformation.

X 0 1 2 3 4 5 6 7 8 9
f(x) 6 3 8 9 14 10 21 43 8 6

g(x) 11 13 5 6 3 8 9 14 10 21

X 0 1 2 3 4 5 6 7 8 9
f(x) 6 3 8 9 14 10 21 43 8 6
-3 5 1 5 -4
g(x) 11 13 5 6 3 8 9 14 10 21
-3 5 1 5 -4
We want to write g(x) as a transformation of
f(x), so g(x) f(x 3)
46
Function Transformations using Sequences
  • Preliminary questions
  • A. When a transformation such as f(x a) is
    used, what happens to the y values?
  • B. When a transformation such as f(x) a is
    used, what happens to the y values?
  • C. When a transformation such as af(x) is
    used, what happens to the y values?

X 0 1 2 3 4 5 6 7 8 9
f(x) 6 3 8 9 14 10 21 43 8 6
-3 5 1 5
g(x) 4 8 -1 0 6 7 4 9 10 15
-3 5 1 5
X 0 1 2 3 4 5 6 7 8 9
f(x) 6 3 8 9 14 10 21 43 8 6

g(x) 4 8 -1 0 6 7 4 9 10 15

g(x) f(x 5) 1
47
Function Transformations using Sequences
x 0 1 2 3 4 5 6 7 8 9
f(x) 6 3 8 9 14 10 21 43 8 6

g(x) -8 3 12 6 16 18 28 20 42 86

x 0 1 2 3 4 5 6 7 8 9
f(x) 6 3 8 9 14 10 21 43 8 6
-3 5 1 5 -4
g(x) -8 3 12 6 16 18 28 20 42 86
-6 1- 2 1- -8
g(x) 2f(x 2)
48
Infinite Sequences, Series and Convergence
There are some really good opportunities to lead
students to important conclusions, as well as to
challenge their intuition with some sophisticated
ideas with infinite sequences and series. We
can extend their numerical sense as well as
exploiting their graphical skills to help
generate conclusions.
49
Infinite Sequences, Series and Convergence
Suppose an 1, 3, 5, 7, 9, Where does an go
as n gets large? Suppose bn 1, 1.01, 1.02 ,
1.03 , 1.04 , Where does bn go as n gets
large? Suppose cn 1, 2, 4, 8, 16, Where
does cn go as n gets large?
50
Infinite Sequences, Series and Convergence
Suppose dn Where does dn go as n gets
large? Since this is the ratio of two
sequences, each of which approaches infinity,
explain your answer to this question.
51
Infinite Sequences, Series and Convergence
Suppose en 1, 0, -1 , 1 , 0 , -1 , 1 , 0 , -1
, Where does en go as n gets large? Suppose
fn 40 , 32, 25.6, 20.48, 16.384, Where does
fn go as n gets large? Type 40 on the
calculator and hit ENTER. Type .8 and hit
ENTER The screen will read ANS .8 Repeatedly
hit enter to generate the sequence.
52
Infinite Sequences, Series and Convergence
Suppose gn 60, 90, 108, 120, ,
135, Where does gn go as n gets large? What
sequence is this? Suppose hn 120, 90, 72, 60,
, 45, Where does hn go as n gets
large? What sequence is this?
53
Infinite Sequences, Series and Convergence
Suppose in Where does in go as n gets
large?
54
Intermediate Level ApplicationsSequence Mode on
the Calculator
Suppose we want to generate the sequence as an
iterated function (recursive function). So 2,
5, 8, 11, 14, 17, could be an 2 3n or an
an-1 3
55
Intermediate Level ApplicationsSequences and
Series on the Calculator
To generate sequences on the HOME screen, go to
LIST (2nd STAT)/OPS/ltOption 5gt which will give
seq( The inputs required for seq(
are seq(expression, variable, begin, end
increment) Example an 2n1 ? 1, 3, 5,
7, 9,
56
Intermediate Level ApplicationsSequence and
Series on the Calculator
Example an 2n1 ? 1, 3, 5, 7, 9,
Notice that the name of the variable does not
matter, as long as it is specified.
57
Intermediate Level ApplicationsSequence and
Series on the Calculator
If the series is desired, the sum( function
is used. Example an 2n1 ? 1, 3, 5, 7,
9 sn sum(an) ? 13579 25
Sum( is found in LIST (2nd STAT)/MATH/ltOption
5gt
58
Intermediate Level ApplicationsSequence and
Series on the Calculator
Partial sums can also be generated, and this is
helpful if there is an application where the sums
should be considered as making a sequence,
perhaps if their convergence is being
considered. The function cumSum( is found under
LIST (2nd STAT)/OPS/ltOption 6gt Example an
2n1 ? 1, 3, 5, 7, 9 cumSum(an) ? 1, 4, 9,
16, 25
59
Intermediate Level ApplicationsSequence and
Series on the Calculator
On the calculator, cumSum( 1, 3, 5, 7, 9)
or cumSum(seq(2N1, N, 0, 4))
If a list is already in the calculator, perhaps
in L1, then cumSum(L1) or sum(L1) will give
series results.
60
Intermediate Level ApplicationsSequence Mode on
the Calculator
First term
First term value
Recursive function
61
Intermediate Level ApplicationsSequence Mode on
the Calculator
Using the same recursive function an an-1 3
or u(n) u(n-1) 3, suppose that we want
to build a sequence in a list on the calculator.
62
Intermediate Level ApplicationsSequence Mode on
the Calculator
It is also possible to use the sequence mode to
graph some more complicated ideas. Suppose that
we are trying to convince a student that the
geometric sequence 100, 80, 64, 51.2,
converges. Set the Window to
63
Intermediate Level ApplicationsSequence Mode on
the Calculator
Go to the 2nd ZOOM Format key, and make sure
that TIME is selected at the top. Hit GRAPH.
u(n)0.8u(n-1)
This is a scatterplot of the (n, an)
64
Intermediate Level ApplicationsSequence Mode on
the Calculator
Instead, select 2nd Zoom Format and choose
Web. Set the Window to 0 lt xlt 105 and 0 lt y lt
105 Hit Graph and Trace. Hit the right arrow to
iterate the web.
65
Intermediate Level ApplicationsSequence Mode on
the Calculator
Lets look at an 10(-.8)n, which becomes u(n)
-.8u(n-1) You have to think about the WINDOW,
but it has a web which looks like
66
Intermediate Level ApplicationsSequence Mode on
the Calculator
An application which is stretching toward the
advanced is the idea of a predator-prey model .
The populations of the two populations depend on
the size of the other population. Depending on
various parameters, the populations will either
die out, grow without bound (!), or move into an
equilibrium. Two sequence functions can be
used Rn Rn-1(10.05 -.001Wn-1)
rabbits Wn Wn-1(10.0002Rn-1 0.03)
wolves (This example is from the TI-83 manual,
page 6-13) There is a long document on my
website about predator-prey models that I
co-wrote as a NSA sponsored project in June 2004.
67
Intermediate Level ApplicationsSequence Mode on
the Calculator
68
Intermediate Level ApplicationsSequence Mode on
the Calculator
Using a WINDOW of nMin 0 and nMax
400 PlotStart 1 PlotStep 1 XMin 0 XMax
400 Xscl 100 YMin 0 YMax 300
YScl 100 Under FORMAT, use the TIME choice.
It makes for great classroom discussion to
interpret these graphs.
69
Intermediate Level ApplicationsSequence Mode on
the Calculator
With the sequence mode, we can do something quite
interesting on the calculator. The first graph
was showing the separate rabbit and wolf
populations as time progressed. But what if we
want to see how the graphs of the two populations
look relative to each other, i.e., (rabbits,
wolves). To do this select the FORMAT key and
then find the uv choice at the top.
70
Intermediate Level ApplicationsSequence Mode on
the Calculator
Experimentation with the data suggests that the
new WINDOW be XMin 80 XMax 250 Xscl
50 YMin 0 YMax 100 YScl 10
71
Infinite Sequences, Series and Convergence
When series go on forever, we call them infinite
series. Lets look at arithmetic series
first. Sn 4 7 10 13 What is the
sum s100? s1000 ? s 1,000,000 ?
72
Infinite Sequences, Series and Convergence
  • Compare the results for each of these arithmetic
    series
  • Sn 4 7 10 13
  • Sn 1 1.1 1.2 1.3 1.4
  • Sn 5 5.001 5.000001
  • Sn 4 3.5 3 2.5 2
  • Conclusion ?

73
Infinite Sequences, Series and Convergence
  • Moving onto geometric series, consider the
    behavior of these sums by taking the number of
    terms to be higher and higher.
  • sn 2 4 8 16
  • sn
  • s10 s100 s1000

74
Infinite Sequences, Series and Convergence
  1. sn 2 2(1.02) 2(1.02)2 2(1.02)3
  2. sn2 2(0.98) 2(.98)2 2(.98)3
  3. sn 1 3 9 27 81
  4. sn 1
  5. sn 1

75
Infinite Sequences, Series and Convergence
It eventually becomes obvious that there are
geometric series which converge and others which
diverge. The idea is that convergence depends on
the value of r (the common ratio). Conclusion
An infinite geometric series converges when 1 lt r
lt 1 or r lt 1
76
Infinite Sequences, Series and Convergence
Looking at the sequences graphically makes some
strong connections with algebra, and the visual
impact helps with understanding about convergence
and divergence. Lets look at some ideas about
series first (because the graphs of sequences vs.
the graphs of series is also an important
distinction).
77
Graphs of Sequences and Series
Examples an bn
cn dn
78
Graphs of Sequences and Series
79
Graphs of Sequences and Series
80
Intermediate Level ApplicationsDeer Populations
In this application, the various quantities
affect each other. This is part of a discrete
mathematics topic. The sequences involved (and
note why they are not series!) affect each other.
Whether or not they converge is the important
point, since this involves whether the
populations remain stable, or whether they
explode or become extinct. There are intuitive
ideas of limits here.
81
Intermediate Level ApplicationsDeer Populations
Newborn Yearling Adult Male Adult Female TOTAL
N Y AM AF
N 0.20 AF Y 0.90 N AM 0.90 AM 0.45 Y AF 0.90 AF 0.48 Y

1 20 16 90 65 191
2 13 18 88 66 185
3 13 11 87 68 179
4 13 11 83 66 173
5 13 11 79 64 167
6 12 11 76 62 161
7 12 10 73 61 156
8 12 10 70 59 151
9 11 10 67 57 145
10 11 9 64 56 140
82
Intermediate Level ApplicationsDeer Populations
Newborn Yearling Adult Male Adult Female TOTAL
N Y AM AF
N 0.20 AF Y 0.90 N AM 0.90 AM 0.45 Y AF 0.90 AF 0.48 Y

11 11 9 61 54 135
12 10 9 58 52 129
13 10 9 56 51 126
14 10 9 54 50 123
15 10 9 52 49 120
16 9 9 50 48 116
17 9 8 49 47 113
18 9 8 47 46 110
19 9 8 45 45 107
20 9 8 44 44 105
21 8 8 43 43 102
22 8 7 42 42 99
83
Intermediate Level Applications
Suppose that we earn simple interest on a bank
account. Lets say that the interest rate is 5
on a principal of 1,000. a0 1000 a1 1000
1000(.05) 1050 a2 1050 1000(.05) 1100 a3
1100 1000(.05) 1150 an 1000, 1050 , 1100
, 1150 , 1200 ,
Compound Interest
84
Intermediate Level Applications
Instead, suppose that we earn 5 interest on a
1,000 principal, compounded annually. a0
1000 a1 1000 1000(.05) 1000(1.05) 1050 a2
1050 1050(.05) 1050(1.05)
1000(1.05)2 a3 1000(1.05)3 ? at
1000(1.05)t
Compound Interest
85
Intermediate Level ApplicationsCompound Interest
Most banks and financial institutions offer
compound interest which is awarded more
frequently than annually, and it is important for
students to realize that there is an advantage to
getting a fraction of the annual interest more
frequently so that more compounding can occur
earlier in time. If yo is the initial principal,
r the annual percentage rate, t the number of
years for the money to be invested, n the
number of times per year that compounding will
occur, yt yo(1 )nt
86
Intermediate Level Applications
If the number of compoundings is discrete, then
this formula is fine. But what if the number of
compoundings each year becomes more and more
frequent? Investigate the sequence of (1
)n as n increases.
Compound Interest
87
Intermediate Level ApplicationsCompound Interest
N ( 1 ) n
100 2.7048
200 2.7115
1000 2.7169
10,000 2.7181
1,000,000 2.7183
88
Intermediate Level Applications
Note that if n gt 1012, the calculator will be
subject to some serious roundoff errors. This is
because the memory of the calculator only holds
about 12 digits, and larger numbers than that
overwhelm the capabilities of the machine. The
sequence is (for n 1, 2, 3, 4, 5) 2, 2.25,
2.370, 2.441, 2.448, 2.522, 2.545, , 2.7048 ,
2.7115 , 2.7169 , , 2.7181 , 2.7183 ? e
Compound Interest
89
Intermediate Level ApplicationsCompound Interest
There are some wonderful problems for students to
solve with interest, and their interest (bad
pun) is piqued with some challenges, such as
Two people each have 10,000. One invests the
money at a 5.1 interest rate, compounded
monthly. The other invests at 5 compounded
daily. Which investment is better after 8 years?
When will they be equal? Which is better after
many years?
90
Intermediate Level ApplicationsLinear and
Exponential Functions Compared
  • Consider two scenarios
  • Invest 5000 with 5 compound interest earned
    annually.
  • Invest 5000 and add 500 each year to the
    account. No interest is earned.
  • Which investment is better?

91
Intermediate Level ApplicationsLinear and
Exponential Functions Compared
The first situation is modeled with an
exponential function, since it is geometric
sequence. The second situation is modeled with a
linear function, since it is an arithmetic
sequence. Eventually..if both sequences
increase, a geometric sequence will exceed an
arithmetic sequence.
92
Intermediate Level ApplicationsLinear and
Exponential Functions Compared
Suppose that person has a debt obligation which
is subject to a compound annual interest rate of
18 (such as a credit card). The amount owed is
50,000. If the minimum monthly payment is 2.5
of the remaining balance, and the minimum payment
is what is made each month, what happens to the
debt? Question Is a (geometric sequence
arithmetic sequence) a good strategy to pay back
a debt? Could it be fine if the minimum payment
is high enough?
93
Intermediate Level ApplicationsAstronomy and
Sequences
In the middle of the 19th century, data
concerning the distance of the planets in our
solar system from the sun indicated that there
was a remarkable sequence with a missing number
Planet Mercury Venus Earth Mars Jupiter
Dist sun 36 67.2 92.9 141.6 483.7
A.U. 0.3875 0.7234 1.0000 1.5242 5.2067
Planet Saturn Uranus Pluto
Dist sun 890.6 1777 2654.4
A.U. 9.5867 19.1281 39.3369
(in millions of miles) astronomical units
94
Intermediate Level ApplicationsAstronomy and
Sequences
It seemed that there were two holes in the
location of the planets, and the location even
the existence (ah, such a word for a
mathematician) of a possible planet was
discovered by calculation rather than by
observation. The conclusion was that there was
another body pulling Uranus out of the orbit
predicted by Bodes Law, so Adams (England) and
Leverrier (France) solved g to
calculate the place where another planet ought
to be found.
95
Intermediate Level ApplicationsAstronomy and
Sequences
Bodes Law
A B SUM SUM/10
4 4 0.4
4 3 7 0.7
4 6 10 1.0
4 12 16 1.6
4 24 28 2.8
4 48 52 5.2
4 96 100 10.0
4 192 196 19.6
4 384 388 38.8
4 768 772 77.2

96
Intermediate Level ApplicationsAstronomy and
Sequences
Planet A.U. Bodes Law
A B SUM SUM/10
Mercury 0.3875 4 4 0.4
Venus 0.7234 4 3 7 0.7
Earth 1.0000 4 6 10 1.0
Mars 1.5242 4 12 16 1.6
4 24 28 2.8
Jupiter 5.2067 4 48 52 5.2
Saturn 9.5867 4 96 100 10.0
Uranus 19.1281 4 192 196 19.6
30.1335 4 384 388 38.8
Pluto 39.3369 4 768 772 77.2
97
Intermediate Level ApplicationsAstronomy and
Sequences
On September 23, 1846, astronomers had their
telescopes trained on the piece of the night sky
where Adams and Leverrier had predicted that a
missing planet might be located. A mere half
hour after they began looking, Neptune was
observed, only 52 minutes of arc (less than one
degree) off from Leverriers prediction. It was
2.8 billion miles from earth. Viva les
mathematiques!
98
Intermediate Level ApplicationsChemistry, Data
Analysis and Sequences
Looks for patterns in atomic weight, specific
heat or boiling points across rows or down
columns.
NB The TI-84 has a built in periodic table, and
there are graphical displays included!
99
Intermediate Level ApplicationsBiological Growth
and Sequences
If a virus grows from a population of 200 at 8 AM
to a population of 1000 by noon, how many virus
will there be at 4 PM? 6 PM? midnight? Answer
y0 200 (8 AM) y4 1000 (noon, which is 4
hours later) The sequence is t 0, 4,
8, 12, 16, 20, at 200, 1000, 5000, 25000,
125000, 625000
100
Intermediate Level ApplicationsBiological Growth
and Sequences
If we are only interested in the virus counts at
whole number of hours, we need the geometric
means, and the multiplier becomes So the
sequence is at 500 (1.4963t-1)
101
Intermediate Level ApplicationsBiological Growth
and Sequences
No wonder healthy people at 8 AM are not feeling
well at the ned of a day!
102
Intermediate Level ApplicationsChemical
Half-Life Radioactivity
The half-life of the chemical element technetium
is about 6 hours. This element is used in
medicine when tracing body functions, especially
renal function or failure in patients receiving
chemotherapy. Given the short half-life, what
percentage of Tc injected into the body remains
after 2 hours? 3 hours? 4 hours? This is done
just as the biological (population) growth, and
the hurly percentages can be thought of as a
sequence which converges to some value.
103
Intermediate Level ApplicationsAntibiotic
Medications Sequences and Series
Suppose that an antibiotic medication dissipates
in the body so that 20 of the amount currently
in the body is gone after 4 hours (or 80 of the
medication remains after 4 hours). A patient is
given a 600 mg bolus (a large initial dosage) to
begin the treatment. Then the dosage is an
additional 100 mg every 4 hours. It is dangerous
for the body to have more than 700 mg at any one
time, and at least 500 mg is needed to fight the
illness (e.g., strep throat).
104
Intermediate Level ApplicationsAntibiotic
Medications Sequences and Series
This is a good example of a problem which can be
considered as both a sequence and as a
series. Sequence a t amount of medication
given at each 4 hour interval a1 600 a2 100
600(.8) 580 mg a3 100 100(.8) 600(.82)
564 mg a4 100 100(.8) 100(.82)
600(.83) 551.2 mg a5 100 100(.8)
100(.82) 100(.83) 600(.84) 540.96 mg
105
Intermediate Level ApplicationsAntibiotic
Medications Sequences and Series
The medication after the bolus forms a geometric
sequence which decreases to zero, and the
repeated medications form a geometric
series sum 100 100(.8) 100(.82)
100(.83)
500
The combined dosages (which are a series) form a
sequence which needs to stay between the
effective and the dangerous drug levels. (What
happens to the original bolus?)
106
Intermediate Level ApplicationsCooling of Liquids
A hot cup of coffee ( of cocoa, tea, ) fresh
from the coffeepot has a temperature of 140o F.
a) How does it cool? b) This can be simulated
with a CBL and TI-83/84. c) Use appropriate
data analysis and regressions.
107
Intermediate Level ApplicationsCooling of Liquids
Time (min) Temp1 Temp2 Temp3 Temp4
0 140 140 140 140
1 135 126 133 130
2 130 113.4 126.7 121
3 125 102.1 121 113
4 120 91.9 115.9 106
5 115 82.67 111.3 100
10 90 48.8 94.4 85
20 65 17 78.5 ??
Which sequence of temperatures makes the most
sense?
How are each of the sequences calculated?
108
Intermediate Level ApplicationsCooling of Liquids
It seems to be good, authentic mathematics and
science to guess which of the sequences is most
reasonable, and then try to fit a function to
that sequence. Following such intuition with a
data collection with a CBL on cooling water will
give data to verify or refute the earlier guess.
Newtons Law of Cooling k(T-Tambient)
or T-Ta (T0-Ta)e -kt
109
Intermediate Level ApplicationsEconomics
  • When a yearbook is printed, suppose it costs
    9000 to print one copy, because of the set-up
    costs for the press, type-setting, importing
    photographs, binding, cover set-up, and artwork.
    It costs as additional 8 for each book, since
    the press is already set up, and only paper,
    binding, and some ink are needed for the second
    copy.
  • What is the cost of 5 books? 10 books? 100 books?
    n books?
  • What is the average cost of n books?
  • What is the difference in average costs for
    printing n to (n1) books for various values of n?

110
Intermediate Level ApplicationsEconomics
As a sequence or series problem, b1 9000 b1
average 9000/1 9000 b2 9000 8 b2
average 9008/2 4504 b3 9000 8 8 b3
average 9016 / 3 3005.33 bn 9000 8(n-1)
bn average (8992 8n) / n 8992/n
8
If 500 yearbooks are ordered, it costs 12,992 to
print them, and the average cost is 25.98
This can be taught as a sequence problem or as a
rational function problem.
111
Advanced LevelCalculus Examples, especially AP
Calculus
Newtons Method uses the definition of derivative
to provide a method to locate the roots of a
function. (It differs from the algorithm ROOT
FINDER in the TI-83/84 calculators which uses the
IVT) x n1 xn This is an
iterative algorithm, where the results (output)
of each stage become the input of the next stage.
If we look at each xn and its subsequent xn1,
then the fraction which is subtracted can be
considered as the correction factor, which
(hopefully) sends us closer, via a sequence, to
the exact location of the root of a function.
112
Advanced LevelCalculus Examples, especially AP
Calculus
An example of Newtons method Suppose we want
to approximate This is a root of f(x) x2
3 The sequence of values from Newtons Method
looks like x0 1 (our choice for a
guess)
The sequence seems to converge.
113
Advanced LevelCalculus Examples, especially AP
Calculus
An example of Newtons method Suppose we want
to approximate the roots of f(x) x2
The sequence of values from Newtons Method
looks like x0 1 (our choice for a guess)
This time, there is no convergence, and we cannot
locate a root.
114
Advanced LevelCalculus Examples, especially AP
Calculus
Riemann sums are the basis for evaluating the
area under a function, as sums of the areas of
rectangles are used to approximate the exact
area. It is probably a good idea to mention the
words sequence and series in the explanation for
the strategy. After all, the C part of the BC
Calculus concerns the ideas of series and
convergence, but the ideas of the convergence of
sequences and series can appear very early in the
A part of the differential calculus when
limits are discussed and when early ideas about
areas under functions are introduced.
115
Advanced LevelCalculus Examples, especially AP
Calculus
There is a sequence of the areas of each
rectangle, and there is a sequence of the partial
sums of the rectangles. Convergence of each of
these is an important idea.
116
Advanced LevelCalculus Examples, especially AP
Calculus
and as the number of partitions goes from 4 to 8
to 16 to , there is a sequence of estimates on
the area, and the idea for calculus students is
to believe that the sequence of series
convergesto the exact area.
117
Advanced LevelCalculus Examples, especially AP
Calculus
We usually consider the Trapezoid Rule and
Simpsons Rule as series, but if we repeat them
with more and more partitions, then sequence of
the series should converge. Trap Simpson

n an even number of partitions required for
Simpsons rule
118
Advanced LevelCalculus Examples, especially AP
Calculus
Evaluate using different
algorithms.
Upper Lower Trapezoids Simpson
n4 30 14 22 21.3333
N 8 25.5 17.5 21.5 21.3333
n 20 22.96 19.76 21.36 21.3333
n 100 21.9776 20.6976 21.3376 21.3333
n 50 21.56544 21.0144 21.3344 21.3333
n 1000 21.365 21.301344 21.33334 21.3333
119
Advanced LevelCalculus Examples, especially AP
Calculus
Eulers method is an iterative algorithm to give
approximate solutions to differential equations.
It is really just a linearization method that is
used repeatedly to give a sequence of points
which serve as a numerical function.
120
Advanced LevelCalculus Examples, especially AP
Calculus
Example Solve with
the initial condition (1, 2) Use
0.1 answer The first point on the solution is
(1,1) because x0 2 and y0 1
2 (1)(22) (0.1) 2.4 ? the
second point is (1.1, 2.4)
121
Advanced LevelCalculus Examples, especially AP
Calculus
So x1 1.1 and y1 2.4 2.4
(1.1)(2.42) (0.1) 3.0336 ? the
third point is (1.2, 3.0336) We continue the
process, generating a sequence of approximate
solutions to the differential equation. (If is
smaller, then the theory says that the sequence
should more closely match the function which is
the solution to the differential equation.)
122
Advanced LevelCalculus Examples, especially AP
Calculus
The exact solution, using separable differential
equation methods, is It is not always the case
that an exact solution can be found, and those
are the examples for which the approximate
solutions algorithms are important.
There are also some algorithms which provide more
accurate approximations. Two of them are called
the Improved Euler method and the Runge-Kutta
Method.
123
Advanced LevelCalculus Examples, especially AP
Calculus
A summary of the results of these algorithms is
X Y (exact) Y (Euler) Y (ImprovedEuler) Y (Runge-Kutta)
1 2 2 2 2
1.1 2.5316 2.4 2.5168 2.5316
1.2 3.5714 3.0336 3.4848 3.5706
1.3 6.4516 4.1379 5.80101 6.4304
perfect! good
better best
124
Advanced LevelCalculus Examples, especially AP
Calculus
Maclaurin and Taylor polynomials are a series of
polynomial (power) terms, and they are typically
taught near the end of a BC Calculus course. A
suggestion is to introduce them much earlier in
the course, since students only need to be able
to do derivatives to calculate these series.
Then the approximation methods that they provide
with polynomials simulating other function can
be used, for example, when an indefinite or
definite integral is to be done, and students
have not yet learned the antiderivative of that
function. We want to convince them that the
polynomial (or power) series is a sequence of
series that converges.
125
Advanced LevelCalculus Examples, especially AP
Calculus
  • A couple of ideas to emphasize the ideas of
    sequences of series with Maclaurin and Taylor
    polynomials.
  • Show simultaneously the graphs of
  • y sin x
  • y

Put increasingly more terms of the series in the
calculator to see how the original function and
its Maclaurin series match.
126
Advanced LevelCalculus Examples, especially AP
Calculus
  • Show, graphically, the limited convergence of
  • y ln (x 1)
  • y

This will provide a good foundation for
understanding the convergence tests and
intervals of convergence ideas which follow.
127
Advanced LevelCalculus Examples, especially AP
Calculus
  • Evaluate using a series for
    y
  • centered at x 9.
  • Show that a Taylor series is easier to evaluate
    (easier uses
  • only simple arithmetic) and can almost be done
    without a
  • calculator at all.

128
Advanced LevelCalculus Examples, especially AP
Calculus
  • James Gregorys method for estimating
    (1671)
  • Since tan -1 (1) which equals the
    value of
  • then a Maclaurin series for the integrand can be
    antidifferentiated and an approximate value can
    be done with ordinary arithmetic.

129
Advanced LevelCalculus Examples, especially AP
Calculus
The sequence of operations that is useful here
is 1.
This last series is accomplished by replacing the
xs by x2s in the first series. This is a very
helpful (and yet unique) feature of Maclaurin
series!
130
Advanced LevelCalculus Examples, especially AP
Calculus
  • Then
  • 0.8349206349 (ok, so I did use a calculator
    to do some of this!!)
  • Then 4 (0.8349206349) 3.33968254

131
Advanced LevelCalculus Examples, especially AP
Calculus
  • We can update what James Gregory did, using
    technology to see whether his series converges.
  • The antiderivative series can be written
    as

On a TI-83/84, put the counters n is L1 (as
far as you want to go) the terms of the
sequence in L2 as (-1)(L11) /(2L1 -1)
the accumulated sum in to L3 as cumSum(L2)
The series does not converge very quickly, so it
is not useful, but it is a valuable method to
teach.
132
Sequences and SeriesFrom Simple Patterns to
Elegant and Profound Mathematics
  • Mathematics is all about expressing patterns,
    numerically and graphically.
  • Patterns can indicate some interesting, usual,
    unusual, and sometimes complicated simulations of
    real phenomena.
  • So sequences and series ought to be as much a
    part of our mathematical language as functions,
    formulas, equations, expressions, and shapes.
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