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Infinite Series
Section 9.10
Taylor and Maclaurin Series
• Find a Taylor or Maclaurin series for a function.
• Find a binomial series.
• Use a basic list of Taylor series to find other Taylor series.
Taylor Series and Maclaurin Series
Bettmann/Corbis
In Section 9.9, you derived power series for several functions using geometric series
with term-by-term differentiation or integration. In this section you will study a
general procedure for deriving the power series for a function that has derivatives of
all orders. The following theorem gives the form that every convergent power series
must take.
THEOREM 9.22
COLIN MACLAURIN (1698–1746)
The development of power series to represent
functions is credited to the combined work of
many seventeenth and eighteenth century
mathematicians. Gregory, Newton, John and
James Bernoulli, Leibniz, Euler, Lagrange,
Wallis, and Fourier all contributed to this
work. However, the two names that are most
commonly associated with power series are
Brook Taylor (1685–1731) and Colin
Maclaurin.
The Form of a Convergent Power Series
If f is represented by a power series f x anx cn for all x in an open
interval I containing c, then an f ncn! and
f x f c fcx c f c
f nc
x c2 . . . x cn . . . .
2!
n!
Proof Suppose the power series anx cn has a radius of convergence R. Then,
by Theorem 9.21, you know that the nth derivative of f exists for x c < R, and
by successive differentiation you obtain the following.
f 0x a0 a1x c a2x c2 a3x c3 a4x c4 . . .
f 1x a1 2a2x c 3a3x c2 4a4x c3 . . .
f 2x 2a2 3!a3x c 4 3a4x c2 . . .
f 3x 3!a3 4!a4x c . . .
f nx
n!an n 1!an1x c . . .
Evaluating each of these derivatives at x c yields
f 0c 0!a0
f 1c 1!a1
f 2c 2!a2
f 3c 3!a3
and, in general, f nc n!an. By solving for an, you find that the coefficients of the
power series representation of f x are
NOTE Be sure you understand Theorem
9.22. The theorem says that if a power
series converges to f x, the series must
be a Taylor series. The theorem does not
say that every series formed with the
Taylor coefficients an f ncn! will
converge to f x.
an f nc
.
n!
Notice that the coefficients of the power series in Theorem 9.22 are precisely the
coefficients of the Taylor polynomials for f x at c as defined in Section 9.7. For this
reason, the series is called the Taylor series for f x at c.
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SECTION 9.10
Taylor and Maclaurin Series
677
Definitions of Taylor and Maclaurin Series
If a function f has derivatives of all orders at x c, then the series
n0
f nc
f nc
x cn f c fcx c . . . x cn . . .
n!
n!
is called the Taylor series for f x at c. Moreover, if c 0, then the series is
the Maclaurin series for f.
If you know the pattern for the coefficients of the Taylor polynomials for a
function, you can extend the pattern easily to form the corresponding Taylor series.
For instance, in Example 4 in Section 9.7, you found the fourth Taylor polynomial for
ln x, centered at 1, to be
1
1
1
P4x x 1 x 12 x 13 x 14.
2
3
4
From this pattern, you can obtain the Taylor series for ln x centered at c 1,
1
1n1
x 1 x 12 . . . x 1n . . . .
2
n
EXAMPLE 1
Forming a Power Series
Use the function f x sin x to form the Maclaurin series
n0
f n0 n
f 0 2
f 30 3
f 40 4 . . .
x f 0 f0x x x x n!
2!
3!
4!
and determine the interval of convergence.
Solution Successive differentiation of f x yields
f x sin x
fx cos x
f x sin x
f 3x cos x
f 4x sin x
f 5x cos x
f 0 sin 0 0
f0 cos 0 1
f 0 sin 0 0
f 30 cos 0 1
f 40 sin 0 0
f 50 cos 0 1
and so on. The pattern repeats after the third derivative. So, the power series is as
follows.
f n0 n
f 0 2 f 30 3
f 40 4
x f 0 f0x x x x . . .
n!
2!
3!
4!
n0
1n x2n1
0
1 3
0
1
0
0 1x x 2 x x 4 x5 x6
2n
1
!
2!
3!
4!
5!
6!
n0
1 7 . . .
x 7!
x3
x5
x7
x . . .
3! 5! 7!
By the Ratio Test, you can conclude that this series converges for all x.
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Infinite Series
x<−π
2
π
x  ≤ 2
x> π
2
−1,
sin x,
f (x) =
Page 678
1,
Notice that in Example 1 you cannot conclude that the power series converges to
sin x for all x. You can simply conclude that the power series converges to some
function, but you are not sure what function it is. This is a subtle, but important, point
in dealing with Taylor or Maclaurin series. To persuade yourself that the series
y
f c fcx c f c
f nc
x c2 . . . x cn . . .
2!
n!
1
−π
2
π
2
−1
Figure 9.23
π
x
might converge to a function other than f, remember that the derivatives are being
evaluated at a single point. It can easily happen that another function will agree with
the values of f nx when x c and disagree at other x-values. For instance, if you
formed the power series (centered at 0) for the function shown in Figure 9.23, you
would obtain the same series as in Example 1. You know that the series converges for
all x, and yet it obviously cannot converge to both f x and sin x for all x.
Let f have derivatives of all orders in an open interval I centered at c. The Taylor
series for f may fail to converge for some x in I. Or, even if it is convergent, it may
fail to have f x as its sum. Nevertheless, Theorem 9.19 tells us that for each n,
f x f c fcx c f c
f nc
x c2 . . . x cn Rnx,
2!
n!
where
Rnx f n1z
x cn1.
n 1!
Note that in this remainder formula the particular value of z that makes the
remainder formula true depends on the values of x and n. If Rn → 0, then the following
theorem tells us that the Taylor series for f actually converges to f x for all x in I.
THEOREM 9.23
Convergence of Taylor Series
If lim Rn 0 for all x in the interval I, then the Taylor series for f converges
n→ and equals f x,
f x f nc
x cn.
n0 n!
Proof For a Taylor series, the nth partial sum coincides with the nth Taylor polynomial. That is, Snx Pnx. Moreover, because
Pnx f x Rnx
it follows that
lim Snx lim Pnx
n→ n→ lim f x Rnx
n→ f x lim Rnx.
n→ So, for a given x, the Taylor series (the sequence of partial sums) converges to f x
if and only if Rnx → 0 as n → .
NOTE Stated another way, Theorem 9.23 says that a power series formed with Taylor
coefficients an f ncn! converges to the function from which it was derived at precisely
those values for which the remainder approaches 0 as n → .
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SECTION 9.10
Taylor and Maclaurin Series
679
In Example 1, you derived the power series from the sine function and you also
concluded that the series converges to some function on the entire real line. In
Example 2, you will see that the series actually converges to sin x. The key observation is that although the value of z is not known, it is possible to obtain an upper
bound for f n1z .
A Convergent Maclaurin Series
EXAMPLE 2
Show that the Maclaurin series for f x sin x converges to sin x for all x.
Solution Using the result in Example 1, you need to show that
x5
x7
1n x 2n1 . . .
x3
. . .
3! 5! 7!
2n 1!
sin x x is true for all x. Because
f n1x ± sin x
or
f n1x ± cos x
you know that f n1z ≤ 1 for every real number z. Therefore, for any fixed x, you
can apply Taylor’s Theorem (Theorem 9.19) to conclude that
0 ≤ Rnx x n1
f n1z n1
x
≤
.
n 1!
n 1!
From the discussion in Section 9.1 regarding the relative rates of convergence of
exponential and factorial sequences, it follows that for a fixed x
lim
n→ xn1
n 1!
0.
Finally, by the Squeeze Theorem, it follows that for all x, Rnx → 0 as n → . So, by
Theorem 9.23, the Maclaurin series for sin x converges to sin x for all x.
Figure 9.24 visually illustrates the convergence of the Maclaurin series for sin x
by comparing the graphs of the Maclaurin polynomials P1x, P3x, P5x, and P7x
with the graph of the sine function. Notice that as the degree of the polynomial
increases, its graph more closely resembles that of the sine function.
y
y
4
3
2
1
−π
4
3
2
1
π
−2
−3
−4
P1(x) = x
2π
y = sin x
x
y
y = sin x
−π
2π
x
−2
−3
−4
4
3
2
1
π
−2
−3
−4
3
P3(x) = x − x
3!
As n increases, the graph of Pn more closely resembles the sine function.
Figure 9.24
y
4
3
2
1
3
5
P5 (x) = x − x + x
3! 5!
2π
y = sin x
x
−π
−2
−3
−4
y = sin x
π
3
7
5
P7(x) = x − x + x − x7!
3! 5!
2π
x
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Infinite Series
The guidelines for finding a Taylor series for f x at c are summarized below.
Guidelines for Finding a Taylor Series
1. Differentiate f x several times and evaluate each derivative at c.
f c, fc, f c, fc, . . . , f n c, . . .
Try to recognize a pattern in these numbers.
2. Use the sequence developed in the first step to form the Taylor coefficients
an f ncn!, and determine the interval of convergence for the resulting
power series
f c fcx c f c
f nc
x c2 . . . x cn . . . .
2!
n!
3. Within this interval of convergence, determine whether or not the series
converges to f x.
The direct determination of Taylor or Maclaurin coefficients using successive
differentiation can be difficult, and the next example illustrates a shortcut for finding
the coefficients indirectly—using the coefficients of a known Taylor or Maclaurin
series.
EXAMPLE 3
Maclaurin Series for a Composite Function
Find the Maclaurin series for f x sin x2.
Solution To find the coefficients for this Maclaurin series directly, you must
calculate successive derivatives of f x sin x 2. By calculating just the first two,
fx 2x cos x 2 and
f x 4x 2 sin x 2 2 cos x 2
you can see that this task would be quite cumbersome. Fortunately, there is an
alternative. First consider the Maclaurin series for sin x found in Example 1.
gx sin x
x3
x5
x7
. . .
3! 5! 7!
Now, because sin x 2 gx 2, you can substitute x 2 for x in the series for sin x to
obtain
x
sin x 2 gx 2
x2 x 6 x10 x14 . . .
.
3!
5!
7!
Be sure to understand the point illustrated in Example 3. Because direct
computation of Taylor or Maclaurin coefficients can be tedious, the most practical
way to find a Taylor or Maclaurin series is to develop power series for a basic list of
elementary functions. From this list, you can determine power series for other
functions by the operations of addition, subtraction, multiplication, division, differentiation, integration, or composition with known power series.
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SECTION 9.10
Taylor and Maclaurin Series
681
Binomial Series
Before presenting the basic list for elementary functions, you wll develop one more
series—for a function of the form f x 1 xk. This produces the binomial series.
EXAMPLE 4
Binomial Series
Find the Maclaurin series for f x 1 xk and determine its radius of convergence.
Assume that k is not a positive integer.
Solution By successive differentiation, you have
f x 1 xk
fx k1 xk1
f x kk 11 xk2
fx kk 1k 21 xk3
f 0 1
f0 k
f 0 kk 1
f0 kk 1k 2
f nx k . . . k n 11 xkn
f n0 kk 1 . . . k n 1
which produces the series
1 kx kk 1x 2 . . . kk 1 . . . k n 1xn . . .
.
2
n!
Because an1an → 1, you can apply the Ratio Test to conclude that the radius of
convergence is R 1. So, the series converges to some function in the interval
1, 1.
Note that Example 4 shows that the Taylor series for 1 xk converges to some
function in the interval 1, 1. However, the example does not show that the series
actually converges to 1 xk. To do this, you could show that the remainder Rnx
converges to 0, as illustrated in Example 2.
EXAMPLE 5
Finding a Binomial Series
3 1 x.
Find the power series for f x Solution Using the binomial series
1 xk 1 kx kk 1x 2 kk 1k 2x3 . . .
2!
3!
let k 13 and write
1 x13 1 2
x
2x 2
2 5x3 2 5 8x 4 . . .
2 3 3 2!
333!
344!
which converges for 1 ≤ x ≤ 1.
P4
−2
2
f(x) =
3
1+x
Figure 9.25
−1
TECHNOLOGY Use a graphing utility to confirm the result in Example 5.
When you graph the functions
f x 1 x13 and P4x 1 x 2 5x3 10x 4
x
3
9
81
243
in the same viewing window, you should obtain the result shown in Figure 9.25.
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Infinite Series
Deriving Taylor Series from a Basic List
The following list provides the power series for several elementary functions with the
corresponding intervals of convergence.
Power Series for Elementary Functions
Interval of
Convergence
Function
1
1 x 1 x 12 x 13 x 14 . . . 1n x 1n . . .
x
1
1 x x 2 x3 x 4 x5 . . . 1n xn . . .
1x
ln x x 1 x 12 x 13 x 14 . . . 1n1x 1n . . .
2
3
4
n
0 < x < 2
1 < x < 1
0 < x ≤ 2
x2
x3
x4
x5
xn
. . . . . .
2! 3!
4!
5!
n!
< x <
sin x x x3
x5
x7
x9
1n x 2n1 . . .
. . .
3! 5! 7! 9!
2n 1!
< x <
cos x 1 x 2 x 4 x 6 x 8 . . . 1n x 2n . . .
2!
4!
6! 8!
2n!
< x <
x3 x5 x7 x9 . . . 1n x 2n1 . . .
3
5
7
9
2n 1
1 ≤ x ≤ 1
ex 1 x arctan x x arcsin x x x3
2
3
1 xk 1 kx 1 3x5
1 3 5x7
2n!x 2n1
. . . n 2
. . .
245 2467
2 n! 2n 1
kk 1x 2 kk 1k 2x3 kk 1k 2k 3x 4 . . .
2!
3!
4!
1 ≤ x ≤ 1
1 < x < 1*
* The convergence at x ± 1 depends on the value of k.
NOTE The binomial series is valid for noninteger values of k. Moreover, if k happens to be a
positive integer, the binomial series reduces to a simple binomial expansion.
EXAMPLE 6
Deriving a Power Series from a Basic List
Find the power series for f x cos
x.
Solution Using the power series
cos x 1 x2 x 4 x6 x8 . . .
2!
4!
6! 8!
you can replace x by x to obtain the series
cos
x 1 x
x2 x3 x 4 . . .
.
2! 4! 6! 8!
This series converges for all x in the domain of cos
x—that is, for x ≥ 0.
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SECTION 9.10
683
Taylor and Maclaurin Series
Power series can be multiplied and divided like polynomials. After finding the
first few terms of the product (or quotient), you may be able to recognize a pattern.
EXAMPLE 7
Multiplication and Division of Power Series
Find the first three nonzero terms in each of the Maclaurin series.
a. e x arctan x
b. tan x
Solution
a. Using the Maclaurin series for e x and arctan x in the table, you have
e x arctan x 1 x
x2 x3 x 4 . . .
1! 2! 3!
4!
x
x3 x5
. . . .
3
5
Multiply these expressions and collect like terms as you would for multiplying
polynomials.
1 4
1 x 12 x 2 16 x3 24
x . . .
x
13 x3
1 5
5x
. . .
x x 2 12 x3 1 4
6x
1 5
24 x
. . .
13 x3 1 4
3x
1 5
6x
. . .
1 5
5x
. . .
x x 2 16 x3 1 4
6x
3 5
40
x . . .
So, e x arctan x x x 2 16 x 3 . . . .
b. Using the Maclaurin series for sin x and cos x in the table, you have
x3 x5 . . .
sin x
3!
5!
tan x .
cos x
x2 x 4 . . .
1 2!
4!
x
Divide using long division.
1
1
1 x2 x 4 . . .
2
24
1
x x3
3
1
x x3
6
1
x x3
2
1 3
x
3
1 3
x
3
2 5
So, tan x x 13 x 3 15
x . . ..
2 5
x
15
1 5
x
120
1 5
x
24
1 5
x
30
1 5
x
6
2 5
x
15
. . .
. . .
. . .
. . .
. . .
. . .
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Infinite Series
A Power Series for sin2 x
EXAMPLE 8
Find the power series for f x sin2 x.
Solution Consider rewriting sin2 x as follows.
sin2 x 1 cos 2x 1 cos 2x
2
2
2
Now, use the series for cos x.
x2 x 4 x6 x8 . . .
2!
4!
6! 8!
2
4
2
2
26
28
cos 2x 1 x 2 x 4 x 6 x 8 . . .
2!
4!
6!
8!
3
5
1
1
2
2
2
27
cos 2x x 2 x 4 x 6 x 8 . . .
2
2 2!
4!
6!
8!
3
5
1 1
1 1
2
2
2
27
sin2 x cos 2x x 2 x 4 x 6 x 8 . . .
2 2
2 2 2!
4!
6!
8!
cos x 1 2 2 23 4 25 6 27 8 . . .
x x x x 2!
4!
6!
8!
This series converges for < x <
.
As mentioned in the preceding section, power series can be used to obtain tables
of values of transcendental functions. They are also useful for estimating the values of
definite integrals for which antiderivatives cannot be found. The next example demonstrates this use.
Power Series Approximation of a Definite Integral
EXAMPLE 9
Use a power series to approximate
1
ex dx
2
0
with an error of less than 0.01.
Solution Replacing x with x 2 in the series for ex produces the following.
x 4 x6 x8 . . .
2!
3! 4!
1
3
5
x
x
x7
x9
2
ex dx x . . .
3
5 2! 7 3! 9 4!
0
1
1
1
1
1 . . .
3 10 42 216
ex 1 x 2 2
1
0
Summing the first four terms, you have
1
ex dx 0.74
2
0
1
which, by the Alternating Series Test, has an error of less than 216
0.005.
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SECTION 9.10
Exercises for Section 9.10
In Exercises 1–10, use the definition to find the Taylor series
(centered at c) for the function.
See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
In Exercises 31–34, find the Maclaurin series for the function.
(See Example 7.)
1. f x e2x, c 0
31. f x x sin x
2. f x e 3x,
sin x ,
x
33. gx 1,
c0
3. f x cos x, c 4
4. f x sin x,
c
4
c1
35. gx 7. f x sin 2x, c 0
8. f x lnx 2 1,
9. f x sec x,
c 0 (first three nonzero terms)
In Exercises 11–14, prove that the Maclaurin series for the
function converges to the function for all x.
11. f x cos x
12. f x e2x
13. f x sinh x
14. f x cosh x
In Exercises 15–20, use the binomial series to find the
Maclaurin series for the function.
1
1 x2
1
f x 1 x
1
f x 4 x 2
4
f x 1x
f x 1 x 2
f x 1 x3
15. f x 17.
18.
19.
20.
arcsin x ,
x
x0
1,
x0
1 ix
e eix sin x
2i
37. f x e x sin x
38. gx e x cos x
39. hx cos x ln1 x
40. f x e x ln1 x
41. gx sin x
1x
ex
1x
42. f x In Exercises 43–46, match the polynomial with its graph. [The
graphs are labeled (a), (b), (c), and (d).] Factor a common
factor from each polynomial and identify the function approximated by the remaining Taylor polynomial.
y
(a)
y
(b)
4
2
x
x
−4
2
4
23. gx sin 3x
2
25. f x cos x32
4
2
2
−2
4
1
27. f x 2e x ex sinh x
28. f x e x ex 2 cosh x
43. y x 2 29. f x cos2 x
30. f x sinh1 x lnx x 2 1 Hint: Integrate the series for x 1 1.
2
x
−4
−2
4
−4
−4
3
4
y
(d)
x
−4
2
−4
y
22. gx e3x
24. f x cos 4x
−4 −2
4
−4
(c)
22
26. gx 2 sin x
x0
34. f x In Exercises 37–42, find the first four nonzero terms of the
Maclaurin series for the function by multiplying or dividing the
appropriate power series. Use the table of power series for
elementary functions on page 682. Use a graphing utility to
graph the function and its corresponding polynomial approximation.
In Exercises 21–30, find the Maclaurin series for the function.
(Use the table of power series for elementary functions.)
21. f x e x
x0
1
36. gx 2 eix eix cos x
c0
10. f x tan x, c 0 (first three nonzero terms)
16.
32. hx x cos x
In Exercises 35 and 36, use a power series and the fact that
i 2 1 to verify the formula.
5. f x ln x, c 1
6. f x e x,
685
Taylor and Maclaurin Series
x4
3!
45. y x x2 44. y x x3
2!
x3
x5
2! 4!
46. y x 2 x3 x 4
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Infinite Series
In Exercises 47 and 48, find a Maclaurin series for f x.
Probability In Exercises 61 and 62, approximate the normal
probability with an error of less than 0.0001, where the probability is given by
x
47. f x et 1 dt
2
0
x
48. f x 1 t3
Pa < x < b dt
0
1
2
b
ex
2
2
dx.
a
y
In Exercises 49 – 52, verify the sum. Then use a graphing utility
to approximate the sum with an error of less than 0.0001.
49.
1
n1
n1
50.
1
ln 2
n
1 2n 1! sin 1
1
n
n0
a
2n
e2
51.
n0 n!
52.
1n1
n1
62. P1 < x < 2
x→0
1 cos x
53. f x x
1
64. f x sin
sin x
dx
x
0
12
56.
57.
c0
65. gx x ln x, c 1
3 x arctan x,
66. hx c1
67. State the guidelines for finding a Taylor series.
68. If f is an even function, what must be true about the
coefficients an in the Maclaurin series
1 x3 dx
0.1
14
58.
x
ln1 x,
2
Writing About Concepts
arctan x
dx
x
0
0.3
In Exercises 63–66, use a computer algebra system to find the
fifth-degree Taylor polynomial (centered at c) for the function.
Graph the function and the polynomial. Use the graph to
determine the largest interval on which the polynomial is a
reasonable approximation of the function.
63. f x x cos 2x, c 0
sin x
x
In Exercises 55–58, use a power series to approximate the value
of the integral with an error of less than 0.0001. (In Exercises 55
and 56, assume that the integrand is defined as 1 when x 0.)
55.
x
b
61. P0 < x < 1
e1
1
n!
e
In Exercises 53 and 54, use the series representation of the function f to find lim f x (if it exists).
54. f x 1 e− x 2/2
2π
f(x) =
f x x lnx 1 dx
a x ?
n
n
n0
Explain your reasoning.
0
69. Explain how to use the series
Area In Exercises 59 and 60, use a power series to approximate
the area of the region. Use a graphing utility to verify the result.
2
59.
60.
0
3
4
1
2
1
4
xn
cos
x dx
to find the series for each function. Do not find the series.
y
(a) f x e x
0.5
y
n!
n0
1
x cos x dx
gx ex (b) f x e3x
1.5
(c) f x xe x
(d) f x e 2x e2x
1.0
70. Define the binomial series. What is its radius of
convergence?
0.5
x
π
8
π 3π
4 8
5π
8
x
0.5
1
1.5
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SECTION 9.10
71. Projectile Motion A projectile fired from the ground
follows the trajectory given by
g
g
kx
y tan x 2 ln 1 kv0 cos k
v0 cos x2 x3 x 4 . . .
,
ln1 x x 2
3
4
1 < x < 1
verify that the trajectory can be rewritten as
y tan x gx 2
kgx3
k 2 gx4
3
4
. . ..
2
3
cos 3v0 cos 4v0 cos4 2v02
72. Projectile Motion Use the result of Exercise 71 to determine
the series for the path of a projectile launched from ground
level at an angle of 60, with an initial speed of v0 64
1
feet per second and a drag factor of k 16
.
73. Investigation Consider the function f defined by
f x 0,e
x0
x 0.
1x 2,
(b) Use the alternative form of the definition of the derivative
(Section 2.1) and L’Hôpital’s Rule to show that f0 0.
[By continuing this process, it can be shown that f n0 0 for n > 1.]
(c) Using the result in part (b), find the Maclaurin series for f.
Does the series converge to f ?
and determine its radius of convergence. Use the first four
terms of the series to approximate ln 3.
In Exercises 77–80, evaluate the binomial coefficient using the
formula
nk kk 1k 2k n!3
k0 1.
82. Prove that e is irrational. Hint: Assume that e pq is
rational ( p and q are integers) and consider
e11
lnt 2 1
dt
t2
0.25
0.50
x
1 x x2
Fx
n
(c) Complete the table, where
x
1
1
. . . . . ..
2!
n!
83. Show that the Maclaurin series of the function
(b) Use a graphing utility to graph f and the eighth-degree
Taylor polynomial P8x for f.
0
78.
is
lnx 2 1
.
x2
22
13
80. 5 53
0.5
79. 4
77.
(a) Find the power series centered at 0 for the function
Fx . . . k n 1
where k is a real number, n is a positive integer, and
gx 74. Investigation
x
1x
1x
81. Write the power series for 1 xk in terms of binomial
coefficients.
(a) Sketch a graph of the function.
f x 687
76. Find the Maclaurin series for
f x ln
where v0 is the initial speed, is the angle of projection, g is the
acceleration due to gravity, and k is the drag factor caused by
air resistance. Using the power series representation
Taylor and Maclaurin Series
0.75
Gx P8t dt.
0
1.00
1.50
where Fn is the nth Fibonacci number with F1 F2 1 and
Fn Fn2 Fn1, for n ≥ 3.
Hint: Write
x
and
n
n1
2.00
x
a0 a1x a2 x 2 . . .
1 x x2
and multiply each side of this equation by 1 x x 2.
Fx
Gx
(d) Describe the relationship between the graphs of f and P8
and the results given in the table in part (c).
xn
75. Prove that lim
0 for any real x.
n→ n!
Putnam Exam Challenge
84. Assume that f x ≤ 1 and f x ≤ 1 for all x on an interval
of length at least 2. Show that fx ≤ 2 on the interval.
This problem was composed by the Committee on the Putnam Prize Competition.
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