Section 11.9 Representations of Functions as Power Series
12. Find a power series representation for the function and determine the interval of convergence. f (x) = xx+a2+a2, a > 0 Solution: SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES ¤ 1121 12. () = +
2+ 2 [ 0] =
2
1
1 − (−22)
+
2
1
1 − (−22)
=
2
∞
=0
−2
2
+ 1
∞
=0
−2
2
= ∞
=0
(−1)2+1
2+1 + ∞
=0
(−1)2
2+1 . The geometric series ∞
=0
−2
2
converges when
−2
2
1 ⇔ || , so = and = (− ).
13. () = 2 − 4
2− 4 + 3 = 2 − 4
( − 1)( − 3) =
− 1 +
− 3 ⇒ 2 − 4 = ( − 3) + ( − 1). Let = 1 to get
−2 = −2 ⇔ = 1 and = 3 to get 2 = 2 ⇔ = 1. Thus, 2 − 4
2− 4 + 3 = 1
− 1+ 1
− 3 = −1 1 − + 1
−3
1
1 − (3)
= − ∞
=0
− 1 3
∞
=0
3
= ∞
=0
−1 − 1 3+1
. We represented as the sum of two geometric series; the first converges for ∈ (−1 1) and the second converges for
∈ (−3 3). Thus, the sum converges for ∈ (−1 1) =
14. () = 2 + 3
2+ 3 + 2= 2 + 3
( + 1)( + 2) =
+ 1+
+ 2 ⇒ 2 + 3 = ( + 2) + ( + 1). Let = −1 to get 1 = and = −2 to get −1 = − ⇔ = 1. Thus,
2 + 3
2+ 3 + 2= 1
+ 1 + 1
+ 2 = 1
1 − (−)+1 2
1
1 − (−2)
= ∞
=0(−)+ 1 2
∞
=0
− 2
= ∞
=0
(−1)
1 + 1
2+1
We represented as the sum of two geometric series; the first converges for ∈ (−1 1) and the second converges for
∈ (−2 2). Thus, the sum converges for ∈ (−1 1) =
15. (a) () = 1
(1 + )2 =
−1 1 +
= −
∞
=0(−1)
[from Exercise 3]
= ∞
=1(−1)+1−1 [from Theorem 2(i)] = ∞
=0(−1)( + 1)with = 1.
In the last step, note that we decreased the initial value of the summation variable by 1, and then increased each occurrence of in the term by 1 [also note that (−1)+2= (−1)].
(b) () = 1
(1 + )3 = −1 2
1
(1 + )2
= −1 2
∞
=0(−1)( + 1)
[from part (a)]
= −12
∞
=1(−1)( + 1)−1= 12 ∞
=0(−1)( + 2)( + 1)with = 1.
(c) () = 2
(1 + )3 = 2· 1
(1 + )3 = 2· 1 2
∞
=0(−1)( + 2)( + 1) [from part (b)]
= 1 2
∞
=0(−1)( + 2)( + 1)+2
To write the power series with rather than +2, we will decrease each occurrence of in the term by 2 and increase the initial value of the summation variable by 2. This gives us1
2
∞
=2(−1)()( − 1)with = 1.
° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
15. (a) Use differentiation to find a power series representation for f (x) = (1+x)1 2. What is the radius of convergence?
(b) Use part (a) to find a power series for f (x) = (1+x)1 3. (c) Use part (b) to find a power series for f (x) = (1+x)x2 3. Solution:
170 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES 9. () = − 1
+ 2 = + 2 − 3
+ 2 = 1 − 3
+ 2 = 1 − 32
2 + 1 = 1 − 3
2· 1
1 − (−2)
= 1 − 3 2
∞
=0
− 2
= 1 −3 2− 3
2
∞
=1
− 2
= −1 2− ∞
=1
(−1)3 2+1 . The geometric series ∞
=0
− 2
converges when− 2
1 ⇔ || 2, so = 2 and = (−2 2).
Alternatively, you could write () = 1 − 3
1
+ 2
and use the series for 1
+ 2found in Example 2.
10. () =
2+ 2 [ 0] =
2
1
1 − (−22)
= 1
∞
=0
−2
2
= ∞
=0
(−1)2
2+1 . The geometric series
∞
=0
−2
2
converges when−2
2
1 ⇔ || , so = and = (− ).
11. () = 2 − 4
2− 4 + 3 = 2 − 4
( − 1)( − 3) =
− 1+
− 3 ⇒ 2 − 4 = ( − 3) + ( − 1). Let = 1 to get
−2 = −2 ⇔ = 1 and = 3 to get 2 = 2 ⇔ = 1. Thus, 2 − 4
2− 4 + 3 = 1
− 1+ 1
− 3 = −1 1 − + 1
−3
1
1 − (3)
= − ∞
=0
−1 3
∞
=0
3
= ∞
=0
−1 − 1 3+1
. We represented as the sum of two geometric series; the first converges for ∈ (−1 1) and the second converges for
∈ (−3 3). Thus, the sum converges for ∈ (−1 1) =
12. () = 2 + 3
2+ 3 + 2= 2 + 3
( + 1)( + 2) =
+ 1+
+ 2 ⇒ 2 + 3 = ( + 2) + ( + 1). Let = −1 to get 1 = and = −2 to get −1 = − ⇔ = 1. Thus,
2 + 3
2+ 3 + 2= 1
+ 1+ 1
+ 2 = 1
1 − (−)+1 2
1
1 − (−2)
= ∞
=0(−)+1 2
∞
=0
− 2
= ∞
=0
(−1)
1 + 1
2+1
We represented as the sum of two geometric series; the first converges for ∈ (−1 1) and the second converges for
∈ (−2 2). Thus, the sum converges for ∈ (−1 1) =
13. (a) () = 1
(1 + )2 =
−1 1 +
= −
∞
=0(−1)
[from Exercise 3]
= ∞
=1(−1)+1−1 [from Theorem 2(i)] = ∞
=0(−1)( + 1)with = 1.
In the last step, note that we decreased the initial value of the summation variable by 1, and then increased each occurrence of in the term by 1 [also note that (−1)+2= (−1)].
(b) () = 1
(1 + )3 = −1 2
1
(1 + )2
= −1 2
∞
=0(−1)( + 1)
[from part (a)]
= −12
∞
=1(−1)( + 1)−1= 12 ∞
=0(−1)( + 2)( + 1)with = 1.
(c) () = 2
(1 + )3 = 2· 1
(1 + )3 = 2· 1 2
∞
=0(−1)( + 2)( + 1) [from part (b)]
= 1 2
∞
=0(−1)( + 2)( + 1)+2 [continued]
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES ¤ 171
To write the power series with rather than +2, we will decrease each occurrence of in the term by 2 and increase the initial value of the summation variable by 2. This gives us1
2
∞
=2(−1)()( − 1)with = 1.
14. (a) 1
1 − = − ln(1 − ) + and
1
1 − =
(1 + + 2+ · · · ) =
+2 2 +3
3 + · · ·
+ = ∞
=1
+ for || 1.
So − ln(1 − ) = ∞
=1
+ and letting = 0 gives 0 = . Thus, () = ln(1 − ) = − ∞
=1
with = 1.
(b) () = ln(1 − ) = − ∞
=1
= −∞
=1
+1
.
(c) Letting = 1
2gives ln1 2 = −
∞
=1
(12)
⇒ ln 1 − ln 2 = − ∞
=1
1
2 ⇒ ln 2 = ∞
=1
1
2.
15. () = ln(5 − ) = −
5 − = −1 5
1 − 5 = −1 5
∞
=0
5
= − 1 5
∞
=0
+1
5( + 1) = − ∞
=1
5 Putting = 0, we get = ln 5. The series converges for |5| 1 ⇔ || 5, so = 5.
16. () = 2tan−1(3) = 2 ∞
=0(−1)(3)2+1
2 + 1 [by Example 7] = ∞
=0(−1)6+3+2 2 + 1 = ∞
=0(−1)6+5 2 + 1for
3 1 ⇔ || 1, so = 1.
17. We know that 1
1 + 4 = 1
1 − (−4) = ∞
=0(−4). Differentiating, we get
−4
(1 + 4)2 = ∞
=1(−4)−1= ∞
=0(−4)+1( + 1), so
() =
(1 + 4)2 = −
4 · −4
(1 + 4)2 = − 4
∞
=0(−4)+1( + 1)= ∞
=0(−1)4( + 1)+1 for |−4| 1 ⇔ || 14, so = 14.
18. 1
2 − = 1
2(1 − 2) = 1 2
∞
=0
2
= ∞
=0
1
2+1. Now
1
2 −
=
∞
=0
1 2+1
⇒
1
(2 − )2 = ∞
=1
1
2+1−1and
1
(2 − )2
=
∞
=1
1
2+1−1
⇒ 2
(2 − )3 = ∞
=2
1
2+1( − 1)−2= ∞
=0
( + 2)( + 1) 2+3 . Thus, () =
2 −
3
= 3
(2 − )3 = 3
2 · 2
(2 − )3 = 3 2
∞
=0
( + 2)( + 1)
2+3 = ∞
=0
( + 2)( + 1) 2+4 +3 for
2
1 ⇔ || 2, so = 2.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
16. (a) Use Equation 1 to find a power series representation for f (x) = ln(1 − x). What is the radius of convergence?
(b) Use part (a) to find a power series for f (x) = x ln(1 − x).
(c) By putting x = 12 in your result from part (a), express ln 2 as the sum of an infinite series.
Solution:
SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES ¤ 171
To write the power series with rather than +2, we will decrease each occurrence of in the term by 2 and increase the initial value of the summation variable by 2. This gives us 1
2
∞
=2(−1)()( − 1)with = 1.
14. (a) 1
1 − = − ln(1 − ) + and
1
1 − =
(1 + + 2+ · · · ) =
+ 2 2 + 3
3 + · · ·
+ = ∞
=1
+ for || 1.
So − ln(1 − ) = ∞
=1
+ and letting = 0 gives 0 = . Thus, () = ln(1 − ) = − ∞
=1
with = 1.
(b) () = ln(1 − ) = − ∞
=1
= − ∞
=1
+1
.
(c) Letting = 1
2gives ln1 2= −
∞
=1
(12)
⇒ ln 1 − ln 2 = − ∞
=1
1
2 ⇒ ln 2 = ∞
=1
1
2.
15. () = ln(5 − ) = −
5 − = −1 5
1 − 5 = −1 5
∞
=0
5
= −1 5
∞
=0
+1
5( + 1) = − ∞
=1
5 Putting = 0, we get = ln 5. The series converges for |5| 1 ⇔ || 5, so = 5.
16. () = 2tan−1(3) = 2 ∞
=0(−1)(3)2+1
2 + 1 [by Example 7] = ∞
=0(−1)6+3+2 2 + 1 = ∞
=0(−1)6+5 2 + 1 for
3 1 ⇔ || 1, so = 1.
17. We know that 1
1 + 4 = 1
1 − (−4) = ∞
=0(−4). Differentiating, we get
−4
(1 + 4)2 = ∞
=1(−4)−1 = ∞
=0(−4)+1( + 1), so
() =
(1 + 4)2 = −
4 · −4
(1 + 4)2 = − 4
∞
=0(−4)+1( + 1)= ∞
=0(−1)4( + 1)+1 for |−4| 1 ⇔ || 14, so = 14.
18. 1
2 − = 1
2(1 − 2) = 1 2
∞
=0
2
= ∞
=0
1
2+1. Now
1 2 −
=
∞
=0
1 2+1
⇒
1
(2 − )2 = ∞
=1
1
2+1−1and
1
(2 − )2
=
∞
=1
1
2+1−1
⇒ 2
(2 − )3 = ∞
=2
1
2+1( − 1)−2= ∞
=0
( + 2)( + 1) 2+3 . Thus, () =
2 −
3
= 3
(2 − )3 = 3
2 · 2
(2 − )3 = 3 2
∞
=0
( + 2)( + 1)
2+3 = ∞
=0
( + 2)( + 1) 2+4 +3 for
2
1 ⇔ || 2, so = 2.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
1
SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES ¤ 171
To write the power series with rather than +2, we will decrease each occurrence of in the term by 2 and increase the initial value of the summation variable by 2. This gives us1
2
∞
=2(−1)()( − 1)with = 1.
14. (a) 1
1 − = − ln(1 − ) + and
1
1 − =
(1 + + 2+ · · · ) =
+ 2 2 + 3
3 + · · ·
+ = ∞
=1
+ for || 1.
So − ln(1 − ) = ∞
=1
+ and letting = 0 gives 0 = . Thus, () = ln(1 − ) = − ∞
=1
with = 1.
(b) () = ln(1 − ) = − ∞
=1
= − ∞
=1
+1
.
(c) Letting = 1
2gives ln1 2 = −
∞
=1
(12)
⇒ ln 1 − ln 2 = − ∞
=1
1
2 ⇒ ln 2 = ∞
=1
1
2.
15. () = ln(5 − ) = −
5 − = −1 5
1 − 5 = −1 5
∞
=0
5
= −1 5
∞
=0
+1
5( + 1) = − ∞
=1
5 Putting = 0, we get = ln 5. The series converges for |5| 1 ⇔ || 5, so = 5.
16. () = 2tan−1(3) = 2 ∞
=0(−1)(3)2+1
2 + 1 [by Example 7] = ∞
=0(−1)6+3+2 2 + 1 = ∞
=0(−1)6+5 2 + 1 for
3 1 ⇔ || 1, so = 1.
17. We know that 1
1 + 4 = 1
1 − (−4) = ∞
=0(−4). Differentiating, we get
−4
(1 + 4)2 = ∞
=1(−4)−1 = ∞
=0(−4)+1( + 1), so
() =
(1 + 4)2 = −
4 · −4
(1 + 4)2 = − 4
∞
=0(−4)+1( + 1)= ∞
=0(−1)4( + 1)+1 for |−4| 1 ⇔ || 14, so = 14.
18. 1
2 − = 1
2(1 − 2) = 1 2
∞
=0
2
= ∞
=0
1
2+1. Now
1 2 −
=
∞
=0
1 2+1
⇒ 1
(2 − )2 = ∞
=1
1
2+1−1and
1
(2 − )2
=
∞
=1
1
2+1−1
⇒ 2
(2 − )3 = ∞
=2
1
2+1( − 1)−2= ∞
=0
( + 2)( + 1) 2+3 . Thus, () =
2 −
3
= 3
(2 − )3 = 3
2 · 2
(2 − )3 = 3 2
∞
=0
( + 2)( + 1)
2+3 = ∞
=0
( + 2)( + 1) 2+4 +3 for
2
1 ⇔ || 2, so = 2.
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
28. Evaluate the indefinite integral as a power series. What is the radius of convergence? R t 1+t3dt Solution:
174 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES 26.
1 + 3 = · 1
1 − (−3) = ∞
=0(−3)= ∞
=0(−1)3+1 ⇒
1 + 3 = + ∞
=0(−1) 3+2
3 + 2. The series for 1
1 + 3 converges when
−3
1 ⇔ || 1, so = 1 for that series and also for the series
1 + 3. By Theorem 2, the series for
1 + 3 also has = 1.
27. From Example 6, ln(1 + ) = ∞
=1(−1)−1
for || 1, so 2ln(1 + ) = ∞
=1(−1)−1+2
and
2ln(1 + ) = + ∞
=1(−1)−1 +3
( + 3). = 1 for the series for ln(1 + ), so = 1 for the series representing
2ln(1 + )as well. By Theorem 2, the series for
2ln(1 + ) also has = 1.
28. From Example 7, tan−1 = ∞
=0(−1)2+1
2 + 1 for || 1, so tan−1
= ∞
=0(−1) 2
2 + 1and
tan−1
= + ∞
=0(−1) 2+1
(2 + 1)2. = 1 for the series for tan−1, so = 1 for the series representing tan−1
as well. By Theorem 2, the series for
tan−1
also has = 1.
29. 1 + 3 =
1
1 − (−3)
= ∞
=0(−3) = ∞
=0(−1)3+1 ⇒
1 + 3 =
∞
=0(−1)3+1 = + ∞
=0(−1)3+2 3 + 2. Thus,
=
03 0
1 + 3 =
2 2 −5
5 +8 8 −11
11 + · · ·
03
0
= (03)2
2 −(03)5
5 +(03)8
8 −(03)11 11 + · · · . The series is alternating, so if we use the first three terms, the error is at most (03)1111 ≈ 16 × 10−7. So
≈ (03)22 − (03)55 + (03)88 ≈ 0044 522 to six decimal places.
30. We substitute 2 for in Example 7, and find that
arctan(2) =
∞
=0(−1)(2)2+1 2 + 1 =
∞
=0(−1) 2+1 22+1(2 + 1)
= + ∞
=0(−1) 2+2
22+1(2 + 1)(2 + 2) Thus,
=
12 0
arctan(2) =
2
2(1)(2)− 4
23(3)(4)+ 6
25(5)(6)− 8
27(7)(8)+ 10
29(9)(10)− · · ·
12
0
= 1
23(1)(2)− 1
27(3)(4)+ 1
211(5)(6)− 1
215(7)(8)+ 1
219(9)(10)− · · ·
[continued]
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
40. The Bessel function of order 1 is defined by J1(x) =
∞
P
n=0 (−1)n
x 2n+1
n!(n + 1)!22n+1 (a) Show that J1 satisfies the differential equation x2J100(x) + xJ10(x) + (x2− 1)J1(x) = 0 (b) Show that J00(x) = −J1(x).
(c) Show that J00(x) = −J1(x).
Solution:
176 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES 35. (a) 0() = ∞
=0
(−1)2
22(!)2 00() = ∞
=1
(−1)22−1
22(!)2 , and 000() = ∞
=1
(−1)2(2 − 1)2−2 22(!)2 , so
2000() + 00() + 20() = ∞
=1
(−1)2(2 − 1)2
22(!)2 + ∞
=1
(−1)22
22(!)2 + ∞
=0
(−1)2+2 22(!)2
= ∞
=1
(−1)2(2 − 1)2
22(!)2 + ∞
=1
(−1)22
22(!)2 + ∞
=1
(−1)−12
22−2[( − 1)!]2
= ∞
=1
(−1)2(2 − 1)2
22(!)2 + ∞
=1
(−1)22
22(!)2 + ∞
=1
(−1)(−1)−12222
22(!)2
= ∞
=1(−1)
2(2 − 1) + 2 − 222 22(!)2
2
= ∞
=1(−1)
42− 2 + 2 − 42 22(!)2
2= 0
(b) 1 0
0() =
1 0
∞
=0
(−1)2
22(!)2
=
1 0
1 −2
4 +4 64 − 6
2304 + · · ·
=
− 3
3 · 4+ 5
5 · 64− 7
7 · 2304+ · · ·
1
0
= 1 − 1 12+ 1
320− 1
16,128+ · · ·
Since 16,1281 ≈ 0000062, it follows from The Alternating Series Estimation Theorem that, correct to three decimal places,
1
0 0() ≈ 1 −121 +3201 ≈ 0920.
36. (a) 1() = ∞
=0
(−1)2+1
! ( + 1)! 22+1, 10() = ∞
=0
(−1)(2 + 1) 2
! ( + 1)! 22+1 , and 100() = ∞
=1
(−1)(2 + 1) (2) 2−1
! ( + 1)! 22+1 .
2100() + 10() +
2− 1
1()
= ∞
=1
(−1)(2 + 1)(2)2+1
! ( + 1)! 22+1 + ∞
=0
(−1)(2 + 1)2+1
! ( + 1)! 22+1 +∞
=0
(−1)2+3
! ( + 1)! 22+1 − ∞
=0
(−1)2+1
! ( + 1)! 22+1
= ∞
=1
(−1)(2 + 1)(2)2+1
! ( + 1)! 22+1 + ∞
=0
(−1)(2 + 1)2+1
! ( + 1)! 22+1
−∞
=1
(−1)2+1
( − 1)! ! 22−1 − ∞
=0
(−1)2+1
! ( + 1)! 22+1
Replace with − 1 in the third term
= 2 −
2 + ∞
=1(−1)
(2 + 1)(2) + (2 + 1) − ()( + 1)22− 1
! ( + 1)! 22+1
2+1= 0
(b) 0() = ∞
=0
(−1)2
22(!)2 ⇒
00() = ∞
=1
(−1)(2)2−1 22(!)2 = ∞
=0
(−1)+12( + 1)2+1
22+2[( + 1)!]2 [Replace with + 1]
= − ∞
=0
(−1)2+1
22+1( + 1)! ! [cancel 2 and + 1; take −1 outside sum] = −1()
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
2
1128 ¤ CHAPTER 11 SEQUENCES, SERIES, AND POWER SERIES (c) 0() = ∞
=0
(−1)2
22(!)2 ⇒
00() = ∞
=1
(−1)(2)2−1 22(!)2 = ∞
=0
(−1)+12( + 1)2+1
22+2[( + 1)!]2 [Replace with + 1]
= − ∞
=0
(−1)2+1
22+1( + 1)! ! [cancel 2 and + 1; take −1 outside sum] = −1()
41. (a) () = 1 + ∞
=1
, where = 3
2 · 3 · 5 · 6 · · · (3 − 1)(3), so lim
→∞
+1
= ||3 lim
→∞
1
(3 + 2)(3 + 3) = 0 for all , so the domain is .
(b), (c) 0= 1has been omitted from the graph. The
partial sums seem to approximate () well near the origin, but as || increases, we need to take a large number of terms to get a good approximation.
To plot , we must first define () for the CAS. Note that for ≥ 1, the denominator of is
2 · 3 · 5 · 6 · · · (3 − 1) · 3 = (3)!
1 · 4 · 7 · · · (3 − 2) = (3)!
=1(3 − 2), so =
=1(3 − 2)
(3)! 3and thus
() = 1 + ∞
=1
=1(3 − 2)
(3)! 3. Both Maple and Mathematica are able to plot if we define it this way.
Maple and Mathematica have two initially known Airy functions, called AI·SERIES(z,m) and BI·SERIES(z,m) from AiryAi and AiryBi in Maple and Mathematica (just Ai and Bi in older versions of Maple). However, it is very
difficult to solve for in terms of the CAS’s Airy functions, although in fact () =
√3AiryAi() + AiryBi()
√3AiryAi(0) + AiryBi(0).
42. () =∞
=0 , where +4= for all ≥ 0. So
4−1= 0+ 1 + 22+ 33+ 04+ 15+ 26+ 37+ · · · + 34−1
=
0+ 1 + 22+ 33
1 + 4+ 8+ · · · + 4−4
→ 0+ 1 + 22+ 33
1 − 4 as → ∞ [by (11.2.4) with = 4] for4 1 ⇔ || 1. Also 4, 4+1, 4+2have the same limits (for example,
4= 4−1+ 04and 4→ 0 for || 1). So if at least one of 0, 1, 2, and 3is nonzero, then the interval of convergence is (−1 1) and () = 0+ 1 + 22+ 33
1 − 4 .
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