answers to the Power Series Worksheet

Math 113, Calculus II
Power Series - Answers
1. Find the radius of convergence and the interval of convergence of the following series.
(a) ∑ ∞
n=0
(b) ∑ ∞
n=1
(c) ∑ ∞
n=0
(d) ∑ ∞
n=0
x n
(2n)!
2 n
(x − 1)n
n2 5 n
n! xn
2. Use the fact that
(−1) n−1
(x + 3) n
n
tan −1 x =
∫ x
0
1
1 + t 2 dt
to find a power series expansion for tan −1 x centered at x = 0.
(a) Use this result to get a formula for π/4. [Hint: what should you set x equal to?]
tan −1 x
(b) Use the series to evaluate lim .
x→0 x
3. Use the Taylor series centered at 0 for e x , sin x, and
at 0 for the following functions.
1
to derive the Taylor series centered
1 − x
(a) e x2 (b) sin(x 3 ) (c) cos x (d) cos 5x
(e)
1
4 + x
(f)
6x
1 − x 2 (g) sin x + cos x (h)
sin x
x
4. Use the Taylor series centered at 0 to determine if sin x is even, odd, or neither. Also determine
the result for cos x and e x .
5. Find the Taylor series centered at 0 for the function sin 2 x. [Hint: Begin by differentiating
the function, then apply a famous trig. identity.] Use your answer to get a power series
representation for cos 2 x.
6. Give the first four terms in the Taylor series centered at 2 for the functions
1
(a)
1
5 x + 1
(b) e −x
7. Give an example of a power series whose interval of convergence is precisely −1 < x ≤ 1.
8. Use power series rather than l’Hôpital’s Rule to calculate the following limits.
(a) lim
x→0
sin(3x 2 )
x 2
(b) lim
x→0
1 + x − e x
x 2

Math 113, Calculus II
Power Series - Answers
Answers to the Power Series Worksheet
1. (a) radius: R = ∞; interval: (−∞, ∞)
(b) radius: R = 1 2
; interval: [1/2, 3/2]
(c) radius: R = ∞; interval: (−∞, ∞)
(d) radius: R = 1; interval: (−4, −2]
2. tan −1 x = ∑ ∞ x2n+1
n=0
(−1)n
2n + 1 = x − x3
3 + x5
5 − x7
7 + · · ·
(a) π/4 = tan −1 (1) = ∑ ∞ (1)2n+1
n=0
(−1)n
2n + 1 = ∑ ∞
n=0
(b) lim x→0
tan −1 x
x
[ (
∑∞
= lim x→0 x −1 x2n+1
n=0
(−1)n
2n + 1
lim x→0
[
1 − x2
3 + x4
5 − x6
7 + · · · ]
= 1
3. (a) e x2 = ∑ ∞ (x 2 ) n
n=0
n!
= ∑ ∞ x 2n
n=0
n!
(b) sin(x 3 ) = ∑ ∞
n=0 (−1)n (x3 ) 2n+1
(2n + 1)! = ∑ ∞ x6n+3
n=0
(−1)n
(2n + 1)!
]
(c) cos x = d d
[sin x] =
dx dx
(d) cos 5x = ∑ ∞ (5x)2n
n=0
(−1)n
(2n)!
(e)
(f)
[
∑∞ x2n+1
n=0
(−1)n
(2n + 1)!
= ∑ ∞
n=0 (−1)n 25n x 2n
(2n)!
(−1) n
2n + 1 = 1 − 1 3 + 1 5 − 1 7 + · · ·
)] [
∑∞ x2n
= lim x→0 n=0
(−1)n
2n + 1
= ∑ ∞ (2n + 1)x2n
n=0
(−1)n
(2n + 1)!
1
4 + x = 1 4 · 1
1 + (x/4) = 1 4 · 1
1 − (−x/4) = 1 4 [∑ ∞
n=0 (−x/4)n ] = 1 4
∑ ∞ (−1) n x n
n=0
4 n+1
[ ]
6x
1 − x 2 = 6x 1
1 − (x 2 = 6x [∑ ∞
n=0
)
(x2 ) n] = ∑ ∞
n=0 6x2n+1
(g) sin x + cos x = ∑ ∞
k=0 c kx k where
⎧
⎪⎨
(−1) k/2
c k = k!
⎪⎩
k!
(−1) (k−1)/2
if k is even
if k is odd
[
∑∞
n=0
]
=
= ∑ ∞ x2n
n=0
(−1)n
(2n)!
(−1) n x n ]
4 n =
(h) sin x
x
= ∑ x−1 ∞ x2n+1
n=0
(−1)n
(2n + 1)! = ∑ ∞ x 2n
n=0 (−1)n (2n + 1)!
4. sin x = ∑ ∞ x2n+1
n=0
(−1)n
(2n + 1)!
only involves odd exponents on x, so it is an odd function;
cos x = ∑ ∞ x2n
n=0
(−1)n
(2n)! only involves even exponents on x, so it is an even function; ex =
∑ ∞ x n
n=0
involves both even and odd exponents on x, so it is neither an even nor an odd
n!
function.

Math 113, Calculus II
Power Series - Answers
5. We know d [
sin 2 x ] = 2(sin x) cos x = sin(2x) = ∑ ∞ (2x)2n+1
n=0
(−1)n
dx
so sin 2 x = ∫ sin(2x) dx = ∫ ∑ ∞
n=0 (−1)n 22n+1 x 2n+1
(2n + 1)!
C + ∑ ∞ (−1) n 2 2n+1
n=0
(2n + 1)!
we find C = 0, so
Thus
cos 2 x = 1 − sin 2 x = 1 −
x 2n+2
(2n + 2) = C + ∑ ∞
sin 2 x =
dx = C+ ∑ ∞
n=0
(2n + 1)! = ∑ ∞
n=0 (−1)n 22n+1 x 2n+1
,
(2n + 1)!
(−1) n 2 2n+1 ∫
x 2n+1 dx =
(2n + 1)!
n=0 (−1)n 22n+1 x 2n+2
. Setting x = 0 on both sides,
(2n + 2)!
∞∑
(−1) n 22n+1 x 2n+2
(2n + 2)!
n=0
∞∑
(−1) n 22n+1 x 2n+2
(2n + 2)!
n=0
= 1 − x 2 + 8 4! x4 − 32
6! x6 + · · ·
1
6. (a) Letting f(x) =
1
5 x + 1 = (1 + 0.2x)−1 , we have f ′ (x) = −(1 + 0.2x) −2 [0.2] = −0.2(1 +
0.2x) −2 , f ′′ (x) = 0.4(1 + 0.2x) −3 [0.2] = 0.08(1 + 0.2x) −3 , and f ′′′ (x) = −0.24(1 +
0.2x) −4 [0.2] = −0.048(1+0.2x) −4 . Thus the first four terms of the Taylor series centered
at 2 for f(x) is
T 3 (x) = f(2) + f ′ (2)(x − 2) + f ′′ (2)
(x − 2) 2 + f ′′′ (2)
(x − 2) 3
2!
3!
= [5/7] + [−5/49](x − 2) + [10/343](x − 2) 2 + [−30/2401](x − 2) 3
(b) Beginning with g(x) = e −x we find f ′ (x) = −e −x , f ′′ (x) = e −x and f ′′′ (x) = −e −x .
Hence the first four terms of the Taylor series centered at x = 2 for g(x) is
T 3 (x) = g(2) + g ′ (2)(x − 2) + g′′ (2)
(x − 2) 2 + g′′′ (2)
(x − 2) 3
2!
3!
= [e −2 ] + [−e −2 ](x − 2) + [e −2 ](x − 2) 2 + [−e −2 ](x − 2) 3
7. An example of a power series that converges precisely for −1 < x ≤ 1 is ∑ ∞ xn
n=1
(−1)n−1
n .
8. (a) limx −2 ∑ ∞
x→0
n=0 (−1)n (3x2 ) 2n+1
(2n + 1)! = lim ∑ x→0 x−2 ∞
n=0 (−1)n 32n+1 x 4n+2
(2n + 1)!
27x4
lim(3 − + · · · ) = 3
x→0 3!
(b) lim
x→0
1 + x − e x
x 2
[
= limx −2 1 + x − ∑ ∞
x→0
n=0
x n ] [
= limx −2
n! x→0
= lim
x→0
∑ ∞
n=0 (−1)n 32n+1 x 4n
(2n + 1)! =
]
1 + x − 1 − x − x2
2! − x3
3! − · · · =
]
= −1/2
[
] [
lim
x→0 x−2 − x2
2! − x3
3! − · · · − xn
n! − · · · = lim − 1
x→0 2! − x 3! − · · · − xn−2
− · · ·
n!