Chapter 1
Integration Techniques
1.1
Integration Techniques
The Riemann integral is defined to be
Z
n
X
b
f (x) dx =
a
lim
k∆xk→0
i=0
f (c0 )(xi+1 − xi )
where the xi partition (a, b). However the Fundamental Theorem of Calculus allows integrals to be calculated by
Z
a
b
f (x) dx = F (b) − F (a)
where dFdx(x) = f (x). This technique of finding an anti-derivative works well if an anti-derivative exists and can be
found. Consider the following problems.
R
1. Integrate the following by finding an anti-derivative. sin(x2 )2x dx
2. Why will the
R technique for finding the anti-derivative above not work on this integral? Note this function is
integrable. xex dx
Because there are no simple rules for finding all anti-derivatives many techniques must be learned.
1.2
1.2.1
Integration by Parts
Derivation
Consider the following use of the derivative product property and be prepared to explain the steps.
f (x)
= g(x)h(x).
0
= g(x)h0 (x) + g 0 (x)h(x).
Z
0
f (x) dx =
g(x)h0 (x) + g 0 (x)h(x) dx.
Z
Z
g(x)h(x) =
g(x)h0 (x) dx + g 0 (x)h(x) dx.
Z
Z
0
g(x)h (x) dx = g(x)h(x) − g 0 (x)h(x) dx.
Z
Z
udv = uv − v du.
f (x)
Z
1
2
CHAPTER 1. INTEGRATION TECHNIQUES
1.2.2
Example
Consider the following technique called integration by parts. Remember
R
R
Equation 1 udv = uv − v du.
Z
Z
Z
xex dx
=
xex dx
=
=
1.2.3
xex dx
u = x
dv
v
du = dx
Z
xex − ex dx
= ex dx
= ex
xex − ex + C.
Practice
Use integration by parts to find the following anti-derivatives.
Z
1.
x sin x dx
Z
2.
x2 sin x dx
Z
x ln x dx
3.
Z
4.
ln x dx
5. Complete the WebAssign problems.
1.2.4
Example
The practice problems required using integration by parts in different ways. The following examples demonstrates
another use of this technique.
Z
Z
ex sin x dx =
ex sin x dx
=
=
u = ex
dv = sin x dx
v = − cos x
du = ex dx
Z
ex (− cos x) − − cos xex dx
Z
x
−e cos x + ex cos x dx
u = ex
dv
du = ex dx
v
=
=
=
cos x dx
=sin x
Z
−ex cos x + ex sin x − sin xex dx
Z
x
x
−e cos x + e sin x − ex sin x dx.
1.3. INTEGRATION OF TRIGONOMETRIC FUNCTIONS
3
Compare the remaining integral on the right to the original integral on the left of the equation.
Z
ex sin x dx =
Glob =
2Glob =
Glob =
Z
1.2.5
ex sin x dx =
−ex cos x + ex sin x −
Z
Glob = ex sin x dx.
Z
ex sin x dx.
−ex cos x + ex sin x − Glob.
−ex cos x + ex sin x.
1
(−ex cos x + ex sin x).
2
1
(−ex cos x + ex sin x) + C.
2
Practice
Find the following anti-derivatives. Try using integration by parts and another method.
Z
1.
ex sin x dx
Z
2.
ex sinh x dx
3. How many types of problems like this exist?
4. Complete the WebAssign problems.
1.3
1.3.1
Integration of Trigonometric Functions
Motivation
• Integrate
R
sin2 y cos y dy
• Integrate
R
cos57 θ sin θ dθ.
• Note that both were easy to integrate using a u-substitution.
R
• Note that sin2 y cos3 y dy cannot be directly integrated using a u-substitution.
• Identities, however, can be used to convert such products of trigonometric functions into a u-substitution form.
1.3.2
Example
Z
Z
sin2 y cos3 y dy
=
sin2 y cos2 y cos y dy
=
Remember: sin2 y + cos2 y = 1.
Z
sin2 y(1 − sin2 y) cos y dy =
Z
u2 (1 − u2 ) du =
Z
u2 − u4 du =
Look at the u-substitution to see why cos y is left.
Explain how this is used in the next step.
u = sin y.
du = cos y.
Make the substitution.
4
CHAPTER 1. INTEGRATION TECHNIQUES
1 3 1 5
u − u +C
3
5
1
1
sin3 y − sin5 y + C.
3
5
1.3.3
=
Undo the substitution.
Example
Z
sin2 y cos2 y dy
sin2 y = 12 [1 − cos(2y)],
Remember:
cos2 y = 12 [1 + cos(2y)].
Z
1
1
[1 − cos(2y)] [1 + cos(2y)] dy
2
2
Z
1
1 − cos2 (2y) dy
4
Z
1
1
1 − [1 + cos(4y)] dy
4
2
Z
1
1 1
− cos(4y) dy
4
2 2
Z
1
1 − cos(4y) dy
8
1
1
y − sin(4y) + C.
8
4
1.3.4
=
Note nothing is left this time.
=
Which identities were used?
=
Note the identity is needed again.
=
=
=
Motivation
• Integrate
R
tan θ sec2 θ dθ.
• Integrate
R
tan θ sec3 θ dθ.
• Note that both were easy to integrate using a u-substitution.
R
• Note that tan3 θ sec4 θ dθ cannot be directly integrated using a u-substitution.
• Identities, however, can be used to convert such products of trigonometric functions into a u-substitution form.
1.3.5
Example
Z
tan3 θ sec4 θ dθ
=
2
2
Remember:
Z sec θ = 1 + tan θ.
tan3 θ sec2 θ sec2 θ dθ =
Z
tan3 θ(1 + tan2 θ) sec2 θ dθ =
Z
u3 (1 + u2 ) du =
Z
u3 + u5 du =
1 4 1 6
u + u +C
4
6
1
1
tan4 θ + tan6 θ + C.
4
6
Look at the u-substitution to see why sec2 θ is left.
u =
du =
tan θ,
sec2 θ dθ.
Make the substitution.
=
Undo the substitution.
1.4. TRIGONOMETRIC SUBSTITUTION
1.3.6
5
Example
Note the different approach in this example.
Z
tan2 θ csc θ dθ
Z 1.3.7
2 1
dθ
sin θ
Z
sin θ
dθ
cos2 θ
Z
1
sin θ dθ
cos2 θ
Z
1
− 2 du
u
1
+C
u
1
+C
cos θ
sec θ + C
sin θ
cos θ
=
=
What identities were used here?
=
=
u =
du =
cos θ,
− sin θ dθ.
=
=
=
Practice
Z
1. Integrate
Z
2. Integrate
3 tan θ sec6 θ dθ.
tan2 θ sec θ dθ as far as possible.
3. Complete the WebAssign problems.
1.4
1.4.1
Trigonometric Substitution
Motivation
1. Identify the anti-derivative for each of the following. At most one might not be known.
Z
Z
1
1
√
dx
dx
1 + x2
1 − x2
Z
Z
x
1
dx
dx
2
1+x
1 − x2
2. All of these have what type (group) of function as anti-derivative?
3. What is similar about each of the integrands above?
1.4.2
Example
Z
y
p
dy
1 − y2
Since
Let
or
and
=
1 − y 2 looks like 1 − sin2 θ
1 − y 2 = 1 − sin2 θ,
y = sin θ
dy = cos θ dθ.
6
CHAPTER 1. INTEGRATION TECHNIQUES
sin θ
p
cos θ dθ
1 − sin2 θ
Z
y
p
dy
1 − y2
Z
y and dy are substituted everywhere.
sin θ
p
cos θ dθ
1 − sin2 θ
Remember cos2 θ = 1 − sin2 θ.
Z
sin θ
√
cos θ dθ.
cos2 θ
Z
cos θ
sin θ
dθ.
cos θ
Z
sin θ dθ.
Z
=
=
=
=
=
=
sin(θ) =
− cos θ + C
p
− 1 − y 2 + C. See Figure 1.1 below.
y
1
1
y
θ
p
1 − y2
Figure 1.1: Reference for trig substitution
1.4.3
Example
The following looks more complicated, but algebra enables the same technique to work.
Z
1
p
Z
q
1
√
5
2y 2 − 5
1
5( 25 y 2 − 1)
Z
1
q
2 2
5y
dy
=
dy
=
Force the constant to be 1.
dy
=
Factor out the constant.
−1
2 2
y −1
5
2
Let y 2 − 1
5
2 2
y
r5
2
y
5
looks like sec2 θ − 1.
Since
=
sec2 θ − 1.
=
sec2 θ.
=
sec θ.
r
5
sec θ.
2
r
5
sec θ tan θ dθ.
2
y
=
dy
=
1.5. INTEGRATION USING PARTIAL FRACTION DECOMPOSITION
Z
1
p
2y 2 − 5
dy
=
=
=
=
=
=
=
7
r
Z
1
1
5
√
√
sec θ tan θ dθ.
2
5
sec θ − 1 2
Z
1
1
√
√
sec θ tan θ dθ.
2
2
sec θ − 1
Z
1
1
√
√
sec θ tan θ dθ.
2
tan2 θ
Z
1
1
√
sec θ tan θ dθ.
tan θ
2
Z
1
√
sec θ dθ.
2
1
√ ln | sec θ + tan θ| + C
2
r
r
2
1
2y 2 − 5 √ ln y+
+ C. See Figure 1.2 below.
5 2 5
p
2y
sec(θ) = p
5
p
2y
p
5
θ
p
2y 2 − 5
Figure 1.2: Reference for trig substitution
1.5
1.5.1
Integration Using Partial Fraction Decomposition
Motivation
Partial fraction decomposition is an algebraic technique for breaking up rational expressions into smaller pieces.
Read the following examples and answer the questions.
2 4
+
5 7
=
=
=
=
7 2 5 4
· + ·
7 5 5 7
14 20
+
35 35
14 + 20
35
34
.
35
• Where did the 5 and 7 in the denominator’s end up in the final expression?
• What is the name of this process (from elementary school)?
• What does this tell us about the way
43
30
might have resulted from a sum?
The next example repeats this process on a rational expression.
2
3
+
x+3 x−5
=
x−5
2
x+3
3
·
+
·
x−5 x+3 x+3 x−5
8
CHAPTER 1. INTEGRATION TECHNIQUES
=
=
=
2x − 10
3x + 9
+
(x + 3)(x − 5) (x + 3)(x − 5)
(2x − 10) + (3x + 9)
(x + 3)(x − 5)
5x − 1
.
x2 − 2x − 15
• Where did the x + 3 and x − 5 in the denominator’s end up in the final expression?
• What is the name of this process (from algebra 1)?
• To break apart
4x−71
x2 −13x+22
what should be (un)done first?
• What does this tell us about the way
1.5.2
4x−71
x2 −13x+22
might have resulted from a sum?
Example
Notice how the process of adding fraction is worked backwards in the following example.
4x − 71
− 13x + 22
4x − 71
(x − 2)(x − 11)
4x − 71
(x − 2)(x − 11)
4x − 71
(x − 2)(x − 11) ·
(x − 2)(x − 11)
4x − 71
x2
=
First factor the denominator.
=
Write the atomic fractions.
A
B
+
. Clear the denominators.
x − 2 x − 11
A
B
= (x − 2)(x − 11) ·
+
.
x − 2 x − 11
= A(x − 11) + B(x − 2).
=
Finding Coefficients Trick
Note the equation 4x − 71 = A(x − 11) + B(x − 2) has a function (polynomial) on both sides.
1. If these two functions are equal, what will be true about their outputs for x = 2? x = 11? x = 908?
2. If any two functions are equal, at how many inputs will their outputs be the same?
Notice how we use this concept of equal functions to find the coefficients.
4x − 71
=
4(2) − 71
=
x =
−63
=
A(x − 11) + B(x − 2).
2.
A(2 − 11) + B(2 − 2).
−9A.
A =
7.
x =
11.
4(11) − 71
−27
=
=
A(11 − 11) + B(11 − 2).
9B.
B
=
−3.
4x − 71
(x − 2)(x − 11)
=
7
3
−
.
x − 2 x − 11
1.5. INTEGRATION USING PARTIAL FRACTION DECOMPOSITION
9
Finding Coefficients Method
4x − 71
=
4x − 71
=
4x − 71
=
A(x − 11) + B(x − 2).
Ax − 11A + Bx − 2B.
(A + B)x + (−11A − 2B).
Note the equation 4x − 71 = (A + B)x + (−11A − 2B) has a polynomial on both sides.
1. If 4x − 71 = M x + N (two polynomials are equal), what should M and N be?
2. To prove this try the following. Plug in x = 0 on both sides. What does the result state about N ?
3. Differentiate both sides. What does the result state about M ?
4. If any two polynomials are equal, at how many inputs will their outputs be the same?
5. If 4x − 71 = (A + B)x + (−11A − 2B), what will be true about 4 and A + B and −71 and −11A − 2B?
A+B
=
11(A + B)
=
11(4). Scale both sides by 11.
11A + 11B
=
44.
−11A − 2B
= −71. This is the other original equation.
= −27. Added the previous two equations.
9B
= −3.
B
1.5.3
4. This is the original equation.
A + (−3)
=
4.
A
=
7.
Example
Consider the following decomposition of a fraction.
41
28
=
=
=
3 5
+ .
4 7
3
5
+
2
2
7
1
1
5
+ 2+ .
2 2
7
Why the 34 is broken into two fractions will be explained in a later course.
This example breaks up a complex rational expression into atomic rational expressions using this repeated term
technique.
−2x2 − 9x − 16
(x + 2)2 (x − 1)
=
−2x2 − 9x − 16
=
A
B
C
+
.
+
2
x + 2 (x + 2)
x−1
−2x2 − 9x − 16
(x + 2)2 (x − 1)
=
(x + 2)2 (x − 1)
A
B
C
2
.(x + 2) (x − 1)
+
+
x + 2 (x + 2)2
x−1
A(x + 2)(x − 1) + B(x − 1) + C(x + 2)2 .
10
CHAPTER 1. INTEGRATION TECHNIQUES
Use the first trick for finding the coefficients.
−2x2 − 9x − 16
= A(x + 2)(x − 1) + B(x − 1) + C(x + 2)2 .
x = −2.
−6
= −3B.
B
=
2.
x =
1.
−27
=
9C.
C
= −3.
x
=
−16
0.
= A(2)(−1) + B(−1) + C(4).
−16 = −2A − 2 − 12.
A = 1.
−2x2 − 9x − 16
(x + 2)2 (x − 1)
1.5.4
=
1
2
−3
+
+
.
x + 2 (x + 2)2
x−1
Practice
1.
R
7
x−2
3
x−11
dx
2.
R
−2x2 −9x−16
(x+2)2 (x−1)
dx
−
3. Complete the WebAssign problems.
1.5.5
Motivation
• Before starting PFD what must be done with
x2 −x−1
x2 −3x+1 ?
• What is the clue that this needs to be done?
• What happens with
• If we use
1.5.6
A
x−1
3−4x
x2 −4x+13 ?
why does
Ax+B
x2 −4x+13
make sense?
Example
This example demonstrates the use of the quadratic term.
2x2 + 3
(x − 1)(x2 + 4)
=
2x2 + 3
=
2x2 + 3
=
2
2x + 3
=
A
Bx + C
+ 2
.
x−1
x +4
A(x2 + 4) + (Bx + C)(x − 1).
Ax2 + 4A + Bx2 − Bx + Cx − C.
(A + B)x2 + (−B + C)x + (4A − C).
2
= A + B.
(1.1)
0
= −B + C.
(1.2)
3
=
4A − C.
(1.3)
1.6. IMPROPER INTEGRATION
11
B
A+C
−4A − 4C
4A − C
−5C
C
1.5.7
1.
R
= C. From 1.2
=
2. From 1.1
= −8. Scaling above equation.
=
3. This is 1.3
= −5. Adding the above equations.
=
1.
B
=
1. From B = C above.
A+1
=
2. From 1.1
A
=
1.
2x2 + 3
(x − 1)(x2 + 4)
=
1
x+1
+
x − 1 x2 + 4
(1.4)
(1.5)
(1.6)
(1.7)
Practice
2x2 +3
(x−1)(x2 +4)
dx
2. Complete the WebAssign problems.
1.6
1.6.1
Improper Integration
Derivation
Review
For each of the following partitions graph the function, draw rectangles for the associated lower sum, and calculate
the value of the lower sum for f (x) = 1 − x2 .
• {−1, 0, 1},
• {−1, − 12 , 0, 12 , 1},
• {−1, − 34 , − 12 , − 14 , 0, 14 , 12 , 34 , 1},
• {−1, − 78 , − 34 , − 58 , − 12 , − 38 , − 14 , − 18 , 0, 18 , 41 , 38 , 12 , 58 , 34 , 78 , 1}.
1. Describe the difference between the areas represented by your lower sums and the actual area between f (x)
and the x-axis.
2. How are the partitions changing? Consider them in order from the first listed to the last listed.
3. How does the area estimate change as the partitions change?
4. How is the exact area obtained from the area estimates (the sums)?
New Application
g(x) =
1. Begin calculating
R1
0
g(x) dx.
2. Why can’t this calculation be finished?
1
x3/2
.
12
CHAPTER 1. INTEGRATION TECHNIQUES
3. REvaluate each
R 1of the following
R 1 integrals. R 1
1
g(x)
dx,
g(x)
dx,
g(x) dx, .0001 g(x) dx.
0.1
0.01
.001
4. Describe the difference between the areas represented by your integrals and the total area between the curve
and the x-axis.
5. How are the integrals changing? Consider them in order from the first listed to the last listed.
6. How does the area estimate change as the limits of integration change?
7. How might you calculate the actual area from the area estimates?
Another New Application
h(x) = e−x .
1. Begin calculating
R∞
0
h(x) dx.
2. Why can’t this calculation be finished?
3. REvaluate each
the following
R 10 of−x
R 100 −xintegrals.
R 1000 −x
1 −x
e
dx,
e
dx,
e dx, 0
e dx.
0
0
0
4. Describe the difference between the areas represented by your integrals and the total area between the curve
and the x-axis (to the right of the y-axis).
5. How are the integrals changing? Consider them in order from the first listed to the last listed.
6. How does the area estimate change as the limits of integration change?
7. How might you calculate the actual area from the area estimates?
1.6.2
Example
1
Z
0
Z
0
1
1
√ dx
x
1
√ dx
x
Z
lim
a→0+
a→0
=
1.6.3
Practice
Z
1
1.
0
Z
∞
2.
1
Z
1
3.
−1
Z
1
dx
x2
1
4.
−1
1
dx
x2
1
dx
x2
1
dx
x
5. Complete the WebAssign problems.
1
1
√ dx
x
a
√ 1
= lim+ 2 xa
a→0
√
√
= lim+ 2 1 − 2 a
=
2.
Chapter 2
Series
2.1
Sequences
The limits and integrals of previous parts of calculus are defined on functions defined on every real number. There
are also limit and sum problems for which functions are only defined on every integer.
2.1.1
Motivation
Complete the following problems to see examples of sequences.
The Story of Bakbuk the Traveler
Bakbuk the traveler has a bushel of grain. Each night he offers half of the grain he has to a person in exchange for
a room for the night.
1. How much grain does he have when he departs on his voyage?
2. How much grain does he have after the 1st night, 2nd, 3rd, and 4th nights?
3. How much grain does he have after the nth night?
4. As he continues to travel what will happen with the amount of grain he carries? Assume he does not buy more.
The Story of the Inheritance of Bigvai
Bigvai is given a bar of gold that is one unit wide. At the end of his life he cuts out the middle third to pay remaining
debts and passes on the remaining gold to his children. His children at the ends of their lives remove the middle
thirds of each piece of gold to pay remaining debts and pass on the remaining gold to their children. This is repeated
each generation.
Bigvai
Generation 1 Generation 2 Generation 3
1. How many bars does Bigvai have?
2. How many bars do the first, second, and third generations have?
3. How many bars do the nth generation have?
4. How wide is the Bigvai’s bar?
5. How wide are the first, second, and third generation’s bars?
6. How wide are the nth generation’s bars?
13
14
CHAPTER 2. SERIES
7. How much of a bar of gold does the nth generation possess?
8. As the generations pass what is the number of bars approaching?
9. As the generations pass what is the width of the bars approaching?
10. As the generations pass what is the amount of gold approaching?
2.1.2
Definition
Definition 1 (Sequence) A function is a sequence if and only if its domain is a subset of the integers.
Below are four notations for sequences. The second is most common and will be used primarily in this text.
1. {0, 2, 4, 6, 8, . . .}
2. an = 2n, n = 0, 1, 2, . . .
3. f (n) = 2n, n = 0, 1, 2, . . .
4. {2n}∞
n=0
2.1.3
Uses
Because sequences are integer value, the only limit of interest is to infinity.
Definition 2 (Limit of a Sequence) lim an = L if and only if for all > 0 there exists an integer N > 0 such
n→∞
that n > N implies |an − L| < .
Using notes from Calculus 1 if needed, write the formal definition for the following sequence limit.
lim an = ∞.
n→∞
Note that the only difference between these definitions and the limit definitions of other functions is that N has
to be an integer. Because the integers are a subset of the reals and due to a theorem presented in analysis, all the
limit techniques learned previously still apply.
1
an 2, 1, 21 , 14 , 18 , 16
,...
2 4 8 16 32 64
bn 1 , 2 , 3 , 4 , 5 , 6 , . . .
Find the limits of each of these sequences. cn 1e , − e12 , e13 , − e14 , e15 , − e16 , . . .
16 37 79 163
dn 53 , 11
5 , 5 , 5 , 5 , 5 ,...
1
3 7
15 31
fn 2 , − 4 , 8 , − 16
, 32 , − 63
64 , . . .
Why can lim an (the limit of a sequence as the input approaches any number) not exist?
n→k
2.1.4
Examples
Algebra
n2 + 2
n→∞
n3
2
n
2
+ 3
lim
n→∞ n3
n
1
2
lim
+ 3
n→∞ n
n
1
2
lim
+ lim 3
n→∞ n
n→∞ n
lim
=
=
=
=
0.
2.1. SEQUENCES
15
L’Hôpital’s Rule
r
2n + n2
n→∞
2n
r
n2
lim
1+ n
n→∞
2
r
n2
lim 1 + n
n→∞
2
r
n2
lim 1 + lim n .
n→∞
n→∞ 2
=
lim
=
, because radicals are continuous
=
Aside
n2
lim n
n→∞ 2
n2
lim n ln 2
n→∞ e
2n
lim
n→∞ (ln 2)en ln 2
2
lim
n→∞ (ln 2)2 en ln 2
r
2n + n2
lim
n→∞
2n
r
n2
lim 1 + lim n
n→∞
n→∞ 2
=
∞/∞
=
L’Hôpital’s Rule
∞/∞
=
L’Hôpital’s Rule
=
0.
=
=
√
=
1.
1+0
Squeeze
Where is Ω(n) =
sin(nπ/4)
n
going? To begin to understand this sequence try the following.
1. Write the first 8 terms.
2. What appears to be happening?
3. Graph the points (n, Ω(n)).
4. What appears to be happening?
5. Speculate where Ω(n) is going.
−1
≤
sin(nπ/4)
≤ 1.
− n1
≤
sin(nπ/4)
n
≤
1
n.
1
n
≤
n→∞
n→∞
0
≤
n→∞
lim −
n→∞
lim
sin(nπ/4)
n
≤
lim
sin(nπ/4)
n
≤ 0.
lim
n→∞
Now complete the WebAssign problems.
sin(nπ/4)
= 0.
n
lim
1
.
n
16
CHAPTER 2. SERIES
Monotonic Bounded
This example illustrates a concept about sequences which provides
√ a method
√ for determining if a limit exists. Note
this concept does not tell us what the limit value is. Given a0 = 2, an = 2 + an−1 . The goal is to find lim an .
n→∞
1. Evaluate the first 8 terms.
2. Approximate each term using a computer or calculator. Record as many decimal places as possible.
3. Looking at these terms what is the relation between a element and the element before it?
4. Guess the limit.
5. Prove that your guess is correct using the following ideas.
Show that an < 2 for all n, that is, an is bounded above.
1. a0 < 2
√
2
<
2.
√
2+ 2
q
√
2+ 2
<
=
2+2
√
2+2
√
4
=
2.
<
2.
2. a1 < 2
a1
<
3. Show a2 and a3 are less than 2.
4. Show an < 2.
Show that an > an−1 for all n, that is, an is strictly increasing.
1. a1 > a0 .
√
2+ 2
q
√
2+ 2
2. Show a2 > a1
3. Show an > an−1
1. In what direction is an tending?
2. Is there a boundary an does not cross in this direction?
3. Because of this, does an converge?
Now complete the WebAssign problems.
>
>
2, so
√
2.
2.2. SERIES
2.2
2.2.1
17
Series
Definition
Remember, a function is a sequence if and only if its domain is a subset of the integers. A series is a sum of the
elements of a sequence. Below are a sequence and its associated series.
1 1 1 1
Sequence: ak = 21k
2 , 4 , 8 , 16 , . . .
∞
X
1
2k
Series:
1
2
+
1
4
+
1
8
+
1
16
+ ...
k=1
The following steps illustrate the definition of a series.
n
1
an =
2
S1 = a1 . S2 = a1 + a2 . Likewise Sn is the sum of the first n elements of the sequence.
1. Evaluate S1 , S2 , S3 , S4 , S5 .
2. What is the relationship between S1 , S2 , S3 , S4 , S5 and the sum of all (infinite) elements of the sequence.
3. What technique could we use to find the total of the infinite sequence?
4. Find a pattern to the sums you calculated (i.e., find a formula for Sn ).
5. Find S∞ using the pattern.
Definition 3 (Series)
∞
X
k=1
2.2.2
ak = lim Sn where Sn =
n→∞
∞
X
ak .
k=1
Example
bn = 2n for n = 0, 1, 2, . . . Calculate
∞
X
bn .
k=0
First find calculate a few values for Sn .
n
0
1
2
3
4
Sn
1
= 1.
1+2
= 3.
1+2+4
= 7.
1+2+4+8
=15
1 + 2 + 4 + 8 + 16=31
The Sn values are each one less than the next value of bn , i.e., 2 − 1, 4 − 1, 8 − 1, 16 − 1, 32 − 1, . . . Thus
∞
X
bn
=
2n
=
−1
=
k=0
∞
X
k=0
n+1
lim 2
n→∞
∞.
If the series value exists and is finite, the series is called convergent otherwise it is called divergent.
18
CHAPTER 2. SERIES
2.2.3
Warning
an = (−1)n , n = 0, 1, 2, . . .
Attempt to find this sum using the same technique as the previous example. The steps are below.
1. Calculate Sn for n = 1, 2, 3, 4, 5, 6.
2. Write a pattern for Sn
3. What kind of sequence is this (describe its action)?
4. Does this limit exist?
5. Does the series value exist?
2.3
2.3.1
Series Techniques
Lower Limit
1. What is
∞
X
1?
k=1
2. What is
∞
X
1?
k=2
3. What is
∞
X
1?
k=3
4. What is
∞
X
1?
k=50
5. If we only care whether a series is finite or infinite do we care about the first 2 terms? 5 terms? 500 terms?
2.3.2
Algebra
1. Write out
P3
2. Write out
P3
i=1
i and
i=1 (i
P3
i=1
i2 . Do not add the terms.
+ i2 ).
3. Compare this to the previous two.
P∞
4. What might we do with i=1 1i −
1
i2
5. What condition would this require?
2.4
2.4.1
nth Term Divergence
Derivation
Suppose an ≥ 2 for all k.
1. What is the
∞
X
2?
n=1
2. As a result what is the
∞
X
n=1
an ?
?
2.5. GEOMETRIC SERIES
19
We will now discover a test for infinite (divergent) series. Suppose lim ak = 3.
k→∞
1. Because the limit is 3, the distance |ak − 3| approach what?
2. Will all the sequence values ak eventually be larger than 2?
∞
X
3. What is the
2?
n=1
4. Therefore what must
∞
X
ak be?
n=1
2.4.2
nth term divergence test
If lim an 6= 0, then
n→∞
2.4.3
∞
X
an diverges.
k=0
Practice
Use the nth term divergence test to determine if the series is definitely divergent or not.
1. an = 1 −
1
2n .
2. bn =
1
1+n .
3. cn =
1
2+3n .
4. Complete the WebAssign problems.
2.5
Geometric Series
Definition 4 (Geometric) A series is geometric if it can be written
P∞
k=0
ark where r is a constant.
The following steps calculate the partial sum Sn .
sn
= a + ar + ar2 + . . . + arn−1 .
rsn
= ar + ar2 + ar3 + . . . + arn .
sn − rsn
=
(ar + ar2 + ar3 + . . . + arn )
sn (1 − r)
sn
(a + ar + ar2 + . . . + arn−1 ) −
= a − arn .
= a(1 − rn ).
a(1 − rn )
.
=
1−r
Thus the sum of a geometric series is given by the following.
∞
X
ark
=
k=0
lim
n→∞
n
X
ark
k=0
a(1 − rn )
n→∞
1−r
a
=
.
1−r
=
However, there are restrictions.
lim
20
CHAPTER 2. SERIES
1. The following are all geometric series.
n
∞ X
27
(a)
133
n=0
∞
X 27 n
(b)
27
n=0
n
∞
X
133
(c)
27
n=0
2. Which of the above will converge?
3. Based on the derivation provided what is the sum of the convergent series above?
Now complete the WebAssign problems.
2.6
2.6.1
Integral Series
Derivation
• (Review) If ak ≤ 5 and ak ≥ ak−1 does the sequence converge?
• Suppose ak = 1/k 2 Note ak = 1/k 2 ≥ 0 for all k ≥ 1.
– Which is bigger a1 or a1 + a2 ? a1 + a2 or a1 + a2 + a3 ?
– Remember Sn is the sum of the first n terms of ak .
– Which is bigger Sn or Sn−1 ?
– Thus the sequence of Sn is doing what?
• We need one more piece to determine convergence or divergence.
1
1/4
1/9
1
2
3
4
5
6
7
8
9
10
Figure 2.1: Integrals and Series (part 1)
• What in Figure 2.1 represents
P∞
• What in Figure 2.1 represents
R∞
1
k=2 k2 ?
1
• Which is bigger?
• Using this compare
R∞
• What is 1 x12 dx?
P∞
1
k=2 k2
and
1
x2
R∞
1
dx?
1
x2
dx.
2.7. COMPARISON TESTS
21
1
1/2
1/3
1/4
1
2
3
4
5
6
7
8
9
10
Figure 2.2: Integrals and Series (part 2)
• Does this series converge or diverge?
P∞
• What in Figure 2.2 represents k=1 k1 ?
R∞
• What in Figure 2.2 represents 1 x1 dx?
• Which is bigger?
R∞
• What is 1 x1 dx?
• Using this compare
P∞
1
k=1 k
and
R∞
1
1
x
dx.
• Does this series converge or diverge?
2.6.2
Summary
∞
X
Z
ak and
k=1
2.6.3
•
•
•
∞
a(x) dx converge or diverge alike.
1
Practice
Z
∞
1
dx
x3/2
∞
1
dx
x1
∞
1
dx
x1/2
1
Z
1
Z
1
• Based on your calculations conjecture for what values of p the following series converges and diverges
• Complete the WebAssign assignment.
2.7
2.7.1
Comparison Tests
Motivation
1. Decide which of the series types in Figure 2.3 the series below look like.
(a)
∞
X
7n
3n − 1
n=1
∞
X
1
.
np
n=1
22
CHAPTER 2. SERIES
(b)
(c)
(d)
∞
X
1
2+1
n
n=1
∞
X
7n
3n + 1
n=1
∞
X
1
3−1
n
n=2
2. Do the series they look like converge or diverge?
3. Based on this information conjecture whether these series converge or diverge?
2.7.2
Examples
Example 1
∞
X
1
1
looks
like
. Note n2 + 1 > n2 so n21+1 <
2+1
2
n
n
n=1
n=1
convergent p-series. Thus this series must converge as well.
∞
X
1
n2 .
Thus
∞
∞
X
X
1
1
1
<
.
Also
note
is a
2+1
2
2
n
n
n
n=1
n=1
n=1
∞
X
Example 2
∞
X
∞ n
∞
∞ n
X
X
X
7n
7n
7
7
7n
7n
n
n
looks like
>
. Note 3 − 1 < 3 so 3n −1 > 3n . Thus
.Also note that
n−1
n−1
3
3
3
3
n=1
n=1
n=1
n=1
P∞
7 n
is a divergent geometric series. Thus this series must diverge as well.
n=1 3
2.8
2.8.1
Limit Comparison
Motivation
1. Attempt to determine if the series
P∞
1
n=2 n3 −1
converges or diverges by comparing it to
P∞
1
n=2 n3 .
2. What difficulty do you encounter.
3. The next method addresses this issue.
2.8.2
Derivation
∞
X
∞
X
an
= c. According to the definition of a limit abnn − c < n→∞ bn
n=1
n=1
an
= c.
for large enough n. We will re-arrange this statement to compare an and bn . Suppose lim
n→∞ bn
Consider the positive series
an and
bn . Suppose lim
− <
c−<
(c − )bn <
an
bn −
an
bn
an
Geometric
∞
X
arn
n=0
c
<
<c+
< (c + )bn
p-series
∞
X
1
np
n=1
Figure 2.3: Known Series
2.9. ROOT & RATIO TEST
23
∞
X
n=1
∞
X
an
<
an
<
(c + )bn .
∞
X
(c + )bn .
n=1
an
<
(c + )
n=1
So if
P∞
n=1 bn
converges, then
an
= c.
Again suppose lim
n→∞ bn
P∞
n=1
an also converges.
an
bn −
an
bn
c−<
(c − )bn <
∞
X
n=1
∞
X
c
n=1 bn
2.8.3
diverges, then
P∞
n=1
<
<c+
an
< (c + )bn
an
>
an
>
(c − )bn .
∞
X
(c − )bn .
n=1
an
>
n=1
P∞
bn .
n=1
− <
So if
∞
X
(c − )
∞
X
bn .
n=1
an also diverges.
Example
Compare
∞
X
∞
X
1
1
to
.
3
n −1
n3
n=2
n=2
1
n3 −1
1
n→∞
n3
3
lim
lim
n
n→∞ n3 − 1
1 3
3n
lim 1 n 3
n→∞ 3 (n − 1)
n
lim
n→∞
1
1 − n13
=
=
=
=
1.
∞
X
1
1
converges
like
.
3
n −1
n3
n=2
n=2
Now complete the WebAssign assignment.
Thus the series
2.9
2.9.1
∞
X
Root & Ratio Test
Notation
1. Find the first 10 derivatives of x10 . Do not expand the coefficient (you will miss the point).
2. With each succeeding derivative how does the coefficient change (what do you do to it)?
3. The tenth derivative is just a coefficient produced by doing what?
4. Because this is a common operation we call it ‘factorial.’
24
CHAPTER 2. SERIES
1. Write
5!
3!
n! = n(n − 1)(n − 2)(n − 3) . . . (2)(1).
using the definition of factorial above.
2. Divide the obvious parts of the expression to simplify.
3. Do the same with
4. What is
2.9.2
7!
5! .
n!
(n−5)! ?
Derivation
1. Suppose an ≥ 0 for all n.
2. If
an+1
an
< 1 what is the sequence doing?
3. If
an+1
an
> 1 what is the sequence doing?
an+1
< 1 what is the sequence doing?
an
an+1
5. If lim
> 1 what is the sequence doing?
n→∞ an
4. If lim
n→∞
6. Since the sequence is positive, lim an > (what number)?
n→∞
7. Because of these facts we can apply what sequence technique to prove convergence?
2.9.3
Ratio Test
If
∞
X
an is a positive series, the series
n=1
2.9.4
converges
when
diverges
when
confuses us
when
lim
an+1
<1
an
lim
an+1
>1
an
n→∞
n→∞
an+1
=1
n→∞ an
lim
Example
∞
X
n
en
n=1
lim
n→∞
an+1
an
n+1
en+1
n→∞ nn
e
n
lim
(n + 1)e
n→∞ nen+1
n+1
lim
n→∞ ne
1
lim
n→∞ e
lim
=
=
=
=
=
<
L’Hôpital’s Rule
1
e
1.
2.10. NON-POSITIVE SERIES
25
Thus the series converges.
2.9.5
Root Test
If
∞
X
an is a positive series, the series
n=1
2.9.6
converges
when
diverges
when
confuses us
when
lim
√
n
an < 1
lim
√
n
an > 1
lim
√
n
an = 1
n→∞
n→∞
n→∞
Example
∞ n
X
1
n=1
n
s n
1
lim n
n→∞
n
n 1/n
1
lim
n→∞
n
1
lim
n→∞ n
=
=
=
0
<
1.
Convergent
2.10
Non-Positive Series
2.10.1
Case 1: Review
an = (1/2)n , n = 1, 2, 3, . . .
15
7
31 32
15 16
3
7 8
16
8
4
3 4
1
1 2
2
0
Figure 2.4: Case 1
31
32
26
CHAPTER 2. SERIES
1. Complete the following table. Note Sn =
an
Sn
Pn
i=1
ai .
2. Compare this sequence to cases 2-4. All the terms an are what kind of numbers?
3. Because of this each new sum is what in comparison to the previous one?
4. In order to converge each term must be what size in comparison to previous ones?
5. How do the an values correspond to the arrows above?
6. How do the Sn values correspond to the axis labels above?
2.10.2
Case 2
bn = −
1
, n = 1, 2, 3, . . .
2n
1. Complete the following table.
bn
Sn
2. All the terms bn are what kind of numbers? Compare to cases 1, 3, and 4.
3. Because of this each new sum is what in comparison to the previous one?
4. How is this different from the previous case?
5. Sketch an arrow diagram for this example.
2.10.3
Case 3
cn =
(−1)n+1
, n = 1, 2, 3, . . .
n
1. Complete the following table.
cn
Sn
2. What do the terms cn do? Compare to cases 1, 2, and 4.
3. Sketch an arrow diagram for this example.
4. What size is each term in comparison to the previous?
5. How is that seen in the arrow diagram?
6. What is lim |cn |?
n→∞
7. Do the sums appear to be converging?
8. How is this seen in the arrow diagram?
2.11. ALTERNATING SERIES
2.10.4
27
Case 4
1
dn = (−1)n+1 1 +
, n = 1, 2, 3, . . .
n
1. Complete the following table.
dn
Sn
2. What do the terms dn do? Compare to cases 1-3.
3. Sketch an arrow diagram for this example.
4. What size is each term in comparison to the previous?
5. What is lim |dn |?
n→∞
6. To what do all the downward pointing arrows seem to converge?
7. To what do all the upward pointing arrows seem to converge?
8. What is the difference between these two values?
9. Compare this answer to the limit.
10. Do the sums appear to be converging?
2.11
Alternating Series
Use the examples in the previous section to answer these questions.
1. What was different between the last two series examples (Case 3 and Case 4)?
2. Why did one converge when the other did not?
3. Use these ideas to determine if en =
A series
∞
X
bn is alternating if
n=a
The alternating series
∞
X
∞
X
n=a
(−1)n
ln n
bn =
∞
X
converges.
(−1)n an where an > 0 for all n.
n=a
n
(−1) an is convergent if
n=a
lim an
n→∞
an+1
2.12
Absolute Convergence
2.12.1
Motivation
=
0.
≤ an .
∞
X
cos n
n2
n=1
1. Write out the first 5 terms (calculators might be helpful).
2. Is this series positive?
3. Is this series alternating?
4. How do we handle this series?
(2.1)
(2.2)
28
CHAPTER 2. SERIES
2.12.2
Derivation
1. Which is bigger 1 +
1
2
+
1
4
+
1
8
or 1 −
1
2
1
4
+
− 18 ?
2. Why?
1
2
3. If 1 +
2.12.3
+
1
4
+
1
8
+ . . . converges should 1 −
1
2
+
1
4
−
1
8
+ . . . converge?
Absolute Convergence Test
∞
X
cos n
n2
n=1
A series
∞
X
an is absolutely convergent if
n=a
∞
X
n=a
|an | is convergent.
Consider
∞ X
cos n 2 =
n
n=1
∞
X
| cos n|
n2
n=1
<
∞
X
1
2
n
n=1
which is finite.
The series is absolutely convergent
2.12.4
Terminology
• If a series is absolutely convergent, it is convergent.
• A series
• A series
∞
X
n=a
∞
X
an is absolutely convergent if
∞
X
n=a
|an | is convergent.
an is conditionally convergent if it is convergent but not absolutely convergent.
n=a
Absolute convergence is a series test not a type of series as this terminology may seem to imply.
Chapter 3
Power Series
3.1
Definition
Definition 5 (Power Series) A series is a power series iff it looks like
∞
X
cn xn = c0 + c1 x + c2 x2 + c3 x3 + . . . .
n=0
The following questions connect the concept and calculations of power series with simple, previous concepts.
1. What are power series?
(a) What type of function is 1 − 3x + 7x2 ?
(b) What type of function is 1 − 3x + 7x2 − 11x3 ?
(c) What type of function is 1 − 3x + 7x2 − 11x3 + 15x4 + 17x7 − 3x10 + 4x11 − 256x13 −
311 27
55 x ?
(d) Look at the definition of power series again. What type of function is a power series?
2. What do we calculate?
(a) Test if the following converge or diverge.
∞ n
X
3
i.
5
n=0
∞
X 7 n
ii.
5
n=0
∞
X x n
iii.
5
n=0
(b) For what values of x will this series converge?
n
P∞
(c) For what values of x is f (x) = n=0 x5 defined?
(d) What do we call this (algebra terminology)?
3.2
Power Series Domains
To determine the values at which a particular power series converges, the following theorem will be needed in addition
to the previously learned series tests.
n
n
1. Compare 13 and 23 .
P∞
x n
2.
converges for x = 2. Why?
n=1 3
n
P∞
3. What does the previous line imply (directly) about n=1 13 converging or diverging?
29
30
CHAPTER 3. POWER SERIES
4. What does this imply (directly) about
P∞
n=1
2.5 n
3
5. Note the use of limits in the following proof.
Theorem 1 (Radius of Convergence) If
P
cn xn converges for x = b, then it converges for all |x| < b.
Proof:
P
Suppose
cn xn converges for x = b.
n
lim cn b = 0.
Which series test tells us this?
n→∞
n
By definition
nofn a limit there exists N such that n > N implies |cn b | < 1.
n
|cn x | = cn bbnx n
|cn xn | = |cnbn | xb n
|cn xn | < xb .
If
P|x| <n b then
P this is convergent.
cn xP< cn bn .
cn xn is convergent.
Thus
3.2.1
Why?
Next follows some algebra.
Which test is this?
Why?
Which test is this?
Example
Using the ratio test
xn+1
xn
÷
n→∞ (n + 1)2
n2
n+1
n2
x
·
lim
n→∞ (n + 1)2 xn
n2
lim x ·
n→∞
(n + 1)2
2n
lim x ·
n→∞
2(n + 1)
lim x
lim
n→∞
=
=
∞/∞
=
L’Hôpital’s Rule
∞/∞
=
L’Hôpital’s Rule
=
x.
Thus by the ratio test p(x) converges when |x| < 1, that is the domain contains (radius of convergence) (−1, 1).
However, this does not determine whether −1 or 1 are in the domain. These must be checked separately.
p(1)
p(−1)
=
∞
X
1n
n2
n=1
=
∞
X
1
.
n2
n=1
=
∞
X
(−1)n
.
n2
n=1
The first converges, because it is a p-series. The second converges by the absolute convergence test. Thus the domain
(called the interval of convergence) is [−1, 1].
3.3
Generating Power Series
One of the uses of power series is as an alternate representation of known functions. The following sections demonstrate how such a power series can be generated.
3.3. GENERATING POWER SERIES
3.3.1
31
Derivation
x
Z
f 0 (t) dt
c
Z
= f (x) − f (c). Note how this definite integral is rearranged.
x
sin x − sin 0.
Z x
cos t dt. Note how integration by parts is used below to continue this integral.
= sin 0 +
cos t dt =
0
sin x
0
u = cos t
dv = dt
du = − sin t dt v = t − x
Z x
x
= 0 + cos t(t − x)|0 −
− sin t(t − x) dt
0
Z x
sin t(t − x) dt
= [cos x(x − x)] − [cos 0(0 − x)] +
0
Z x
sin t(t − x) dt Note the specific choice for u and dv made to continue integration by parts.
= x+
0
u = sin t
dv = (t − x) dt
du = cos t dt v = 12 (t − x)2
x Z x
1
1
= x + sin t (t − x)2 −
cos t (t − x)2 dt
2
2
0
0
Z x
1
1
1
= x + sin x (x − x)2 − sin 0 (0 − x)2 −
cos t (t − x)2 dt
2
2
2
0
Z x
1
2
= x−
cos t (t − x) dt
2
0
Complete the following steps to understand how this process works.
1. Continue this integration by parts process for four more steps. You may not guess and skip.
2. How long could this integration by parts problem continue?
3. If this process could be “finished,” what would the result be?
4. What is sin(1) approximately?
3.3.2
Calculation
Definition 6 A function is smooth of order n if its nth derivative exists and is continuous.
1. sin x is infinitely smooth. What other functions are infinitely smooth (at least where they are defined)?
2. The coefficient of each term (power of x) came from what calculation?
For a sufficiently smooth function
f (x)
(x − c)2 00
(x − c)3 000
f (c) +
f (c) + . . .
2!Z
3!
x
n
n
(x − c) (n)
(t − x) (n+1)
+
f (c) + (−1)n
f
(t)dt.
n!
n!
c
Z x
n
X
(t − x)n (n+1)
(x − c)j (j)
=
f (c) + (−1)n
f
(t)dt
j!
n!
c
j=0
= f (c) + (x − c)f 0 (c) +
= Pn (x) + Rn (x)
Note the following.
32
CHAPTER 3. POWER SERIES
• The polynomial portion Pn (x) is called the Taylor polynomial about c of degree n.
• Rn (x) is referred to as the remainder.
• If c = 0 these are also called Maclaurin series.
• If the process is (can be) continued infinitely it is called a Taylor series.
1. Calculate the Taylor series for sin x with c = 0.
2. Calculate the Taylor series for cos x with c = 0.
3. Calculate the Taylor series for ex with c = 0.
4. Calculate the Taylor series for ln x with c = 1. Why is c = 1 instead of c = 0?
5. Calculate the following Taylor series.
(a) x1/2 with c = 1.
(b) x1/3 with c = 1.
(c) (1 + x)−1 with c = 0.
(d) xr with c = 1.
6. Complete the WebAssign assignment.
3.4
3.4.1
Generating Power Series Efficiently
Motivation
The following problem illustrates why directly using the Taylor method to generate power series is not always the
most efficient means to do so. Generate the Taylor series (infinite) for f (x) = sin(x2 ) about a = 0. List at least the
first two, non-zero terms. Use of a computer is recommended for the steps. You may not let the computer generate
the whole series, however.
3.4.2
Techniques
Memorize these four series for future use.
sin x
=
x
x3
x5
x7
x9
x11
−
+
−
+
−
+ ...
1!
3!
5!
7!
9!
11!
cos x
=
1−
x2
x4
x6
x8
x10
+
−
+
−
+ ...
2!
4!
6!
8!
10!
ex
=
1+
x2
x3
x4
x5
x
+
+
+
+
+ ...
1!
2!
3!
4!
5!
ln x =
(x − 1) (x − 1)2
(x − 1)3
(x − 1)4
−
+
−
+ ...
1
2
3
4
3.4. GENERATING POWER SERIES EFFICIENTLY
Algebra
1. Evaluate sin(x2 ) using the power series provided above.
2. Compare this to your previous result.
3. Generate the power series for sin(x3 )/x using a similar method.
4. Generate the power series for (1 + x)ex .
5. Generate the power series for
ex
1+x .
Derivatives
1. What is the derivative of sin x?
2. Copy the Taylor series for sin x.
3. Take the derivative of this Taylor series.
4. Identify this Taylor series in the provided list.
5. Does the result make sense?
Integrals
1. What is the anti-derivative of ex ?
2. Integrate the power series for ex .
3. Use e0 = 1 to calculate +C in the integral.
4. Integrate sin(x2 ).
5. Write the Taylor series for (1 + x)−1 about c = 0.
6. Write the Taylor series for (1 + x2 )−1 about c = 0.
7. Integrate this Taylor series.
8. Write the Taylor series for arctan x about c = 0.
Complex Series
Calculate the following using their Taylor series, then compare.
1. eix
2. cos x
3. i sin x
33
34
CHAPTER 3. POWER SERIES
3.5
Taylor Polynomial Error
When a Taylor polynomial is used instead of a Taylor series, the value produced is not exact. Consider the following
using T7 (x), a 7th degree Taylor polynomial for sin x with c = 0.
f (x)
T7 (x)
=
=
sin x
3
x − x3! +
x5
5!
−
x7
7!
1. What is T7 (1)?
2. How close is this value to sin 1? Note your calculator does not know, so don’t ask it.
3. But wait, your calculator claims to know. What is it actually giving you?
2
1
-Π
-
3Π
4
-
Π
-
2
Π
Π
Π
3Π
4
4
2
4
Π
-1
-2
f (x)
T3 (x)
T5 (x)
T7 (x)
=
=
=
=
sin x
3
x − x3!
3
x − x3! +
3
x − x3! +
x5
5!
x5
5!
x7
7!
−
black
red
blue
green
Use the examples above to answer the following.
1. Which Taylor polynomial looks the least like sin x?
2. Which Taylor polynomial looks the most like sin x?
3. Which Taylor approximation would be better than all three of these?
4. Where is the red Taylor polynomial (T3 ) closest to sin x?
5. Where is the blue Taylor polynomial (T5 ) closest to sin x?
6. All of these Taylor polynomials were about c = 0.
7. If we wanted to estimate sin(π + 1) what might we do?
3.5.1
Error Estimation
Remember
f (x)
=
n
X
(x − c)j
j=0
|
f (x) ≈
f (j) (c) + (−1)n
j!
|
{z
}
Pn (x)
n
X
(x − c)j
j=0
|
j!
{z
Pn (x)
f (j) (c) .
}
Z
c
x
(t − x)n (n+1)
f
(t)dt
n!
{z
}
Rn (x)
3.5. TAYLOR POLYNOMIAL ERROR
35
For a sufficiently smooth function f (x)
f (x) =
n
X
(x − c)j
f (j) (c) +
j!
j=0
{z
}
|
Taylor Polynomial
f (n+1) (z)
(x − c)n+1
(n + 1)!
{z
}
|
Remainder Term
for some z between c and x.
2
3
4
T4 (x) = 1 + x + x2! + x3! + x4! is the 4th degree Taylor polynomial for ex with c = 0. The procedure below estimates
the error for T4 (2).
According to the remainder theorem for some z between c = 0 and x = 2,
f (n+1) (z)
(x − c)n+1
(n + 1)!
f (5) (z)
(2 − 0)5
5!
ez 5
2
5!
= Rn (x).
=
, because n = 4 and x = 2
≤
e2 5
2 , because ex is increasing
5!
1.97041
=
Thus the error in our estimate of e2 is less than or equal to 1.97041.
3.5.2
Error control
If error is calculated after the fact, it may be too large to be useful. The following example determines what degree
Taylor polynomial is required to approximate e2 with an error less than 0.1.
Again according to the remainder theorem for some z between c = 0 and x = 2,
f (n+1) (z)
(x − c)n+1
(n + 1)!
f (n+1) (z)
(2 − 0)n+1
(n + 1)!
ez
2n+1
(n + 1)!
e2
2n+1
(n + 1)!
< 0.1.
< 0.1, because x = 2.
< 0.1,
< 0.1, because ex is increasing.
• Since we don’t know an exact value for e we will replace it with a nearby integer, 3.
• Note
e2
n+1
(n+1)! 2
• Also if
<
32
n+1
(n+1)! 2
32
n+1
,
(n+1)! 2
< 0.1, then
because e < 3.
e2
n+1
(n+1)! 2
< 0.1 since it is even smaller.
• Thus we need to find a value for n that satisfies
32
n+1
(n+1)! 2
< 0.1.
• Use a computational device to plug in numbers n until the result is less than 0.1.
36
CHAPTER 3. POWER SERIES
Chapter 4
Analytic Geometry
4.1
4.1.1
Parametric Form
Motivation
This example will demonstrate limitations of the expressing curves as functions with scalar inputs and outputs (e.g.,
y = f (x).). Consider the flight in the provided animation.
1. What is the shape?
2. What is the perimeter of the shape?
3. How much distance did the plane cover?
4. How far did it get?
5. What is an equation for this shape?
6. Is this a function (remember formal definition)?
7. From the equation can you tell how many times the plane flew around the circle?
8. From the equation can you tell what direction the plane flew?
9. Do we need a better model?
4.1.2
New Notation
Parametric form expresses curves as functions of one variable with vector outputs (e.g., P (t) = (x, y)). Use the
following example to practice graphing in parametric form.
x = cos t.
y = sin t.
t ∈ [0, 2π].
Plot the curve by first completing the table below, then plotting the points ((x, y) coordinates).
t 0
x
y
π
4
π
2
3π
4
π
37
5π
4
3π
2
7π
4
2π
38
CHAPTER 4. ANALYTIC GEOMETRY
4.1.3
Converting
While not frequently useful, it is possible to convert from parametric form to standard function form in some cases.
This is used below to demonstrate the greater flexibility of parametric form. Consider x = t2 + 5, y = t4 + 10t2 + 25,
t ∈ [−10, 10]. Suppose we wanted to go back to standard functional notation.
Note
y
= t4 + 10t2 + 25
(t2 + 5)(t2 + 5)
=
= x2 . For,
∈
t
[0, 10],
∈
x
[5, 105].
Consider x = cos θ, y = sec θ, θ ∈ (−π/2, π/2). Suppose we wanted to go back to standard functional notation.
Note
y
=
=
=
∈
θ
∈
x
4.1.4
A
D
G
sec θ
1
cos θ
1
. For
x
(−π/2, 0],
(0, 1].
Flexibility
x
y
t
x
y
t
x
y
t
=
=
∈
=
=
∈
=
=
∈
2t + 3,
10t + 18,
(−∞, ∞)
3t2 ,
15t2 + 3,
(−∞, ∞)
sin t,
sin2 t + 4,
(−∞, ∞)
B
E
x
y
t
x
y
t
=
=
∈
=
=
∈
2t,
4t2 + 4,
(−∞, ∞)
t + 1,
t2 + 2t + 5,
(−∞, ∞)
C
F
x
y
t
x
y
t
= cos(2t),
= cos2 (t),
∈ (−∞, ∞)
= 5 − t,
= 28 − 5t,
∈ [0, ∞)
1. Graph each function.
2. Determine the direction for each function.
3. Determine the number of times the curve is traversed.
4. Identify the shape.
5. Convert from parametric form to standard function notation.
4.2
4.2.1
Parametric Derivatives
Piecewise Derivatives
For curves expressed as scalar functions the first derivative indicates where the curve is increasing (going up) and
decreasing (going down). Because the x and y portions are separate in parametric the (two) derivatives provide
different information. Use the following example to discover what information they provide.
x = cos(2t).
y = sin(3t).
t ∈ [−π, π].
4.2. PARAMETRIC DERIVATIVES
39
1. Graph this curve. Use technology.
2. Calculate
dx
dt
and
3. Calculate where
dy
dt .
dx
dt
= 0.
4. Label these points on the graph.
5. Repeat these two steps with
dy
dt .
6. What do these derivatives indicate about the curve?
4.2.2
Slope in Parametric
Derivation
1. Remember the following mantras about slope
• Slope is rise over run.
• Rise is change in height (y).
• Run is change in distance (x).
• Thus slope is change in y over change in x.
2. What calculus concept expresses “change in”?
3. Using this, how can we calculate “change in x” for x = cos(2t)?
4. Using this, how can we calculate “change in y” for y = sin(3t)?
5. If slope is change in y over change in x how do we use the previous two parts to calculate slope?
Example
x = cos(2t).
y = sin(3t).
t ∈ [−π, π].
What is the slope at t = π/4?
dy
dt
dx
dt
dy
dx
=
= −2 sin(2t).
=
=
3 cos(3t) −2 sin(2t) 3 cos(3t).
=
t=π/4
=
dy/dt
dx/dt
3 cos(3t)
.
−2 sin(2t)
√
−1/ 2
−2
1
√ .
2 2
40
CHAPTER 4. ANALYTIC GEOMETRY
4.2.3
Practice
x = cos(2t).
y = sin(3t).
t ∈ [−π, π].
1. Calculate the slope at t = −π/6, π/6, and π/3.
2. Evaluate
dx
dt
and
dy
dt
at t = −π/2 and π/2.
3. What is the slope of the tangent at these points?
4. What is happening at these points?
5. Complete the WebAssign assignment.
4.3
Modeling
Parametric form is convenient for modeling a variety of physical motions. The following is a classic example. First
view the illustration provided then read and answer the questions. Do not adjust the parameters yet. The model
(parametric function) will be completed in class.
1. Why does the light (red dot) move?
2. What is the shape of the wheel?
3. What is the parametric form of this shape? You know this from trigonometry.
4. This parameterization expresses points by giving the x and y distance from what part of the wheel?
5. What is the y value of this part at all times (suppose this is a 27 in diameter wheel)?
6. We want to move this part of the wheel up by this value. How can we do this arithmetically?
7. Why does the hub move forward?
8. How far does the hub move forward if the light has gone from the ground all the way around back to the
ground?
9. How far does the hub move forward if the light moves from the ground to straight above?
10. In general how far does the hub move as the wheel turns?
11. What is the x value of this part at all times?
12. We want to move this part of the wheel right by this value. How can we do this arithmetically?
13. In this animation where did the light begin?
14. Does our representation need to be modified for this starting point?
15. How can the starting point be adjusted?
4.4
4.4.1
Polar Coordinates
Illustrated
Use the provided illustrations to answer the following.
1. Use the Cartesian illustration to move the skidsteer from (0, 0) to (4, 4).
2. Note how the skidsteer drives like it is on roads.
3. Use the Polar illustration to move the skidsteer from (0, 0) to (5, π/4).
4. Experiment to find more ways to get to the same location.
4.4. POLAR COORDINATES
4.4.2
41
Graphing
Old Problem
y = 1 + sin x. x ∈ [0, 2π].
1. Fill out the table below.
2. Graph the points (Cartesian coordinates).
3. In what direction (order) did you plot the points.
x
y
0
π/4
π/2
3π/4
π
5π/4
3π/2
7π/4
2π
New Problem
r = 1 + sin θ.
1. Fill out the table below.
2. Graph the points (polar coordinates).
3. In what direction (order) did you plot the points.
θ
r
0
π/4
π/2
3π/4
π
5π/4
3π/2
7π/4
2π
Another New Problem
r = sin 2θ.
1. Fill out the table below.
2. Graph the points (polar coordinates).
3. In what direction (order) did you plot the points.
θ
r
4.4.3
0
π/4
π/2
3π/4
π
5π/4
3π/2
7π/4
2π
Advanced Polar Graphing
r = 1 − cos θ.
1. Find the roots.
2. Where is any point of the form (r; θ) = (0, θ)?
3. Find where the function is increasing and decreasing.
4. What does it mean for r to increase?
5. What would you expect to happen between roots in terms of increasing/decreasing?
6. Graph this curve using this information.
42
CHAPTER 4. ANALYTIC GEOMETRY
4.5
Polar Slopes
Slope is described as rise over run. In Section 4.2 this was calculated using parametric form. In this section it will
be calculated from polar coordinates.
1. Graph the polar point (1; π/4).
2. Sketch the line segment from the origin to this point.
3. Calculate the x and y coordinates of this point.
4. Calculate the x and y coordinates of r = cos(2θ) using the same technique.
5. Calculate
dx
dθ
6. Calculate
dy
dx .
and
dy
dθ .
7. Determine where the slope of the tangents are 0.
4.6
Practice
1. Convert the following points from polar to Cartesian form. (0; π/2), (1; π/2), (2; 3π/4).
2. Convert the following points from Cartesian to polar form. (1, 1), (2, 0), (3, 4).
3. Graph the following curves expressed in polar coordinates.
(a) r = cos(θ).
(b) r = cos(3θ).
(c) r = 1 + sin(θ).
4. Complete the WebAssign assignment.
4.7
Polar Area
4.7.1
Derivation
Area in Cartesian
f (x) = 1 − x2 , x ∈ [−1, 1].
1. Graph the function on this domain.
2. Draw rectangles to approximate the area as done in Calculus I.
3. In Cartesian which variable is the input (domain)?
4. How is this used in calculating the area of the rectangles?
5. Where does this show up in the area integral?
6. In Cartesian which variable is the output (codomain)?
7. How is this used in calculating the area of the rectangles?
8. Where does this show up in the area integral?
4.7. POLAR AREA
43
Area in Polar
r(θ) = 1 + cos θ.
1. Graph the function.
2. What are the domain values in polar?
3. Draw lines to show divisions in the domain.
4. What shape is produced by two of these divisions and the function?
5. What is the area of a whole circle of radius r?
6. What is the area of 1/2, 1/4, 1/8 of a circle of radius r?
7. What is the angle for a 1/2, 1/4, 1/8 circle?
8. Use this to write the area of a circular wedge in terms of the radius and angle.
4.7.2
Example
Find the area inside r = cos θ for θ ∈ [0, π].
Z
Z
0
π
π
1
cos2 θ dθ
0 2
1 1
[1 + cos(2θ)] dθ
2 2
Z π
1
[1 + cos(2θ)] dθ
0 4
π
1
1
θ + sin(2θ) 4
2
=
=
=
=
0
=
4.7.3
Advanced Polar Area
Find the area inside both r = sin θ and r = cos(2θ).
1. Graph the functions.
2. Identify the region(s) enclosed between the two curves.
3. Determine at what angle these regions begin and end.
4. Setup the integrals.
5. Integrate to find the area.
6. Complete the WebAssign assignment.
1
([π + 0] − [0 + 0])
4
π
.
4
44
CHAPTER 4. ANALYTIC GEOMETRY
4.8
Arclength
4.8.1
Derivation
1. Sketch y = cos x for x ∈ [0, π].
2. Sketch a line segment from (0, 1) to (π, −1).
3. Calculate the length of the line segment.
4. What is the line segment (length thereof) with respect to length of the curve cos x?
5. Sketch a line segment from (0, 1) to (π/2, 0) and another line segment from (π/2, 0) to (π, −1).
6. Calculate the total length of these line segments.
7. What is the total length with respect to length of the curve cos x?
8. How can we obtain a more accurate length for cos x?
9. How can we obtain the exact length?
length ≈
length
=
=
n p
X
(xi+1 − xi )2 + (yi+1 − yi )2
i=0
lim
n→∞
n p
X
(xi+1 − xi )2 + (yi+1 − yi )2
i=0
n p
X
lim
(x0 (c))2 + (y 0 (d))2 by Mean Value Theorem
n→∞
Z
=
i=0
b
p
(x0 (t))2 + (y 0 (t))2 dt.
a
1. The formula above is written in which form?
2. Can we directly use the above to find the length of r = cos(2θ)?
3. Can we calculate x and y for r = cos(2θ)?
4. Now can we calculate the arclength?
5. How could we handle y = x2 for x ∈ [0, 4]?
6. Complete the WebAssign assignment.
4.9
Volumes of Objects of Revolution
If an object has sufficient regularity its volume and surface area can be calculated. Note how the formulas are derived
in the same fashion as was the arclength formula.
1. Pyramids
(a) Long, long ago, in a land far, far away, a ruler built a pyramid in his own honor. It consisted of 8 levels.
The bottom level measured 10 × 10 cubits; the height of each level was 5/4 cubit. Each level leaves 5/9
cubits of the previous level exposed (size of the ledges). What was the volume of this pyramid? Suppose
each level is solid.
4.9. VOLUMES OF OBJECTS OF REVOLUTION
45
(b) Almost as long ago, in the same land far, far away, the ruler’s son built a pyramid in his own honor. He
could not afford to make it taller than his father’s pyramid, but he could make one with more levels. It
consisted of 16 levels. The bottom level measured 10 × 10 cubits; the height of each level was 5/8 cubit.
Each level leaves 5/17 cubits of the previous level exposed (size of the ledges). What was the volume of
this pyramid? Suppose each level is solid.
(c) How could these be used to compute the volume of a smooth sided pyramid (think Egypt)?
2. Cones
(a) Guido is making a conical cake. He begins by baking five cake rounds. Each is 2 inches tall. The bottom
cake has a radius of 20/3 inches. Each cake is 4/3 inches smaller. What is the total volume of cake?
(b) Guido makes a second conical cake. To waste less cake when it is cut to form a cone, he begins by baking
ten cake rounds. Each is 1 inch tall. The bottom cake has a radius of 80/11 inches. Each cake is 8/11
inches smaller. What is the total volume of cake?
(c) How could these be used to compute the volume of a cone?
4.9.1
Disc Method
If the object can be produced by continuously rotating a curve (see the provided illustration), then this method of
calculating volumes produces a simple formula.
Derivation
1. Sketch the curve y = 2x − x2 .
2. Draw one of the rectangles that would be used to estimate area.
3. Imagine rotating this curve around the x-axis. (See the provided illustration.)
4. What does the 3D object look like?
5. What shape does the rectangle form?
6. What is the volume of this shape?
7. How can we use this in an integral?
Example
Find the volume of the object formed by revolving y = 2x − x2 around the x-axis.
Z
π(2x − x2 )2 dx
=
4x2 − 4x3 + x4 dx
=
0
2
Z
π
0
2
1
4 3
x − x4 + x5 =
3
5
0
32
32
π
− 16 +
.
3
5
π
2
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CHAPTER 4. ANALYTIC GEOMETRY
Practice
1. Calculate the volume of the solid produced by rotating the following region around the y-axis. The region is
enclosed by y = x2 , y = 2, x = 0.
2. Calculate the volume of the solid produced by rotating the following region around the y-axis. The region is
enclosed by y = x2 , y = 0, x = 3.
3. Calculate the volume of the solid produced by rotating the following region around the x-axis. The region is
enclosed by y = x2 , y = 2, x = 0.
4. Complete the WebAssign assignment.
4.9.2
Shell Method
Sometimes calculating volumes using discs is overly difficult. Note the problem in the following example.
Derivation
1. Attempt to find the volume of the object generated by rotating the following region about the y-axis. The
region is enclosed between y = (x − 1)2 and y = 1.
2. What is the difficulty?
3. Sketch the region enclosed between y = (x − 1)2 and y = 1.
4. Sketch one of the rectangles that would estimate area.
5. If this rectangle is rotated about the y-axis what shape would it form? See the provided illustration.
6. If the rectangle had width 0.1 and height 2, what would its volume be?
Example
Find the volume formed by rotating the following region around the y-axis. The region is enclosed between y = (x−1)2
and y = 1.
Z
0
2
(2πx)[1 − (x − 1)2 ] dx
2π
0
2π
=
2
Z
2x2 − x3 dx
=
2
2 3 1 4 x − x =
3
4
0
16
2π
−4 .
3
Practice
1. Calculate the volume of the solid produced by rotating the following region around the x-axis. The region is
enclosed by y = x2 , y = 2, x = 0.
2. Calculate the volume of the solid produced by rotating the following region around the x = 3. The region is
enclosed by y = x2 , y = 0, x = 2.
3. Complete the WebAssign assignment.
4.9. VOLUMES OF OBJECTS OF REVOLUTION
4.9.3
47
Surface Area
Derivation
1. Sketch f (x) = 2 + sin x for x ∈ [0, 2π].
2. Sketch a line segment between the points (π/2, f (π/2)) and (π, f (π)).
3. Sketch the rotation of the curve and the line segment.
4. Use the provided video to identify the shape the line segment forms.
√
2
2
5. The surface area of a right, circular cone
√ is πr r + h where r is the radius of the base and h is the height
(orthogonal from point to base). Note r2 + h2 is just the length of the hypotenuse.
6. Calculate the surface area for the frustrum (partial) cone with radii R and r in Figure 4.1.
7. Note r and R form bases of similar triangles. Thus
the surface area formula.
lr
r
=
lr +lR
R .
Clear the denominators and use this to simplify
8. As the side length approaches zero, what will happen to r and R?
9. Use this to derive the integral formula for surface area.
lr
r
lR
R
Figure 4.1: Frustrum of a Cone
The surface area of an object of revolution can be calculated by
Z
b
2πr(arclength) dx
a
Examples
Find the surface area of the object produced by revolving the region enclosed by y = 2x − x2 and the x-axis about
the x-axis.
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CHAPTER 4. ANALYTIC GEOMETRY
First find the curve segment.
2x − x2
=
0.
x(2 − x)
=
0.
P (t)
=
x =
0
P (t)
arclength
0, 2.
(x, 2x − x2 ).
(1, 2 − 2x).
p
=
12 + (2 − 2x)2 .
=
1.0
0.5
0.5
1.0
1.5
2.0
-0.5
-1.0
Because the curve is rotated around the x-axis, the radius is given by the function. The surface area is
Z
1
0
p
2π(2x − x2 ) 1 + (2 − 2x)2 dx.
Find the surface area of the object produced by revolving y = 2x − x2 about the y-axis.
1.0
0.8
0.6
0.4
0.2
-2
-1
1
2
Because the curve is rotated around the y-axis, the radius is given by the x-value. The surface area is
Z
0
1
p
2πx 1 + (2 − 2x)2 dx.
Practice
Use technology if needed.
1. Calculate the surface area of the solid produced by rotating y = x2 for x ∈ [0, 2] about the x-axis.
2. Calculate the surface area of the solid produced by rotating y = x2 for x ∈ [0, 2] about the y-axis.
3. Calculate the surface area of the solid produced by rotating the following region around the x = 3. The region
is enclosed by y = x2 , y = 0, x = 2.
4. Complete the WebAssign assignment.
4.9. VOLUMES OF OBJECTS OF REVOLUTION
4.9.4
Generalizing
1. The first two illustrations in this lesson were constructed by doing what?
2. As the number of levels was increased each object became more like what?
3. How is this similar to the arclength or area (rectangles) calculations?
49