MATH 152, SPRING 2012
COMMON EXAM III - VERSION B Solutions
Last Name:
First Name:
Signature:
Section No:
PART I: Multiple Choice (4 pts each)
1. Which of the following is a unit vector in the direction of b − 3a, where a = h1, −2, 2i and b = h5, 0, −3i?
2
6 9
(a) − , − ,
11 11 11
9
2 6
Correct Answer
, ,−
(b)
11 11 11
(c) h2, 6, −9i
2
18
14
, −√
,√
(d) − √
524
524 524
14
2
18
(e) √
,√
, −√
524
524
524
Solution: b − 3a = h5, 0, −3i − 3 h1, −2, 2i = h2, 6, −9i. Divide
by the magnitude
to make this vector a unit vector.
√
9
2 6
h2, 6, −9i
.
=
, ,−
Since | h2, 6, −9i | = 4 + 36 + 81 = 11, u =
11
11 11 11
2. Find the sum of the series
(a)
(b)
(c)
(d)
(e)
∞ (−2)n
P
.
n!
n=0
0
e−2 Correct Answer
e2
sin(−2)
cos(−2)
Solution: Since ex =
3. Represent
∞ xn
∞ (−2)n
P
P
, it follows that e−2 =
.
n!
n=0 n!
n=0
1
as a power series about 0 and give the interval of convergence.
1 + 4x2
(a)
∞
X
1
=
(−4)n x2n , where |x| < 4.
1 + 4x2
n=0
(b)
∞
X
1
1
=
4n x2n , where |x| < .
2
1 + 4x
4
n=0
(c)
∞
X
1
1
4n x2n , where |x| < .
=
1 + 4x2
2
n=0
(d)
∞
X
1
1
=
(−4)n x2n , where |x| < .
1 + 4x2
4
n=0
∞
X
1
1
=
(−4)n x2n , where |x| < . Correct Answer
(e)
1 + 4x2
2
n=0
Solution:
∞
∞
X
X
1
1
2 n
=
(−4x
)
=
(−4)n x2n , where | − 4x2 | < 1 , hence |x| < .
2
1 + 4x
2
n=0
n=0
1
4. Find the cosine of the angle, θ, between the vectors a = h2, 0, 1i and b = h1, −2, 1i.
5
(a) cos(θ) = √
30
4
(b) cos(θ) = √
30
5
(c) cos(θ) = √
12
3
(d) cos(θ) = √ Correct Answer
30
3
(e) cos(θ) = √
12
Solution: cos(θ) =
h2, 0, 1i · h1, −2, 1i
3
a·b
=
=√
|a||b|
| h2, 0, 1i || h1, −2, 1i |
30
5. Find the 15th derivative at x = 0 for f (x) =
(−1)n
xn .
n
n=0 7 (n + 5)
∞
P
(a) f (15) (0) = 0
15
715 (20)
15
(c) f (15) (0) = − 15
7 (20)
15!
Correct Answer
(d) f (15) (0) = − 15
7 (20)
15!
(e) f (15) (0) = 15
7 (20)
(b) f (15) (0) =
(−1)n
1
1
1
1
1
1
xn = −
x + 2 x2 − 3 x3 + ... − 15
x15 + 16
x16 + .... Now, if
n (n + 5)
7
5
7(6)
7
(7)
7
(8)
7
(20)
7
(21)
n=0
the degree on x is less than 15, the 15th derivative of these terms will be zero. If the degree on x is greater than 15,
the 15th derivative of these terms will still contain a power of x, and hence will equal 0 when x = 0. If the degree
15!
on x is 15, then f (15) (x15 ) = 15! Thus f 15 (0) = − 15
7 (20)
Solution: f (x) =
6. The series
∞
P
n=1
en
∞
P
1
√ is
+ n
∞ 1
P
1
1
√ < n and
is a convergent geometric series. Correct Answer
n
e
+ n
n=1 e
∞
P
1
1
1
√ is a divergent p series.
(b) divergent because n √ > √ and
e + n
n
n
n=1
∞
P
1
1
1
√ is a divergent p series.
(c) divergent because n √ < √ and
e + n
n
n
n=1
∞
P
1
1
1
√ is a convergent p series.
(d) convergent because n √ < √ and
e + n
n
n
n=1
∞ 1
P
1
1
(e) divergent because n √ > n and
is a divergent geometric series.
n
e
e + n
n=1 e
∞ n
X
∞ 1
P
1
1
1
1
is a convergent geometric series (r = ), and n √ < n , it follows
=
Solution: Since
n
e
e
e
e
+
n
e
n=1
n=1
(a) convergent because
that
en
∞
X
1
1
√
, thus by the comparison test, since the larger series converges, so does the smaller series.
<
n
n n=1 en
n=1 e +
∞
P
2
7. Find the Taylor Polynomial of degree 2, T2 (x), for f (x) =
1
(x − 9)2
108
1
T2 (x) = 3 + (x − 9) −
6
1
T2 (x) = 3 + (x − 9) −
6
1
(x − 9)2
T2 (x) = −
216
None of these.
√
x centered at a = 9.
(a) T2 (x) = −
(b)
(c)
(d)
(e)
1
(x − 9)2
108
1
(x − 9)2 Correct Answer
216
Solution: T2 (x) = f (9) + f ′ (9)(x − 9) +
f ′′ (9)
(x − 9)2 .
2
1
1
T2 (x) = 3 + (x − 9) −
(x − 9)2 .
6
216
8. The series
∞ (−1)n n
P
2
n=1 n + 1
(a) is convergent but not absolutely convergent. Correct Answer
(b) is absolutely convergent.
(c) is divergent.
(d) is absolutely divergent but not divergent.
(e) converges absolutely by the ratio test.
∞ (−1)n n
P
n
converges
by
the
Alternating
Series
Test
since
since
the
sequence
2
n2 + 1
n=1 n + 1
∞
P
n
for convergence. Using the Limit
decreases to zero . To test for absolute convergence, test the series
2+1
n
n=1
n
∞
∞ 1
2+1
P
P
1
n
an
n
Comparison Test, with bn = , lim
= lim
diverges, so does
.
= 1 > 0. Since
2+1
1
n→∞
n n→∞ bn
n
n
n=1
n=1
n
∞ (−1)n n
P
x
is converget but not absolutely convergent. Another solution: Another solution: Since 2
is
Hence
2
x +1
n=1 n + 1
∞
P
n
continuous, positive and decreasing on [1, ∞), we may use the integral test to show
diverges:
2
n=1 n + 1
Z ∞
∞
1
x
dx = ln(x2 + 1)1 = ∞.
2
x +1
2
1
Solution: The series
9. What does it mean to say
∞
P
an is convergent?
n=1
(a) lim an = 0.
n→∞
(b) lim an exists.
n→∞
(c) There are numbers A and A′ such that A ≤ an ≤ A′ for all n.
(d) There are numbers B and B ′ such that B ≤ a1 + a2 + ... + an ≤ B ′ for all n.
(e) lim (a1 + a2 + ... + an ) exists. Correct Answer
n→∞
Solution: The partial sum is sn = a1 + a2 + ... + an and
∞
P
n=1
3
an = lim sn .
n→∞
10. The series
∞
P
2
ne−n
n=1
(a) converges by the Comparison Test with
∞
P
2
e−n .
n=1
(b) diverges by the Ratio Test.
(c) converges by the Integral Test. Correct Answer
(d) diverges by the Test for Divergence.
(e) diverges by the p-series test.
2
Solution: Since xe−x is a continuous, positive, decreasing function on [1, ∞), we may apply the integral test:
Z
∞
∞
2
2
1
1
xe−x dx = − e−x = . Since the improper integral converges, so does the series.
2
2e
1
1
11. If we apply the Ratio Test to the series
∞ (−4)n n2
P
, which of the following statements is true?
5n+1
n=1
an+1 4
= < 1, therefore the series converges. Correct Answer
(a) lim n→∞
an 5
an+1 = − 4 < 1, therefore the series converges.
(b) lim n→∞
an 5
an+1 = 4 < 1, therefore the series converges.
(c) lim n→∞
an 25
an+1 = ∞ > 1, therefore the series diverges.
(d) lim n→∞
an an+1 = 0 < 1, therefore the series converges.
(e) lim n→∞
an an+1 (−4)n+1 (n + 1)2 5n+1 (−4)n (−4)(n + 1)2 5n+1 Solution: lim = lim
= lim
n→∞
an n→∞ 5n+2
(−4)n (n2 ) n→∞ 5n+1 (5)
(−4)n (n2 ) (−4)(n + 1)2 4
= < 1, therefore the series converges absolutely, therefore converges.
= lim 5
n→∞
5n2
12. What is the intersection of the sphere (x − 2)2 + (y − 1)2 + (z − 3)2 = 16 with the xy-plane?
(a) The circle (x − 2)2 + (y − 1)2 = 7. Correct Answer
(b) The circle (x − 2)2 + (y − 1)2 = 1.
(c) The point (2, 1, 0).
(d) The circle (x − 2)2 + (y − 1)2 = 16.
(e) The sphere does not intersect the xy-plane.
Solution: The sphere intersects the xy-plane when z = 0. Substituting z = 0 into the equation
(x − 2)2 + (y − 1)2 + (z − 3)2 = 16 yields (x − 2)2 + (y − 1)2 + (0 − 3)2 = 16, hence (x − 2)2 + (y − 1)2 = 7
13. In three-dimensional space, R3 , the equation x2 + y 2 = 9 describes
(a) a plane.
(b) a circle.
(c) a parabola.
(d) a cylinder. Correct Answer
(e) a sphere.
Solution: In two-dimension, x2 + y 2 = 9 describes a circle. Thus, in three-dimension, as z varies, x2 + y 2 = 9
describes a cylinder.
4
PART II: WORK OUT (48 points total)
Directions: Present your solutions in the space provided. Show all your work neatly and concisely and box your
final answer. You will be graded not merely on the final answer, but also on the quality and correctness of the work
leading up to it.
14. a.) (5 pts) Using the known Maclaurin series for sin x, find the Maclaurin series for sin(x3 ).
∞ (−1)n x2n+1
∞ (−1)n x6n+3
P
P
Solution: sin x =
, thus sin(x3 ) =
.
n=0 (2n + 1)!
n=0 (2n + 1)!
sin(x3 ) − x3
.
x→0
x9
b.) (5 pts) Using the Maclaurin series for sin(x3 ), compute lim
∞ (−1)n x6n+3
P
x15
x21
x9
3
− x3
+
−
+ ... − x3
x
−
sin(x ) − x
n=0 (2n + 1)!
3!
5!
7!
= lim
= lim
Solution: lim
x→0
x→0
x→0
x9
x9
x9
6
12
x
x
1
−
+ ...
x9 − +
1
1
3!
5!
7!
= lim
=− =−
x→0
x9
3!
6
3
3
2
15. (5 pts) Using the known Maclaurin series for ex , find a Maclaurin series for e−x .
∞ xn
∞ (−1)n x2n
P
P
2
Solution: ex =
, thus e−x =
.
n!
n=0 n!
n=0
2
b.) (5 pts) Expand the Maclaurin series for e−x found above to degree 4. Use this 4th degree Maclaurin polynomial
Z 1
2
e−x dx. Do not simplify.
to estimate
0
∞ (−1)n x2n
P
x4
≈ 1 − x2 + .
n!
2
n=0
1
Z 1
Z 1
4
x3
1
x5 1
23
x
2
−x2
dx = x −
+
1−x +
Thus
e
dx ≈
= 1 − 3 + 10 = 30 .
2
3
10
0
0
0
2
Solution: e−x =
16. (10 pts) Find a power series about 0 for f (x) = ln(5x + 4). What is the associated radius of convergence?
n+1
n X
∞
∞ 5x 5x
5
d
5
5
5X
=
−
=
(−1)n
xn , where < 1.
Solution:
(ln(5x + 4)) =
= 5x
dx
5x + 4
4 n=0
4
4
4
n=0
4 1+
4
n+1
n+1 n+1
Z X
∞
∞
X
5
5
x
4
(−1)n
Thus, ln(5x + 4) =
xn dx = C +
(−1)n
, where |x| < . Choosing x = 0 to
4
4
n
+
1
5
n=0
n=0
n+1
n+1
∞
P
x
4
5
,R= .
solve for C yields C = ln 4. Thus ln(5x + 4) = ln 4 +
(−1)n
4
n+1
5
n=0
5
∞ (3x − 1)n
P
. Be sure to test the endpoints for convergence.
n6n
n=1
an+1 (3x − 1)(n) 3x − 1 (3x − 1)n+1
n6n .
Solution: Using the Ratio Test, lim = lim
= lim
=
n→∞
an n→∞ (n + 1)6n+1 (3x − 1)n n→∞ (n + 1)6 6 ∞
an+1 P
< 1, then the series
The Ratio Test states if lim an converges. Hence
n→∞
an
n=1
3x − 1 < 1, so |3x − 1| < 6, thus x − 1 < 2. Solve this inequality for x: − 5 < x < 7 .
6 3
3
3
Test the endpoints of the interval for convergence :
∞
X
∞ (−6)n
(−1)n
5 P
=
which converges by the Alternating Series Test since the sequence
Test x = − :
3 n=1 n6n
n
n=1
1
decreases to zero.
n
∞
X
∞ (6)n
1
7 P
=
which diverges by the p-series test. Thus the interval of convergence is
Test x = :
3 n=1 n6n
n
n=1
17. (10 pts) Find the interval of convergence of the series
−
5
7
≤x< .
3
3
18. Consider the series
∞ (−1)n
P
.
2
n=1 3n + 1
a.) (4 pts) Prove the series is absolutely convergent.
∞
∞
∞
X
∞ 1
∞ (−1)n P
P
1
1
1X 1
= P
=
. Since
is a convergent p-series, it follows
Solution:
<
3n2 + 1 2
2
2
2
3n
3 n=1 n
n=1 3n + 1
n=1 n
n=1
n=1
∞
∞ (−1)n
∞
P
P
P
1
1
converges, thus
also converges by the comparison test. Hence
is absolutely
that
2
2
2
n=1 3n + 1
n=1 3n + 1
n=1 3n
convergent.
∞ (−1)n
P
b.) (2 pts) Find s2 , the sum of the first 2 terms, to approximate
.
2
n=1 3n + 1
Solution: S2 =
2 (−1)i
P
1
1
9
=− +
=−
2
4 13
52
i=1 3i + 1
c.) (2 pts) Find an upper bound on |R2 |, the absolute value of the remainder, in using s2 to approximate
Solution: |R2 | ≤ |a3 |, where an =
(−1)n
1
. Thus |R2 | ≤
.
3n2 + 1
28
6
∞ (−1)n
P
.
2
n=1 3n + 1