Series 369- Begin with the known seriesf(x)=
1
1 −x
=
∑∞n= 0xn= 1 +x+x^2 +x^3 +···and replacexwith−x^2. Then1
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 seriesconverges tof(x)=1
1 +x^2
whennlim→∞∣∣
∣∣x2 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, theinterval 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= 183
102 m= 1 + 83
∑∞n= 11
102 m.
The geometric series∑∞n= 11
102 m
convergestos=
a
1 −r=
1 / 100
1 − 1 / 100
=
1
99
.
Therefore, 1. 83 = 1 + 83∑∞n= 11
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