Series 369
- Begin with the known seriesf(x)=
1
1 −x
=
∑∞
n= 0
xn= 1 +x+x^2 +x^3 +···
and replacexwith−x^2. Then
1
1 +x^2
=
1
1 −(−x^2 )
= 1 +(−x^2 )+(−x^2 )^2 +
(−x^2 )^3 +···= 1 −x^2 +x^4 −x^6 +···=
∑∞
n= 0
(−x^2 )n=
∑∞
n= 0
(−1)nx^2 n. The series
converges tof(x)=
1
1 +x^2
when
nlim→∞
∣∣
∣∣x
2 n+ 1
x^2 n
∣∣
∣∣=lim
n→∞|x|<1. The series
converges on (−1, 1). Whenx=1, the
series becomes
∑∞
n= 0
(−1)n, which diverges.
Whenx=−1, the series becomes
∑∞
n= 0
(−1)^3 n, which diverges. Therefore, the
interval of convergence is (−1, 1).
- Use the MacLaurin seriesf(x)=
sinx=
∑∞
n= 0
(−1)nx^2 n+^1
(2n+1)!
=x−
x^3
3!
+
x^5
5!
−
x^7
7!
+···with 9◦=
π
20
. Then
sin 9◦=sin
π
20
=
∑∞
n= 0
(−1)n(π/ 20 )^2 n+^1
(2n+1)!
=
π
20
−
(π/ 20 )^3
3!
+
(π/ 20 )^5
5!
−
(π/ 20 )^7
7!
+··· ≈ 0. 156.
20. 1. 83 = 1 +
83
102
+
83
104
+
83
106
+···
= 1 +
∑∞
n= 1
83
102 m
= 1 + 83
∑∞
n= 1
1
102 m
.
The geometric series
∑∞
n= 1
1
102 m
converges
tos=
a
1 −r
=
1 / 100
1 − 1 / 100
=
1
99
.
Therefore, 1. 83 = 1 + 83
∑∞
n= 1
1
102 m
= 1 + 83
(
1
99
)
=
182
99
.
14.12 Solutions to Cumulative Review Problems
- x′(t)=
1
t
x′′(t)=
− 1
t^2
y′(t)= 2 t
y〈′′(t)=2. The acceleration
− 1
t^2
,2
〉
=〈−1, 2〉whent=1.
The speed of the object at time√ t=1is
(
1
t
) 2
+(2t)^2 =
√
5.
- x=5cos 3θcosθandy=5cos 3θsinθ.
dx
dθ
=−5sinθcos 3θ−15cosθsin 3θ, and
dy
dθ
=5cosθcos 3θ −15sinθsin 3θ.
dy
dx
=
5cosθcos 3θ −15sinθsin 3θ
−5sinθcos 3θ−15cosθsin 3θ
=
3sinθsin 3θ−cosθcos 3θ
3cosθsin 3θ+sinθcos 3θ
.
Evaluated atθ=
2 π
3
,
dy
dx
=
3sin
( 2 π
3
)
sin 3
( 2 π
3
)
−cos
( 2 π
3
)
cos 3
( 2 π
3
)
3cos
( 2 π
3
)
sin 3
( 2 π
3
)
+sin
( 2 π
3
)
cos 3
( 2 π
3
)
=
1
√
3
=
√
3
3
. The slope of the tangent is
√
3
3