Use
10-6 Functions as Infinite Series
Use
to find a power series representation of g(x). Indicate the interval on
which the series converges. Use a graphing
calculator to graph g(x) and the sixth partial
sum of its power series.
1. g(x) =
to find a power series representation of g(x). Indicate the interval on
which the series converges. Use a graphing
calculator to graph g(x) and the sixth partial
sum of its power series.
1. g(x) =
SOLUTION: To find the transformation that relates f (x) to g(x),
use u-substitution. Substitute u for x in f (x), equate
the two functions, and solve for u.
SOLUTION: To find the transformation that relates f (x) to g(x),
use u-substitution. Substitute u for x in f (x), equate
the two functions, and solve for u.
. Therefore, g(x) =
Replace x with
.
. Therefore, g(x) =
Replace x with
Therefore, g(x) =
.
can be represented by the power series
This series converges for
Therefore, g(x) =
power series
can be represented by the < 1, which is
equivalent to –7 < x < 1. Find the sixth partial sum S 6(x) of this series.
This series converges for
< 1, which is
Graph g(x) and S 6(x).
equivalent to –7 < x < 1. Find the sixth partial sum S 6(x) of this series.
Graph g(x) and S 6(x).
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2. g(x) =
SOLUTION: Page 1
The sixth partial sum S 6(x) of this series is
10-6 Functions as Infinite Series
Graph g(x) and S 6(x) .
2. g(x) =
SOLUTION: To find the transformation that relates f (x) to g(x),
use u-substitution. Substitute u for x in f (x), equate
the two functions, and solve for u.
3. g(x) =
SOLUTION: To find the transformation that relates f (x) to g(x),
use u-substitution. Substitute u for x in f (x), equate
the two functions, and solve for u.
. Therefore, g(x) =
.
Replace x with
f(x) =
= for |x| < 1
=
for
< 1. . Therefore, g(x) =
Therefore, g(x) =
can be represented by the Replace x with
.
power series
.
This series converges for
for |x| < 1 f(x) =
< 1, which is
=
equivalent to –1 <
< 1 or
The sixth partial sum S 6(x) of this series is
. Therefore, g(x) =
can be represented by the power series
Graph g(x) and S 6(x) .
< 1. for
This series converges for
equivalent to –1 <
< 1, which is
< 1 or –1 < x < 1. The
sixth partial sum S 6(x) of this series is
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Graph g(x) and S 6(x) .
Page 2
This series converges for
equivalent to –1 <
< 1 or –1 < x < 1. The
10-6 Functions as Infinite Series
equivalent to –1 <
< 1, which is
< 1 or –4 < x < 2. sixth partial sum S 6(x) of this series is
The sixth partial sum S 6(x) of this series is
Graph g(x) and S 6(x) .
Graph g(x) and S 6(x) .
4. g(x) =
5. g(x) =
SOLUTION: To find the transformation that relates f (x) to g(x),
use u-substitution. Substitute u for x in f (x), equate
the two functions, and solve for u.
To find the transformation that relates f (x) to g(x),
use u-substitution. Substitute u for x in f (x), equate
the two functions, and solve for u.
. Therefore, g(x) =
Replace x with
SOLUTION: .
. Therefore, g(x) =
= f(x) =
for |x| < 1 Replace x with
=
Therefore, g(x) =
for
can be represented by the power series
This series converges for
equivalent to –1 <
< 1. = f(x) =
for |x| < 1 =
Therefore, g(x) =
< 1, which is
.
for
< 1. can be represented by the power series
< 1 or –4 < x < 2. This series converges for
< 1, which is
The sixth partial sum S 6(x) of this series is
equivalent to –1 <
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Graph g(x) and S 6(x) .
< 1 or < x <
The sixth partial sum S 6(x) of this series is
. Page 3
This series converges for
< 1, which is
10-6 Functions as Infinite Series
equivalent to –1 <
< 1 or < x <
. The sixth partial sum S 6(x) of this series is
Therefore, g(x) =
the power series
This series converges for
can be represented by equivalent to –1 <
< 1, which is
< 1 or −
<x<
Graph g(x) and S 6(x) .
. The sixth partial sum S 6(x) of this series is
or
.
Graph g(x) and S 6(x) .
6. g(x) =
SOLUTION: To find the transformation that relates f (x) to g(x),
use u-substitution. Substitute u for x in f (x), equate
the two functions, and solve for u.
Use the fifth partial sum of the exponential
series to approximate each value. Round to
three decimal places.
7. e0.5
SOLUTION: . Therefore, g(x) =
Replacing x with
f(x) =
= .
for |x| < 1 =
< 1. for
8. e−0.25
Therefore, g(x) =
can be represented by the power series
ThisManual
series- Powered
converges
for
eSolutions
by Cognero
equivalent to –1 <
SOLUTION: < 1, which is
< 1 or −
<x<
Page 4
9. e−2.5
10-6 Functions as Infinite Series
8. e−0.25
SOLUTION: 12. e3.5
SOLUTION: 9. e−2.5
SOLUTION: 13. ECOLOGY The population density P per square
meter of zebra mussels in the Upper Mississippi
0.08t
River can be modeled by P = 3.5e
, where t is
measured in weeks. Use the fifth partial sum of the
exponential series to estimate the zebra mussel
population density after 4 weeks, 12 weeks, and 1
year.
e
0.8
SOLUTION: SOLUTION: x
Use the power series representation of e to find the
0.08t
fifth partial sum of P = 3.5e
and evaluate the
expression for t = 4, t = 12, and t = 52.
t=4
11. e−0.3
t = 12
SOLUTION: t = 52
12. e3.5
SOLUTION: Therefore, the zebra mussel population density after
4 weeks, 12 weeks, and 1 year will be 4.8
2
2
2
mussels/m , 9.1 mussels/m , and 134.0 mussels/m ,
respectively.
Use the fifth partial sum of the power series for
cosine or sine to approximate each value.
Round to three decimal places.
14. sin
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13. ECOLOGY The population density P per square
Page 5
Therefore, the zebra mussel population density after
4 weeks, 12 weeks, and 1 year will be 4.8
2
2
2
, 9.1 mussels/m
, and
134.0 mussels/m ,
10-6mussels/m
Functions
as Infinite
Series
respectively.
Use the fifth partial sum of the power series for
cosine or sine to approximate each value.
Round to three decimal places.
17. cos
SOLUTION: 14. sin
SOLUTION: Substitute
Substitute
;
.
18. cos
SOLUTION: 15. cos
SOLUTION: Substitute
.
Substitute
.
19. sin
16. sin
SOLUTION: SOLUTION: Substitute
.
Substitute
20. AMUSEMENT PARK A ride at an amusement
park is in the shape of a giant pendulum that swings
riders back and forth in a 240º arc to a maximum
height of 137 feet. The pendulum is supported by a
tower that is 85 feet tall and dips below ground-level
into a pit when swinging below the tower. Use the
fifth partial sum of the power series for cosine or
sine to approximate the length of the pendulum.
Refer to the photo on Page 642.
17. cos
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Substitute
;
SOLUTION: Page 6
Draw a diagram in which the ride is swinging in a
240º arc to a maximum height of 137 feet. Include a
into a pit when swinging below the tower. Use the
fifth partial sum of the power series for cosine or
sine to approximate the length of the pendulum.
10-6Refer
Functions
asonInfinite
to the photo
Page 642.Series
{Note: We should already know that this value
is 0.5)
Find r.
SOLUTION: Draw a diagram in which the ride is swinging in a
240º arc to a maximum height of 137 feet. Include a
horizontal in the diagram, as shown.
Therefore, the length of the pendulum is 104 feet.
Write each complex number in exponential
form.
21. + i
SOLUTION: Write the polar form of
Because the tower is 85 feet tall, the distance from
the maximum height of the ride to the horizontal is
137 – 85 = 52 feet. So, the length of the side opposit
the 30º angle in the right triangle shown is 52 feet.
=
+ i. In this expression, a
, b = 1, a > 0.
Find r.
An acute angle measure and the opposite side length
are known, so the sine function can be used to write
an expression for the length of the pendulum r.
Find
.
Use the fifth partial sum of the power series for sine
to find sin 30º. Converting to radians, 30º =
Now write
+ i in exponential form.
= 22. –i
SOLUTION: Write the polar form of
{Note: We should already know that this value
is 0.5)
=
– i. In this expression, a
, b = –1, a > 0.
Find r.
Find r.
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Find
Page 7
.
+ i in exponential form.
Now write
– i in exponential form.
Now write
10-6 Functions as Infinite Series
22. –i
23. SOLUTION: SOLUTION: Write the polar form of
– i. In this expression, a
, b = –1, a > 0.
=
i
–
Write the polar form of
Find r.
expression, a =
Find r.
Find
Find
.
– i in exponential form.
Now write
iθ
24. −
–
Write the polar form of
Find
i. In this
–
,b =–
, a > 0.
.
–i
Write the polar form of −
a=–
i. In this expression,
, b = –1, a < 0.
Find r.
Find
.
iθ
Therefore, because a + b i = re , the exponential
form of
–
i is
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24. −
i is
SOLUTION: SOLUTION: expression, a =
Find r.
, a > 0.
Therefore, because a + b i = re , the exponential
i
–
,b =–
.
form of
23. i. In this
–
–i
.
Now write i in exponential form.
.
Page 8
iθ
Therefore, because a + b i = re , the exponential
10-6 Functions as Infinite Series
–
form of
24. −
i is
.
25. 1 –
–i
SOLUTION: SOLUTION: Write the polar form of −
a=–
i. In this expression,
, b = –1, a < 0.
a = 1, b = –
Find r.
Find r.
i. In this expression,
Write the polar form of 1 –
Find
i
, a > 0.
.
Find
Now write i in exponential form.
.
Now write 1 –
i in exponential form. 25. 1 –
i
26. −1 +
SOLUTION: SOLUTION: Write the polar form of 1 –
a = 1, b = –
i. In this expression,
, a > 0.
Find r.
Find
i
.
i in exponential form. NowManual
write- 1Powered
–
eSolutions
by
Cognero
Write the polar form of −1 +
expression, a = –1, b =
Find r.
Find
i. In this
, a < 0.
.
Write −1 +
i in exponential form.
Page 9
i in exponential form. Now write 1 –
i in exponential form.
Write −1 +
10-6 Functions as Infinite Series
26. −1 +
i
27. −
SOLUTION: expression, a = –1, b =
Find r.
i. In this
, a < 0.
Write the polar form of −
expression, a = –
Find r.
Find
.
i in exponential form.
Write −1 +
i
+
SOLUTION: Write the polar form of −1 +
Find
,b =
i. In this
+
, a < 0.
.
Write −
+
i in exponential form.
27. −
i
+
28. −1 –
SOLUTION: SOLUTION: Write the polar form of −
expression, a = –
Find r.
Find
,b =
+
i. In this
, a < 0.
Write the polar form of −1 –
expression, a = –1, b = –
Find r.
Find
.
i. In this
, a < 0.
.
Write −
i
+
i in exponential form.
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Write −1 –
i in exponential form.
Page 10
Write −
+
i in exponential form.
31. ln (−2.45)
SOLUTION: 10-6 Functions as Infinite Series
28. −1 –
32. ln (−7)
i
SOLUTION: SOLUTION: Write the polar form of −1 –
expression, a = –1, b = –
Find r.
i. In this
, a < 0.
33. ln (−4.36)
SOLUTION: 34. ln (−9.12)
Find
SOLUTION: .
35. POWER SERIES Use the power series
Write −1 –
i in exponential form.
Find the value of each natural logarithm in the
complex number system.
29. ln (−6)
SOLUTION: representations of sin x and cos x to answer each of
the following questions.
a. Graph f (x) = sin x and the third partial sum of the
power series representing sin x. Repeat for the
fourth and fifth partial sums. Describe the interval of
convergence for each.
b. Repeat part a for f (x) = cos x and the third,
fourth, and fifth partial sums of the power series
representing cos x. Describe the interval of
convergence for each.
c. Describe how the interval of convergence
changes as n increases. Then make a conjecture as
to the relationship between each trigonometric
function and its related power series as n → .
SOLUTION: a.
30. ln (−3.5)
Graph y = sin x and
.
SOLUTION: 31. ln (−2.45)
SOLUTION: 32. ln (−7)
Sample answer: convergence, (–1.5, 1.5) SOLUTION: Graph y = sin x and
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33. ln (−4.36)
SOLUTION: .
Page 11
Sample answer: convergence, (–1.5, 1.5) 10-6 Functions as Infinite Series
Graph y = sin x and
.
Sample answer: convergence, (–2.5, 2.5)
Graph y = cos x and
.
Sample answer: convergence, (–2.5, 2.5)
Graph y = sin x and
Sample answer: convergence, (–3, 3)
c. Sample answer: The interval of convergence
widens as n increases. As n approaches infinity, the
power series equals the trigonometric function that it
represents.
Solve for z over the complex numbers. Round
to three decimal places.
36. 2ez + 5 = 0
SOLUTION: Sample answer: convergence, (–3.5, 3.5)
b. Graph y = cos x and
.
37. e2z + 12 = 0
Sample answer: convergence, (–1.5, 1.5)
Graph y = cos x and
SOLUTION: .
Sample answer: convergence, (–2.5, 2.5)
38. 4e2z + 7 = 6
SOLUTION: Graph
y = -cos
x andby Cognero
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10-6 Functions as Infinite Series
38. 4e2z + 7 = 6
SOLUTION: Therefore, the solutions are iπ or about 0.693.
41. 10e2z + 17ez = –3
SOLUTION: 42. ECONOMICS The total value of an investment of
z
39. 3(e – 1) + 5 = –2
SOLUTION: P dollars compounded continuously at an interest
rt
rate of r over t years is Pe . Use the first five terms
of the exponential series to approximate the value of
an investment of $10,000 compounded continuously
at 5.25% for 5 years.
SOLUTION: 43. RELATIVE ERROR Relative error is the
2z
z
40. e – e = 2
SOLUTION: absolute error in estimating a quantity divided by its
true value. The relative error of an approximation a
of a quantity b is given by
. Find the relative
2.1
error in approximating e using two, three, and six
terms of the exponential series.
SOLUTION: two terms: Therefore, the solutions are iπ or about 0.693.
41. 10e2z + 17ez = –3
SOLUTION: three terms:
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Page 13
10-6 Functions as Infinite Series
= 2%
43. RELATIVE ERROR Relative error is the
absolute error in estimating a quantity divided by its
true value. The relative error of an approximation a
of a quantity b is given by
. Find the relative
2.1
error in approximating e using two, three, and six
terms of the exponential series.
Approximate the value of each expression
using the first four terms of the power series
for sine and cosine. Then find the expected
value of each.
44. sin2
2
+ cos
SOLUTION: SOLUTION: two terms: sin
≈ 1 –
cos
≈ 1 –
+ –
+ –
or 0.4794
or 0.8776
2
2
0.4794 + 0.8776 ≈ 1
three terms:
The expected value is exactly 1 because sin x +
2
cos x = 1.
2
45. sec2 1 – tan2 1
SOLUTION: = 35%
sin 1 ≈ 1 –
six terms:
+ –
≈ 0.8415
cos 1 ≈ 1 –
+ –
≈ 0.5403
tan 1 =
≈
≈ 1.5575
= 2%
sec 1 =
Approximate the value of each expression
using the first four terms of the power series
for sine and cosine. Then find the expected
value of each.
44. sin2
2
+ cos
≈ 1 –
cos
≈ 1 –
2
≈ 1.8508
2
2
1.8508 – 1.5575 ≈ 0.9997
2
The expected value is exactly 1 because sec x –
2
tan x = 1.
46. RAINBOWS Airy's equation, which is used in
physics to model the diffraction of light, can also be
used to explain how a light wave front is converted
into a curved wave front in forming rainbows.
SOLUTION: sin
= + –
+ –
or 0.4794
or 0.8776
2
0.4794
+ 0.8776
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Page 14
2
The expected value is exactly 1 because sin x +
2
cos x = 1.
This equation can be represented by the power
. Find a power series representing E(x)
2
2
1.8508 – 1.5575 ≈ 0.9997
2
expected value is exactly 1 because sec x –
10-6The
Functions
as Infinite Series
2
tan x = 1.
if k is a constant and d = 1.
SOLUTION: First, rewrite E(x) in terms of one fraction.
46. RAINBOWS Airy's equation, which is used in
physics to model the diffraction of light, can also be
used to explain how a light wave front is converted
into a curved wave front in forming rainbows.
To find the transformation that relates E
This equation can be represented by the power
series shown below.
f (x) = 1 +
(x)
to f (x) =
, use u-substitution.
Substitute u for x in f (x), equate the two functions,
and solve for u.
Use the fifth partial sum of the series to find f (3).
Round to the nearest hundredth.
SOLUTION: 47. ELECTRICITY When an electric charge is
accompanied by an equal and opposite charge
nearby, such an object is called an electric dipole. It
consists of charge q at the point x = d and charge –q
at x = –d, as shown below.
. Therefore, g(x) =
.
Replace x with
f(x) =
= Along the x-axis, the electric field strength at x is the
sum of the electric fields from each of the two
charges. This is given by E(x) =
for |x| < 1 =
–
for
. Find a power series representing E(x)
< 1. if k is a constant and d = 1.
SOLUTION: Therefore, g(x) =
can be First, rewrite E(x) in terms of one fraction.
represented by the power series
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Page 15
< 1. g(x) =as Infinite Series
can be 10-6Therefore,
Functions
represented by the power series
50. cos (–x) = cos x
SOLUTION: 48. SOUND The Fourier Series represents a periodic
function of time f (t) as a summation of sine waves
and cosine waves with frequencies that start at 0
and increase by integer multiples. The series below
represents a sound wave from the digital data fed
from a CD into a CD player.
f (t) = 0.7 +
51. APPROXIMATIONS The infinite series for the
–1
inverse tangent function f (x) = tan
Graph the series for n = 4. Then analyze the graph.
SOLUTION: Graph S 4(t).
x is given by
. However, this series is only valid for
values of x on the interval (–1, 1).
a. Write the first five terms of the infinite series
–1
representation for f (x) = tan x.
b. Use the first five terms of the series to
–1
approximate tan 0.1.
–1
c. On the same coordinate plane, graph f (x) = tan
x and the third partial sum of the power series
–1
representing f (x) = tan x. On another coordinate
plane, graph f (x) and the fourth partial sum. Then
graph f (x) and the fifth partial sum.
d. Describe what happens on the interval (–1, 1) and
in the regions x ≥ 1 or x ≤ –1.
SOLUTION: a. tan−1 x ≈ x –
Sample answer: The graph oscillates in a pattern that
repeats about every 0.0232 second.
IDENTITIES Use power series
representations from this lesson to verify each
trigonometric identity.
49. sin (–x) = –sin x
SOLUTION: + –
+ b.
c.
−1
Graph tan
x and S 3(x).
50. cos (–x) = cos x
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Page 16
−1
more closely resemble the graph of f (x) = tan x on
the interval (–1, 1). Outside of the interval (–1, 1),
the end behavior of the polynomial approximations
causes the graphs of the partial sums to diverge from
−1
the graph of f (x) = tan x.
c.
−1
Graph tan
x and S 3(x).
10-6 Functions as Infinite Series
52. WRITING IN MATH Describe how using
additional terms in the approximating series for e
affects the outcome.
x
SOLUTION: Sample answer: In general, using additional terms
provides an approximation that is closer to the actual
x
value of e . Consider the fourth, fifth, and sixth
2
partial sums of e .
−1
Graph tan
x and S 4(x).
2
2
Using a calculator to evaluate e results in e ≈ 7.39.
Therefore, as the number of terms increases, the
x
approximation approaches the actual value of e .
−1
Graph tan
53. REASONING Use the power series for sine to
x and S 5(x).
explain why, for x-values on the interval [–0.1, 0.1],
a close approximation of sin x is x.
SOLUTION: Sample answer: The power series representation for
sine is given by sin x
x –
+ –
+ –
. . .. For x- values on the interval [−0.1, 0.1], the
cubic term and those of higher degree have values
less than or equal to
d. As n increases, the graphs of the partial sums
−1
more closely resemble the graph of f (x) = tan x on
the interval (–1, 1). Outside of the interval (–1, 1),
the end behavior of the polynomial approximations
causes the graphs of the partial sums to diverge from
−1
the graph of f (x) = tan x.
. This represents less than
one-tenth of 1 percent of the value. Therefore, for
the values of x on the interval [–0.1, 0.1], the first
term of the series, x, is a close approximation of sin
x.
54. CHALLENGE Prove that
SOLUTION: 52. WRITING IN MATH Describe how using
additional terms in the approximating series for e
affects the outcome.
x
SOLUTION: Sample
answer:
In general,
using
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additional terms
provides an approximation that is closer to the actual
x
value of e . Consider the fourth, fifth, and sixth
Page 17
iα
found by adding integer multiples of 2 to θ. For e
iβ
and e , α and β represent θ. Thus, if α and β differ
by an integer multiple of 2 , then the complex
one-tenth of 1 percent of the value. Therefore, for
the values of x on the interval [–0.1, 0.1], the first
of the series,
is a close approximation
of sin
10-6term
Functions
asx,Infinite
Series
x.
iα
iβ
numbers represented by e and e will be the same.
PROOF Show that for all real numbers x, the
following are true.
54. CHALLENGE Prove that
56. SOLUTION: SOLUTION: First, find an expression for e
−ix
ix
. Since e = cos x +
i sin x, .
55. REASONING For what values of α and β does eiα
iβ
= e ? Explain.
SOLUTION: The exponential form of a complex number a + b i is
iθ
given by a + b i = re , where θ is the measure of an
angle in radians. Since θ is the measure of an angle,
θ has infinitely many coterminal angles that can be
Thus, e
iα
= cos x − i sin x.
found by adding integer multiples of 2 to θ. For e
iβ
and e , α and β represent θ. Thus, if α and β differ
by an integer multiple of 2 , then the complex
iα
−ix
iβ
numbers represented by e and e will be the same.
PROOF Show that for all real numbers x, the
following are true.
57. 56. SOLUTION: SOLUTION: First, find an expression for e
i sin x, First, find an expression for e
−ix
ix
. Since e = cos x +
i sin x,
Thus, e
ix
. Since e = cos x +
.
.
Thus, e
−ix
−ix
= cos x − i sin x.
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−ix
= cos x − i sin x.
Page 18
10-6 Functions as Infinite Series
58. CHALLENGE The hyperbolic sine and hyperbolic
57. SOLUTION: First, find an expression for e
i sin x,
−ix
ix
. Since e = cos x +
.
cosine functions are analogs of the trigonometric
functions that you studied in Chapters 4 and 5. Just
as the points (cos x, sin x) form a unit circle, the
points (cosh t, sinh t) form the right half of an
equilateral hyperbola. An equilateral hyperbola has
perpendicular asymptotes. The hyperbolic sine (sinh)
and hyperbolic cosine (cosh) functions are defined
below. Find the power series representations for
these functions.
cosh x =
sinh x =
SOLUTION: Begin with the right side of sinh x =
Thus, e
−ix
.
= cos x − i sin x.
The expanded form of the series, x +
58. CHALLENGE The hyperbolic sine and hyperbolic
cosine functions are analogs of the trigonometric
functions that you studied in Chapters 4 and 5. Just
as the points (cos x, sin x) form a unit circle, the
points (cosh t, sinh t) form the right half of an
equilateral hyperbola. An equilateral hyperbola has
perpendicular asymptotes. The hyperbolic sine (sinh)
and hyperbolic cosine (cosh) functions are defined
below. Find the power series representations for
these functions.
sinh x =
+ … ,
resembles the power series representation for sin x
except that the terms are all positive. Therefore, the
power series representation for sinh x is sinh x =
=x+
+ + + … .
Begin with the right side of cosh x =
.
cosh x =
SOLUTION: Begin with the right side of sinh x =
.
The expanded form of the series, 1 +
+ … ,
resembles the power series representation for cos x
except that the terms are all positive. Therefore, the
power series representation for cosh x is cosh x =
+ + …
Use Pascal's triangle to expand each binomial.
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The expanded form of the series, x +
+ … ,
resembles the power series representation for sin x
59. (3m +
)
4
SOLUTION: Page 19
except that the terms are all positive. Therefore, the
power series representation for cosh x is cosh x =
+ Series
+ …
10-6 Functions as Infinite
Use Pascal's triangle to expand each binomial.
59. (3m +
)
1, 8, 28, 56, 70, 56, 28, 8, and 1. Use these numbers
as the coefficients of the terms in the series. Then
simplify.
62. Prove that 4 + 7 + 10 + … + (3n + 1) =
4
for all positive integers n.
SOLUTION: 4
Write a series for (3m +
coefficients.
) omitting the
SOLUTION: Let Pn be the statement 4 + 7 + 10 + … + (3n + 1) =
The numbers in the 4th row of Pascal’s triangle are
1, 4, 6, 4, and 1. Use these numbers as the
coefficients of the terms in the series. Then simplify.
is a true . Because 4 =
statement, Pn is true for n = 1. Assume that Pk : 4 +
7 + 10 + … + (3k + 1) =
is true for a positive integer k. Show that Pk + 1 must be
60. SOLUTION: Write a series for
omitting the
true.
coefficients.
The numbers in the 5th row of Pascal’s triangle are
1, 5, 10, 10, 5, and 1. Use these numbers as the
coefficients of the terms in the series. Then simplify.
This final statement is exactly Pk + 1, so Pk + 1 is
true. Because Pn is true for n = 1 and Pk implies Pk
+ 1, Pn
is true for n = 2, n = 3, and so on. That is, by
the principle of mathematical induction, 4 + 7 + 10 +
… + (3n + 1) =
is true for all positive integers n.
61. (p 2 + q)8
Find each power, and express it in rectangular
form.
63. (−2 + 2i)3
SOLUTION: 2
8
Write a series for (p + q) omitting the coefficients.
SOLUTION: First, write –2 + 2i in polar form.
The numbers in the 8th row of Pascal’s triangle are
1, 8, 28, 56, 70, 56, 28, 8, and 1. Use these numbers
as the coefficients of the terms in the series. Then
simplify.
62. Prove that 4 + 7 + 10 + … + (3n + 1) =
The polar form of –2 + 2i
is
. Now use De Moivre’s
Theorem to find the third power.
for all positive integers n.
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Page 20
Let Pn be the statement 4 + 7 + 10 + … + (3n + 1) =
. Because 4 =
is a true Therefore,
.
+1
n
the principle of mathematical induction, 4 + 7 + 10 +
… + (3n + 1) =
is true for all positive 10-6 Functions as Infinite Series
integers n.
Find each power, and express it in rectangular
form.
63. (−2 + 2i)3
Therefore,
65. (
+
.
i)−2
SOLUTION: First, write
SOLUTION: + i in polar form.
First, write –2 + 2i in polar form.
The polar form of –2 + 2i
is
. Now use De Moivre’s
Theorem to find the third power.
Therefore,
Therefore,
.
.
66. Given t = <−9, −3, c>, u = <8, −4, 3>, v = <2, 5,
i)4
64. (1 +
i
The polar form of
+ is
. Now use De Moivre’s
Theorem to find the negative second power.
−6>, and that the volume of the parallelepiped having
adjacent edges t, u, and v is 93 cubic units, find c.
SOLUTION: i in polar form.
First, write 1 + SOLUTION: The volume is the absolute value of the triple scalar
product, so t · (u × v) can equal 93 or – 93.
The polar form of −1 +
i
is
. Now use De
Moivre’s Theorem to find the fourth power.
Therefore,
65. (
+
.
i)−2
SOLUTION: First, write
+ i in polar form.
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Use an inverse matrix to solve each system
of 21
Page
equations, if possible.
67. x − 8y = −7
10-6 Functions as Infinite Series
Use an inverse matrix to solve each system of
equations, if possible.
67. x − 8y = −7
2x + 5y = 28
The solution is (9, 2).
68. 4x + 7y = 22
−9x + 11y = 4
SOLUTION: Write the system in matrix form AX = B.
4x + 7y = 22
−9x + 11y = 4
Write the system in matrix form AX = B.
Use the formula for the inverse of a 2 × 2 matrix to Use the formula for the inverse of a 2 × 2 matrix to SOLUTION: 1
−1
find A .
Multiply A
find A .
−1
by B to solve the system.
The solution is (9, 2).
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4x + 7y = 22
−9x + 11y = 4
Multiply A
−1
by B to solve the system.
The solution is (2, 2).
69. w + 2x + 3y = 18
4w − 8x + 7y = 41
−w + 9x − 2y = −4
Page 22
10-6 Functions as Infinite Series
The solution is (2, 2).
69. w + 2x + 3y = 18
Multiply A
−1
by B to solve the system.
4w − 8x + 7y = 41
−w + 9x − 2y = −4
SOLUTION: w + 2x + 3y = 18
4w − 8x + 7y = 41
w + 9x − 2y = −4
Write the system in matrix form AX = B.
−1
Use a graphing calculator to find A . Enter the
−1
values for the matrix A. Enter [A] . Select
to get the values in reduced fraction form.
The solution is (7, 1, 3).
Determine whether A and B are inverse
matrices.
70. A =
,B=
SOLUTION: A=
Multiply A
−1
by B to solve the system.
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,B=
If A and B are
inverse matrices, then AB = BA = I.
Page 23
10-6 Functions as Infinite Series
The solution is (7, 1, 3).
Because AB = BA = I, B = A
Determine whether A and B are inverse
matrices.
70. A =
,B=
−1
and A = B .
,B=
SOLUTION: If A and B are inverse matrices, then AB = BA = I.
SOLUTION: A=
72. A =
−1
,B=
If A and B are
inverse matrices, then AB = BA = I.
AB ≠ I, so A and B are not inverses.
73. CONFERENCE A university sponsored a
conference for 680 women. The Venn diagram
shows the number of participants in three of the
activities offered. Suppose women who attended the
conference were randomly selected for a survey.
AB ≠ I, so A and B are not inverses.
71. A =
,B=
SOLUTION: If A and B are inverse matrices, then AB = BA = I.
a. What is the probability that a woman selected
participated in hiking or sculpting?
b. Describe a set of women such that the probability
of their being selected is about 0.39.
SOLUTION: Because AB = BA = I, B = A
72. A =
−1
−1
and A = B .
,B=
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SOLUTION: If A and B are inverse matrices, then AB = BA = I.
a. From the diagram, 98 women participated only in
hiking, 75 women participated in both hiking and
sculpting, 123 women participated only in sculpting,
38 women participated in sculpting and quilting, 15
women participated in all three events, and 21
women participated in hiking and quilting.
Page 24
Therefore, the probability that a randomly selected
woman participated in hiking or sculpting is about
54.41%.
a. From the diagram, 98 women participated only in
hiking, 75 women participated in both hiking and
sculpting, 123 women participated only in sculpting,
women participated in sculpting and quilting, 15
10-638
Functions
as Infinite Series
women participated in all three events, and 21
women participated in hiking and quilting.
So, the probability of a selecting a woman who
participated in hiking or quilting but not sculpting is
about 0.39.
74. SAT/ACT PQRS is a square. What is the ratio of
the length of diagonal
Therefore, the probability that a randomly selected
woman participated in hiking or sculpting is about
54.41%.
to the length of side ?
b. Sample answer: First, find the size of a set with a
probability of 0.39 or 39%.
39% of 680 = 0.39(680) or about 265
Look at the Venn diagram and find a set of numbers
whose sum is close to 265.
148 + 21 + 98 = 266
These numbers would represent the set of women
that participated in hiking or quilting but not
sculpting. Check:
A2
B
C1
D
E
SOLUTION: So, the probability of a selecting a woman who
participated in hiking or quilting but not sculpting is
about 0.39.
74. SAT/ACT PQRS is a square. What is the ratio of
the length of diagonal
to the length of side ?
Since PQRS is a square
and all
angles are right angles. Let this side length be
represented by x. Consider triangle QRS. This is a
right triangle with 2 side lengths labeled x. Apply the
Pythagorean Theorem to assign a value for the
hypotenuse.
A2
B
C1
D
The ratio of
The correct answer is B.
.
75. REVIEW What is the sum of the infinite geometric
series
+ + + + …?
E
F
SOLUTION: Since PQRS is a square
and all
angles are right angles. Let this side length be
represented by x. Consider triangle QRS. This is a
right triangle with 2 side lengths labeled x. Apply the
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Pythagorean
Theorem
to assign a value for the
hypotenuse.
G1
H1
J1
Page 25
SOLUTION: The sum S of an infinite geometric series for which |
The ratio of
.
10-6 Functions as Infinite Series
The correct answer is B.
75. REVIEW What is the sum of the infinite geometric
series
+ + + d. Find the expression for the number of dots in the
nth figure in the original sequence.
e . Prove, through mathematical induction, that the
sum of the sequence found in part b is equal to the
expression found in part d.
+ …?
F
G1
SOLUTION: H1
a.
J1
SOLUTION: The sum S of an infinite geometric series for which |
r | < 1 is given by S =
. So, first find the
common ratio.
The common ratio r is
< 1. This infinite geometric series has a sum. Use the formula the
sum of an infinite geometric series.
b. There is 1 dot in the first figure, 4 dots in the
second figure, 9 dots in the third figure and 16 dots in
the fourth figure.
a1 = 1
a 2 = 4 – 1 or 3
a 3 = 9 – 4 or 5
a 4 = 16 – 9 or 7
Therefore, the sequence is 1, 3, 5, 7, … .
c. Each term a n in this sequence can be found by
The correct answer is F.
76. FREE RESPONSE Consider the pattern of dots
shown.
a. Draw the next figure in this sequence
b. Write the sequence, starting with 1, that
represents the number of dots that must be added to
each figure in the sequence to get the number of
dots in the next figure.
c. Find the expression for the nth term of the
sequence found in part b.
d. Find the expression for the number of dots in the
nth figure in the original sequence.
e . Prove, through mathematical induction, that the
sum of the sequence found in part b is equal to the
expression found in part d.
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eSolutions
subtracting 1 from 2n.
2(1) – 1 = 1
2(2) – 1 = 3
2(3) – 1 = 5
2(4) – 1 = 7
So, an explicit formula for this sequence is a n = 2n –
1.
d. The sequence of dots is 1, 4, 9, 16, … , which can
2
2
2
2
also be written as 1 , 2 , 3 , 4 , … . Therefore an
expression for the number of dots in the nth figure in
2
the original sequence is n .
e . Let Pn be the statement 1 + 3 + 5 + 7 + . . . + (2n
2
2
– 1) = n . Because 1 = 1 is a true statement, Pn is
true for n = 1. Assume that 1 + 3 + 5 + 7 + . . .Page
+ 26
2
(2k – 1) = k is true for a positive integer k. Show
that P
must be true.
2
the original sequence is n .
Let Pn be the as
statement
1 + 3Series
+ 5 + 7 + . . . + (2n
10-6e .Functions
Infinite
2
2
– 1) = n . Because 1 = 1 is a true statement, Pn is
true for n = 1. Assume that 1 + 3 + 5 + 7 + . . . +
2
(2k – 1) = k is true for a positive integer k. Show
that Pk + 1 must be true.
This final statement is exactly Pk + 1, so Pk + 1 is
true. Because Pn is true for n = 1 and Pk implies Pk
+ 1, Pn
is true for n = 2, n = 3, and so on. That is, by
the principle of mathematical induction, 1 + 3 + 5 + 7
2
+ . . . + (2n – 1) = n is true for all positive integers
n.
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