Quiz 7 for MATH 105 SECTION 205
March 30, 2015
Family Name
Given Name
Student Number
1. (1 point) Is the series
∞
X
k 10 10k (k!)2
(2k)!
k=1
convergent or divergent?
divergent
1.
Answer.
Let ak =
k 10 10k (k!)2
, then
(2k)!
ak+1
ak
=
=
(k+1)10 10k+1 [(k+1)!]2
[2(k+1)]!
k10 10k (k!)2
(2k)!
10
(k + 1) 10k+1 [(k +
1)!]2
[2(k + 1)]!
·
(2k)!
k 10 10k (k!)2
(k +
10k+1
(k + 1)! 2
(2k)!
=
·
·
·
10
k
k
k!
(2k + 2)!
10
10
1
1
=
1+
· 10 · (k + 1)2 ·
k
(2k + 2)(2k + 1)
10
(k + 1)2
1
·
= 10 1 +
k
(2k + 2)(2k + 1)
5
1
→ 10 · 1 · = , as k → ∞.
4
2
1)10
By the ratio test, we know that the series
∞
X
k 10 10k (k!)2
k=1
2. Consider the power series
∞
X
2k (x − 3)k
k=1
k+1
(2k)!
diverges.
.
(a) (1 point) Find the radius of convergence of this series.
(a)
Answer.
Let ck =
1
2
2k
, then
k+1
ck+1
2k+1 k + 1
2k+1 k + 1
k+1
=
· k = k ·
=2·
→ 2,
ck
k+2
k+2
k+2
2
2
as k → ∞.
1
So the radius of convergence is R = .
2
(b) (1 point) Find the interval of convergence of this series.
(b)
(5/2, 7/2)
1
and center is a = 3, then the interval of convergence
2
is (a − R, a + R) = (3 − 1/2, 3 + 1/2) = (5/2, 7/2).
Answer.
Since the radius of convergence is R =
3. (1 point) Find the value of x such that
∞
X
ekx = 1.
k=1
− ln 2
3.
Answer.
∞
∞
X
X
x k
Since
(e ) =
ekx = 1, then |ex | < 1 and
k=1
k=1
∞
X
ekx =
k=1
ex
= 1.
1 − ex
1
So we get ex = 1 − ex , that is, 2ex = 1. So we get ex = , that is,
2
x = ln
4. (1 point) Find the power series of f (x) =
1
= − ln 2.
2
4
centered at 0.
4 + x2
4.
∞ X
1 k 2k
−
x
4
k=0
Answer.
In fact, we have
f (x) =
=
4
4 + x2
4
1
·
4 1 + x2
4
=
1
1+
x2
4
1
2
1 − − x4
k
∞ X
x2
x2
=
−
if −
< 1, that is, |x| < 2
4
4
k=0
∞ X
1 k 2k
=
−
x .
4
=
k=0
5. (2 points) Find the power series of f (x) =
1
centered at 1.
4 − 2x
5.
∞
X
(x − 1)k
k=0
2
Answer.
In fact, we have
1
4 − 2x
f (x) =
1
4 − 2(x − 1) − 2
1
2 − 2(x − 1)
1
1
·
2 1 − (x − 1)
∞
1X
(x − 1)k if |x − 1| < 1
2
=
=
=
=
k=0
∞
X
=
k=0
(x − 1)k
.
2
6. (1 point) Find the function represented by the series
k
∞ 2
X
x −1
k=0
3
.
6.
Answer.
In fact, we have
k
∞ 2
X
x −1
=
3
k=0
=
=
1
1−
x2 − 1
< 1, that is, |x2 − 1| < 3, then |x| < 2
3
if
x2 −1
3
3
3 − x2 + 1
3
.
4 − x2
7. (2 points) *Find the function represented by the series
∞
X
x2k
k=1
4k
.
7.
Answer.
3
4 − x2
Approach I: Let f (x) =
∞
X
x2k
k=1
4k
, then
f 0 (x) =
=
d
dx
∞
X
x2k
k=1
∞
X
x2k−1
k=1
x
2
!
4k
2
if |x2 | < 1
1 − x2
1
x
= − · 2
.
2 x −1
=
So we get
1
f (x) = −
2
Z
x2
x
1
dx = − ln |1 − x2 | + C.
−1
4
1
− ln(1 − x2 )
4
Taking x = 0 in the above identity, we get C = 0. So we get
∞
X
x2k
k=1
1
= − ln(1 − x2 ).
4k
4
Approach II: In fact, we have
∞
X
x2k
k=1
4k
∞
=
=
=
=
=
=
1 X yk
Let y = x2
4
k
k=1
∞ Z
X
1
y k−1 dy
4
k=1
!
Z X
∞
1
y k−1 dy
4
Z k=1
1
1
dy
4
1−y
1
− ln |1 − y| + C if |y| < 1, that is, |x| < 1
4
1
− ln(1 − x2 ) + C.
4
Taking x = 0 in the above identity, we get C = 0. So we get
∞
X
x2k
k=1
8. *Consider the power series
∞
X
4k
1
= − ln(1 − x2 ).
4
(−1)k 32k+1 x3k+1 .
k=0
(a) (1 point) Find the radius of convergence of this series.
2
3− 3
(a)
Answer.
Notice that
∞
X
(−1)k 32k+1 x3k+1 =
k=0
∞
∞
X
X
(−1)k 32k+1 x3k · x = x
(−1)k 32k+1 x3k .
k=0
Let y = x3 , then
k=0
∞
∞
X
X
(−1)k 32k+1 x3k+1 = x
(−1)k 32k+1 y k .
k=0
k=0
Let ck = (−1)k 32k+1 , then
|ck+1 |
32k+3
= 2k+1 = 9,
|ck |
3
Then the radius of convergence of the series
∞
X
(−1)k 32k+1 y k is
k=0
r
∞
X
convergence of the series
(−1)k 32k+1 x3k+1 is
k=0
for all k ≥ 0.
3
2
1
= 3− 3 .
9
1
. Since y = x3 , then the radius of
9
(b) (1 point) Evaluate the sum of this series.
3x
1 + 9x3
(b)
Answer.
In fact, we have
∞
X
k 2k+1 3k+1
(−1) 3
x
∞
X
=
(−1)k 32k · 3 · x3k · x
k=0
k=0
= 3x
∞
X
(−1 · 32 · x3 )k
k=0
= 3x
∞
X
(−9x3 )k
k=0
=
3x
1 + 9x3
if | − 9x3 | < 1.
Your Score:
/12