Quiz 8 for MATH 105 SECTION 205
April 01, 2015
Given Name
Family Name
Student Number
1. (1 point) Is the series
∞
X
sin(k 2 + 1)
√
convergent or divergent?
k3 − 1
k=1
convergent
1.
Answer.
For
∞
X
k=1
Notice that
√
1
k3
−1
sin(k 2 + 1) ≤ √ 1
√
.
k3 − 1 k3 − 1
, since
√ 1
k3 −1
1
3
k2
√
k3
=√
=
k3 − 1
r
k3
→ 1,
−1
as k → ∞.
k3
∞
X
1
∞
X
1
3
Since
> 1, then
By the limit comparison test, then
3 is a p-series with p =
3 converges.
2
2
2
k=1 k
k=1 k
∞
∞ ∞
X
X
X
sin(k 2 + 1) sin(k 2 + 1)
1
converges. So
√
√
√
converges. By the comparison test, then
converges.
k3 − 1
k3 − 1 k3 − 1
k=1
k=1
k=1
2. (1 point) Find the Taylor series of e2x centered at 1.
2.
∞ 2 k
X
e 2
k=0
Answer.
k!
(x − 1)k
In fact, we have
e2x = e2(x−1)+2)
= e2 · e2(x−1)
∞
X
[2(x − 1)]k
= e2
k!
= e2
k=0
∞
X
k=0
=
2k
(x − 1)k
k!
∞ 2 k
X
e 2
k=0
k!
(x − 1)k .
3. (1 point) Find the Maclaurin series of log3 (x + 1).
3.
∞
X
(−1)k+1
k=1
k ln 3
xk
Answer.
Notice that
log3 (x + 1) =
Since ln(1 − x) = −
∞
X
xk
k=1
k
ln(x + 1)
.
ln 3
, then
ln(x + 1) = −
∞
X
(−x)k
k=1
So we get
log3 (x + 1) = −
4. (1 point) Evaluate lim
x→0
k
.
∞
∞
k=1
k=1
X (−1)k+1
1 X (−x)k
=
xk .
ln 3
k
k ln 3
2 cos(2x) − 2 + 4x2
.
2x4
4.
Answer.
Since cos(x) =
∞
X
(−1)k
k=0
(2k)!
2
3
x2k , then
cos(2x) =
∞
X
(−1)k
k=0
=
(2k)!
∞
X
(−4)k
k=0
(2k)!
(2x)2k
x2k
4 2 16 4 64 6
x + x − x + ···
2!
4!
6!
4 2 16 4 64 6
= 1 − x + x − x + ···
2
24
6!
2 4 64 6
2
= 1 − 2x + x − x + · · · .
3
6!
= 1−
So we get
2 cos(2x) − 2 + 4x2
2x4
=
=
=
=
So we have
lim
x→0
6
2
2 1 − 2x2 + 32 x4 − 64
6! x + · · · − 2 + 4x
2x4
4
6
2
2 − 4x2 + 3 x4 + 2 − 64
6! x + · · · − 2 + 4x
2x4
64 6
4 4
3 x + 2 − 6! x + · · ·
2x4
2
64 2
2
+ − x + · · · → , as x → 0.
3
6!
3
2 cos(2x) − 2 + 4x2
2
= .
2x4
3
2
5. (1 point) Evaluate lim x sin
.
x→∞
x
5.
2
Answer.
Let t =
Since sin(t) =
2
2
, then x = and t → 0 as x → ∞. So we get
x
t
2
2 sin(t)
lim x sin
= lim
.
x→∞
t→0
x
t
∞
X
t3
(−1)k 2k+1
t
= t − + · · · , then
(2k + 1)!
3!
k=0
2 sin(t)
=
t
h
2 t−
t3
3!
+ ···
i
2
t
= 2 + 2 − + · · · → 2,
3!
t
as t → 0.
So we get
2 sin(t)
2
= lim
lim x sin
= 2.
x→∞
t→0
x
t
6. (1 point) Evaluate lim
x→4
x2 − 16
.
ln(x − 3)
8
6.
Answer.
Let t = x − 4, then x = t + 4 and t → 0 as x → 4 and
x2 − 16
(t + 4)2 − 16
t2 + 8t
t
=
=
=
· (t + 8).
ln(x − 3)
ln(t + 1)
ln(t + 1)
ln(t + 1)
Since ln(1 − t) = −
∞ k
X
t
k=1
k
, then
ln(t + 1) = −
∞
X
(−t)k
k=1
Then
k
=t−
t2
+ ··· .
2
2
t − t2 + · · ·
t
ln(t + 1)
=
= 1 + − + · · · → 1,
t
t
2
as t → 0.
So we get
lim
t→0
Then
ln(t + 1)
= 1.
t
(t + 4)2 − 16
t2 + 8t
t
x2 − 16
=
=
=
· (t + 8) → 1 · 8 = 8,
ln(x − 3)
ln(t + 1)
ln(t + 1)
ln(t + 1)
as t → 0.
So we get
lim
x→4
x2 − 16
=8
ln(x − 3)
7. (1 point) Identify the functions represented by the power series
∞
X
xk
k=1
7.
k
.
− ln |1 − x|
Answer.
Approach I: In fact, we have
!
Z X
Z
∞
∞ Z
∞
X
X
xk
k−1
k−1
=
x
dx =
x
dx =
k
k=1
k=1
k=1
Z
1
dx if |x| < 1
=
1−x
= − ln |1 − x| + C.
∞
X
!
xk
dx
k=0
Plug x = 0 into the above identities, then C = 0. So we get
∞
X
xk
k=1
Approach II: Let f (x) =
∞
X
xk
k=1
k
= − ln |1 − x|.
k
, then
∞
X
xk
d
f 0 (x) =
dx
k=1
So we have
Z
!
k
∞
X
xk−1 =
k=1
1
,
1−x
if |x| < 1.
1
dx = − ln |1 − x| + C.
1−x
f (x) =
Since f (0) = 0, then C = 0. Hence we get
∞
X
xk
k=1
k
= − ln |1 − x|.
8. (1 point) Identify the functions represented by the power series
∞
X
k(k − 1)xk
3k
k=2
6x2
(3 − x)3
8.
Answer.
.
In fact, we have
∞
X
k(k − 1)xk
k=2
3k
=
=
=
∞
X
k=2
∞
X
k=2
∞
X
x k
k(k − 1)
3
k(k − 1)y k
Let y =
x
3
y 2 · k(k − 1)y k−2
k=2
= y2
∞
X
k=2
k
d k−1 y
dy
∞
X
d2 k
2
(y )
= y
dy 2
k=2
∞
X
d2
= y2 2
dy
= y
2
d2
dy 2
!
yk
k=2
y2
1−y
if |y| < 1, that is, |x| < 3
=
=
=
=
=
=
=
=
=
d 2y(1 − y) + y 2
y
dy
(1 − y)2
2y − y 2
2 d
y
dy (1 − y)2
(2 − 2y)(1 − y)2 + (2y − y 2 ) · 2(1 − y)
y2 ·
(1 − y)4
(2 − 2y)(1 − y) + 2(2y − y 2 )
y2 ·
(1 − y)3
2 − 2y − 2y + 2y 2 + 4y − 2y 2
y2 ·
(1 − y)3
2
y2 ·
(1 − y)3
2y 2
(1 − y)3
2
2 · x3
x
Since y =
x 3
3
1−
2
3
2
6x
(3 − x)3
.
9. (2 points) *Identify the functions represented by the power series
∞
X
k=2
xk
.
k(k − 1)
(1 − x) ln(1 − x) + x
9.
Answer.
Let f (x) =
∞
X
k=2
xk
, then
k(k − 1)
0
f (x) =
d
dx
∞
X
k=2
xk
k(k − 1)
!
∞
X
xk−1
=
k−1
k=2
f 00 (x) =
=
d
dx
∞
X
∞
X
xk−1
k−1
!
k=2
xk−2
k=2
=
Then
0
f (x) =
Z
1
1−x
if |x| < 1.
1
dx = − ln |1 − x| + C.
1−x
Since f 0 (0) = 0, then C = 0. So we get
f 0 (x) = − ln |1 − x|.
Then
Z
f (x) = − ln(1 − x) dx
Z
=
ln u du Let u = 1 − x
Z
1
= u ln u − u · du Use the integration by parts
u
= u ln u − u + C
= (1 − x) ln(1 − x) − (1 − x) + C
Since u = 1 − x
= (1 − x) ln(1 − x) + x + C.
Since f (0) = 0, then C = 0. So we get
∞
X
k=2
xk
= (1 − x) ln(1 − x) + x.
k(k − 1)
10. (2 points) *Consider differential equation y 0 (x) = 2xy with initial condition y(0) = 1, find the power series
solution to the this differential equation.
∞
X
x2k
10.
k=0
Answer.
Let y(x) =
∞
X
k!
ck xk , since y(0) = 1, then c0 = 1. Since y 0 (x) = 2xy, then
k=0
0
y (x) =
d
dx
= 2x ·
∞
X
!
k
ck x
k=0
∞
X
=
∞
X
kck xk−1 =
k=0
ck xk =
∞
X
k=0
(2ck )xk+1
∞
X
(2ck−1 )xk .
=
k=1
k=0
k=0
∞
X
(k + 1)ck+1 xk
So we have c1 = 0 and (k + 1)ck+1 = 2ck−1 for all k ≥ 1, that is,
ck =
2
ck−2 ,
k
for all k ≥ 2,
and
c0 = 1,
c1 = 0.
So we know that ck = 0 if k is odd. If k = 2m is even, then
c2m =
2
2
2
2
2
2 2
1
c2m−2 =
·
c2m−4 =
·
· · · · c0 =
.
2m
2m 2m − 2
2m 2m − 2
4 2
m!
So we have
y(x) =
∞
X
m=0
c2m x2m =
∞
X
x2m
2
= ex .
m!
m=0
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